Preview Fundamentals of Inorganic Chemistry For Competitive Exams by Ananya Ganguly (2010) Preview Fundamentals of Inorganic Chemistry For Competitive Exams by Ananya Ganguly (2010) Preview Fundamentals of Inorganic Chemistry For Competitive Exams by Ananya Ganguly (2010) Preview Fundamentals of Inorganic Chemistry For Competitive Exams by Ananya Ganguly (2010) Preview Fundamentals of Inorganic Chemistry For Competitive Exams by Ananya Ganguly (2010)
Fundamentals of Inorganic Chemisty Ananya Ganguly Chandigarh • Delhi • Chennai The aim of this publication is to supply information taken from sources believed to be valid and reliable This is not an attempt to render any type of professional advice or analysis, nor is it to be treated as such While much care has been taken to ensure the veracity and currency of the information presented within, neither the publisher, nor its authors bear any responsibility for any damage arising from inadvertent omissions, negligence or inaccuracies (typographical or factual) that may have found their way into this book Copyright © 2011 Dorling Kindersley (India) Pvt Ltd Licensees of Pearson Education in South Asia No part of this eBook may be used or reproduced in any manner whatsoever without the publisher’s prior written consent This eBook may or may not include all assets that were part of the print version The publisher reserves the right to remove any material present in this eBook at any time ISBN 9788131759691 eISBN 9789332511682 Head Office: A-8(A), Sector 62, Knowledge Boulevard, 7th Floor, NOIDA 201 309, India Registered Office: 11 Local Shopping Centre, Panchsheel Park, New Delhi 110 017, India Contents Preface v Chapter Periodic Properties and Chemical bonding Chapter Alkali Metals (+2) 2.1—2.33 Chapter Group-II Alkaline Earth Metals 3.1—3.28 Chapter Group13-Boron Family 4.1—4.56 Chapter Group14-Carbon Family 5.1—5.54 Chapter Group15-Nitrogen Family 6.1—6.54 Chapter Group16-Oxygen Family 7.1—7.44 Chapter Group17-Halogens 8.1—8.36 Chapter Group18-Noble Gases 9.1—9.16 1.1—1.110 Chapter 10 Hydrogen 10.1—10.23 Chapter 11 11.1—11.55 Co-ordination Chemistry Chapter 12 Transition Elements (d-block) and Their Compounds 12.1—12.57 Chapter 13 Principles of Metallurgy 13.1—13.48 Chapter 14 Salt Analysis 14.1—14.65 Question Bank QB.3—QB.94 This page is intentionally left blank Preface In recent years, the question format of non-medical examinations like IIT-JEE and AIEEE has been restructured with greater emphasis on the theoretical and conceptual intricacies and the application of the underlying basic concepts and principles The poor performance of students in these examinations is partly due to non-availability of a comprehensive text book which lays adequate stress on the authentic logical theoretical concept building, numericals and related problems The present book grew out of my experience of classroom teaching through lectures, notes and assignments Effort has been made to maintain lucid style and simplicity of expression Apart from this, the book has numerous carefully selected examples and solved illustrations which include almost all the previous years’ questions asked in IIT-JEE and AIEEE examinations Selected questions of different formats keeping in mind the recent pattern of examinations have been listed at the end of each chapter Solutions to almost all advanced questions have been incorporated I would like to thank Showick Thorpe and Sanjay Sharma for their constant encouragement during the publication of this book I extend my thanks to Nitkiran Bedi for her contribution in bringing out this book in record time I am also thankful to the staff members of my academy for their sincere help I will appreciate comments, suggestions and criticism from the readers and will incorporate them in the subsequent editions Ananya Ganguly This page is intentionally left blank c h a p t e r 1 Periodic Properties and Chemical bonding INTRODUCTION Chemists have long tried to find patterns in the properties of the elements Some were discovered fairly readily; for example, elements were classified as metals, non-metals, and many of their compounds as acids, alkalis, or salts Table 1.1 Typical Properties of Metals and Non-metals Appearance and properties Metals Non-metals Solids, some with high melting points; lustrous, malleable and ductile Gases, or solids with low melting points Conduction of heat and electricity Very good Poor Compounds Ionic compounds with non-metals; alloys with other metals Ionic compounds with metals; covalent compounds with other non-metals Charge on ions Positive Negative Chemical nature Reducing agents Oxidizing agents Electronegativity Low High However, it was widely believed that there had to be an underlying reason for the patterns Prout’s Hypothesis One of the first suggestions was due to Prout Prout’s hypothesis was that all elements were made from a whole number of hydrogen atoms (Be careful here: an atom’ in Prout’s time was a very different thing to our understanding of the word.) According to him the atomic masses of the elements should be a whole number of times that of hydrogen, i.e., they should be integers Unfortunately, from Prout’s point of view, the results of experiments showed that the atomic masses of many elements were not integers Dobereiner’s Triads According to Dobereiner, when elements of same properties are kept in the increasing order of their atomic weights, the atomic weight of middle element is equal to the mean atomic weight of remaining two elements Such a group of elements is called Dobereiner’s triad 1.2 Periodic Properties and Chemical Bonding Triad of atoms Mean of first and last element Li Na K 23 39 Be Mg Ca 24 40 + 39 = 23 + 40 = 24 Dobereiner could arrange only a few elements as triads and there are some such elements present in a triad, whose atomic weights are approximately equal, e.g., Fe Ru Co Rh Ni Pd Therefore, this hypothesis was not acceptable for all elements The Telluric Helix It was in 1862, that a periodic classification of the elements was developed that approached the idea we have today At that time A.E de Chancourtois, a professor of Geology at the Ecole des Mines in Paris presented an account of his telluric helix in which he indicated the relative properties of elements and their atomic weights He used a vertical cylinder with 16 equidistant lines on its surface, the lines lying parallel to the axes Then he drew a helix at 45 degree to the axis and arranged the elements on the spiral in the order of their increasing atomic weights In this manner, elements that differed from each other in atomic weight by 16 or multiples of 16 fell very nearly on the same vertical line In addition to the 16 vertical lines, de Chancourtois felt that other connecting lines could be drawn, and that all elements lying on such lines were related in some manner His arrangement resulted in the proposal by de Chancourtois that the properties of the elements are the properties of numbers Newland’s Rule of Octave A major attempt at making a link was made by new lands in 1864 (Table 1.2) He grouped elements into sets of eight and claimed that every eighth element in the pattern was chemically similar Newlands’ law of octaves was largely ignored, or at best treated with mild amusement The first thorough attempt at relating chemical properties to atomic masses was made by the Russian Table 1.2 Examples of Newlands’ Octaves* No H No F 8 No Cl 15 No Co and Ni 22 No Br 29 Li Na 9 K 16 Cu 23 Rb 30 G Mg 10 Ca 17 Zn 24 Sr 31 Bo Al 11 Cr 18 Y 25 Ce and Le 32 C Si 12 Ti 19 In 26 Zr 33 N P 13 Mn 20 As 27 Di and Mo 33 O S 14 Fe 21 Se 28 Ro and Ru 35 *This is part of a table that John Newlands presented in a talk he gave to the Chemical Society on March 1866 The talk was entitled The Law of Octaves, and the Causes of Numerical Relations among the Atomic Weights’ (You might like to work out which elements G, Bo, etc., stand for.) Here is part of an account of the talk: The author claims the discovery of a law according to which the elements analogous in their properties exhibit peculiar relationships, similar to those subsisting in music between a note and its octave Professor G F Foster humorously enquired of Mr Newlands whether he had ever examined the elements according to the order of their initial letters? Newlands was not at all happy about the credit that went to Mendeleeff over the discovery of the periodic law In 1884 Newlands wrote: Having been the first to publish the existence of the periodic law more than nineteen years ago, I feel, under existing circumstances, compelled to assert my priority in this matter As a matter of simple justice, and in the interest of all true workers in science, both theoretical and practical, it is right that the originator of any proposal or discovery should have the credit of his labour Periodic Properties and Chemical Bonding 1.3 80 70 60 87 81 53 49 35 31 17 13 50 40 30 20 10 Atomiv Volume (c.c.) Lothar Meyer’s Volume Curves The graphs of atomic volumes against atomic weights are known as Lothar Meyer volume curves (i) Lothar Meyer plotted a graph between atomic weight and atomic volume (i.e., atomic weight in solid state/ density) (ii) Elements with similar properties occupied the similar positions on the graph (iii) Strong electropositive elements of IA except Li, all others Na, K, Rb, Cs etc., occupied the top position on the graph Atomic Number Fig 1.1 Atomic volume versus atomic number curve (iv) IIa group elements Be, Mg, Ca, Sr, Ba etc., occupied the positions on the ascending part of the graph (v) Inert gases except He occupied the positions on the descending part of the graph (vi) Halogens also occupied the descending part of the graph nn Transition elements have very small volumes and therefore these are present at the bottoms of the curve metals have highest atomic volumes 6.(a) Mendeleef’s Periodic Table nn Periodic table is based on atomic weight nn In the periodic table, the horizontal lines are called periods and the vertical lines are called groups nn The periodic table consists of seven periods and nine groups (The earlier periodic table had only groups The noble gases were added later in the zero group because these were not discovered when Mendeleef put forward his periodic table nn All the groups (except VIII and Zero groups) are divided into subgroups A and B nn 2, 8, 18 and 32 are called magic numbers (b) Merits of Mendeleef’s Periodic Table nn Classification of elements then known, was done for the first time and the elements having similar properties were kept in the same group nn It encouraged research and led to discovery of newer elements nn Mendeleef had even predicted the properties of many elements not discovered at that time This helped in the discovery of these elements For example, Mendeleef predicted the properties of the following elements: (a) ka-boron—This was later called scandium (Sc) (b) Eka-aluminium—This was later called gallium (Ga) (c) Eka-silicon—This was later called germanium (Ge) nn Atomic weight of elements were corrected Atomic weight of Be was calculated to be × 4.5 = 13.5 by considering its valency Mendeleef calculated it × 4.5 = Periodic Properties and Chemical Bonding 1.59 The bond order of CO, CN– and NO+ is 3.0; Bond order of CO+, CN and NO is 2.5 and Bond order of NO– is 2.0 nn In molecules like PCl (sp d-hybridization), the two axial bonds are slightly elongated and hence slightly weaker than equatorial bonds This makes PCl5 quite reactive nn Coordinate Bond It is a special type of covalent bond in which both the shared electrons are contributed by one atom only It may be defined as “a covalent bond in which both electrons of the shared pair are contributed by one of the two atoms” Such a bond is also called as dative bond A coordinate or a dative bond is established between two such atoms, one of which has a complete octet and possesses a pair of valence electrons while the other is short of a pair of electrons [[ $ %[ [[ [ [[ %[ [[ [ $ % RU $ This bond is represented by an arrow (→) pointing towards acceptor atom The atom which contributes electron pair is called the donor while the atom which accepts it is called acceptor The compound consisting of the coordinate bond is termed coordinate compound Some examples of coordinate bond formation are given below: Note: Coordinate bond after formation is indistinguishable from a covalent bond The formation of a coordinate bond can be looked upon as a combination of electrovalent and covalent bonds The formation may be assumed to have taken place in two steps: (i) The donor atom loses one electron and transferred to acceptor atom As a result donor atom acquires a positive charge and the acceptor atom acquires a negative charge $ % $ ± % (ii) These two charged particles now contribute one electron each and this pair is shared by both the atoms $ ± % $ % ± As the coordinate bond is a combination of one electrovalent bond and one covalent bond, it is also termed as semi polar bond The compound consisting of the coordinate bond is termed coordinate compound Some examples of coordinate bond formation are given below: (i) Combination of ammonia and boron trifiuoride: Although the nitrogen atom has completed its octet in ammonia, it still has a lone pair of electrons in the valency shell which it can donate The boron atom in boron trifiuoride is short of two electrons which it accepts and completes its octet + + ) % + ) ) 'RQRU $FFHSWRU + + ) % + ) ) &RRUGLQDWH &RPSRXQG Any atom or ion or molecule which has one unshared electron pair which it can donate is termed as Lewis base while those which are capable of accepting the lone pair are termed as Lewis acids In above example, ammonia is a Lewis base while boron trifiuoride is a Lewis acid Note: H+ ion and cations of transition metals such as Cu2+, Co2+, Fe2+, Mn2+, Cr3+, Ni2+, etc., act as Lewis acids The donors are also called as ligands 1.60 Periodic Properties and Chemical Bonding (ii) Formation of ammonium ion: Hydrogen ion (H+) has no electron and thus accepts a lone pair donated by nitrogen H H | | → H–N → H H–N : + H + | | H H + (iii) Formation of H2O2: Oxygen in water has two lone pairs of electrons, out of which one is donated to another oxygen atom forming hydrogen peroxide + 2 + + + 'RQRU $FFHSWRU (iv) Formation of ozone: Oxygen molecule consists of two oxygen atoms linked by a double covalent bond Each oxygen atom has two lone pairs of electrons When one lone pair of electrons is donated to a third oxygen atom which has six electrons, a coordinate bond is formed 2 'RQRU 2 $FFHSWRU (v) Formation of CO: Carbon has four valency electrons and oxygen has six Two combine to form a double bond and a coordinate bond as to achieve their octet completed & $FFHSWRU & 'RQRU Characteristics of Coordinate Compounds: The properties of coordinate compounds are intermediate between the properties of electrovalent compounds and covalent compounds The main properties are described below: (i) Melting and Boiling Points: Their melting and boiling points are higher than purely covalent compounds and lower than ionic compounds (ii) Solubility: These are sparingly soluble in polar solvents like water but readily soluble in non-polar (organic) solvents (iii) Conductivity: Like covalent compounds, these are also bad conductors of electricity The solutions or fused mass not allow the passage of electricity In the previous section, we discussed about those compound which deviate from fully ionic to some degree of covalency A similar trend can also be observed with pure covalent molecules which can change to a partially ionic bond This happens when the electronegativities of the two atoms which form the covalent bond are not the same The atom having higher electronegativity will draw the bonded electron pair more towards itself resulting in a partial charge separation The distribution of the electron cloud in the bond does not remain uniform and shifts towards the more eletronegative one Such bonds are called polar covalent bonds For example, the bond formed between hydrogen and chlorine or between hydrogen and oxygen in water is of this type δ+ δ– H – Cl δ+ δ– δ+ H− O − H Molecules like HCl, H2O, NH3 i.e., molecules of the type H – X having two polar ends (positive and negative) are known as polar molecules The extent of polar character or the degree of polarity in a compound is given by it’s dipole moment which is defined as the product of the net positive or negative charge and the distance of separation of the charges i.e., the bond length The symbol of dipole moment is m m = electronic charge (e) × distance The unit of dipole moment is Debye (D) 1D = 3.33 × 10–30 cm = 10–18 esu cm Periodic Properties and Chemical Bonding 1.61 Dipole moment is indicated by an arrow having a symbol ( ) pointing towards the negative end Dipole moment has both magnitude and direction and therefore it is a vector quantity To calculate the dipole moment of a molecule we should calculate the net dipole due to all bonds and for lone pair if any Diatomic molecules like HCl, HF have the dipole moment of the bond (called bond dipole) equal to the molecular dipole as the structure has one bond only But for poly atomic molecules the net dipole is the resultant of the individual bond dipoles A compound having a zero dipole moment indicates that the compound is a symmetrical one Symmetrical molecule is non-polar even though it contains polar bonds For example, CO2, BF3, CH4, CCl4 being symmetrical molecules have zero resultant dipole moments and hence are non-polar since dipole moments summation of all the bonds present in the molecule cancel each other Dipole moment is usually indicated by an arrow having + on the tail (+→), above the polar bond and pointing towards the negative end ) & % ) 8QV\PPHWULFDO + ) & + + + 1RQOLQHDU 3RO\DWRPLF 0ROHFXOHV Worked-out Examples 75 Dipole moment of CCl4 is zero while that of CHCl3 is non zero Solution: Both CCl4 and CHCl3 have tetrahedral structure but CCl4 is symmetrical while CHCl3 is nonsymmetrical &O &O + & & &O &O 6\PPHWULFDO &O &O 1RQ6\PPHWULFDO &O Due to the symmetrical structure of CCl4 the resultant of bond dipoles comes out to be zero But in case of CHCl3 it is not possible as the presence of hydrogen introduces some dissymmetry 76 Both CO2 and N2O are linear, but dipole moment of CO2 is zero but N2O is non-zero Why? Solution: The answer lies in the structure of these molecules While CO2 is symmetrical, N2O is not, for which the bond dipoles not cancel each other leaving the molecule with a resultant dipole N≡N→O Exercise: Compare the dipole moment of NH3 and NF3 Dipole Moment in Aromatic Ring System The dipole moments of the aromatic compounds present a very good illustration of dipole moment We all know when a substituted benzene is treated with a reagent different products namely ortho, meta and para products are formed The dipole moments of these products are different since the orientation of the groups are different at ortho, meta and para position Let us take an example which will make it easily digestive for you Suppose we have three isomers of o-nitrophenol, m-nitrophenol and p-nitrophenol We have also the e.g., o-aminophenol, m-aminophenol and p-aminophenol 1.62 Periodic Properties and Chemical Bonding ; ; ; < < 0HWD 2UWKR < 3DUD In the case X = Y, the para isomer becomes symmetrical and have zero dipole Now the obvious question that is peeping through your mind is that which isomer in which case has got higher dipole moment The answer lies in the nature of the groups linked to the benzene ring In nitrophenol groups one group, is electron pushing and the other is electron withdrawing while in the second case both the groups attached are electron pushing So depending on the nature of the groups attached one of the isomer, o, m or p has the largest dipole moment Case (i): Now when X and Y are both electron pushing or electron withdrawing Suppose, bond dipole of C – × = m1 And that of C – Y = m2 Here, we have assumed a sign of (+) when groups are electron pushing and (–) when groups are electron withdrawing The net dipole is the resultant of two bond dipoles at different orientations When both X and Y are electron pushing or electron withdrawing mortho = m12 + m 22 + 2m1m cos 60° = m12 + m 22 + 2m1m ⋅ ∴ m0 = m12 + m 22 + m1m mmeta = m12 + m 22 + 2m1m cos120° ∴mm = m12 + m 22 − m1m mpara = m12 + m 22 + 2m1m cos180 ° = m12 + m 22 − 2m1m ∴ mp = m1 – m2 From the above expression of m0, mm and mp it is clear that when both X and Y are of the same nature i.e., both are electron withdrawing or both are electron pushing the para product has the least dipole moment and ortho product has the highest Now, when X = Y, mp = m1 – m2 = m1 – m1 = Which we have already discussed Case (ii): When X is electron pushing and Y is electron withdrawing or vice versa Let C – X dipole = m1 and C – Y dipole = m2 \ m0 = m12 + (−m ) + 2m1 (−m ) cos 60° = m12 + m 22 − m1m = (m1 + m ) − 3m1m mmeta = m12 + (−m ) + 2m1 (−m ) cos120° = m12 + m 22 + m1m = (m1 + m ) − m1m mpara = m12 + (−m ) + 2m1 (−m ) cos180° = m12 + m 22 + 2m1m =m1 + m2 Now if you see the expressions, it is very clear that the para isomer has the highest dipole moment and ortho is the least So to calculate the dipole moments of disubstituted benzene one should consider about the nature of the groups linked and then only one can predict the dipole moment of the molecule Periodic Properties and Chemical Bonding 1.63 Exercise: Arrange the following compounds in order of increasing dipole moments &O &O &O &+ &O &O &O , ,, ,,, &O ,9 Percentage of Ionic Character Every ionic compound having some percentage of covalent character according to Fajan’s rule The percentage of ionic character in a compound having some covalent character can be calculated by the following equation The percent ionic character = Observed dipole moment × 100 Calculated dipole moment assuming 100% ionic bond Worked-out Examples 77 Dipole moment of KCl is 3.336 × 10–29 coulomb metre which indicates that it is highly polar molecule The interatomic distance between k+ and Cl– is 2.6 × 10–10m Calculate the dipole moment of KCl molecule if there were opposite charges of one fundamental unit located at each nucleus Calculate the percentage ionic character of KCl Ans Dipole moment m = e × d coulomb metre For KCl d = 2.6 × 10–10 m For complete separation of unit charge e = 1.602 × 10–19 C Hence m = 1.602 × 10–19 × 2.6 × 10–10 = 4.1652 × 10–29 cm mKCl = 3.336 × 10–29 Cm ∴ % ionic character of KCl = 3.336 × 10−29 × 100 = 80.09% 4.165 × 10−29 78 Dipole moment of HX is 2.59 × 10–30 coulomb-metre Bond length of HX is 1.39Å Calculate percentage ionic character of molecule Ans dipole moment = q × d mcal = q × 1.39 × 10–10 coulomb – metre mcal = 1.6 × 10–19 × 1.39 × 10–10 µ exp 2.59 × 10−30 × 100 = 11.65% \ % age ionic character = × 100 = µ cal 1.6 × 10−19 × 1.36 × 10−10 79 Calculate the percentage of covalent character of HI having bond length = 1.62Å and observed dipole moment = 0.39 D 46 In water, the H—O—H bond angle is 105° Using the dipole moment of water and the covalent radii of the atoms, determine the magnitude of the charge on the oxygen atom in the water molecule 1.64 Periodic Properties and Chemical Bonding Solution: m = 1.85 D = 1.85 × 10–18 esu.cm = δd cos 52.5° = d/0.94 Å + G + d = (0.609)(0.94 Å) = 0.572 Å δ= or 2G± c G m 1.85D 1.8510 –18 esu cm = = = 3.2 ×10 –10 esu d 0.572 Å 0.572 ×10 –8 cm 0.67 times the electronic charge (4.8 × 10‑10 esu) Some Noteworthy Points nn nn The dipole moment helps to predict whether a molecule is polar or non polar A molecule may contain polar covalent bonds but its dipole moment may be zero if it is a symmetrical molecule The dipole moment helps to predict geometry of molecules The percentage of ionic character can be calculated For example, (i) A diatomic molecule having a dipole moment of 1.92 Debye and bond length of 2.0 Å shall have 20% ionic character This can be calculated as, m=e×d or 1.92 × 10–18 esu cm = e × × 10–8 cm nn nn or change (e) = 1.92 × 10−18 − 0.96 × 10−10 esu × 10−8 Since charge on isolated ion = 4.8 × 10–10 esu \ % age ionic character = 0.96 × 10−10 × 100 = 20% 4.8 × 10−10 (ii) The % age ionic character can also be calculate as follows; % age ionic character = Experimental value of dipole moment × 100’ Theoretical value of dipole moment e.g., when the dipole moment of HX is 1.92 D and bond distance is 1.2 Å, the % age ionic character can be calculate as: Theoretical value of dipole moment (m) when ionic character is 100% is = q × d = 4.8 × 10–10 esu × 1.2 × 10–8 cm (Θ Å = 10–8 cm) m = 5.76 × 10–18 esu cm m = 5.76 D nn % age ionic character = Experimental dipole moment 1.92 × 100 = × 100 = 33.3% Theoretical dipole moment 5.76 Odd Electron bonds: Luder (in 1916) postulated that there are a number of stable molecular in which the bonds are formed by sharing of an odd number of electrons (usually one or three The bond of this type involving odd number of electrons is called odd electron bond This type of bond is found in H +2 , He +2 , O2, NO, NO2, C1O2 molecule etc Periodic Properties and Chemical Bonding 1.65 nn Dipole moment values can be used to distinguish between the cis and trans isomers, Usually, cis isomers have higher dipole moment and are more polar than the trans isomers, e.g., + &O & & & & &O + nn + &O + &O Among the ortho, meta and para isomers, dipole moment is greatest for ortho isomer for para isomer (same substltuents) In general, dipole moment of o > m > p &O &O &O &O &O 2UWKR P ' 3DUD P 0HWD P ' If a molecule of the type MX2 has zero dipole moment, the σ-bonding orbitals used by M (Z < 21) must be sp-hybridized (e.g., BeF2) nn If a molecule of the type MX has zero dipole moment, the σ-bonding orbitals uses by M (Z < 21) must be sp -hybridized (e.g., BF3) nn If a molecule of the type MX has zero dipole moment, the σ-bonding orbitals used by M (Z < 21) must be sp3-hybridized (e.g., CCl4) nn Worked-out Examples 80 The dipole moment of NH3 is more than that of NF3 Explain Ans The dipole moment of NH3 acts in the directions H ——» N and thus moment due to unshared pair of electron will naturally increase the moment of the NH3 molecule while in the case of NF3, the dipole moment acts in the direction N ——> F and thus unshared electron pair will partially neutralize the dipole moment, causing a lower moment of NF3 relative of NH3 + : : + + 7RWDO 7RWDO ) ) ) 81 o-Hydroxy benzaldehyde is a liquid at room temperature while p-hydroxy benzaldehyde is a high melting solid [IIT 1999] Ans o-Hydroxy benzaldehyde show intramolecular H-bonding or chelation, a weaker one than intermolecular H-bonding in p-hydroxy benzaldehyde 1.66 Periodic Properties and Chemical Bonding + & 2 + 2+& &+2 + + 82 The dipole moment of Kcl is 3.336 × 10–29 coulomb metre which indicates that it is a highly polar molecule The interatomic distance between K+ and Cl– in this molecule is 2.6 × 10–10 m Calculate the dipole moment of KCI molecule, if there were opposite charges of one fundamental unit located at each nucleus Calculate percentage ionic character of Kcl [IIT 1993] Ans Dipole moment m = δ × d ∴ 3.336 × 10–29 = × 2.6 × 10–10 ∴ δ = ∴ 1.602 × 10–19 charge on each, % character = 100 ∴ 1.283 × 10–19 charge on each, % character = 3.336 × 10−29 = 1.283 × 10–29 coulomb 2.6 × 10−10 1.283 × 10−19 × 100 = 80.09% 1.602 × 10−19 ∴ If one unit charge, then δ = 1.602 × 10–19 C ∴ m = 1.602 × 10–19 × 2.6 × 10–10 = 4.1652 × 10–29 coulomb metre Hydrogen Bonding In 1920, Latimer and Rodebush introduced the idea of “hydrogen bond” to explain the nature of association in liquid state of substance like water, hydrogen fluoride, ammonia, formic acid etc In a hydrogen compound, when hydrogen is bonded to highly electronegative atom (such as F, O, N) by a covalent bond, the electron pair is attracted towards electronegative atom so strongly that a dipole results i.e., one end carries a positive charge (H-end) and other end carries a negative charge (X-end) G± ; G + G± G± ; + RU (OHFWURQHJDWLYH DWRP If a number of such molecules are brought nearer to each other, the positive end of one molecule and negative end of the other molecule will attract each other and weak electrostatic force will develop Thus, these molecules will associate together to form a cluster of molecules δ− δ+ δ− δ+ δ− δ+ δ− δ+ δ− δ+ X − H X − H X − H X − H X − H The attractive force that binds hydrogen atom of one molecule with electronegative atom of the other molecule of the same or different substance is known as hydrogen bond Periodic Properties and Chemical Bonding 1.67 Hydrogen bonding is of two types: (a) Intermolecular hydrogen bonding: This type of bonding results between the positive and negative ends of different molecules of the same or different substances Example (i) Ammonia G G G G G + + + + + G± G± G± G± G± G + + + + + + G (ii) Water + G + G + G + G + G + G± + G + G± + G + G± + G + G± (iii) Acetic acid +& + + &+ This type of hydrogen bonding increases the boiling point of the compound and also its solubility in water The increase in boiling point is due to association of several molecules of the compound (b) Intramolecular hydrogen bonding: This type of bonding results between hydrogen and an electronegative element both present in the same molecule This type of bonding is generally present in organic compounds Examples are o-nitro-phenol, o-hydroxy benzoic acid, etc +± 2 R1LWURSKHQRO +± 2 R+\GUR[\EHQ]RLFDFLG This type of bonding decreases the boiling point of the compound The solubility of the compound also decreases Hence, compound becomes more volatile Properties Explained by Hydrogen Bonding (a) Strength of certain acids and bases can be explained on the basis of hydrogen bonding (b) Solubility: An organic substance is said to be insoluble in water if it does not form hydrogen bonding with water The organic compound like alkanes, alkenes, ethers, etc., are insoluble in water as they not form hydrogen bonding with water, while alcohols and acids are soluble because they readily form hydrogen bonds with water (i) Melting and boiling points of hydrides of N, O and F: If the melting points and boiling points of the hydrides of the elements of IVA, VA, VIA and VIIA groups are plotted against the molecular weights of these hydrides, we shall get the plots as shown in figure (a) and (b) From these plots it may be seen that although in case of SbH3, AsH3, PH3 (VA group elements hydrides), H2Te, H2Se, H2S (VI A group elements hydrides) and HI, HBr, HCl (VIII group elements hydrides) there is a progressive decrease in their m.p’s and b.p’s with the decrease in their molecular weights, the m.p’s and b.p’s of NH3, H2O and HF hydrides suddenly increase with a further decrease of their molecular weights The sudden increase in m.p’s and b.p’s in these hydrides is due to the inter-molecular H-bonding in between H and F in case of HF, in between 1.68 Periodic Properties and Chemical Bonding H and O in case of H2O and in between H and N in case of NH3 respectively The existence of H-bonding in these molecules gives polymerized molecules (NH3)n Thus, m.p’s and b.p’s of these molecules are suddenly raised Having no power to form H-bonds, the simple carbon family hydrides (SnH4, GeH4, SiH4 and CH4) show a decrease in their bp’s and mp’s with the decrease in their molecular weights + 2 1+ ± +6H + 6 +3 +%U 3+ 6Q+ 9,$ 9,,,$ 9$ 6E+ +&O +) &+ ± +6H $V+ 6Q+ *H+ ,9$ %RLOLQJSRLQWV & LQFUHDVLQJ 0HOWLQJSRLQWV & LQFUHDVLQJ +2 +) +6H 1+ +6 +&O ± 3+ 6Q+ +6H 6E+ +, +%U *H+ 6Q+ 9,$ 9$ 9,,,$ ,9$ &+ ± 0ROHFXODUZHLJKWLQFUHDVLQJ D 0ROHFXODUZHLJKWLQFUHDVLQJ E Fig 1.11 (ii) Ice has less density than water: The explanation of this fact is as follows: In the crystal structure of ice, the Oatom is surrounded by four H-atoms Two H-atoms are linked to O-atom by covalent bonds as shown (by normal covalent bond) and the remaining two H-atoms are linked to O-atom by two H-bonds shown by dotted lines Thus in ice every water molecule is associated with four other water molecules by H-bonding in a tetrahedral fashion Ice has an open cage like structure with a large empty space due to the existence of H-bonds As ice melts at 0°C, a number of H-bonds are broken down and the space between water molecules decreases so that water molecules move closer together The density of water increases, from 0° to 4°C, and at 4°C it is maximum Above 4°C the increase in kinetic energy of the molecules is sufficient to cause the molecules to begin to disperse and the result is that the density decrease with increasing temperature + + c c ± + c ZDWHUPROHFXOH + + + + + + + Fig 1.12 2SHQFDJHOLNHWHWUDKHGUDOFU\VWDOVWUXFWXUH RILFH&LUFOHVLQGLFDWHR[\JHQDWRPV %RQGVUHSUHVHQWHGE\VROLGOLQHDUHQRUPDO FRYDOHQWERQGVZKLOHWKRVHUHSUHVHQWHGE\ GRWWHGOLQHVDUHK\GURJHQERQGV Periodic Properties and Chemical Bonding 1.69 Worked-out Examples 83 Which of the following hydrogen bonds is the strongest? (a) O—H—N (b) F—H—F (c) O—H—O (d) O—H—F [AIEEE 2007] Ans (b) Greater the difference between electronegativity of bonded atoms, stronger will be bond Since F is most electronegative hence F–H … F is the strongest bond 84 An ether is more volatile than alcohol having same molecular formula This is due to: (a) inter molecular H-bonding in ethers (b) inter molecular H-bonding in alcohols (c) dipolar character of ethers (d) resonance structure in alcohols [AIEEE 2003] Ans (b) In ether, there is no H-bonding while alcohols have intermolecular H-bonding (d) The C=O bond is weaker than the C—O bond Rules for linear combination of atomic orbitals In deciding which atomic orbitals may be combined to form molecular orbitals, three rules must be considered: The atomic orbitals must be roughly of the same energy This is important when considering overlap between two different types of atoms The orbitals must overlap one another as much as possible This implies that the atoms must be close enough for effective overlap and that the radial distribution functions of the two atoms must be similar at this distance In order to produce bonding and anti-bonding MOs, either the symmetry of the two atomic orbitals must remain unchanged when rotated about the internuclear line, or both atomic orbitals must change symmetry in an identical manner In the same way that each atomic orbital has a particular energy, and may be defined by four quantum numbers, each molecular orbital has a definite energy, and is also defined by four quantum numbers The principal quantum number n has the same significance as in atomic orbitals The subsidiary quantum number l also has the same significance as in atomic orbitals The magnetic quantum number of atomic orbitals is replaced by a new quantum number l In a diatomic molecule, the line joining the nuclei is taken as a reference direction and l, represents the quantization of angular momentum in h/2p units with respect to this axis l takes the same values as m takes for atoms, i.e., l = – l, , –3, –2, –1, 0, +1, +2, +3, , + l When l = 0, the orbitals are symmetrical around the axis and are called o orbitals When l = ±1 they are called p orbitals and when l = ±2 they are called δ orbitals The spin quantum number is the same as for atomic orbitals and may have values of ± The Pauli exclusion principle states that in a given atom no two electrons can have all four quantum numbers the same The Pauli principle also applies to molecular orbitals: No two electrons in the same molecule can have all four quantum numbers the same ... of Formation of Ionic Substances The energy included in the formation of an ionic compound from its constituent elements may be considered as shown by the Born-Haber Cycle for the formation of. .. negative value of the heat of formation, greater would be the stability of the ionic compound produced Thus on the basis of the above equation, formation of an ionic compound is favoured by, (a) low.. .Fundamentals of Inorganic Chemisty Ananya Ganguly Chandigarh • Delhi • Chennai The aim of this publication is to supply information taken from sources believed