CH 111: General Chemistry

Dr. Ananda M

The University of Dodoma, Tanzania

Teaching Compendium on

General Chemistry
(CH 111)

by

Dr. AnandaMurthyM.Sc., M.Phil., Ph.D.
Department of Chemistry, School of Physical Sciences, College of Natural &Mathematical Sciences, The University of Dodoma.

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CH 111: General Chemistry

Dr. Ananda M

Preface
The importance of learning chemistry cannot be overemphasized. In this regard, all branches of chemistry contribute to the overall role that chemistry plays in daily life. The course content of CH 111 is designed to bridge the gap between the high school and the undergraduate chemistry, with respect to some fundamental topics in chemistry for first year chemistry students. This compendium will be helpful to the students, due to its relevance to the course content and will promote better understanding of the subject matter. It is intended to enable students to achieve the learning objectives and learning outcomes of CH 111 by being a quick reference to learners. The author of the compendium has presented the contents in a simplified manner by using various illustrations, including structures, tables, figures, and other relevant information to help the learner understand easily. This compendium will, to greater extent, help in understanding the basic concepts in chemistry.

Dr. J.J. Makangara Senior Lecturer, Department of Chemistry, Former Dean, School of Physical Sciences, College of Natural and Mathematical Sciences, The University of Dodoma, Dodoma, Tanzania.

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CH 111: General Chemistry

Dr. Ananda M

Dedicated to My Students of The University of Dodoma

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CH 111: General Chemistry

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Acknowledgements
I express my profound gratitude and regards to the management of the University of Dodoma for providing me this opportunity to serve at CNMS, UDOM. It is a great pleasure to express my gratitude to the Head, Department of Chemistry, the Dean, School of Physical Sciences and The Principal, CNMS for their constant support and encouragement in bringing out this compendium.I also thank Dr. J. J. Makangara for his cooperation towards publication of this compendium.I also place on record my appreciation to my fellow colleagues at CNMS, UDOM for their cooperation. I wish to present this simple compendium on General chemistry course (CH 111) for the benefit of the students of first year B.Sc. Chemistry of the University of Dodoma. This compendium is believed to be helpful to the students, as it is prepared to cater to the needs of the students, keeping in mind the relevance of each contents of the course toward better understanding of the subject and withillustration including suitable figures, tables, and examples of structures and configurations of the complexes with simple presentation formats. I will be happy, if my little work, serves the purpose of student‘s better understanding of the concepts of Chemistry and superior performance in the examinations.

Author Dr. Ananda Murthy H C

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CH 111: General Chemistry

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Contents

Sl. Nos. * * * * 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12 1.13 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12 2.13

Contents Preface Dedication Acknowledgements Contents Course Description Electronic Configurations of Elements Rules for Electronic Configurations The electronic configuration of the elements Shapes of Atomic orbitals- s, p, d & f Modern Periodic law s, p, d and f block elements Periodic trends in properties of elements Theories of Chemical Bonding Hybridization Valence Shell Electron Pair Repulsion Theory Polar and Non-polar covalent bonds Metallic Bond Introduction to Symmetry and Point Groups Theories of acids and bases Mole concept applications Methods of expressing concentration of the solutions Red-ox reactions Balancing chemical equations Basics of electrochemistry Properties of solutions Solutions of gases in liquids - Henry‘s law, Solution of liquids in liquids - Raoult's law Binary liquid mixtures Chemical equilibrium Phase rule and applications References

Page 2 3 4 5 6 7 8 10 13 15 17 21 36 45 49 52 54 59 70 79 81 88 90 92 113 115 116 120 128 141 149

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CH 111: General Chemistry

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CH 111: General Chemistry
10 Credits Course Description This course is designed to bridge the gap between the high school and the undergraduate programs, with respect to some fundamental topics in chemistry. They are the structure, chemical bonding, acid-base and redox aspects of elements and compounds and also updates in the areas of solutions, chemical equilibrium and thermo chemistry. Objective To provide the students with an ability to explain the experimentally measured properties and behaviour of matter on molecular terms. Learning outcome After completing with the course, the students will gain basic knowledge on the electronic structure of the atoms, periodicity of physical and chemical properties of the elements, Chemical equilibrium, redox reactions and electrochemistry. Students acquire skills in solving problems of different types – theoretical explanations and numerical calculations. Course Content Electronic configurations of elements, orbital shapes of s, p, d and f orbitals, modern periodic table- s, p, d and f block elements and trends in physical and chemical properties, theories of chemical bonding- ionic, covalent and metallic bonding- properties and applications, introduction to symmetry and point groups, mole concept applications, methods of expressing concentration of the solutions, theories of acids and bases, properties of solutions: solutions of gases in liquids, Henry‘s law, solution of liquids in liquids, Raoult's law, binary liquid mixtures, ideal solutions, chemical equilibrium- heterogeneous equilibrium, Phase rule and applications, redox reactions, balancing chemical equations and basics of electrochemistry. Contact Hours: 40 hours of lectures, 20 hours of tutorials Mode of Assessment:40 % Coursework (at least 2 tests); 60 % final examination

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CH 111: General Chemistry 1.1 Electronic Configurations of Elements

Dr. Ananda M

The scientific enquiry to understand the distribution of electrons in atoms started since the discovery of electrons by J J Thomson in 1897. Rutherford and Bohr proposed the stable atomic model with discrete orbits for electrons around the atom. The great minds suchasSchrodinger and Heisenberg have made significant contributions in evolving the probability density for electrons, defined by mathematical functions called atomic orbitals.The region in space around the nucleus of an atom where there is a maximum probability of finding an electron is called the atomic orbital. The understanding of distribution of electrons around the nucleus of an atom in orbitals resulted in the evolution of a method of representation of electrons by the way called electronic configuration of elements.In general electronic configuration describes the distribution of electrons around the nucleus of atoms in various shells and sub shells (orbitals). The two important ways of representations of electronic configurations includes, 1. spdf notation: Thespdf notation uses a number to designate a Principal shell and a letter-s,p,d,f- for the identification of sub shell. The number of electrons in the orbital is represented by a superscript number.

2s1
Thus the electronic configuration for carbon 1 1 is given as1s2 2s2 2p2 or 1s22s2 2Px 2Py .For the heavier elements, electronic configurations include the reference of nearest noble gas elements, for example electronic

configuration of calcium is written as 1s2 2s 2 2p6 3s 2 3p6 4s 2 or [Ar] 4s 2 . 2. Orbital diagram: In this method, electronic configuration is represented by using either box or circle and arrows to represent electrons. Upward arrow represents an electron spinning in the clockwise directions and the downward arrow for anticlockwise electron. The orbital diagram for carbon is

OR

In general both the above electronic configurations are used for the illustrations of atomic and molecular bonding. UDOM- Study Material 2013-14 Page 7

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1.2 Rules for Electronic Configurations The best representation of electronic configuration for elements is achieved through learning few important rules developed by Wolf-gang Pauli and Friedrich Hund. a. Pauli’s Exclusion principle Pauli’s exclusion principle states that “It is impossible for any two electrons in the same atom to have all the four quantum numbers identical” This means each electron in an atom has only one set of values n, l,ml and s. Any two electrons in the same orbital will have different values for at least one quantum numbers. An atomic orbital can accommodate only two electrons with opposite spins. For example 4s-orbital 4s n=4 l=0 m=0 s=+½ - clockwise electron n=4 l=0 m=0 s=-½ - Anticlockwise electron Same values Different s values The Pauli‘s exclusion principle limits the number of electrons that can beaccommodated in principal shells and sub shells (orbitals). These are illustrated in the table given below, Main shell Principal quantum number [n] Azimuthal q.no [l] [ values between 0 and( n-1)] 0 0 1 0 1 2 0 1 2 3 Subshell Magnetic q. no.[ml] designation (2l+1) no. of values. These values vary between –l to +l. 1s 0 2s 0 2p -1,0,+1 3s 0 3p -1,0,+1 3d -2,-1,0,+1,+2 4s 0 4p -1,0,+1 4d -2,-1,0,+1,+2 4f -3,-2,-1,0,+1,+2,+3 No. of orbitals in sub shells Principal shell capacity [2n2] 2 2+ 6=8 2+ 6+ 10=18 2+ 6+ 10+ 14=32

K L M

1 2 3

N

4

1 1 3 1 3 5 1 3 5 7

b. Hund’s rule of maximum multiplicity According to this rule electron pairing in any of the s, p, d or f orbital is not possible until all the available orbitals of a given set contain one electron each. For example, electronic configuration of N is 1s2 2s2 2p3. This indicates that 1s and 2s orbitals are completely filled with two electrons each. But according to Hund‘s rule, electrons occupy all UDOM- Study Material 2013-14 Page 8

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the available orbitals singly with parallel spins, before they pair up in any orbital. Thus the three ' ' ' electrons of the 2p-orbital will have the configuration 2p x , 2p y , 2p z . Thus the new electronic
' configuration of nitrogen is 1s2 2s2 2 p x 2 p 'y 2 p ' z

In case of oxygen 2p orbital has four electrons, first three electrons in px, py and pz orbitals singly, fourth electron occupies px orbital and pairs with the electron already present in it. Thus 2 1 1 the electronic configuration of oxygen is 1s2 2s2 2p x 2p y 2p z Hunds rule also states that the total energy of a many electron atom with more than one electron occupying a set of degenerate orbitals is lowest it, as far as possible, electrons occupy different atomic orbitals and have parallel spins. The lowest energy configuration for an atom is the one with maximum number of unpaired electrons.

c. (n+l ) rule In neutral isolated atoms, the most stable orbitals is the one for which the sum (n+l) is least. When two orbitals have the same (n+l) value the orbital with lower value of n is the stabler one. For example for 4s sublevel n=4 l=0 n+l=4 for 3d sublevel n=3 l=2 n+l=5 For 4s sublevel (4+l) has lower value than for the 3d sublevel. So 4s level is occupied by electrons before the 3d level. d. Aufbau principle Aufbau principle states that the electrons enter the various orbital in the order of increasing energy. Aufbau principle describes the method which involves building up of electronic configuration for atom by adding on electron to the outer shell and one proton to the nucleus of that atom. Aufbau means building up in German language. Pauli‘s exclusion principle, Hund‘s rule, (n+l) rule and Afbau principle are employed in building the electronic configuration of atoms.The ascending order of energy of the sublevels, in accordance with aufbau principle is 1s, 2s,2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s,4p, 5d, 6p, 7s, 5p, 6d, 7p Many of the principles or rules discussed earlier, namely UDOM- Study Material 2013-14 Page 9

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Afbau principle, Hunds rule, (n+l) ruleand Pauli‘s exclusion principle provide us enough information in filling of electrons to various orbitals of different elements. The first element, hydrogen atom (atomic number z = 1) having only one electron in s orbital, is half filled i.e., the configuration is 1s1. The electronic configuration is 1s2 for the next element helium, He (z = 2). Maximum capacity of s-orbital is 2 in accordance with the formula 2n2 (where n-principal quantum no.) Now k-shell is completely filled. Next the third electron in Lithium enters 2s orbital of L-shell. The configuration of Li (z = 3) is 1s2 2s1. In beryllium (z = 4), 2s orbital is completely filled, so the configuration of Be is 1s2 2s2.Boron, the next element has five electrons of which 5th electron occupies 2p orbital of L-shell so the configuration of B (z =5) is 1s2 2s2 2px1.In carbon, 6th electron, does not get paired up with electron of 2px orbital but enters 2py [orbital] or subshell in accordance with Hund‘srule.So the configuration of C(z = 6) is1s2 2s2 2px12py1.Nitrogen has electronicnic configuration with three of its electron in p three 2p orbitals1s2 2s2 2px12py12pz1. In oxygen, O (z = 8) the 8th electron gets paired with 2px1orbital and has the configuration 1s2 2s2 2px22py12pz1. Similarly in fluorine and neon, 9th and 10th electrons pair up with 2py1and 2pz1orbitals respectively. Hence the configurations of fluorine (z =9) is 1s2 2s2 2px22py22pz1and that of neon (z =10) is 1s2 2s2 2px22py22pz2.The 2nd shell (L-Shell) is completed in neon. 1.3 The electronic configuration of the first 10 elements Electronic configuration At No. 1 2 3 4 5 6 7 8 9 10 Element Hydrogen Helium Lithium Beryllium Boron Carbon Nitrogen Oxygen Fluorine Neon Symbol H He Li Be B C N O F Ne Spdf-notation 1s1 1s2 1s2 2s1 1s2 2s2 1s2 2s2 2px1 1s2 2s2 2px12py1 1s2 2s2 2px12py12pz1 1s2 2s2 2px22py12pz1 1s2 2s2 2px22py22pz1 1s2 2s2 2px22py22pz2 Page 10 Orbital diagram 1s 2s 2p

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The third shell -M-shell gets its quota of electrons from sodium to Argon. The noble gas core abbreviated electron configuration for the elements is given below, 11 12 13 14 15 16 17 18 Sodium Magnesium Aluminium Silicon Phosphorous Sulphur Chlorine Argon Na Mg Al Si P S Cl Ar 1s2 2s2 2p6 3s1 or [Ne] 3s1 [Ne] 3s2 [Ne] 3s23p1 [Ne] 3s23p2 [Ne] 3s23p3 [Ne] 3s23p4 [Ne] 3s23p5 [Ne] 3s23p6

The fourth shell, electron filling starts from K, where 19th electron moves to 4s orbital rather to 3rd orbital in accordance with (n+l) rule. 19 20 21 22 23 24 25 26 27 28 29 30 Potassium Calcium Scandium Titanium Vanadium Chromium* Manganese Iron Cobalt Nickel Copper* Zinc K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn
1s2 2s2 2p6 3s2 3p6 4s1 or [Ar] 4s1 1s 2 2s 2 2p6 3s 2 3p6 4s 2 or [Ar] 4s 2 [Ar] 4s 2 3d1

[Ar] 4s2 3d2 [Ar] 4s2 3d3 [Ar] 4s1 3d5 [Ar] 4s2 3d5 [Ar] 4s2 3d6 [Ar] 4s1 3d7 [Ar] 4s2 3d8 [Ar] 4s1 3d10 [Ar] 4s2 3d10

Both chromium and Copper have unusual electronic configuration deviating form normal electron filling process because of a general rule, which states that ―completely filled and completely half filled configurations are more stable than other configuration‖. In the absence of this rule Cr and Cu should have the following configuration respectively, [Ar] 4s2 3d4 and [Ar] 4s2 3d1.

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CH 111: General Chemistry 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Gallium Germanium Arsenic Selenium Bromine Krypton Rubidium Strontium Yttrium Zirconium Niobium* Molybdenum* Technetium Ruthenium* Rhodium* Palladium* Silver * Cadmium Indium Tin Antimony Tellurium Iodine Xenon Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe [Ar] 4s2 3d104p1 [Ar] 4s2 3d104p2 [Ar] 4s1 3d104p3 [Ar] 4s2 3d104p4 [Ar] 4s1 3d104p5 [Ar] 4s2 3d104p6 [Kr] 5s1 [Kr] 5s2 [Kr] 5s24d1 [Kr] 5s24d2 [Kr] 5s14d4 [Kr] 5s14d5 [Kr] 5s24d5 [Kr] 5s14d7 [Kr] 5s14d8 [Kr] 5s04d10 [Kr] 5s14d10 [Kr] 5s24d10 [Kr] 5s24d105p1 [Kr] 5s24d105p2 [Kr] 5s24d105p3 [Kr] 5s24d105p4 [Kr] 5s24d105p5 [Kr] 5s24d105p6

Dr. Ananda M

Practice questions 1. Identify the specific element that corresponds to each of the following electron configurations: a) 1s2 2s2 2p6 3s2 b) [Ne] 3s23p4 c) [Ar] 4s1 3d10 d) [Kr] 5s24d105p3

2. The following electron configurations represent excited states. Identify the element, and write its ground state condensed configuration, a) 1s2 2s2 3p2 3p1 b) [Ar] 4s1 3d10 4p4 5s1 d) [Kr] 5s24d65p1

3. Scientists have speculated that element 126 might have a moderate stability, allowing it to be synthesized and characterized. Predict what the condensed electron configuration of this element might be.

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CH 111: General Chemistry 1.4Shapes of Atomic orbitals- s, p, d & f

Dr. Ananda M

Solving the Schrödinger wave equation results in a set of wave functions called atomic orbitals. Each atomic orbital contains information about the region of space where an electron of a given energy is most likely located because Ѱ 2 represents electron probability density.Thewave function, Ѱ 2, can be represented in graphical form as a picture of the three-dimensional region of the atom where an electron with a given energy state is most likely to be found. That is, Ѱ 2is related to the probability of finding the electron in a given region of space. This probability, known as the electron density, can be visualized as an array of dots, each dot representing the possible location of the electron with a given energy in relation to the nucleus. The region in space around the nucleus of an atom of an element where there is a maximum probability of finding an electron is called the atomic orbital.Atomic orbitals are of different kinds and each orbital has a definite energy. Solving of wave equations corresponding to different electrons in an atom shows that the atomic orbitals are mainly of four types namely s-orbitals, p-orbitals, d-orbitals and f-orbitals. 1. S-orbital: In the s-orbital, the electron density is distributed symmetrically about the nucleus, s-orbital spherical in shape.The s-orbital corresponding to the k-shell has the lowest energy. 2s orbital is associated with more energy and so on.

2.

P-orbitals: p-orbitals have dumb bell shape with two lobes. For a p-orbital the regions of high charge density lie on both sides of the nucleus. A sub shell with l=1, there are three porbitals, designated as Px, Py and Pz which are oriented mutually at right angles to one another and along the x,y and z axes in space. These orbitals have three-fold degeneracy corresponding to the quantum numbers ml = +1, 0, -1.

3.

d-orbitals: d-orbitals have double dumb bell shape. A sub shell with l=2, there can be five d-orbitals. These are designated as dxy,dxz, dyz, dx2-y2anddz2. Each d- orbital has four lobes, except for dz2. The first three orbitals lie in the xy,xz and yz planes respectively. They are Page 13

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oriented at 45o to the axes. These orbitals have five-fold degeneracy corresponding to the quantum numbers ml = +2, +1, 0, -1, -2.

4.

f-orbital: f-orbital have complex shapes. There are seven f-orbitals, f-orbitals have sevenfold degeneracy corresponding to the quantum numbers ml = +3, +2, +1, 0, -1, -2, -3.

Note that an atomic orbital (quantum mechanical model) is not the same as an orbit (Bohr model). In the quantum mechanical model, the principal quantum number, n, is a measure of the most probable distance of the electron from the nucleus, not the radius of a well-defined orbit.

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CH 111: General Chemistry The origin of labels s,p,d and f for orbitals

Dr. Ananda M

Spectroscopists associated transitions involving energy levels with different values with different groups of lines in the line spectra of the alkali metals. The line groups were called sharp, principal, diffuse, and fundamental. When the angular momentum quantum number was used to describe and explain these groups of lines, s became an abbreviation for = 0, p meant = 1, d meant = 2, and f meant = 3. For consistency, higher values of the angular momentum quantum numbers are designated alphabetically (g means = 4, h means = 5, and so on). The orbital labels s, p, d, and f originate from a now-discredited system of categorizing spectral lines as sharp, principal, diffuse, and fundamental, based on their observed fine structure. 1.5Modern Periodic law and the present form of the periodic table In the year 1913, Henry Moseley discovered a new property of elements, called Atomic number, which provided a better basis for the periodic arrangement of the elements. Mendeleev periodic law was modified and called as Modern periodic law. The last major changes to the periodic table resulted from Glenn Seaborg's work in the middle of the 20th Century. Starting with his discovery of plutonium in 1940, he discovered all the transuranic elements from 94 to 102. He reconfigured the periodic table by placing the actinide series below the lanthanide series. In 1951, Seaborg was awarded the Nobel Prize in chemistry for his work. Element 106 has been named seaborgium (Sg) in his honor. The Modern periodic law states that ―The physical and chemical properties of elements are periodic function of their atomic numbers”. The periodic table: A table in which elements are arranged in the order of increasing atomic numbers in the manner that the elements with similar properties fall in the same vertical column is known as the periodic table. The elements, which fall in the same column and resemble one another in their properties are said to belong to the same group or family of elements. Importance of the periodic table The objective of the periodic table is to organize and systematize the chemistry of the elements. It helps us to understand the characteristic behaviour of each element and resemblance of group elements. It helps us to understand the cause of periodicity of properties and the reason why similar properties recur at certain regular intervals, i.e., after 2,8,18 and 32 elements. These numbers namely 2,8,18 and 32 sometimes referred to as magic numbers.

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CH 111: General Chemistry The long form of the periodic table

Dr. Ananda M

It is an arrangement in which the elements are arranged in the increasing order of atomic number such that the elements with similar properties fall in vertical columns called groups: Horizontal rows are called periods. Periodic table consists of 18 groups and 7 periods. Instead of listing the 103 elements as one long list, the periodic table arranges them into several horizontal rows or periods, in such a way that each row begins with an alkali metal and ends with a noble gas. The sequence in which the various energy levels are filled determines the number of elements in each period and the periodic table can be divided into four main regions according to whether the s, p, d or f-levels are being filled.

Characteristics of groups All the group elements have same valence electronic configuration and thus exhibits similar chemical properties. The elements in a group are separated numbers.(2,8,8,18,18,32- Magic numbers) by definite gaps of atomic

In general physical properties of group elements (melting point, boiling point etc) follow a systematic pattern down the group. The atomic sizes of the elements in a group always decrease down the group. s-block elements belong to group 1 and 2, p-block elements belong to groups 13 to 18, dblock elements belong to groups 3 to 12 and f-block elements are considered to be in group 3. UDOM- Study Material 2013-14 Page 16

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Characteristics of periods In all the elements of period, electrons are successfully filled to the same valence shell and thus all the period elements have different electronic configuration. In general elements exhibit different chemical properties along a period. The metallic character of elements decreases across a period, while their non metallic character increases. Cause of periodicity The cause of periodicity in properties appears to lie in the recurrence of similar outer electronic configurations at certain regular intervals. For example, the general electronic configuration of first group alkali metals is ns1. The recurrence of the same electronic configuration for alkali metals, starts from Na[3s1 after Li [2s1]] then continues for K[4s1], Rb [5s1] Cs[6s1] and Fr [7s1]. The similar outer electronic configurations recur at definite intervals of atomic numbers 2, 8, 8, 18, 18, 32 and 32. Hence periodicity in properties appears. 1.6 s, p, d and f block elements The aufbau (build up) principle and the electronic configuration of atoms provide a theoretical foundation for the periodic classification. The elements in a vertical column of the Periodic Table constitute a group or family and exhibit similar chemical behaviour. This similarity arises UDOM- Study Material 2013-14 Page 17

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because these elements have the same number and same distribution of electrons in their outermost orbitals. The periodic table is divided into four blocks namely s, p, d and fblocks depending on the type of atomic orbitals that are being filled with electrons.

The element in any block is linked with the sub-shell or orbital in which the last electron in its atom is filled. If the valence electron is present in s-orbital then it belongs to s-block. Same is the case with the other elements. The S-block elements: The general electronic configuration of elements belonging to this block is ns1-2. n is principal quantum number. S blocks include 1 and 2 groups. Group 1 elements (Alkali metals): Li, Na, K, Rb, Cs & Fr. Group 2 elements (Alkaline earth metals): Be, Mg, Ca, Sr, Ba and Ra. The atoms of elements of the two groups 1 and 2 receive the last electron in the S-orbital of their outermost energy level. The general characteristics of s- block elements are 1. They are soft metals. 2. They are highly reactive and form ionic compounds. 3. They have low ionization enthalpies and are highly electropositive. 4. They have +1 and +2 oxidation states. 5. They act as good reducing agents as they lose electron readily. Note* Hydrogen which is placed above Group 1 alkali metals is not a member of that group because it is neither metallic nor a solid. UDOM- Study Material 2013-14 Page 18

CH 111: General Chemistry 3.6.2 The P-block elements:

Dr. Ananda M

The general electronic configuration of elements belonging to this block is ns2 np1-6. P-block elements include 13, 14, 15, 16, 17 and 18 group elements. The Atoms of the elements of p-block receive their last electron in their p-orbitals. Helium contains no p orbital electron but is included in group 18. This is because helium completes k- Shell electronic configuration and exhibit chemical inertness. The outermost electronic configuration varies from ns2np1 to ns2np6 in each period. At the end of each period is a noble gas element with a closed valence shell ns2np6 configuration. Preceding the noble gas family are two chemically important groups of non-metals. They are the halogens (Group 17) and the chalcogens (Group 16).

The general Characteristics of p-block elements are 1. p-block elements are representative elements which include metals, metalloids and non metals. 2. Compounds of these elements are mostly covalent in nature. 3. They exhibit variable oxidation states 4. They are highly electronegative elements. 5. Most of them form acidic oxides. 6. The nonmetallic character increases left to right across the periods and metallic character increases down the groups. 7. The reactivity of the elements decreases down the group. The d-block (Transition) elements The general electronic configuration of elements of d block is (n-1)d1-10 ns1-2. The d-orbitals in the penultimate [inner to the outermost] shell, however, are still empty. The last electron in these elements therefore, enters into the d-orbital, which thereby, get progressively filled as we move across a period. There are four transition series in d-block. Thus the elements from Sc to Zn - 3d series Page 19

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CH 111: General Chemistry Y to Cd La to Hg - 4d series - 5d series,

Dr. Ananda M

Ac to Uub - 6d series are called d-block elements. The d-block elements are known as transition elements because the properties of d- block elements are transitional (Intermediate) between s-block elements and p-block elements.

The general characteristics of d-block elements are 1. They are all metals with high melting and boiling points. 2. They form coordinate complexes which give coloured solutions. 3. They exhibit variable oxidation states 4. Most of the compounds of elements exhibit paramagnetic nature. 5. They possess good catalytic properties. Elements of group 12 (Zn, Cd and Hg) donot show transition properties, because d- orbital is fully filled. The f-block elements General electronic configuration is (n-2) f1-14 (n-1) d 0-1 ns2. The elements in which the last electron enters into the f-orbitals of their atoms are called f-block elements. f-blocks include 14 lanthanides and 14 Actinides.

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In lanthanides, 4-f orbitals and in actinides, 5-f orbitals are being progressively filled. The two rows of elements at the bottom of the Periodic Table, called the Lanthanoids, Ce(Z = 58) – Lu(Z = 71) and Actinoids, Th(Z = 90) – Lr (Z = 103). In this f-block, atoms of the elements receive electrons to their pre-penultimate shell (innermost), therefore f-block elements are also called inner transition elements. The elements after uranium are called Transuranium elements. 1.7 Periodic trends in properties of elements The physical and chemical properties of elements, either across the period or down the group, show a definite periodic variation which is probably due to systematic change in electronic configurations of elements which are repeated after intervals of 2, 8, 8, 18, 18 and 32. This is responsible for periodicity in certain properties of elements such as atomic radius, ionic radius, ionization enthalpy, electron gain enthalpy and electronegativity. Trends in physical properties  a) Atomic radius: In general, the distance between the centre of the nucleus and the outermost shell of electrons is called radius of atom. However, there is no certainty with regard to the exact position occupied by electrons at any time. It may be very close to the nucleus at one time or may be for a way from the nucleus at next moment. There are three important concepts of atomic radius. i) Vander Waals Radius: One half of the distance between the nuclei of two adjacent atoms belonging to two neighbouring molecules of an element in the solid state is usually known as vanderwaals radius. Covalent radius: One half of the distance between the nuclei of two covalently bonded atoms of the same element in a molecule is taken as the covalent radius of the atom of that element.

ii)

Covalent radius of chlorine atom is = 99 pm iii) Metallic radius: Metallic radius is taken as half of the internuclear distance separating the metal cores in the metallic crystal. For example the distance between two adjacent copper atoms in solid copper is 256 pm; hence the metallic radius of copper is assigned a value of 128 pm.

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CH 111: General Chemistry  Variation of atomic radii in a period

Dr. Ananda M

Atomic number increases on moving from left to right across a period. Nuclear charge of atoms also increases across a period. As the atomic number increases, the effective nuclear charge increases. The electrons which are added to the same orbital across a period, experience a least shielding by the electrons already in the orbital and experience more nuclear charge, Hence these electrons are getting attracted more and more towards the nucleus. Therefore size of an atom reduces on moving across a period. The atomic radius decreases on moving left to right across a period. Trends in atomic radius in Periods 2 and 3

Trends in atomic radius across periods The atomic size generally decreases across a period as illustrated in Figure for the elements of the second period. It is because within the period the outer electrons are in the same valence shell and the effective nuclear charge increases as the atomic number increases resulting in the increased attraction of electrons to the nucleus. You have to ignore the noble gas at the end of each period. Because neon and argon don't form bonds, you can only measure their van der Waals radius - a case where the atom is pretty well "unsquashed". All the other atoms are being measured where their atomic radius is being lessened by strong attractions.

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CH 111: General Chemistry  Variation of atomic radii in a group:

Dr. Ananda M

Atomic radius increases on moving down a group. In moving down a group, the number of shells increases and therefore, the size of the atom increases. This effect no doubt, is partly annulled by the drawing in of the electron shell on account of the increasing nuclear charge. But the effect of adding a new shell is so large that it overcomes the contractive effect of the increased nuclear charge. Hence the radius of the atom increases in moving from top to bottom in a group.

Trends in atomic radius down a group Within a family or vertical column of the periodic table, the atomic radius increases regularly with atomic number as illustrated in Figure. For alkali metals and halogens, descending the groups, the principal quantum number (n) increases and the valence electrons are farther from the nucleus. This happens because the inner energy levels are filled with electrons, which serve to shield the outer electrons from the pull of the nucleus. Consequently the size of the atom increases as reflected in the atomic radii. It is fairly obvious that the atoms get bigger as you go down groups. The reason is equally obvious - you are adding extra levels of electrons. b) Ionic radius. Ionic radius may be defined as the distance from the nucleus of an ion up to which it has influence on its electron cloud. The internuclear distance between the nuclei of Na+ and Cl- ions is determined from x-ray technique. This distance is taken as the sum of the radii of Na+ and Cl- ions. The contribution of Na+ and Cl- ions to the internuclear distance is then evaluated. In this way, the radii of Na+ and Clfound out.

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CH 111: General Chemistry Radii of Cations and Anions A positive ion is obtained by the loss of one or more electrons from the outer shell or valence shell of an atom. In the formation of positive ions, the outer shell of electrons is generally removed completely. The cation, therefore is much smaller than the corresponding atom. At the same time, with the elimination of one or more orbital electrons. The effective nuclear charge increases. The electrons are pulled in more towards the nucleus than before. This effect also tends to decrease the radius of the cation. The radius of the cation is always smaller than that of the atom.

Dr. Ananda M

Atomic radius pm Na = 186 A1 = 143 Mg = 160

Ionic radius pm Na+ = 95 A13+= 50 Mg2+= 80 It is evident that radius of a cation is invariably smaller than that of the corresponding atom. Addition of one or more electrons into an outermost shell of a neutral atom leads to the formation of an anion. The effective nuclear charge of anion of atom is reduced and therefore, the electrons acquire greater freedom of getting away from the nucleus. As a result of this, the radius of a negative ion is invariably larger than that of the corresponding atom The radius of the anion is always greater than that of the atom. Atomic radius pm Cl - 99 Ionic radius pm Cl- - 198

Br - 114

Br - - 195

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Isoelectronic ions: Isoelectronic ions are those ions which have the same number of electrons but differ in the charge on their nuclei. Ions No. of electrons Charge on nucleus Radius pm N310 7 171 O210 8 140 F10 9 136 Na+ 10 11 95 Mg2+ 10 12 80 Al3+ 10 13 50

c) Ionisation enthalpy Ionisation enthalpy is the measure of the tendency of an atom of an element to lose electron. The ionisation enthalpy of an element is the minimum amount of energy required to remove an electron from an isolated gaseous atom of that element resulting in the formation of a positive ion. Ionisation enthalpy ( iH) is the measure of the tendency of an atom to lose an electron and to become a cation, as represented below.

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CH 111: General Chemistry Ionisation enthalpies for group I and II elements [KJ mol-1] 1st Li Na K Rb Cs Fr 520 496 419 403 374 -----2nd 7296 4563 3069 2650 2420 -------Be Mg Ca Sr Ba Ra 1st 899 737 590 549 503 509 2nd 1757 1450 1145 1064 965 979 3rd 14847 7731 4910 4207

Dr. Ananda M

The energy required to remove the first electron from an atom is called the first ionisation enthalpy IE1, the energies required to remove the second, third electrons etc are called second IE [IE2] third I.E[IE3] etc. Energy is always required to remove electrons from an atom and hence ionization enthalpies are always positive. The second ionization enthalpy will be higher than the first ionization enthalpy because it is more difficult to remove an electron from a positively charged ion than from a neutral atom. In the same way the third ionization enthalpy will be higher than the second and so on. The term ―ionization enthalpy‖, if not qualified, is taken as the first ionization enthalpy. The factors that influence the ionisation enthalpy are: 1. The size of the atom 2. The charge on the nucleus 3. How effectively the inner electron shells screen the nuclear charge?- shielding effect 4. The type of electron involved (s.p.d or f) In a small atom the electrons are tightly held, whilst in a larger atom, the electrons are loosly held. Thus the I.E decreases as the size of the atoms increases.

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First ionisation energy shows periodicity. That means that it varies in a repetitive way as you move through the Periodic Table. For example, look at the pattern from Li to Ne, and then compare it with the identical pattern from Na to Ar. These variations in first ionisation enthalpy can all be explained in terms of the structures of the atoms involved. The ionisation enthalpy also depends on the type of electron, which is being removed. s,p,d and f electrons have orbitals with different shapes. An s electron penetrates nearer to the nucleus and is therefore more tightly held than a p electron. For similar reasons a p electron is more tightly held than a d electron and d electron is more tightly held than f electron. Other factors being equal the ionisation enthalpies are in the order s>p>d>f. Thus the increase in I.E is not quite smooth on moving from left to right in the periodic table. In general, the IE decreases on descending a group and increases on crossing a period. Removal of successive electrons becomes more difficult and first IE < second IE < third IE. The first ionization enthalpies of elements having atomic numbers up to 60 are plotted in the figure. The periodicity of the graph is quite striking. Maxima is observed at the noble gases which have closed electron shells and very stable electron configurations. On the other hand, minima occur at the alkali metals and their low ionization enthalpies can be correlated with their high reactivity. In general first ionization enthalpy increases as we go across a period and decreases as we descend in a group. To understand these trends, we have to consider two factors: (i) the attraction of electrons towards the nucleus, and (ii) the repulsion of electrons from each other. The effective nuclear charge experienced by a valence electron in an atom will be less than the actual charge on the nucleus because of ―shielding‖ or ―screening‖ of the valence electron from the nucleus by the intervening core electrons. For example, the 2s electron in lithium is shielded from the nucleus by the inner core of 1s electrons. As a result, the valence electron experiences a net positive charge which is less than the actual charge of +3. In general, shielding is effective when the orbitals in the inner shells are completely filled. This situation occurs in the case of alkali metals which have a lone ns-outermost electron preceded by a noble gas electronic configuration. On moving from lithium to fluorine across the UDOM- Study Material 2013-14 Page 27

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second period, successive electrons are added to orbitals in the same principal quantum level and the shielding of the nuclear charge by the inner core of electrons does not increase very much to compensate for the increased attraction of the electron to the nucleus. Thus, across a period, increasing nuclear charge outweighs the shielding. Consequently, the outermost electrons are held more and more tightly and the ionization enthalpy increases across a period. As we go down a group, the outermost electron being increasingly farther from the nucleus, there is an increased shielding of the nuclear charge by the electrons in the inner levels. In this case, increase in shielding outweighs the increasing nuclear charge and the removal of the outermost electron requires less energy down a group.

d) Electron gain enthalpy Electron gain enthalpy is the measure of the tendency of an atom of an element to accept electron. The energy released when an extra electron is added to a neutral gaseous atom to convert it in to its anion is termed the electron gain enthalpy. The electron affinity gives the measure of the tendency of an atom to change in to anion.

Usually only one electron addition gives a uni negative ion. Since energy is evolved, these terms have a negative sign. Electron affinities depend on the size and effective nuclear charge. Negative electron affinity values indicate that energy is given out when the atom accepts an electron. The above values show that the halogens evolve a large amount of energy on forming negative halide ions.Energy is evolved when one electron is added to an O and s atom, forming the species O- and S-, but a substantial amount of energy is absorbed when two electrons UDOM- Study Material 2013-14 Page 28

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are added to form O2- and S2- ions. Thus the electron affinities to O  O2- and SS2- have a positive side. group 17 elements (the halogens) have very high negative electron gain enthalpies because they can attain stable noble gas electronic configurations by picking up an electron. On the other hand, noble gases have large positive electron gain enthalpies because the electron has to enter the next higher principal quantum level leading to a very unstable electronic configuration. In general, the electron affinity decreases in going from top to bottom in a group and increases in going from left to right across a period.

Variation of electron affinity along the period The variation in electron gain enthalpies of elements is less systematic than for ionization enthalpies. As a general rule, electron gain enthalpy becomes more negative with increase in the atomic number across a period. The effective nuclear charge increases from left to right across a period and consequently it will be easier to add an electron to a smaller atom since the added electron on an average would be closer to the positively charged nucleus. However, electron gain enthalpy of O or F is less negative than that of the succeeding element. This is because when an electron is added to O or F, the added electron goes to the smaller n = 2 quantum level and suffers significant repulsion from the other electrons present in this level. For the n = 3 quantum level (S or Cl), the added electron occupies a larger region of space and the electron-electron repulsion is much less. Variation of electron affinity down the group Electron affinity decreases down the group due to increase in the atomic size and decrease in the nuclear attraction for electrons. On moving down a group, electron gain enthalpy UDOM- Study Material 2013-14 Page 29

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become less negative because, the size of the atom increases and the added electron would be farther away from the nucleus. e) Electro negativity Electronegativity is a measure of the tendency of an atom to attract a shared pair of electrons towards itself The Pauling scale is the most commonly used. Fluorine (the most electronegative element) is assigned a value of 4.0, and values range down to cesium and francium which are the least electronegative at 0.7. The electronegativity of any given element is not constant; it varies depending on the element to which it is bound. Electronegativity generally increases across a period from left to right (say from lithium to fluorine) and decrease down a group (say from fluorine to astatine) in the periodic table. The attraction between the outer (or valence) electrons and the nucleus increases as the atomic radius decreases in a period. The electronegativity also increases. On the same account electronegativity values decrease with the increase in atomic radii down a group. The trend is similar to that of ionization enthalpy. Non-metallic elements have strong tendency to gain electrons. Therefore, electronegativity is directly related to that nonmetallic properties of elements. It can be further extended to say that the electronegativity is inversely related to the metallic properties of elements. Thus, the increase in electronegativities across a period is accompanied by an increase in non-metallic properties (or decrease in metallic properties) of elements. Similarly, the decrease in electronegativity down a group is accompanied by a decrease in non-metallic properties (or increase in metallic properties) of elements. Trends in electronegativity across a period As you go across a period the electronegativity increases. The chart shows electronegativities from sodium to chlorine - you have to ignore argon. It doesn't have electronegativity, because it doesn't form bonds.

Trends in electronegativity down a group As you go down a group, electronegativity decreases. (If it increases up to fluorine, it must decrease as you go down.) The chart shows the patterns of electronegativity in Groups 1 and 7.

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CH 111: General Chemistry

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Fluorine is the most electronegative element (electronegativity = 4.0), the least electronegative is Cesium (notice that are at diagonal corners of the periodic chart) General trends: Electronegativity is the relative tendency of a bonded atom to attract electrons to itself. Electro negativity increases from left to right along a period For the representative elements (s and p block) the electro negativity decreasesas you go down a group An atom with extremely low electronegativity, like a Group I metal, is said to be electropositive since its tendency is to lose rather than to gain, or attract, electrons For the commonly encounted elements, the order in decreasing electronegativity is F > O > N ~ Cl > Br > C ~ S ~ I > P ~ H > Si. UDOM- Study Material 2013-14 Page 31

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Periodic trends in chemical properties It has been learnt that the atomic properties like atomic radius, ionization enthalpy, electron gain enthalpy and electronegativity exhibit periodicity. The periodicity of the valence state of the elements and anomalous properties of the second period elements from Li to F is discussed in the following section. a) Periodicity of valence or oxidation states The combining capacity of an element is termed as the valency of that element. Valency of an element is linked with the number of electrons present in the outer most energy shell of its atom. The electrons in the outer most shell (valence shell) are called valence electrons. The valence (Oxidation state) of an element may be defined as the number of hydrogen or halogen atoms or twice the number of oxygen atoms which combine with one atom of the element. For example, the valency of carbon in CH4 is 4, the valency of Nitrogen in NH3 is 3, the valency of oxygen in H2O is 2 and the valency of fluorine in HF is 1. Nowadays the term oxidation state is frequently used for valence. Consider the two oxygen containing compounds: OF2 and Na2O. The order of electronegativity of the three elements involved in these compounds is F > O > Na. Each of the atoms of fluorine, with outer electronic configuration 2s22p5, shares one electron with oxygen in the OF2 molecule. Being highest electronegative element, fluorine is given oxidation state -1. Since there are two fluorine atoms in this molecule, oxygen with outer electronic configuration 2s22p4 shares two electrons with fluorine atoms and thereby exhibits oxidation state +2. In Na2O, oxygen being more electronegative accepts two electrons, one from each of the two sodium atoms and, thus, shows oxidation state -2. On the other hand sodium with electronic configuration 3s1 loses one electron UDOM- Study Material 2013-14 Page 32

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to oxygen and is given oxidation state +1. Thus, the oxidation state of an element in a particular compound can be defined as the charge acquired by its atom on the basis of electronegative consideration from other atoms in the molecule. The variation in the valence of elements is shown in the table given below.

b) Anomalous properties of second period elements The first element of each of the groups 1 (lithium) and 2 (beryllium) and groups 13-17 (boron to fluorine) differs in many respects from the other members of their respective group. For example, lithium unlike other alkali metals, and beryllium unlike other alkaline earth metals, form compounds with pronounced covalent character; the other members of these groups predominantly form ionic compounds. On moving diagonally across the periodic table the elements show certain similarities. These similarities are more pronounced in the following pairs of elements. Li Be B Mg Al Si

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Lithium is similar to Magnesium in many of its properties and beryllium is similar to aluminium. Therefore, diagonal relationship is the similarity between a pair of elements which are diagonally opposite in periodic table. So the two elements Li and Mg having diagonal relationship resemble very much in properties. The anomalous behaviour is attributed to their small size, large charge/ radius ratio and high electronegativity of the elements. In addition, the first member of group has only four valence orbitals (2s and 2p) available for bonding, whereas the second member of the groups have nine valence orbitals (3s, 3p, 3d). As a consequence of this, the maximum covalency of the first member of each group is 4 (e.g., boron can only form [BF4 ] − , whereas the other members of the groups can expand their valence shell to accommodate more than four pairs of electrons e.g., aluminium forms [AlF6 ]3−). Furthermore, the first member of p-block elements displays greater ability to form pπ – pπ multiple bonds to itself (e.g., C = C, C ≡ C, N = N, N ≡ Ν) and to other second period elements (e.g., C = O, C = N, C ≡ N, N = O) compared to subsequent members of the same group. Periodic Trends and Chemical Reactivity Periodic trends in certain fundamental properties such as atomic and ionic radii, ionization enthalpy, electron gain enthalpy and valence has been observed so far. We know by now that the periodicity is related to electronic configuration. That is, all chemical and physical properties are a manifestation of the electronic configuration of elements. We shall now try to explore relationships between these fundamental properties of elements with their chemical reactivity. The atomic and ionic radii, as we know, generally decrease in a period from left to right. As a consequence, the ionization enthalpies generally and electron gain enthalpies become more negative across a period. In other words, the ionization enthalpy of the extreme left element in a period is the least and the electron gain enthalpy of the element on the extreme right is the highest negative (note : noble gases having completely filled shells have rather positive electron gain enthalpy values). This results into high chemical reactivity at the two extremes and the lowest in the centre. Thus, the maximum chemical reactivity at the extreme left (among alkali metals) is exhibited by the loss of an electron leading to the formation of a cation and at the extreme right (among halogens) shown by the gain of an electron forming an anion. This property can be related with the reducing and oxidizing behaviour of the elements which you will learn later. However, here it can be directly related to the metallic and non-metallic character of elements. Thus, the metallic character of an element, which is highest at the extremely left decreases and the non-metallic character increases while moving from left to right across the period. The chemical reactivity of an element can be best shown by its reactions with oxygen and halogens. Here, we shall consider the reaction of the elements with oxygen only. Elements on two extremes of a period easily combine with oxygen to form oxides. The normal oxide formed by the element on extreme left is the most basic (e.g., Na2O), whereas that formed UDOM- Study Material 2013-14 Page 34

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by the element on extreme right is the most acidic (e.g., Cl2O7). Oxides of elements in the centre are amphoteric (e.g., Al2O3, As2O3) or neutral (e.g., CO, NO, N2O). Amphoteric oxides behave as acidic with bases and as basic with acids, whereas neutral oxides have no acidic or basic properties. Practice questions 1. An atom with a valence electronic configuration of 1s22s22p63s23p3 belongs to group A. 3 B. 5 C. 15 D. 17 2. Which orbital is characterized by the quantum numbers n = 3, l = 0, ml = 0? A. 3s B. 3p C. 3d D. 2s 3. Arrange the following ions in the order of decreasing size: Br-, I-, Mg2+, Ca2+ A. I-> Br-> Ca2+> Mg2+ B. Br-> I-> Ca2+ > Mg2+ C. Ca2+> Mg2+> I-> Br- D. I-> Br-> Mg2+> Ca2+ 4. The first ionization energy of beryllium is 899 KJ/mol and the second ionization energy is 1,757 KJ/mol. How much energy would it take to remove the valence electrons from three moles of beryllium? A. 5312 KJ B. 2656 KJ C. 7968 KJ D. 10704 KJ 5. Which one of the following correctly describes the trend in atomic radius? A. Increases across a period and decreases down a group B. decreases period and decreases down a group C. Increases across a period and increases down a group D. decreases period and increases down a group 6. Which one of the following elements has the highest second ionization enthalpy? A. Carbon B. Nitrogen C. Oxygen D. Fluorine 7. The electron affinity of chlorine is 349 KJ/mol. What is the correct equation for the formation of chloride? A. Cl (s) + e- → Cl- (s) + 349 kJ B. Cl (s) + e- → Cl- (s) - 349 kJ C. Cl (s) + 349 kJ + e- → Cl- (s) D. Cl (g) + 349 kJ + e- → Cl- (g) 8. The correct ordering of the electronegativities of the following atoms is A. O C6H5COOH > CH3COOH > C6H5OH

2. Decreasing order of basic character P b values : Bases k k k

k

3.22

3.35

4.22

4.76

9.37

: k (CH3)2 NH > CH3NH2> (CH3)3N

> NH3> C6H5NH2

3. P a + P b = 14 P a of acetic acid = 4.7 . P b of it's conjugate base CH3COO- is calculated as follows: P b = 14 - P a = 14 - 4.7 = 9.3 Ionic Product of Water Water is a weak electrolyte and does not ionize much in the pure state. Water undergoes self ionization as shown below. Applying the law of mass action to this equilibrium or K[H2O] = [H+] [OH-] Since water undergoes ionization to a very small extent, [H2O] may be regarded as constant. K[H2O] is a constant. Thus we have The constant KW is known as ionic product of water. KW = 10-14 mol2/L2 at 298 K. Thus in pure water at 298 K. [H+] [OH-] = 10-14 [H+] = [OH-] = At 25oC UDOM- Study Material 2013-14 Page 77 k k

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[H+] = 10-7 mol/dm3 [OH-] = 10-7 mol/dm3 KW = 10-7.10-7 = 10-14 mol2/(dm3)2 The ionic product of water varies with temperature. Concentration of H+ and OH- ions in aqueous solution of Acids and Bases When an acid or base is added to pure water, the concentration of H+ions and OH- ions changes but the value of ionic product of water i.e., KW remains unchanged so long as the temperature remains constant.{ K W [H ] [OH ] = 10-7.10-7 = 10-14 mol2/(dm3)2} Addition of an acid in water decreases the [OH-] according to the relation. KW [OH ] [H ] Addition of base in water decreases [H+] according to the relation KW [H ] [OH ] + Thus H and OH ions are always present in an aqueous solution no matter whether it is neutral, acidic or basic. However their relative concentrations are different in different types of solutions. In general, In neutral solutions; [H+] = [OH-] In acidic solutions; [H+] > [OH-] In basic solutions; [H+] < [OH-] Practice questions 1. Which of the following acid act both as Arrhenius and Bronsted-Lowry acid? a) CO2 b) BF3 c) HNO3 d) AlCl3 2. The conjugate base of H2PO4– is, a) PO43 – b) P2O5

c) H3PO4

d) HPO42 –

3. Which one of the following is not a conjugate acid – base pair? a) HCl and Clb) H2CO3 and HCO3c) H2SO4 and SO424. The pOH of 0.0001M HCl is, a) 4 b) 5

d) HPO42 – and PO43-

c) 10

d) 12 d) 9.5

5. The pKa value of the strongest acid among the following is, a) 4.7 b) 3.6 c) 5.2

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CH 111: General Chemistry 2.2 Mole concept

Dr. Ananda M

A mole is a unit which is used to express the amount of substance. In general one mole of any substance contains Avogadro number of (6.022 x 1023) atoms or molecules. Mole is defined as the amount of substance which contains Avogadro number of particles. 6.022 x 1023, is called Avogadro‘s number (represented by NA), named in the honour of an Italian scientist Amedeo Avogadro. Atomic mass unit is defined as a mass exactly equal to one-twelfth the mass of one carbon -12 atom. One atomic mass unit also called one Dalton. Mass of one mole C-12 = 12 g = mass of 6.023 × 1023C-12 atom Mass of one C-12 atom = 12/6.023 × 1023 g = 12 × 1.66 × 10–24 = 12 amu(1 amu or 1 dalton = 1.66 × 10–24 g). The mass of a hydrogen atom was designated as 1 amu. Mole in terms of mass The mole is the amount of substance (Elements or compounds) which has a mass equal to its gram atomic mass or gram molecular mass. Ex.1. One mole of oxygen atoms = 16 g (0ne gm. atomic mass). Ex.2. One mole of oxygen molecule = 32 g (gm mass) If weight of a given species is given, then

No. of moles( for atoms)

weight Atomic weight

or 1. Calculate the number of number of moles of oxygen atoms present in 300 g of O2.

No. of moles( for atoms)

300 18.75 16

2. If 1 mole of CaSO4 has a mass of 136.2 g, what is the mass of 0.5 moles of CaSO4? Weight = No. of moles x Mol. Wt. = 136.2 x 0.5 = 68.1 g CaSO4 3. How many moles of CaSO4 are there in 88.55 g of CaSO4?

No. of moles( for molecules )

weight 88.5 0.6501moles CaSO4 Molecularweight 136.2

4. What weight of H2O is equivalent to 10 moles?

Weight = No. of moles x Mol. Wt. = 10 x 18 = 180 g
5. Calculate the no. of moles in: i) 60g of Ca (gram atomic mass of Ca = 40g)

No. of moles of Ca

Mass of Ca Gram atomicmass

60 1.5 mol 40
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Mole in terms of number Mole is a unit of chemists to count atoms, molecules, ions, electrons, protons, neutrons, Chemical bond etc. and one mole of any substance is equal to 6.022×1023 particles of that substance. One mole of substance contain one Avogadro‘s number. 1 gm mole of hydrogen atom contains 6.022 x 1023 hydrogen atoms. 1 gm mole of Hydrogen molecule contains 6.022 x 1023 hydrogen molecule. Molecular mass of water (H2O) is 18. One mole of water (H2O) contain 6.022 x1023 molecules of water. One mole of Na = 6.022×1023atom of Na One mole of H2O = 6.022×1023molecule of H2O One mole of CO32–= 6.022×1023ions of CO32–

No. of moles( for atoms)

No. of atoms NA No. of molecules No. of moles( for ions) NA No. of ions NA

No. of moles( for molecules )

1. Calculate the number of Fe atoms present in 100 g of Fe specimen. No. of atoms of Fe = No. of moles x NA

Mass NA Gram atomic mass 100 6.023 1023 = 10.78 × 1023 = 55.85
= 2. h No. of K2Cr2O7 molecule = No. of moles x NA

Mass NA Gram Molar mass 50 6.023 1023 = 294
= = 1.024× 1023 Therefore, No. of K atoms = 2 x 1.024× 1023 = 2.048 x 1023 No. of Cr atoms = 2 x 1.024× 1023 = 2.048 x 1023 No. of O atoms = 7 x 1.024× 1023 = 7.168 x 1023 UDOM- Study Material 2013-14 Page 80

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Mole in terms of Volume One mole of gas under standard temperature (273K) and Pressure 1 atm (STP) occupies 22.4 dm3 of volume. A mole of gaseous substance can also be defined the amount of substance that can occupy the volume of 22.4 dm3 at STP or 0.0224 m3 No. of moles = given mass in gm gram molecular mass OR

No. of moles = given mass in gm gram atomic mass 1 mole = 6.022 x 1023 particles = 1gm Molecular mass = 22.4 dm3 at STP (i) One mole NH3 = 22.4 litres NH3 at N.T.P. (Molar volume or Mol. Gm. Vol.) = One gram molecule NH3 = 17 g NH3 = 6.022×1023 molecules of NH3. (ii) One mole of He = 22.4 litre He at N.T.P. (Molar volume or gram atomic volume) = One gram atom He = 4 g He = 6.022×1023 atoms of He. (iii) One mole of Hydrogen = 22.4 litres hydrogen at N.T.P. (Molar volume or gram molecular volume) = 2 g hydrogen = 6.022×1023 molecules of hydrogen = 2×6.022×1023 atoms of hydrogen.

2.3 Methods of expressing concentration of the solutions There are a number of ways to express the relative amounts of solute and solvent in a solution. The following five ways of expressing the concentration are the important ones, 1. Molarity (M) 2. Molality (m) 3. Percentage by mass or volume (% wt. or % vol) 4. Mole fraction (X) 5. Parts per million / parts per billion (ppm or ppb) UDOM- Study Material 2013-14 Page 81

CH 111: General Chemistry 1. Molarity (M)

Dr. Ananda M

Molarity (M), or molar concentration, is a common unit for expressing the concentrations of solutions. Molarity is defined as the number of moles of solute per litre of solution:

Molarity

Moles of solute Volume of solutionin litres where weight of solute Molecular weight

Moles of solute

To prepare one litre of a one molar solution, one mole (gram molecular weight) of solute is placed in a one litre volumetric flask, enough solvent is added to dissolve the solute, and solvent is then added until the volume of the solution is exactly one litre. To calculate the molarity of any solution, the following formula is used,

Molarity

Weight of solute 1 Molecular weight Volume of solutionin litres

If you are using 250 ml volumetric flask for the preparation of a standard solution, the above formula can be written as

Molarity

Weight of solute 1000 Molecular weight 250

Molarity

Weight of solute 4 Molecular weight

1. What is the molarity of NaOH in a solution which contains 24.0 gram NaOH dissolved in 300 ml of solution?

Molarity

Weight of solute 1000 Molecular weight Volume of solutionin ml

Molarity

24 1000 40 300

Molarity 2
2. Calculate the molarity of solution obtained by the dissolution of 180 gm of K2Cr2O7 in 0.25 liters of solution.

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Molarity

Weight of solute 4 Molecular weight

Molarity

180 4 294

Molarity 2.44
3. How many ml of water must be added to 200 ml of 0.65 M HCl to dilute the solution to 0.20 M?

Molarity volume 1 0.2 volume 1 volume 1

Molarity volume 2

0.65 200 2

0.65 200 0.2

volume 1

0.65 200 0.2

volume 650 ml 1
Volume of water that must be added = 650 ml – 200 ml = 450 ml 4. What is the molarity of a solution prepared by dissolving 37.94 g of potassium hydroxide in some water and then diluting the solution to a volume of 500 ml?

Molarity

Weight of solute 1000 Molecular weight 250

Molarity

37.94 1000 56 500

Molarity 1.355
5. Calculate the amount of EDTA required for preparing 2 liters of 0.75M EDTA solution.

Molarity

Weight of solute 1 Molecular weight Volume of solutionin litres

Weight of EDTA Molarity Molecular weight Volume of solutionin L Weight of EDTA 0.25 372 2
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Weight of EDTA 186 g
We often express the volume of a solution in milliliters rather than in litres. Likewise, we may express the amount of solute in millimoles (mmol) rather than in moles. Because one milliliter is 1/1000 of a litre and one millimole is 1/1000 of a mole, molarity also may be expressed as the number of millimoles of solute per millilitre of

Molarity

Milli moles of solute Volume of solutionin milli liters where weight of solute 1000 Molecular weight

No. of milliMolesof solute

Water is the solvent in most of the solutions, until other solvent not indicated.

2. Molality (m) Molality is defined as the number of moles of solute per kg of solvent. 1m solution of a solute is prepared by dissolving one mole (Gram molecular weight) of a solute in one Kg of solvent.

Molality

Moles of solute Weight of solvent in Kg where weight of solute Molecular weight

Moles of solute

To prepare one molal solution of a solute, one mole (gram molecular weight) of solute is dissolved in one kg of solvent.

Molality

Weight of solute 1 Molecular weight Weight of solvent in Kg

1. Determine the molal concentration of a solution containing 81.3 g of ethylene glycol dissolved in 166g of water.

Molality

Weight of solute 1 Molecular weight Weight of solvent in Kg

Molality

81.3 1 62.08 166 10 3

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Molality 7.89
2. What is the molality of solution of 560 g of acetone in 620 g of water?

Molality

Weight of solute 1 Molecular weight Weight of solvent in Kg

Molality

560 1 58 620 10

3

Molality 15.6
3. Percentage by mass or Volume (% wt. or % vol) The concentration of a solution is often expressed as percent concentration by mass or percent by volume of solute in solution. Percent by mass is calculated from the mass of solute in a given mass of solution. A 5%-by-mass aqueous solution of sodium chloride contains 5 g sodium chloride and 95 g water in each 100 g solution.

Percent by mass

Mass of solute 100 0 0 Mass of solution

1. How many grams of glucose and of water are in 500 g of a 5.3% by-mass glucose solution Solution We know that 5.3% of the solution is glucose:

The remainder of the 500 g is water: If both solute and solvent are liquids, the concentration may be expressed as percent by volume. Both ethyl alcohol and water are liquids; the concentration of alcohol-water solutions is often given as percent by volume. For example, a 95% solution of ethyl alcohol contains 95 mL ethyl alcohol in each 100 mL solution.

Percent by Volume

Volume of solute 100 0 0 Volume of solution

Because the density of liquids changes slightly as the temperature changes, a concentration given in percent by mass is accurate over a wider range of temperatures than is a concentration given in percent by volume. Sometimes a combination of mass and volume is used to express the UDOM- Study Material 2013-14 Page 85

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concentration--the mass of solute dissolved in each 100 mL solution. Using this method, a 5% (wt/vol) solution of sodium chloride contains 5 g sodium chloride in each 100 mL solution.

Way of expression Symbol of Concentration Molarity M

Meaning

Preparation

Moles solute per 5M: Dissolve 5 mol of liter of solution solute in 1 L of a solution Moles solute per 5m: Dissolve 5 mol of kilogram of solvent solute in 1 kg of solvent Grams solute per 5%: Dissolve 5 g of solute 100 g of solution in 95 g of solvent

Molality

m

Percentage

%

4. Mole Fraction The mole fraction, X, of a component in a solution is the ratio of the number of moles of that component to the total number of moles of all components in the solution. To calculate mole fraction, we need to know: The number of moles of each component present in the solution. The mole fraction of A, XA, in a solution consisting of A, B, C, ... is calculated using the equation:

To calculate the mole fraction of B, XB, use:

5. Concentration in Parts per Million (ppm) and parts per Billion (ppb) The terms (ppm) and parts per billion (ppb) are encountered more and more frequently as we become aware of the effects of substances present in trace amounts in water and air, and as we develop instruments sensitive enough to detect substances present in such low concentrations. UDOM- Study Material 2013-14 Page 86

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In discussing mass, parts per million means concentration in grams per 106 grams, or micrograms per gram. For example, total hardness of water is expressed in ppm of CaCO3. If the hardness of water is 200 ppm of CaCO3, it means that the hard water contains 200 parts of CaCO3 in 106 (million) parts of H2O. In discussing volume, parts per million means milliliters per cubic meter, or the mixed designation of milligrams per cubic meter. For parts per billion, the general trend is toward the use of micrograms per liter when discussing water contaminants, micrograms per cubic meter for air, and micrograms per kilogram for soil concentrations. Practice questions 1. The number of water molecules present in 9 grams of H2O is, a) 6.022×1023 b) 3.011×1023 c) 12.044×1023 d) 6.022×1024

2. The number of moles of oxygen molecules present in 320 grams of oxygen gas is, a) 2 b) 10 c) 20 d) 30 3. The molarity of a solution obtained by dissolving 400g of NaOH in 1 litre of water is, a) 1 b) 10 c) 100 d) None 4. During reduction, the oxidation number of the chemical species, a) increases b) decreases c) remains same

d) None

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2.4 Red-ox reactions Oxidation-Reduction (REDOX) reactions are among the most important chemical reactions. Redox reactions are involved in a wide variety of important natural processes including the browning of foods, respiration of animals and rusting of metals and alloys. Chemical reactions involving transfer of electronsoxidation and reduction- are called Red-ox reactions. Red- ox reactions involve both oxidation and reduction simultaneously. In general, when any reaction occur oxidation of some chemical species takes place and at the same time certain other chemical species are reduced. The joint processoxidation and reduction- is called Red-ox reaction. Oxidation is the process of removal of electrons. During oxidation, electrons are lost from the chemical species. For example, Iron reacts with chlorine to give ferric chloride. In this reaction iron is undergoing oxidation from Fe0 to Fe3+ state. It is also confirmed that, during oxidation the oxidation state of iron increases from 0 to 3+. Iron has lost 3 electrons in this reaction and the same 3 electrons are added to chlorine. Thus iron acts as reducing agent and reduces chlorine to chloride but itself gets oxidized.

2 Fe

3 Cl2

2 FeCl3

On the other hand chlorine involves reduction. Reduction is the process of addition of electrons. During reduction, electrons are added to the chemical species. In the above reaction, chlorine is being reduced by iron form the oxidation state 0 to -1. Chlorine has accepted 1 electron to become chloride anion. Thus chlorine acts as oxidizing agent and oxidizes Fe0 to Fe3+.

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Recognizing Oxidizing and Reducing Agents - I Problem: Identify the oxidizing and reducing agent in each of the Reactions: a) Zn(s) + 2 HCl(aq) → ZnCl2 (aq) + H2 (g) b) S8 (s) + 12 O2 (g) → 8 SO3 (g) c) NiO(s) + CO (g) → Ni(s) + CO2 (g) Step 1: Assign oxidation numbers (O.N.) to each atom (or ion) based on the rules in Table.

Step 2: Find out oxidation reaction and Reducing agent. An atom, ion or molecule which involves oxidation accompanies increase in its oxidation state and thus acts as reducing agent. In

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this case Zn has undergone oxidation and therefore Zn is the oxidizing agent. The reactant is the reducing agent if it contains an atom that is oxidized (O.N. increased in the reaction). Step 3: Find out reduction reaction and oxidising agent. An atom, ion or molecule which involves reduction accompanies decrease in its oxidation state and thus acts as oxidising agent. In this case HCl has undergone reduction and therefore HCl is the reducing agent. The reactant is the reducing agent if it contains an atom that is oxidized (O.N. increased in the reaction).

Zn

2 HCl

ZnCl2

H2

HCl is the oxidizing agent and Zn is the reducing agent. On the similar logic, S8 and CO are reducing agents and O2 and NiO are oxidizing agents. 2.5 Balancing REDOX reactions by the method of half reactions Problem 1: Balance the following equation.

Fe2

Cr2O7 2

2Cr 3

Fe3

1. Oxidation reaction: multiply oxidation reaction equation by 6 as the observed change in the oxidation state of two Cr atoms in a molecule is 3.

Fe2 6Fe2

Fe3 6Fe3

1emultiplyby 6 6eequation 1

2. Reduction reaction: multiply reduction reaction equation by 1 as the observed change in the oxidation state of Fe is 1.

Cr2O7 2

6e

2Cr 3 equation2

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6Fe2
Cr2O7 2 6Fe2

6Fe3
6e Cr2O7 2 Cr2O7 2

6eequation 1
2Cr 3 equation2 2Cr 3 6Fe3

3. Balance oxygen by the addition of 7 H2O molecules to the right of the equation.

6Fe2

2Cr 3

6Fe3

7H 2O

4. Balance hydrogen by the addition of 14 H+ ions to the left of the equation.

6Fe2

Cr2O7 2

14H

2Cr 3

6Fe3

7H 2 O

Problem 2: Balance the equation

Fe2

MnO4 Cu2

Mn2 NO Mn2

Fe3

Cu HNO3 Problem 3:Balance the equation
Problem 4: Balance the equation MnO 4

C2O4 2

CO2

Problem 5: Balancing redox equations in basic solutions: First balance in acidic solution, then add OH- to cancel H+. Cr2O72- (aq) + 2NO (g)+ 6H+ (aq) → 2Cr3+ (aq) + 2NO3- (aq)+3H2O(l) Add OH- equal to number of H+ on both sides 6OH- (aq) + Cr2O72- (aq) + 2NO (g) + 6H+ (aq) → 2Cr3+ (aq) + 2NO3- +3H2O (l)+ 6OH-(aq) Combine OH- and H+ to form water to maximum extent possible 6H2O (l) + Cr2O72- (aq) + 2NO (g) → 2Cr3+(aq)+2NO3-+3H2O(l) +6OH- (aq) Cancel H2O on both sides to max extent possible 3H2O (l) + Cr2O72- (aq) + 2NO(g) →2Cr3+ (aq)+2NO3- +6OH- (aq)

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Balancing redox equation using half-cell method in acidic solutions

2.6 Basics of electrochemistry All chemical interactions are electrical at the atomic level so that in a sense all chemistry is electrochemistry. Electrochemistry provides an important way of utilizing the free energy available in spontaneous chemical reactions to perform useful work. Electrical work can be derived at the cost of chemical energy. Electrical energy can be applied to bring about the desired chemical change. The inter-conversion of chemical energy into electrical energy and vice-versa is extensively studied in a separate branch of chemistry known as electrochemistry. Electrochemistry deals with the conversion of chemical energy into electrical energy and viceversa. Electrochemistry is the area of chemistry dealing with inter-conversion of electrical energy and chemical energy of redox reactions. There are many applications of this in every-day UDOM- Study Material 2013-14 Page 92

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life. Batteries, control of corrosion, metallurgy and electrolysis are just a few examples of the applications of electrochemistry. An electrochemical cell converts chemical energy into electrical energy. Electrical energy is utilized to bring about chemical energy changes in an electrolytic cell.
Electrochemical cell

Chemical Energy
Electrolytic cell

Electrical Energy

Electrolytes and Non-electrolytes Water soluble chemical compounds can be classified into following categories. 1. Electrolytes 2. Non-electrolytes 1. Electrolytes: Electrolytes are the substances which dissociate into ions in aqueous solution and conduct electric current. Ex. NaCl, KCl etc. Electrolytes can further be classified into two types: a) Strong electrolytes Strong electrolytes are the substances which undergo complete ionization into ions in aqueous solution. Ionisation takes place to an extent of 100%. Example:
NaCl
H 2O

Na(aq)

Cl(aq)

KCl HCl
NaOH

H 2O H 2O
H 2O

K(aq) H(aq)

Cl(aq) Cl(aq)
OH(aq)

Na(aq)

b) Weak electrolytes Weak electrolytes are the substances which undergo incomplete [partial] ionization in aqueous solution. Example: CH3COOH .NH4OH etc.
CH3COOH CH3COO H

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NH4OH

NH4 OH

Always there exists an equilibrium between the undissociated molecule and dissociated ions of molecule in aqueous solution. Chemical equilibrium is a characteristic of reversible reaction. 2. Non-electrolytes Non-electrolytes are the substances which dissolve in water but do not dissociate in solution and do not conduct electric current. Examples: Sugar, urea, glucose, glycerine etc. Electrolysis Electrolysis is the process of decomposition of an electrolyte due to the flow of an electric current through the solution of an electrolyte. For example, when an electric current is passed through the solution of sodium chloride, sodium chloride decomposes liberating chlorine at the anode and sodium at the cathode. Chemical reactions that occur during electrolysis are given below,

(aq) 1. 2. At cathode [Reduction]
Na e Na 3. At anode [oxidation] 1 Cl Cl2 e 2

NaCl

H 2O

Na

Cl(aq)

Faraday's Laws of Electrolysis Michael Faraday (1832) discovered the relationship between the quantity of electricity passing through an electrolyte and the amount of material liberated at the electrodes. Faraday's 1st law of electrolysis can be stated as follows: "The amount of substance liberated or deposited at a particular electrode during

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electrolysis is directly proportional to the quantity of electricity passed in the solution"

If 'W'g of a substance get deposited at a particular electrode on passing 'Q' coloumbs electricity , then according to the first law of electrolysis.

of

W Q but Q = I x t W I t

[charge = current x time]

W=ZxIxt Where Z is the electrochemical equivalent of the substance If I = 1 ampere t = 1 second, then W = Z Electrochemical equivalent may be defined as the amount of substance deposited by a current of one ampere passed for one second 1. One mole of electrons (6.23 x 1023) carries a charge of 96500 coulomb. 2. One coulomb of current can deposit one electrochemical equivalent weight of the electrolyte. One equivalent of any substance can be liberated at a particular electrode only when the corresponding species either gains or loses one mole of electrons. At anode [Oxidation]
B(aq)
1 equivalent

B

e

1 equivalent 1 mole of electrons

At cathode [Reduction] A( aq) e A

1 equivalent 1 equivalent Since one equivalent of any substance requires one Faraday of electricity to get liberated at a particular electrodes, one Faraday must be the charge carried by one mole of electrons. 1 Faraday = charge carried by one mole of electrons = charge carried by an electron x Avagadro number = 1.602 x 10-19 x 6.023 x 1023 C mol-1 1 Faraday = 96488.5 C mol-1 If an electrode reaction involves 'n' moles of electrons, the total quantity of electricity (Q) involved in the reaction is given by Q = nF Thus knowing the number of moles of electrons involved in an electrode reaction, the quantity of electricity required for the reaction can be determined. The mass of substance produced in the reaction can be calculated with the help of Faraday's first law of electrolysis.
Faraday's Second Law of Electrolysis It can be stated as follows:

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"When the same quantity of electricity is passed through the solutions of different electrolytes, the masses of the substances liberated or deposited at the electrodes are directly proportional to their equivalent masses".

On passing electric current, electrolysis takes place in these three cells and deposition of Cu and Ag occur. If the same current is passed for the same time, it is found that mass of copper deposited Equivalent mass of copper mass of silver deposited Equivalent mass of silver

If W1 and W2 are the masses of the substance liberated by passing the same quantity of electricity and E1 and E2 are their equivalent masses respectively, then according to Faraday's second law of electrolysis W 1 E1 W2 E2 …(1) If Z1 and Z2 are the electrochemical equivalents of the two substances and I is the current passed for 't' seconds. We have W1 = Z1It and W2 = Z2It Therefore equation (1) can be written as Z1It E1 Z2It E 2 Z1 E1 or E1 Z1 Z2 E 2 E 2 Z2 orE = F x Z F Faraday's constant = 96500 coulombs
Equivalent mass Faraday' s constan t Equivalent mass 96500 g/C

Electrochemical equivalent Z

E F

Electrochemical equivalent is expressed in the units of gC-1 Note :1. Electron reduces and deposits 1M+ ion at an electrode UDOM- Study Material 2013-14 Page 96

CH 111: General Chemistry 2. n mole of electrons reduce 1 mole of Mn+ ions 3. 1 mole of electrons reduces1/n mole of Mn+ ions

Dr. Ananda M

Electrode Potentials and Cells When a metal is placed in a solution containing its own ions a potential difference is established between the metal and its own ions in equilibrium with it at the metal solution interface. This potential difference is termed as the single electrode potential. Single electrode potential (E) Single electrode potential is defined as the potential developed at the interface between the metal (M) and solution, when it is in contact with a solution of its ion. Oxidation Potential Oxidation is a process of removal of electrons from chemical species oxidation potential is a measure of the tendency of the metal to lose electrons and become cation.
M Mn ne Electrode potential is denoted as EM / Mn

Example
Zn Zn2 2e; E
Zn / Zn 2

Higher the oxidation potential of a metal, the greater will be its tendency to oxidize or ionize. Metals with higher oxidation potentials are more reactive and acts as good reducing agents. Reduction Potential Reduction is a process of oxidation of electrons to chemical species. Reduction potential is a measure of the tendency of metal ion to accept electrons and get deposited as metal.
Mn ne M Electrode potential is denoted as EMn / M

Example
Cu 2 2e Cu; E
Cu 2 / Cu

The higher reduction potential of metal is the indication of its lower reactivity.
2 Oxidation potential of Zn / Zn is +0.76V 2 Reduction potential of Zn / Zn is -0.76 V The magnitude of the electrode potential depends up on. 1) The nature of the electrode 2) The nature of the ions in solution 3) The concentration of the ions in solution 4) The temperature and 5) Pressure in the case of gas electrodes

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Origin of Electrode Potential Consider a metal M dipped in a solution containing it's ions Mn+. Here metal M exhibits the following two tendencies 1) The metal shows the tendency to go into the solution as metal ions by loosing electrons.
M Mn Mn ne ; Oxidation reaction M; Reduction reaction

2) The metal ions in the solution show the tendency to get deposited as metal atoms. ne These two opposite tendencies of metal results electrode, when contact with its ions leads to charge separation i.e., electrical double layer formation and results in development of an electrode potential. Helmholtz double layer is visualized as a parallel plate capacitor with a thickness of approximately the radius of metal ion. Outside the HDL there is a thin diffused region of ions called Guoy-champman layer. The potential difference across the HDL is the cause for the electrode potential.

Standard Electrode Potential [Eo] It is the electrode potential development at the metal and solution interface, when metal is in contact with it's ionic solution of unit concentration at 298K. Formation of a Cell A metal in contact with the solution of its own ions constitute a single electrode or half cell. A cell is formed by the combination of two half cells. The difference between the potentials of the two half cells constitutes the electromotive (emf) force of the cell. EMF of a cell causes the flow of current from one half cell (High potential) to other half cell (low potential). EMF of a cell depends on the nature of the electrode temperature and concentration of the electrolyte solution. Electrochemical Cells: Conventions & Notations An electrochemical cell is a device which involves interconversion of chemical energy and electrical energy. There are two types of electrochemical cells. 1) Galvanic cell converts chemical energy into electrical energy. Ex. Daniel cell. 2) Electrolytic cell converts electrical energy into chemical energy. UDOM- Study Material 2013-14 Page 98

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Daniel Cell Daniel Cell consists of two electrodes of dissimilar metals, Zn and Cu. Zinc and copper rods are dipped in 1 molar solution of Zinc sulphate and copper sulphate respectively.

The Zinc and copper half cells are connected internally by a salt bridge and externally by a wire through voltmeter. 1) Sign of the electrode The metal electrode or half cell at which oxidation occurs is given a negative sign and called anode.Anode is a negative electrode at which electrons are generated.
0.76 V Zn The metal electrode or half cell at which reduction occurs is given a positive sign and called cathode.Cathode is positive electrode at which electrons are absorbed by metal cations. Zn Zn2 E0 2e
Zn2

Cu2

E0 2e Cu

Cu2

Cu

0.34V

2) Cell Representation The metal electrode at which oxidation occurs is written at the leftside and reducing metal electrode is written at right side. salt bridge Left side electrode anode 3) Cell reactions At anode: oxidation reaction [Loss of electrons] Zn Right side electrode Zn(s) / Zn2+ (aq) || Cu2+(aq) / Cu (s) cathode

Zn2

2e

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At cathode: Reduction reaction [Gain of electrons] Cu

2

2e

Cu

2 Overall reaction: Zn Cu 4) EMF of a cell

Cu Zn2

EMF of a cell is defined as the algebraic difference between the reduction potentials of the cathode and anode. o Ecell o o o Ecathode Eanode Ecell Or o ERHE o ELHE

Eo 2 cu Cu

Eo 2 Zn Zn

[ 0.34 ( 0.76)] V

= 1.1 V Function of salt bridge The use of salt bridge provides the electrical contact between the two solutions and prevents the establishment of liquid junction potential. In the absence of salt bridge, the emf of the cell includes liquid junction potential also. Salt bridge consists of KCl or NH4NO3, whose ions have almost same migration velocities. The positive and negative ions of the salt in the salt bridge migrate with equal speed into cathode and anode compartments respectively, there by avoiding liquid junction potential. Distinguishing factors between Galvanic cell and Electrolytic cell Galvanic cell Electrolytic cell 1 It does not require a source of external It requires a source of electrical energy energy 2 It converts chemical energy to electrical It converts electrical energy into energy. chemical energy. 3 The redox reaction is spontaneous The redox reaction is non-spontaneous 4 Each electrode is dipped in its own Both the electrodes are immersed in the ionic solution in two separate same electrolyte solution (single compartments. compartment). 5 Salt bridge is required. No salt bridge is required. 6 Cathode is positive electrode and anode Cathode is negative electrode and anode is negative electrode. is positive electrode.

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NERNST Equation for Electrode Potential NERNST equation relates electrode potentials and concentration of ions at a given temperature. The equation is 2.303 RT E Eo log10 [Mn ] nF where,E - electrode potential, Eo - standard electrode potential R - gas constant = 8.314 JK-1 mol-1, T - Temperature n - Number of electrons involved in the reaction. n+ [M ] - Concentration of metal ions, Mn+ F - Faraday = 96500 coulombs Substituting the above values, nernst equation reduces to 0.0591 E Eo log10 [Mn ] n E becomes equal to Eo, when [Mn+] is unity.

Factors affecting electrode potential The electrode potential depends on the following factors. 1. Nature of the metal electrode 2. Concentration of the metal ions in solution and 3. Temperature Nernst equation for galvanic cell is given as follows: [oxidisedspecies ] 0.0591 E cell E o log10 cell n [Re ducedspecies ] For Daniel Cell
E cell Eo cell 0.0591 [Zn 2 ] log10 n [Cu 2 ]

Problems: 1. Calculate the potential of zinc electrode which is in equilibrium with zinc half solution at 25oC. Solution: o EZn

0.76 V ,
Eo Zn

[Zn2 ]

0.01 M eo Zn 0.76
Page 101

E zn

0.0591 log [Zn2 ] n

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0.0591 log10 10 2 2 = [Mn+] = [Zn2+] = 0.01n = -0.76V - 0.0591 T = 25oC EZn = -0.8191 V 2. Calculate the electrode potential of silver electrode in contact with Ag+ ions of concentration 0.76V
0.025 M Solution:

[E0 Ag 0.80V]
0.80V

Eo Ag

E Ag

o E Ag

0.591 log10[ Ag ] 1

[Ag+] = 0.025

EAg

= 0.80 + 0.591 log [0.025] = 0.80 + 0.591 (-0.6021) = 0.80 - 0.036 = 0.764 V

EMF of a cell and its relation to free energy change Thermodynamics demonstrates a fundamentals relationship between the change in free energy G, of a spontaneous chemical reaction at constant temperature and pressure and the maximum electrical work that such a reaction is capable of producing. - G = WElectrical work The maximum electrical work available from the electrode is nFEjoules, hence free energy decrease of the system is given by - G = nFE Under standard conditions - Go = nFEo

ne M For a reaction The equilibrium constant KC is related to change in free energy by the equation.
G Go RT ln KC

Mn

Reference electrode: Standard hydrogen Electrode [SHE] Standard hydrogen electrode is a primary reference electrode. It is used to measure the (electrode) potential of metal electrode. 1. Cell representation: SHE is represented as follows: Pt/H2 (1 atm) / HCl [H+] = 1M 2. Cell construction

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SHE is constructed by immersing a platinised platinum foil, into the solution containing 1M HCl. Platinum foil is sealed into a glass tube by means of platinum wire. External electrical connection is establishment by using mercury and copper wire. Hydrogen gas is bubbled through the HCl solution with the help of jacket tube. 3. Working of hydrogen electrode When a pure hydrogen gas is passed through the solution, platinum foil absorbs hydrogen. The adsorbed film of hydrogen comes into contact with hydrogen ions of solution and an equilibrium is established.

2H

2e

H2

This equilibrium is responsible for the potential of the hydrogen electrode. 4. Potential of hydrogen electrode The potential of S.H.E is taken as zero volt by convention at all temperature. 0.0591 E H 2 Eo log10 [H ] H2 n
Eo H
2

0

n

1

EH2

0.0591log10[H ]

5. Use of SHE Standard H.E is used to determine the standard electrode potentials of other electrodes. Limitations: 1.Construction and working is difficult 2. Poisoning of platinum by impurities in H2 gas 3. It cannot be used in the presence of oxidizing agents Measurement of single electrode potential The potential of a given electrode is measured using the standard hydrogen electrode (SHE) whose potential is arbitrarily taken as zero volts at all temperatures and is the standard reference for all potential measurements. Measurement of electrode potential of zinc electrode involves three steps. 1) The electrode whose potential is to be determined is coupled with standard hydrogen electrode. Zn / Zn2+ || HCl, [H+] = 1 / H2 (1 atm) / Pt Zinc electrode SHE 2) Assignment of signs to electrodes Zn electrode  Anode  -Ve SHE  Cathode  +Ve The potential of the cell is found to be 0.76 V UDOM- Study Material 2013-14 Page 103

CH 111: General Chemistry o Ecell o ERHE o ELHE

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3) EMF of Cell

Eo 2 Eo 2 0.76V 0.76 = 0 - Zn / Zn Zn / Zn Standard hydrogen electrode is primary reference electrode. Example 1: Calculate the electrode potential at a copper electrode dipped in a 0.1 M solution of copper sulphate at 25o C. The standard electrode potential of Cu2+/Cu system is 0.34 volt at 298 K. Solution: We know that Ered = Eored + 0.0591/n log10[ion] Putting the values of Eored =0.34 V, n = 2 and [Cu2+]= 0.1 M Eored= 0.34+0.0591/2 log10[0.1] = 0.34 + 0.02955 × (-1) = 0.34 - 0.02955 = 0.31045 volt
Example 2: Calculate the standard electrode potential of Cu2+/Cu if its electrode potential at 250C is 0.296V when [Cu2+] = 0.015 E = Eo +0.0591 log10 [Mn+] n Eo = E - 0.0591 log10 [Cu2+] n E0 = 0.296 E0= 0.349V
0.0591 2

log100.015

Example 3: What is the single electrode potential of a half-cell for zinc electrode dipping in 0.01 M ZnSO4 solution at 25o C? The standard electrode potential of Zn/Zn2+ system is 0.763 volt at 25o C. Solution: We know that E = Eored+ 0.0591/n log10[ion] substituting Eoox=0.763 V, (Eored = -0.763V), n=2 and [Zn2+]=0.01 M E = -0.763 + 0.0591/2 log [0.01] = -0.763 + 0.02955 × (-2) = (-0.763 - 0.0591) volt = -0.8221 volt Problems on Nernst Equation 1. Calculate the voltage of a cell which consists of a rod iron immersed in a 1M solution of FeSO4 and rod manganese immersed in a 0.1M solution of MnSO4 at 250C.Write the cell reaction. Given EoFe2+/Fe = – 0.44 V and EoMn2+/Mn = – 1.18V. Solution: (i)Cell representation: Mn|Mn2+ (0.1 M) || Fe2+ (1.0 M) | Fe (ii) Electrode reactions: At anode: Mn Mn2+ + 2e¯ At cathode: Fe2+ + 2e¯ Fe Net cell reaction: Mn + Fe2+ Mn2+ + Fe UDOM- Study Material 2013-14 Page 104

CH 111: General Chemistry (iii)The emf of the cell is Ecell = E0cell +0.0591 log [Fe2+] n [Mn2+]
2 Ecell = ( EoFe2+/Fe – EoMn2+/Mn)+ log = [-0.44-(-1.18)] + 0.02955= 0.74+0.02955 Ecell = 0.7696 V 0.0591

Dr. Ananda M

1.0 0.1

2. An electrochemical cell consists of an iron electrode, dipped in 0.1M FeSO4 and silver electrode dipped in 0.05M AgNO3 solution. Write cell representation, cell reaction and calculate the emf of the cell at 298K. Given SRP of Fe and Ag are -0.44 and +0.8V respectively. Solution: (i)Cell representation: Fe|FeSO4 (0.1 M) || AgNO3 (0.05 M) | Ag (ii) Electrode reactions At anode: Fe Fe2+ + 2e¯ At cathode: [Ag+ +e¯ Ag] x 2 + 2+ Net cell reaction: Fe + 2Ag 2Ag + Fe (iii)The emf of the cell is Ecell = E0cell +0.0591 log [Ag+]2 n [Fe2+] Ecell = ( EoAg+/Ag – EoFe2+/Fe) +0.0591 log [0.05]2 2 [0.1] = [0.8-(-0.44)] + 0.02955 log0.025= 1.24-0.0473 Ecell = 1.1927 V. Concentration cells Concentration cell is an electrochemical cell in which two identical electrodes are immersed in the same ionic solution but of different concentration. As a result, the potential difference arises due to transfer of substance from a solution of higher concentration to a solution of lower concentration. Derivation of expression for the cell potential of a concentration cell Concentration cell can be represented as Zn| Zn2+ (C1) || Zn2+ (C2)| Zn The half cell reactions are At anode: Zn  Zn2+ (C1) +2eAt cathode: Zn2+ (C2) +2e- Zn UDOM- Study Material 2013-14 Page 105

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Net cell reaction: Zn2+ (C2)  Zn2+ (C1) C1 and C2 are concentrations of Zn2+ in the two half cells respectively and C2> C1.The left hand electrode is the anode and the right hand electrode is the cathode. Emf of a concentration cell can be calculated using Nernst Equation as follows: Eanode = E0electrode + 2.303RT log C1 nF Ecathode= E0electrode + 2.303RT log C2 nF Ecell = Ecathode – Eanode = (E0 + 2.303RT log C2) _ (E0 + 2.303RT log C1) nF nF Ecell = 0.0591 log C2 n C1 Numerical problems 1. The emf of the cell Cd/CdSO4(0.0093M) // CdSO4(xM) /Cd is 0.086V at 250C. Find the value of x. Ecell = 0.0591 log C2 n C1 0.086 = 0.0591 log x 2 0.0093 log x = 2.910 0.0093 log x- log0.093=2.910 logx + 2.0315=2.910 logx =2.910-1.0315 = 0.8785 x = antilog (0.8785) = 7.55 M 2. Represent the cell formed by the coupling of two Cu electrodes immersed in CuSO4 solutions concentration of cupric ions in one electrode system in 100 times more concentrated than other. Write the cell reaction and calculate the potential at 300K. (i)Cell representation: Cu│CuSO4 (x M || CuSO4 (100x M│Cu (ii)Electrode reactions: At anode: Cu(s) Cu2+ (x M) + 2e¯ At cathode: Cu2+ (100x M) + 2e¯ Cu(s) 2+ Net cell reaction: Cu (100x M) Cu2+ (x M) (iii)The emf of the concentration cell, Ecell = 2.303RT log C2 nF C1 Ecell = 2.303 × 8.314 × 300 log 100x 2 × 96500 x = 0.0297 log 100 UDOM- Study Material 2013-14 Page 106

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= 0.0297 × 2 Ecell = 0.0595V Electrochemical Series The standard reductions potentials of various metals can be determined by using S.H.E. Different electrodes have different standard electrode potentials. The standard reduction potentials (Eo) for some electrodes are negative while for some others Eo values are positive. The arrangement of various electrodes in the order of increasing standard reduction potentials is called electrochemical series.
Standard electrode Electrode reaction Standard potential reduction

Li+/Li
K /K

Li
K

e e Li
K

-3.05V -2.93 V
Ba

Ba2+ / Ba Sr2+/Sr Ca2+/Ca Na+/Na Mg2+/Mg Al3+/Al Zn2+/Zn Fe2+/Fe Ni2+/Ni H+/½ H2.Pt Cu2+/Cu Ag+/Ag Au3+/Au

Ba 2

2e

-2.90 V -2.890V -2.87V -2.71V -2.37 -1.66V -0.76V -0.44V -0.25V 0.0V +0.34V +0.80V +1.50V

Sr 2
Ca 2

2e
2e

Sr
Ca

Na
Mg2

e
2e

Na
Mg

Al3 Zn 2 Fe2 Ni2 H
Cu 2
Ag

3e 2e 2e 2e e
2e
e

Al Zn Fe Ni H
Cu
Ag

Au 3

3e

Au

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Applications of Electrochemical Series The standard electrode potentials given in the electrochemical series are the reduction potentials. Some applications of the electrochemical series are given below. 1. In predicting the reactivity of metals The reactivity of the metal depends upon its tendency to lose valence electrons i.e., on the tendency to get oxidized. Greater the tendency of a metal to get oxidized, greater is its reactivity. Metals with lowest reduction potential exhibits highest oxidation potential i.e., the metals will exhibit highest reactivity. The metal placed at the top of the series, exhibit highest reactivity. [Li, Na, Al, Zn etc.]. The metals placed at the bottom possess high reduction potential but are least reactive [Au, Pt, etc]. 2. In predicting the feasibility of a redox reaction The feasibility of a redox reaction can also be predicted by calculating the value of given cell. o o Eo cell Ecathode Eanode

Eo cell

for the

If

Eo cell

is positive, then the reaction is feasible

Eo 0, If cell then the reaction in is equilibrium Eo If cell =-ve, then the reaction is not feasible.
3. In predicting the displacement of hydrogen by metals from dilute acids. The displacement of hydrogen gas from acid solution is a reduction process,

2H (aq)

2e

H 2 (g)

This reduction process can be made to takes place by the metal whose reducing power is greater than that of hydrogen. The metals with low reduction potential, usually undergo oxidation and reduce H+ ion to liberate H2, gas. Reducing power of the substances decreases in going down the electro-chemical series. Hence any metal placed above hydrogen in the electrochemical series can displace hydrogen from acid solution. Metals with negative electrode potentials such as, Li, Al, Zn, Mg, Fe displace hydrogen from acid solution, but metals like Cu, Ag, Au cannot. 4. In predicting the displacement of metals from its salt solution by another metal. A metal in the electrochemical series can displace another metal below it in the series from its salt solution. Zinc is placed above copper in the series. Zinc = [Eo = -0.76V] displaces Cu [Eo = +0.34V]
Zn(s) Cu 2 (aq)
Cu(s) Zn 2 (aq)

Copper does not displace iron from iron sulphate. UDOM- Study Material 2013-14 Page 108

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5. In predicting electropositive character of metal and thermal stability of their oxides. The electropositive character of a metal is its tendency to form positive ions by losing one or more valence electrons. The metal at the top of the electrochemical series possess low reduction potential and exhibit greatest tendency to undergo oxidation. Consequently the metal at the top possesses highest electropositive character. The thermal stability of a metallic oxide depends upon the electropositive character of the metal. Higher the electropositive character, greater is the stability of its oxides. Hence stability of oxides of metals decreases in going down the series. The oxides of Li, Na, Ketc are very stable and donot decompose easily on heating whereas the oxides of metals like Ag, Hg, Au etc are not stable and decomposes on heating.

Corrosion The corrosion of metals is one of the most significant problems faced by advanced industrial societies. It is estimated that about 20% of iron produced annually is used just to replace the rusted Iron objects. Corrosion is the spontaneous slow chemical interaction of metal or alloy with it's environment resulting in the formation of one of its compounds like oxide, hydrated oxide, carbonate, sulphide sulphate etc. Extraction of metals from their ores is an endothermic process. Pure metals, being highly energetic, have natural tendency to revert back to their combined states.
Metallurgical extraction

Metal

Metal ores

Corrosion

Corrosion is defined as the destruction of metals or alloys by the surrounding environment through chemical or electrochemical changes. The most familiar example of corrosion is rusting of iron exposed to the atmospheric conditions. 3 2 Fe (s) O (g) x H 2O (l) Fe2O3 . x H 2O (s) 2 2 Chemical Composition of rust

Corrosion of metals occurs either by direct chemical attack or by electro chemical attack on the metal by the corrosive environment.Dry corrosion involves direct attack of atmospheric gases on metal (In the absence of moisture). Wet corrosion involves electro chemical attack on metals in aqueous environments. Wet corrosion is more prevalent than dry corrosion. Electrochemical theory of corrosion According to the electrochemical theory of corrosion, the corrosion of a metal, such as Iron, exposed to the environment may be thought of as three steps process. UDOM- Study Material 2013-14 Page 109

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1) Formation of large number of galvanic cells i.e., the Development of anodic and cathodes regions on the same metal. The electrons released at the anodic region are absorbed at cathodic region. 2) Corrosion (oxidation) takes place at the anodic region. 3) Reduction of O2 (air) in presence of water to hydroxyl [OH-] ions takes place at cathodic region.

Corrosion Reactions Anodic reaction: Metal undergoes oxidation (corrosion) with +ve release of electrons

Fe

Fe2

2e

At the cathodic region, electrons are absorbed and cause reduction of the constituents. Cathodic reactions Cathodic reactions are dependent on the constituents of the corrosion medium. There are three possible ways in which reduction takes place. i) If the solution is aerated and almost neutral, oxygen is reduced in presence of H2O to OH- ions.

O2

2H2O 2e 2e

4e H2 H2

4 OH 2 OH

ii) If the solution is deaerated and almost neutral. H2 is liberated along with OH- ions.

2H2O 2H

iii) If the solution is deaerated and acidic, H+ ions are reduced to hydrogen gas

The metal ions formed at the anode combine with hydroxyl ions and form the corresponding metallic hydroxide Fe(OH)2 which further gets oxidized to hydrated ferric oxide [rust].

2Fe2

4 OH
O2 (x

2Fe(OH)2
2) H2O Fe2O3 . x . H2O
Rust

2 Fe (OH)2

In the presence of limited oxygen, Black rust is formed as follows: 1 3Fe(OH)2 O2 Fe3 O4 . 3H2O 2 Black rust UDOM- Study Material 2013-14 Page 110

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Cathode Protection The technique of offering protection to a specimen against corrosion by providing electrons from an external source is called cathodic protection. Cathodic protection can be achieved by the following two methods. a) Sacrificial anode methodb) Impressed current method a) Sacrificial anode method In this method, metal structure is converted into a cathode by connecting into a more active metal. This active metal acts as an auxiliary anode. Zn, Mg and Al are the common auxiliary anodes. These metals being more active, acts as anode and undergo preferential corrosion, protecting the metal structure. Since the anodic metals are sacrificed to protect the metal structure. The method is known as sacrificial anode method. New ones are replacing exhausted sacrificial anodes as and when required. Examples of sacrificial anode methods are A magnesium block connected to a buried pipelines Magnesium bars are fixed to the sides of the ships

b) Impressed current method Another method of providing cathodic protection is by applying a direct current larger than the corrosion current. The protected metal is made cathodic by connecting it to the negative terminal of a d.c. source. Positive terminal is connected to inert anode such as Graphite. The metal structure being cathode, does not undergo corrosion.

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CH 111: General Chemistry Practice questions 1. The oxidizing agent and reducing agent are, a) Electron deficient & Electron rich species respectively b) Electron rich & Electron deficient species respectively c) Only electron deficient species d) Only electron rich species.

Dr. Ananda M

2. The standard electrode potentials of Zn+2 / Zn and Ag+ / Ag are -0.763 V and + 0.799 V respectively. The standard potential of the cell is, a) 1.562 V b) 0.036 V c) -1.562 V d) 0.799 V 3. In a galvanic cell, the electrons flow from, a) Anode to cathode through the solution b) Cathode to anode through the solution c) Anode to cathode through the external circuit d) Cathode to anode through the external circuit 4. The overall net cell reaction for Cu-Ag cell is, a) Cu + Ag+ → Cu2+ + Ag b) Cu + 2Ag+ → Cu2+ + 2Ag c) Cu2+ + Ag → Cu + Ag+ d) Cu2+ + 2Ag → Cu + 2Ag+ 5. Which one of the following metals does not displace hydrogen from acid solution, a) Iron b) Zinc c) Copper d) Aluminium 6. Corrosion of iron takes place at very fast rate in, a) Soft water b) Mineral water c) Sea water

d) Distilled water

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CH 111: General Chemistry 2.7 Properties of solutions

Dr. Ananda M

Solutions are homogeneous (single-phase) mixtures of two or more components. They are extremely important in Chemistry because they allow intimate and varied encounters between molecules of different kinds, a condition that is essential for rapid chemical reactions to occur. Several more explicit reasons can be cited for devoting a significant amount of time to the subject of solutions: 1. For the reason stated above, most chemical reactions that are carried out in the laboratory and in industry, and that occur in living organisms, take place in solution. 2. Solutions are so common; very few pure substances are found in nature. 3. Solutions provide a convenient and accurate means of introducing known small amounts of a substance to a reaction system. Advantage is taken of this in the process of titration. 4. The physical properties of solutions are sensitively influenced by the balance between the intermolecular forces of like and unlike (solvent and solute) molecules. The physical properties of solutions thus serve as useful experimental probes of these intermolecular forces. A solution is a homogenous mixture of two or more substances that exist in a single phase. There are two main parts to any solution. The solute is the component of a solution that is dissolved in the solvent; it is usually present in a smaller amount than the solvent. The solvent is the component into which the solute is dissolved, and it is usually present in greater concentration. For example, in a solution of salt water, salt is the solute and water is the solvent. In solutions where water is the solvent, the solution is referred to as an aqueous solution. A solution does not have to involve liquids. For instance, air is a solution that consists of nitrogen, oxygen, carbon dioxide, and other trace gases, and solder is a solution of lead and tin. The general rule of thumb for solutions is the idea that like dissolves like. Polar, ionic substances are soluble in polar solvents, while nonpolar solutes are soluble in nonpolar solvents. For example, alcohol and water, which are both polar, can form a solution and iodine and carbon tetrachloride, which are both nonpolar, make a solution. However, iodine will not readily dissolve in polar water. In a solution, the particles are really small—anywhere from 0 to 100 nm. They never settle on standing, they cannot be separated by filtering, and light will pass through a solution unchanged. One type of mixture that is not a solution is known as the colloid. In a colloid, particles are between 100 and 1000 nm in size—still too small for our eyes to distinguish, but particles this small will not settle. As is the case in solutions, the particles cannot be filtered, but they do scatter light. Some examples of colloids include gelatin, fog, smoke, and shaving cream. Another type of mixture that is not considered a solution is known as a suspension. Suspensions have much larger particles: usually over 1000 nm. Particles in a suspension will settle on standing, can UDOM- Study Material 2013-14 Page 113

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often be separated by a filter, and may scatter light, but they are usually not transparent. Some examples of suspensions are muddy water, paint, and some medicines, like Pepto-Bismol. The Solution Process In order for a solute to be dissolved in a solvent, the attractive forces between the solute and solvent particles must be great enough to overcome the attractive forces within the pure solvent and pure solute. The solute and the solvent molecules in a solution are expanded compared to their position within the pure substances. The term solubility refers to the maximum amount of material that will dissolve in a given amount of solvent at a given temperature to produce a stable solution. By looking at the plot of solubilities below, you can see that most solids increase in solubility with an increase in temperature. Gases, however, decrease in solubility with an increase in temperature. Degrees of Saturation When referring to solutions, there are three degrees of saturation— unsaturated, saturated, and supersaturated. If a solution is unsaturated, the solvent is capable of dissolving more solute. When the solution is saturated, the solvent has dissolved the maximum amount of solute that it can at the given temperature. At this point we say that the solution is in a state of dynamic equilibrium—the processes of dissolving and precipitation are happening at the same rate. A supersaturated solution is one in which the solvent contains more solute than it can theoretically hold at a given temperature. Supersaturated solutions are often formed by heating a solution and dissolving more solute, then cooling the solution down slowly. These solutions are unstable and crystallize readily.

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CH 111: General Chemistry 2.8 Soutions of Gases in liquids – Henry’s law

Dr. Ananda M

When a gas dissolves in a liquid, the confinement of the gas in the much smaller volume of the liquid causes a loss in molecular disorder that is not usually compensated by the presence of two kinds of molecules in the liquid phase. Such processes are not favored unless there are strong compensating factors; gases therefore tend to be only slightly soluble in liquids. Still, there is always some solubility, as is shown in Table given below for several common gases in water. The greatest solubilities occur when the gas reacts chemically with the solvent (a―compensating factor‖), as happens, for example, with CO2, HCl, SO2, and especially NH3in water. In these cases, the fall in potential energy associated with the reaction helps overcome the unfavorable randomness effect. One important consequence of the reduction in disorder when a gas dissolves in a liquid is that the solubility of a gas decreases at higher temperatures; this is in contrast to most othersituations, where a rise in temperature usually leads to increased solubility. Solute ammonia carbon dioxide nitrogen oxygen Formula NH3 CO2 N2 O2 Solubiliy, mol L−1atm−1 57 0.0308 0.000661 0.00126

Henry's law: Henry's law is one of the gas laws formulated by William Henry in 1803. It states: "At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid." An equivalent way of stating the law is that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. To explain this law, Henry derived the equation: C=k Pgaswhere, C is the solubility of a gas at a fixed temperature in a particular solvent (in units of M or mL gas/L) k is Henry's law constant (often in units of M/atm) Pgas is the partial pressure of the gas (often in units of Atm) UDOM- Study Material 2013-14 Page 115

CH 111: General Chemistry 2.9 Solutions of liquids in liquids – Raoult’s law

Dr. Ananda M

Whereas all gases will mix to form solutions regardless of the proportions, liquids are much more fussy. Some liquids, such as ethyl alcohol and water, are miscible in all proportions. Others, like the proverbial ―oil and water‖, are not; each liquid has only a limited solubility in the other, and once either of these limits is exceeded, the mixture separates into two phases. Mixing of two liquids can be exothermic, endothermic, or without thermal effect, depending on the particular substances. Whatever the case, the energy factors are not usually very large, but neither is the increase in randomness; the two factors are frequently sufficiently balanced to produce limited miscibility. A useful general rule is that liquids are completely miscible when their intermolecular forces are very similar in nature; “like dissolves like”. Thus water is miscible with other liquids that can engage in hydrogen bonding, whereas a hydrocarbon liquid in which London or dispersion forces are the only significant intermolecular effect will only be completely miscible with similar kinds of liquids. Substances such as the alcohols, CH3(CH2)nOH, which are hydrogen-bonding (and thus hydrophilic) at one end and hydrophobic at the other, tend to be at least partially miscible with both kinds of solvents. If n is large, the hydrocarbon properties dominate and the alcohol has only a limited solubility in water. Very small values of n allow the -OH group to dominate, so miscibility in water increases and becomes unlimited in ethanol (n = 1) and methanol (n = 0), but miscibility with hydrocarbons decreases owing to the energy required to break alcohol-alcohol hydrogen bonds when the nonpolar liquid is added.

Raoult's Law In the case of a single volatile liquid (the solvent) and a non-volatile solute, Raoult's Law can be states as follows, ―The vapour pressure of a solution of a non-volatile solute is equal to the vapour pressure of the pure solvent at that temperature multiplied by its mole fraction.‖ In equation form, this reads:

In this equation, Po is the vapour pressure of the pure solvent at a particular temperature. xsolv is the mole fraction of the solvent. That is exactly what it says it is - the fraction of the total number of moles present which is solvent.

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CH 111: General Chemistry You calculate this using:

Dr. Ananda M

Suppose you had a solution containing 10 moles of water and 0.1 moles of sugar. The total number of moles is therefore 10.1 The mole fraction of the water is:

A simple explanation of why Raoult's Law works There are two ways of explaining why Raoult's Law works - a simple visual way, and a more sophisticated way based on entropy. Remember that saturated vapour pressure is what you get when a liquid is in a sealed container. An equilibrium is set up where the number of particles breaking away from the surface is exactly the same as the number sticking on to the surface again. Now suppose you added enough solute so that the solvent molecules only occupied 50% of the surface of the solution. A certain fraction of the solvent molecules will have enough energy to escape from the surface (say, 1 in 1000 or 1 in a million, or whatever). If you reduce the number of solvent molecules on the surface, you are going to reduce the number which can escape in any given time.But it won't make any difference to the ability of molecules in the vapour to stick to the surface again.

If a solvent molecule in the vapour hits a bit of surface occupied by the solute particles, it may well stick. There are obviously attractions between solvent and solute otherwise you wouldn't have a solution in the first place.

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The net effect of this is that when equilibrium is established, there will be fewer solvent molecules in the vapour phase - it is less likely that they are going to break away, but there isn't any problem about them returning. If there are fewer particles in the vapour at equilibrium, the saturated vapour pressure is lower. Limitations on Raoult's Law Raoult's Law only works for ideal solutions. An ideal solution is defined as one which obeys Raoult's Law. Now we are finally in a position to see what effect a non-volatile solute has on the melting and freezing points of the solution. Look at what happens when you draw in the 1 atmosphere pressure line which lets you measure the melting and boiling points. The diagram also includes the melting and boiling points of the pure water from the original phase diagram for pure water (black lines).

Because of the changes to the phase diagram, you can see that: the boiling point of the solvent in a solution is higher than that of the pure solvent; the freezing point (melting point) of the solvent in a solution is lower than that of the pure solvent. We have looked at this with water as the solvent, but using a different solvent would make no difference to the argument or the conclusions. The only difference is in the slope of the solidliquid equilibrium lines. For most solvents, these slope forwards whereas the water line slopes backwards. You could prove to yourself that that doesn't affect what we have been looking at by re-drawing all these diagrams with the slope of that particular line changed.

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CH 111: General Chemistry Boiling Point Elevation

Dr. Ananda M

If addition of a nonvolatile solute lowers the vapor pressure of the solution, then the temperature must be raised to restore the vapor pressure to the value corresponding to the pure solvent. In particular, the temperature at which the vapor pressure is 1 atm will be higher; the boiling point will be raised.There is a simple formula for calculating the change in boiling point. ΔTb = kb x m ΔTb is the change in the boiling temp., added to the liquid‘s original BP. kb is the ‗molal boiling point constant‘which depends on the solvent. For water, kb is 0.51°C/molal, m is the molality of the solution. Freezing Point Depression The freezing point is the temperature at which solid and liquid can simultaneously coexist, meaning that the escaping tendencies of molecules from the two phases are the same. Suppose now that we dilute the solvent by adding some solute. The escaping tendency of solvent molecules from the liquid phase is now reduced owing to their increased disorder in the solution, but in the solid nothing has changed and the escaping tendency of solvent molecules from this phase remains the same. This means that there will be a net movement of solvent molecules from the solid phase to the liquid; the solid melts. In order to keep the solid from melting, the escaping tendency of molecules from the solid must be reduced. This can be accomplished by reducing the temperature; this lowers the escaping tendency of molecules from both phases, but it affects those in the solid more than those in the liquid, so we eventually reach the new, lower freezing point where the two quantities are again in exact balance and both phases can coexist.Solution freezing points are always lower than that of the solvent liquid.A similar formula is used to calculate the change in the freezing point. ΔTf= kf x m ΔTf is the change in the freezing temp., subtracted from the liquid‘s original FP. kf is the ‗molal freezing point constant‘which depends on the solvent. For water, kf is 1.86°C/molal, m is the molality of the solution.

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CH 111: General Chemistry 2.10Binary liquid mixtures IDEAL MIXTURES OF LIQUIDS Raoult’s law

Dr. Ananda M

The partial pressure exerted by solvent vapor above the solution, PA, equals to the product of the mole fraction of the solvent in the solution, Xa, times the vapor pressure of the pure solvent. PA = XA x P°A where • XA is the mole fraction of compound A • P°A is the normal vapor pressure of Aatthat temperature Raoult's law plot for a mixture of hexane and heptane. An ideal mixture is one which obeys Raoult's Law. There are some solutions whose components follow Raoult's law quite closely. An example of such a solution is one composed of hexane C6H14 and heptane C7H16. The total vapor pressure of this solution varies in a straight-line manner with the mole fraction composition of the mixture. Note that the mole fraction scales at the top and bottom run in opposite directions, since by definition, Xhexane = 1 – Xheptane. Ptotal is the sum of the Raoult's law plots for the two pure compounds, indicating that this solution behaves ideally. An ideal solution is one whose vapor pressure follows Raoult's law throughout its range of compositions. Raoult's Law only works for ideal mixtures.Solutions that approximate ideal behavior are composed of molecules having very similar structures. Thus hexane and heptane are both linear hydrocarbons that differ only by a single –CH2group.This provides a direct clue to the underlying cause of non-ideal behavior in solutions of volatile liquids. In an ideal solution, the UDOM- Study Material 2013-14 Page 120

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interactions are there, but they are all energetically identical. Thus in an ideal solution of molecules A and B, A—A and B—B attractions are the same as A—B attractions. This is the case only when the two components are chemically and structurally very similar. Note that this differs in a fundamental way from the definition of an ideal gas, defined as a hypothetical substance that follows the ideal gas law. The kinetic molecular theory that explains ideal gas behavior assumes that the molecules occupy no space and that intermolecular attractions are totally absent.The definition of an ideal gas is clearly inapplicable to liquids, whose volumes directly reflect the volumes of their component molecules. And of course, the very ability of the molecules to form a condensed phase is due to the attractive forces between the molecules. So the most we can say about an ideal solution is that the attractions between its all of its molecules are identical — that is, A-type molecules are as strongly attracted to other A molecules as to B-type molecules. Ideal solutions are perfectly democratic: there are no favorites. NON-IDEAL MIXTURES OF LIQUIDS Raoult's Law only works for ideal mixtures. In these, the forces between the particles in the mixture are exactly the same as those in the pure liquids. The tendency for the particles to escape is the same in the mixture and in the pure liquids. That's not true in non-ideal mixtures. Positive deviations from Raoult's Law In mixtures showing a positive deviation from Raoult's Law, the vapour pressure of the mixture is always higher than expected for an ideal mixture. The deviation can be small - in which case, the straight line in the last graph turns into a slight curve.Notice that the highest vapour pressure anywhere is still the vapour pressure of pure A. Cases like this, where the deviation is small, behave just like ideal mixtures as far as distillation is concerned, and we don't need to say anything more about them.

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CH 111: General Chemistry Explaining the deviations

Dr. Ananda M

The fact that the vapour pressure is higher than ideal in these mixtures means that molecules are breaking away more easily than they do in the pure liquids. That is because the intermolecular forces between molecules of A and B are less than they are in the pure liquids. Less heat is evolved when the new attractions are set up than was absorbed to break the original ones. Heat will therefore be absorbed when the liquids mix. The enthalpy change of mixing is endothermic. The classic example of a mixture of this kind is ethanol and water. This produces a highly distorted curve with a maximum vapour pressure for a mixture containing 95.6% of ethanol by mass. Negative deviations from Raoult's Law In exactly the same way, there are mixtures with vapour pressures which are lesser than expected for ideal mixtures by Raoult's Law. In some cases, the deviations are small, but in others they are much greater giving a minimum value for vapour pressure lower than that of either pure component.

Explaining the deviations These are cases where the molecules break away from the mixture less easily than they do from the pure liquids. New stronger forces must exist in the mixture than in the original liquids. Heat is evolved during the mixing of the liquids - more heat is given out when the new stronger bonds are made than was used in breaking the original weaker ones. Many (although not all) examples of this involve actual reaction between the two liquids. The example of a major negative deviation that we are going to look at is a mixture of nitric acid and water. These two covalent molecules react to give hydroxonium ions and nitrate ions.

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CH 111: General Chemistry Boiling point / composition diagrams for non-ideal mixtures A large positive deviation from Raoult's Law: ethanol and water mixtures

Dr. Ananda M

It is very much obvious that large positive deviation from Raoult's Law produces a vapour pressure curve with a maximum value at some composition other than pure A or B. If a mixture has a high vapour pressure it means that it will have a low boiling point.

The molecules are escaping easily and less heating is required for the mixture to overcome the intermolecular attractions completely. The implication of this is that the boiling point / composition curve will have a minimum value lower than the boiling points of either A or B.In the case of mixtures of ethanol and water, this minimum occurs with 95.6% by mass of ethanol in the mixture. The boiling point of this mixture is 78.2°C, compared with the boiling point of pure ethanol at 78.5°C, and water at 100°C.You might think that this 0.3°C doesn't matter much, but it has huge implications for the separation of ethanol / water mixtures. The next diagram shows the boiling point / composition curve for ethanol / water mixtures.

Using the diagram Suppose you are going to distil a mixture of ethanol and water with composition C1 as shown on the next diagram. It will boil at a temperature given by the liquid curve and produce a vapour with composition C2. When that vapour condenses it will, of course, still have the composition C2.

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If you reboil that, it will produce a new vapour with composition C3.You can see that if you carried on with this boiling-condensing-reboiling sequence, you would eventually end up with a vapour with a composition of 95.6% ethanol. If you condense that you obviously get a liquid with 95.6% ethanol.

What happens if you reboil that liquid? The liquid curve and the vapour curve meet at that point. The vapour produced will have that same composition of 95.6% ethanol. If you condense it again, it will still have that same composition. You have hit a barrier. It is impossible to get pure ethanol by distiling any mixture of ethanol and water containing less than 95.6% of ethanol. This particular mixture of ethanol and water boils as if it were a pure liquid. It has a UDOM- Study Material 2013-14 Page 124

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constant boiling point, and the vapour composition is exactly the same as the liquid. It is known as a constant boiling mixture or an azeotropic mixture or an azeotrope. The implications of this for fractional distillation of dilute solutions of ethanol are obvious. The liquid collected by condensing the vapour from the top of the fractionating column can't be pure ethanol. The best you can produce by simple fractional distillation is 95.6% ethanol.What you can get (although it isn't very useful!) from the mixture is pure water. As ethanol rich vapour is given off from the liquid boiling in the distillation flask, it will eventually lose all the ethanol to leave just water. To summarise:Distilling a mixture of ethanol containing less than 95.6% of ethanol by mass lets you collect: a distillate containing 95.6% of ethanol in the collecting flask (provided you are careful with the temperature control, and the fractionating column is long enough); pure water in the boiling flask.

A large negative deviation from Raoult's Law: nitric acid and water mixtures Nitric acid and water form mixtures in which particles break away to form the vapour with much more difficulty than in either of the pure liquids. That means that mixtures of nitric acid and water can have boiling points higher than either of the pure liquids because it needs extra heat to break the stronger attractions in the mixture. In the case of mixtures of nitric acid and water, there is a maximum boiling point of 120.5°C when the mixture contains 68% by mass of nitric acid. That compares with the boiling point of pure nitric acid at 86°C, and water at 100°C. Notice the much bigger difference this time due to the presence of the new ionic interactions.

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CH 111: General Chemistry Using the diagram: Distilling dilute nitric acid

Dr. Ananda M

Start with a dilute solution of nitric acid with a composition of C1 and trace through what happens. The vapour produced is richer in water than the original acid. If you condense the vapour and reboil it, the new vapour is even richer in water. Fractional distillation of dilute nitric acid will enable you to collect pure water from the top of the fractionating column. As the acid loses water, it becomes more concentrated. Its concentration gradually increases until it gets to 68% by mass of nitric acid. At that point, the vapour produced has exactly the same concentration as the liquid, because the two curves meet You produce a constant boiling mixture (or azeotropic mixture or azeotrope). If you distil dilute nitric acid, that's what you will eventually be left with in the distillation flask. You can't produce pure nitric acid from the dilute acid by distilling it. Distilling nitric acid more concentrated than 68% by mass This time you are starting with a concentration C2 to the right of the azeotropic mixture. The vapour formed is richer in nitric acid. If you condense and reboil this, you will get a still richer vapour. If you continue to do this all the way up the fractionating column, you can get pure nitric acid out of the top.As far as the liquid in the distillation flask is concerned, it is gradually losing nitric acid. Its concentration drifts down towards the azeotropic composition. Once it reaches that, there can't be any further change, because it then boils to give a vapour with the same composition as the liquid.Distilling a nitric acid / water mixture containing more than 68% by mass of nitric acid gives you pure nitric acid from the top of the fractionating column and the azeotropic mixture left in the distillation flask. You now have strong ionic attractions involved.

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Practice questions 1. An aqueous solution of ethanol in water has vapour pressure A) Equal to water B) More than that of water C) Equal to that of ethanol D) Less than that of water 2. Which of the following liquid pairs shows a positive deviation from Raoult‘s law? A) Water – hydrochloric acid B) Acetone – chloroform C) Water – nitric acid D) Benzene – methanol 3. A binary liquid solution is prepared by mixing n-heptane and ethanol. Which one of the following statements is correct regarding the behavior of the solution? A) The solution formed is an ideal solution B) The solution is non-ideal, showing +ve deviation from Raoult‘s Law C) The solution is non-ideal, showing –ve deviation from Raoult‘s Law D) n-heptane shows +ve deviation while ethanol shows –ve deviation from Raoult‘s Law 4. At higher altitudes the boiling point of water decreases because A) The atmospheric pressure is high B) The temperature is low C) The atmospheric pressure is low D) The temperature is high

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CH 111: General Chemistry 2.11 Chemical equilibrium

Dr. Ananda M

Chemical change occurs when the atoms that make up one or more substances rearrange themselves in such a way that new substances are formed. These substances are the components of the chemical reaction system; those components which decrease in quantity are called reactants, while those that increase are products. A given chemical reaction system is defined by a balanced net chemical equation which is conventionally written asReactants → Products When a chemical reaction takes place in a container which prevents the entry or escape of any of the substances involved in the reaction, the quantities of these components change as some are consumed and others are formed. Eventually this change will come to an end, after which the composition will remain unchanged as long as the system remains undisturbed. The system is then said to be in its equilibrium state, or more simply, "at equilibrium". Chemical equilibrium: A state in which the rates of the forward and reverse reactions are equal and the concentrations of the reactants and products remain constant. ⇒ Equilibrium is a dynamic process. The conversions of reactants to products and products to reactants are still going on, although there is no net change in the number of reactant and product molecules. For the reaction: N2O4

g

2NO2 g

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Under certain conditions, one mole of the colourless gas N2O4 will decompose to form two moles of brown NO2 gas: N2O4→ 2 NO2 Colourless brown

Under other conditions, you can take 2 moles of brown NO2 gas and change it into one mole of N2O4 gas: 2 NO2 → N2O4 Brown colourless

From the above, it follows that equilibrium is a dynamic process in which microscopic change (the forward and reverse reactions) continues to occur,butmacroscopic change (changes in the quantities of substances) is absent. In other words, this reaction, as written may go forward or in reverse, depending on the conditions. Chemical equilibrium has been reached in a reaction when the rate of the forward reaction is equal to the rate of the reverse reaction. When a chemical reaction has reached equilibrium, collisions are still occurring: the reaction is now happening in each direction at the same rate. This means that reactants are being formed at the same rate as products are being formed, and this is indicated by double arrows . At equilibrium, the reaction can lie far to the right, meaning that there are more products in existence at equilibrium, or far to the left, meaning that at equilibrium there are more reactants.

The two diagrams below show how the concentrations of the three components of this chemical reaction change with time. Examine the two sets of plots carefully, noting which substances have zero initial concentrations, and are thus "products" of the reaction equations shown. Satisfy yourself that these two sets represent the same chemical reaction system, but with the reactions occurring in opposite directions. Most importantly, note how the final (equilibrium) concentrations of the components are the same in the two cases. UDOM- Study Material 2013-14 Page 129

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Whether we start with an equimolar mixture of H2 and I2 (left) or a pure sample of hydrogen iodide (shown on the right, using twice the initial concentration of HI to keep the number of atoms the same), the composition after equilibrium is attained (shaded regions on the right) will be the same. The equilibrium composition is independent of the direction from which it is approached. Characteristics of a System at Dynamic Equilibrium 1. The rate of the forward reaction = Therateof the reverse reaction 2. Microscopic processes (the forward and reverse reaction) continue in a balance which yields no macroscopic changes. (so nothing appears to be happening.) 3. The system is closed and the temperature is constant and uniform throughout. 4. The equilibrium can be approached from the left (starting with reactants) or from the right (starting with products). The Equilibrium Constant For a reaction: aA + bB cC + dDEquilibrium constant: K c

[C]c D d [ A]a [B]b

The equilibrium constant, Kc, is the ratio of the equilibrium concentrations of products over the equilibrium concentrations of reactants each raised to the power of their stoichiometric coefficients. For example, the equilibrium constant Kc, for N2O4 g

2NO2 g is given as

Kc

[ NO2 ]2 [ N 2O4 ]
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Dr. Ananda M

Law of mass action - Two Norwegian chemists, Cato Maximilian Guldberg (1836-1902) and Peter Waage (1833-1900) proposed in 1864 a general description of an equilibrium condition. The law of mass action states that any chemical change is a competition between a forward reaction (left-to-right in the chemical equation) and a reverse reaction. The rate of each of these processes is governed by the concentrations of the substances reacting; as the reaction proceeds, these rates approach each other and at equilibrium they become identical. The value of the equilibrium constant expression, Kc, is constant for agiven reaction at equilibrium and at a constant temperature. ⇒The equilibrium concentrations of reactants and products may vary, but the value for Kc remains the same. Other Characteristics of Kc 1) Equilibrium can be approached from either direction. 2) Kc does not depend on the initial concentrations of reactants and products. 3) Kc does depend on temperature. Magnitude of Kc ⇒ If the Kc value is large (Kc >> 1), the equilibrium lies to the right and the reaction mixture contains mostly products. ⇒ If the Kc value is small (Kc