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Unit, , 10, , Chemical bonding, Learning Objectives, After studying this unit students will be able to, , Linus Carl Pauling, was an American chemist,, biochemist, peace activist,, author and educator. In, addition to his contribution, to chemistry and he, also worked with many, biologists., He received the Nobel, Prize in Chemistry in 1954, for his research into the, nature of the chemical bond, and its application to the, elucidation of the structure, of complex substances., , •, , describe Kossel – Lewis approach to chemical, bonding, , •, , explain the octet rule, , •, , sketch the Lewis structures of simple, molecules, , •, , describe the formation of different types of, bonds and bond parameters, , •, , sketch the resonance structures for simple, molecules, , •, , apply the concept of electronegativity to, explain the polarity of covalent bonds, , •, , describe VSEPR theory and predict the shapes, of simple molecules, , •, , explain the valence bond approach for the, formation of covalent bonds, , •, , explain the different types of hybridisation, involving s, p & d orbitals and sketch shapes, of simple covalent molecules, , •, , explain the molecular orbital theory, calculate, the bond order and explain the magnetic, properties of H2, O2, N2 CO and NO, , •, , describe metallic bonding briefly., , 10.1 Introduction, Diamond is very hard while its allotrope, graphite is very soft. Gases like hydrogen and oxygen, are diatomic while the inert gases are monoatomic., UpBright Classes, Haldia WhatsApp 9734878427, Unit 10 layout.indd 67, , 67, , Page 1 of 43, 27-08-2018 17:53:02

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or sharing one or more electrons from, their outer shell. For example, sodium loses, one electron to form Na+ ion and chlorine, accepts that electron to give chloride ion (Cl–),, enabling both atoms to attain the nearest, noble gas configuration. The resultant ions,, Na+ and Cl- are held together by electrostatic, attractive forces and the attractive force is, called a chemical bond, more specifically an, electrovalent bond., , Carbon combines with chlorine to form, carbon tetrachloride, which is a liquid and, insoluble (immiscible) in water. Sodium, combines with chlorine atom to form sodium, chloride, a hard and brittle compound that, readily dissolves in water. The possible, reason for these observations lies in the type, of interaction that exists between the atoms, of these molecules and these interactions, are responsible for holding the atoms/ions, together. The interatomic attractive forces, which hold the constituent atoms/ions, together in a molecule are called chemical, bonds., , [Ar], , Why do atoms combine only in, certain combinations to form molecules? For, example oxygen combines with hydrogen to, give water (H2O) and with carbon it gives, carbon dioxide (CO2). The structure of, water is ‘V’ shaped while that of the carbon, dioxide is linear. Such questions can be, answered using the principles of chemical, bonding. In this unit we will analyse the, various theories and their principles, which, were developed over the years to explain the, nature of chemical bonding., , G. N. Lewis proposed that the, attainment of stable electronic configuration, in molecules such as diatomic nitrogen,, oxygen etc… is achieved by mutual sharing, of the electrons. He introduced a simple, scheme to represent the chemical bond and, the electrons present in the outer shell of, an atom, called Lewis dot structure. In this, scheme, the valence electrons (outer shell, electrons) of an element are represented, as small dots around the symbol of the, element. The first four valence electrons, are denoted as single dots around the four, sides of the atomic symbol and then the fifth, onwards, the electrons are denoted as pairs., For example, the electronic configuration, of nitrogen is 1s2, 2s2, 2p3. It has 5 electrons, in its outer shell (valence shell). The Lewis, structure of nitrogen is as follows., , 10.1.1 Kossel – Lewis approach to, chemical bonding, A logical explanation for chemical, bonding was provided by Kossel and, Lewis in 1916. Their approach to chemical, bonding is based on the inertness of the, noble gases which have little or no tendency, to combine with other atoms. They proposed, that the noble gases are stable due to their, completely filled outer shell electronic, configuration. Elements other than noble, gases, try to attain the completely filled, electronic configurations by losing, gaining, UpBright Classes, Haldia WhatsApp 9734878427, Unit 10 layout.indd 68, , N, Fig 10.1 Lewis Structure of Nitrogen atom, , Similarly, Lewis dot structure of carbon,, oxygen can be drawn as shown below., , 68, , Page 2 of 43, 27-08-2018 17:53:02

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configuration by mutually sharing the, electrons available with them. Similarly,, in the case of oxygen molecule, both the, oxygen atoms share two electron pairs, between them and in nitrogen molecule, three electron pairs are shared between, two nitrogen atoms. This type of mutual, sharing of one or more pairs of electrons, between two combining atoms results in, the formation of a chemical bond called a, covalent bond. If two atoms share just one, pair of electron a single covalent bond is, formed as in the case of hydrogen molecule., If two or three electron pairs are shared, between the two combining atoms, then the, covalent bond is called a double bond or a, triple bond, respectively., , O, , C, , Fig 10.2 Lewis Structures of C & O atoms, , Only exception to this is helium which has, only two electrons in its valence shell which, is represented as a pair of dots (duet)., , He, Fig 10.3 Lewis Structures of He atoms, , Octet rule, The idea of Kossel – Lewis approach, to chemical bond lead to the octet rule,, which states that “the atoms transfer or, share electrons so that all atoms involved, in chemical bonding obtain 8 electrons in, their outer shell (valence shell)”., , 10.2 Types of chemical bonds, The chemical bonds can be classified, based on the nature of the interaction, between the bonded atoms. Two major, types of chemical bonds are covalent bonds, and ionic bonds. Generally metals reacts, with non-metals to form ionic compounds,, and the covalent bonds are present in the, compounds formed by nonmetals., , Single Covalent Bond, , (O2), Double Covalent Bond, , 10.2.1 Covalent bonds:, Do you know all elements (except, noble gases) occurs either as compounds or, as polyatomic molecules? Let us consider, hydrogen gas in which two hydrogen, atoms bind to give a dihydrogen molecule., Each hydrogen atom has one electron and, it requires one more electron to attain the, electronic configuration of the nearest, noble gas helium. Lewis suggested that, both hydrogen atoms will attain the stable, , UpBright Classes, Haldia WhatsApp 9734878427, Unit 10 layout.indd 69, , (H2), , (N2), Triple Covalent Bond, , Fig 10. 4 Representation of Lewis Structures, of covalent bonds, , 69, , Page 3 of 43, 27-08-2018 17:53:02

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10.2.2 Representing a covalent bond Lewis structure (Lewis dot structure), , hydrogen)] + [1 × 6 (valence electrons, of oxygen)] = 2 + 6 = 8., , Lewis structure (Lewis dot structure), is a pictorial representation of covalent, bonding between the combining atoms. In, this structure the shared valence electrons, are represented as a pair of dots between, the combining atoms and the unshared, electrons of the atoms are represented as, a pair of dots (lone pair) on the respective, individual atoms., , 3., , The Lewis dot structure for a given, compound can be written by following the, steps given below. Let us understand these, steps by writing the Lewis structure for, water., , 4., , 1., , 2., , H, , Draw the skeletal structure of, the molecule. In general, the most, electronegative atom is placed at the, centre. Hydrogen and fluorine atoms, should be placed at the terminal, positions. For water, the skeletal, structure is, H, , O, , H, , Distribute the remaining valence, electrons as pairs (lone pair), giving, octet (only duet for hydrogen) to, the atoms in the molecule. The, distribution of lone pairs starts with the, most electronegative atoms followed by, other atoms., , H, 5., , Calculate the total number of, valence electrons of all the atoms in, the molecule. In case of polyatomic, ions the charge on ion should also be, considered during the calculation of the, total number of valence electrons. In, case of anions the number of negative, charges should be added to the number, of valence electrons. For positive ions, the total number of positive charges, should be subtracted from the total, number of valence electrons., , O, , H, , Verify weather all the atoms satisfy the, octet rule (for hydrogen duet). If not,, use the lone pairs of electrons to form, additional bond to satisfy the octet rule., In case of water, oxygen has octet and, the hydrogens have duets, hence there is, no need for shifting the lone pairs. The, Lewis structure of water is as follows, H, , O, , H, , Fig 10. 5 Lewis structure of water, , Let us draw the Lewis structure for, nitric acid., , In water, total number of valence, electron =[2×1 (valence electron of, , Unit 10 layout.indd 70, , O, , In case of water, the remaining four, electrons (two lone pairs) are placed, on the most electronegative central, oxygen, giving octet., , H, , UpBright Classes, Haldia WhatsApp 9734878427, , Draw a single bond between the, atoms in the skeletal structure of the, molecule. Each bond will account for, two valence electrons (a bond pair)., For water, we can draw two bonds, accounting for four valence electrons as, follows., , 1., , 70, , Skeletal structure, Page 4 of 43, 27-08-2018 17:53:02

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(three each) to satisfy their octet and, two pairs are distributed to the oxygen, that is connected to hydrogen to satisfy, its octet., , H O N O, O, 2., , Total number of valence electrons in, HNO3, , H, , = [1 × 1(hydrogen)] + [1 × 5(nitrogen)], + [3× 6(oxygen)] = 1+ 5 + 18 = 24, 3., , N O, O, , Draw single bonds between atoms., Four bonds can be drawn as shown in, the figure for HNO3 which account for, eight electrons (4 bond pairs)., H O N O, , 5., , Verify weather all the atoms have octet, configuration. In the above distribution,, the nitrogen has one pair short for, octet. Therefore, move one of the lone, pair from the terminal oxygen to form, another bond with nitrogen., , O, 4., , O, , Distribute the remaining sixteen (24, - 8= 16) electrons as eight lone pairs, starting from most electronegative, atom, the oxygen. Six lone pairs are, distributed to the two terminal oxygens, , The Lewis structure of nitric acid is given as, H, , O, , N O, O, , Fig 10. 6 Lewis structure of Nitric acid, , Table 10.1 : The Lewis dot structures for some molecules, S. No, , Molecule, , 1., , Sulphur trioxide (SO3), , 2., , Ammonia (NH3), , O, O, , Lewis Structure, , O, , OS O, , N H, , H, H N H, , S, H, , H, , H, 3., , Methane, , H, , C H, H, , 4., , Dinitrogen Pentoxide, (N2O5), , O N O, O, , O, , N O, O, , H, H C H, H, , O NON O, O O, , Note, It is to be noted that nearly in all their compounds, certain elements form a fixed, number of bonds. For example, Fluorine forms only one bond. Hydrogen, oxygen,, nitrogen and carbon atoms form one, two, three and four bonds, respectively., , UpBright Classes, Haldia WhatsApp 9734878427, Unit 10 layout.indd 71, , 71, , Page 5 of 43, 27-08-2018 17:53:02

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Evaluate Yourself, , for octet. Therefore, to satisfy the octet, rule two lone pairs from one oxygen, or one pair from each oxygen can be, moved to form multiple bonds, leading, the formation of two possible structures, for carbon dioxide as shown below, , ?, , 1) Draw the lewis structures for, i), , Nitrous acid (HNO2), , ii), , Phosphoric acid, , iii) Sulphur troxide (SO3), , O C O, structure 1, , 10.2.3 Formal charge:, , Fig 10. 7 (a) two possible structures for carbon, dioxide, , Let us draw the Lewis structure for carbon, dioxide., 1., , Similarly, the Lewis structure for, many molecules drawn using the above steps, gives more than one acceptable structure., Let us consider the above mentioned two, structures of carbon dioxide., , Skeletal structure, O C O, , 2., , Which one the above forms, represents the best distribution of electrons, in the molecule. To find an answer, we need, to know the formal charge of each atom in, the Lewis structures. Formal charge of an, atom in a molecule, is the electrical charge, difference between the valence electron in an, isolated atom and the number of electrons, assigned to that atom in the Lewis structure., , Total number of valence electrons in, CO2, , = [1 x 4(carbon)] +[2 x 6(oxygen)] =, 4+ 12 = 16, 3., , Draw single bonds between atoms. Two, bonds can be drawn as shown in the, figure for CO2 which accounts for four, electrons (2 bond pairs)., O, , 4., , C, , Where,, Nv-, , Number of valence electron of atom, in its isolated state., , Nl - Number of electrons present as lone, pairs around the atom in the Lewis, structure, , O, , Nb - Number of electrons present in bonds, around the atom (bond pairs) in the, , Verify weather all the atoms have octet, configuration. In the above distribution,, the central carbon has two pair short, , UpBright Classes, Haldia WhatsApp 9734878427, Unit 10 layout.indd 72, , N , , Formal charge of an atom = Nv -  N l + b , 2 , , , O, , Distribute the remaining twelve, electrons (16 - 4= 12) as six lone pairs, starting from most electronegative, atom, the oxygen. Six lone pairs are, distributed to the two terminal oxygens, (three each) to satisfy their octet., O, , 5., , C, , O C O, structure 2, , 72, , Page 6 of 43, 27-08-2018 17:53:03

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Lewis structure], Now let us calculate the formal charge on all, atoms in both structures,, For Structure 1,, N , , Nv --  N, N l ++ N bb , Formal charge, charge on, on carbon, carbon == N, Formal, v,  l, 22 , , , == 44 -- 00 ++, , , 88  = 0, = 0, 22 , , N , , = Nv -  Nl + b , 2 , , 8, , = 4 - 0 +  = 0, 2, , Formal charge on singly bonded oxygen, 2, , = 6 -  6 +  = −1, 2, , Formal charge on triply bonded oxygen, 6, , = 6 - 2 +  = +1, 2, , , O, , structure 1, , O, , structure 2, , Unit 10 layout.indd 73, , 2., , Molecules containing odd electrons, , 3., , Molecules with expanded valence shells, , F, , After calculating the formal charges,, the best representation of Lewis structure, can be selected by using following guidelines., , electron, , electron, , deficient, , deficient, , B, , F, , F, , A structure in which all formal charges, are zero preferred over the one with, charges., , UpBright Classes, Haldia WhatsApp 9734878427, , Molecules with, central atoms, , Let us consider boron trifluoride,, as an example. The central atom boron has, three valence electron and each fluorine has, seven valence electrons. The Lewis structure, is, , O, , Fig 10. 7 (b) two possible structures for, carbon dioxide (with formal charges), , 1., , 1., , Molecules with, central atoms, , –1, , C, , A structure in which negative formal, charges are placed on the most, electronegative atom is preferred., , The octet rule is useful for writing, Lewis structures for molecules with second, period element as central atoms. In some, molecules, the central atoms have less than, eight electrons around them while some, others have more than eight electrons., Exception to the octet rule can be categorized, into following three types., , Formal charge on carbon, , C, , 3., , 10.2.4 Lewis structures for exceptions, to octet rule, , For structure 2, , O, , A structure with small formal charges, is preferred over the one with higher, formal charges., , In case of CO2 structures, the, structure one is preferred over the structure, 2 as it has zero formal charges for all atoms., , 44 , , Formal charge, charge on, on oxyge, oxygen, n == 66 --  44 ++ 2 , Formal, 2 , , for both, both oxygens, oxygens)), == 00 (( for, , +1, , 2., , Fig 10. 8 (a) Lewis structure of BF3, , In the above structure, only six, , 73, , Page 7 of 43, 27-08-2018 17:53:03

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bonding pair of electrons or twelve electrons., , electrons around boron atom. Moving a, lone pair from one of the fluorine to form, additional bond as shown below., , Cl, Cl, , F B F, F, , Cl, , However, the above structure is, unfavourable as the most electronegative, atom fluorine shows positive formal charge, and hence the structure with incomplete, octet is the favourable one. Molecules such, as BCl3, BeCl2, etc... also have incomplete, octets., , F, F, , S, , F, , F, SF6, Fig 10. 10 Lewis structures for SF6 and PCl5, , Few molecules have a central atom, with an odd number of valence electrons., For example, in nitrogen dioxide and nitric, oxide all the atoms does not have octet, configuration. The lewis structure of the, above molecules are shown in the figure., , Evaluate Yourself, , ?, , 2) Calculate the formal charge on each, atom of carbonyl chloride (COCl2), , 10.3 Ionic or electrovalent bond, , +1 –1, , When the electronegativity difference, between the two combining atoms is large,, the least electronegative atom completely, transfers one or more of its valence, electrons to the other combining atom so, that both atoms can attain the nearest inert, gas electronic configuration. The complete, transfer of electron leads to the formation of, a cation and an anion. Both these ions are, held together by the electrostatic attractive, force which is known as ionic bond., , O N O, , Fig 10. 9 Lewis structures of Nitric oxide and, Nitrogen dioxide (with formal charges), , Molecules with expanded valence shells, In molecules such as sulphur, hexafluoride, (SF6),, phosphorous, pentachloride (PCl5) the central atom, has more than eight valence electrons, around them. Here the central atom can, accommodate additional electron pairs, by using outer vacant d orbitals. In SF6 the, central atom sulphur is surrounded by six, , Unit 10 layout.indd 74, , F, , F, , Molecules containing odd electrons, , UpBright Classes, Haldia WhatsApp 9734878427, , Cl, (PCl5), , Fig 10. 8 (b) Lewis structure of BF3, , N O, , Cl, , P, , Let us consider the formation, potassium chloride. The electronic, configuration of potassium and chlorine are, , 74, , Page 8 of 43, 27-08-2018 17:53:03

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Potassium (K) : [Ar] 4s1, , These type of bonds are called coordinate, covalent bond or coordinate bond. The, combining atom which donates the pair, of electron is called a donor atom and the, other atom an acceptor atom. This bond, is denoted by an arrow starting from the, donor atom pointing towards the acceptor, atom. (Later in coordination compound, we, will refer the donor atom as ligand and the, acceptor atom as central-metal atom/ion., , Chlorine (Cl) : [Ne]3s2, 3p5, Potassium has one electron in its, valence shell and chlorine has seven electron, in its valence shell. By loosing one electron, potassium attains the inert gas electronic, configuration of argon and becomes a, unipositive cation (K+) and chlorine accepts, this electron to become uninegative chloride, ion (Cl–) there by attaining the stable, electronic configuration of argon. These two, ions combine to form an ionic crystal in, which they are held together by electrostatic, attractive force. The energy required for the, formation of one mole of K+ is 418.81 kJ, (ionization energy) and the energy released, during the formation of one mole of Cl- is, -348.56 kJ (electron gain enthalpy). The, sum of these two energies is positive (70.25, kJ) However, during the formation of one, mole potassium chloride crystal from its, constituent ions, 718 kJ energy is released., This favours the formation of KCl and its, stabilises., , Evaluate Yourself, , For Example, in ferrocynide ion, [Fe(CN)6]4–, each cyanide ion (CN–) donates, a pair of electrons to form a coordinate, bond with iron (Fe2+) and these electrons, are shared by Fe­2+ and CN­-­., 4–, , NC, , CN, , CN, , Fig 10. 11 Structure of Ferrocyanide ion, , In certain cases, molecules having, a lone pair of electrons such as ammonia, donates its pair to an electron deficient, molecules such as BF3. to form a coordinate, , 3) Explain the ionic bond formation in, MgO and CaF2, , F, , 10.4 Coordinate covalent bond, , F, , H, , H, , F, F, , H, , N, , B, , F, , In the formation of a covalent bond,, both the combining atoms contribute one, electron each and the these electrons are, mutually shared among them. However,, in certain bond formation, one of the, combining atoms donates a pair of electrons, i.e. two electrons which are necessary for the, covalent bond formation, and these electrons, are shared by both the combining atoms., , Unit 10 layout.indd 75, , CN, , Fe, , ?, , UpBright Classes, Haldia WhatsApp 9734878427, , CN, , NC, , H, N, , B, , F, , H, H, , Fig 10. 12 Structure of BF3 → NH3, , 75, , Page 9 of 43, 27-08-2018 17:53:03

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length. For example, the carbon-carbon, single bond is longer than the carboncarbon double bond (1.33 Å) and the, carbon-carbon triple bond (1.20 Å)., , 10.5 Bond parameters, A covalent bond is characterised by, parameters such as bond length, bond angle,, bond order etc... A brief description of some, of the bond parameters is given below., , 10.5.2 Bond order, The number of bonds formed, between the two bonded atoms in a molecule, is called the bond order. In Lewis theory, the, bond order is equal to the number of shared, pair of electrons between the two bonded, atoms. For example in hydrogen molecules,, there is only one shared pair of electrons, and hence, the bond order is one. Similarly,, in H2O, HCl, Methane, etc the central atom, forms single bonds with bond order of one., , 10.5.1 Bond length, The distance between the nuclei of, the two covalently bonded atoms is called, bond length. Consider a covalent molecule, A-B. The bond length is given by the sum, of the radii of the bonded atoms (rA + rB)., The length of a bond can be determined, by spectroscopic, x-ray diffraction and, electron-diffraction techniques The bond, length depends on the size of the atom and, the number of bonds (multiplicity) between, the combining atoms., , rB, , Bond length, Fig 10. 13 Bond length of covalent molecule, A–B, , Greater the size of the atom, greater, will be the bond length. For example,, carbon-carbon single bond length (1.54 Å), is longer than the carbon-nitrogen single, bond length (1.43 Å)., Increase in the number of bonds, between the two atoms decreases the bond, UpBright Classes, Haldia WhatsApp 9734878427, Unit 10 layout.indd 76, , 76, , S., No., , Bonded atoms, , rA, , B, , Molecule, , A, , Table 10.2 Bond order of some common, bonds:, Bond, order (No., of shared, pair of, electrons, between, bonded, atoms), , 1, , H2, , H-H, , 1, , 2, , O2, , O=O, , 2, , 3, , N2, , N≡N, , 3, , 4, , HCN, , C≡N, , 3, , 5, , HCHO, , C=O, , 2, , 6, , CH4, , C–H, , 1, , 7, , C2H4, , C=C, , 2, , Page 10 of 43, 27-08-2018 17:53:03

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10.5.3 Bond angle, , bonds, the arithmetic mean of the bond, energy values of the same type of bonds is, considered as average bond enthalpy. For, example in water, there are two OH bonds, present and the energy needed to break, them are not same., , Covalent bonds are directional in, nature and are oriented in specific directions, in space. This directional nature creates a, fixed angle between two covalent bonds in, a molecule and this angle is termed as bond, angle. It is usually expressed in degrees., The bond angle can be determined by, spectroscopic methods and it can give some, idea about the shape of the molecule., , H2O(g)→H(g)+OH(g) ΔH1 = 502 kJ mol-1, OH(g)→H(g)+O(g), , The average bond enthalpy of OH, 502+427, bond in water =, = 464.5 kJ mol-1, 2, , S. No., , Molecule, , Atoms, defining, the angle, , Bond, angle (⁰), , Table 10.3 Bond angles for some, common molecules, , 1, , CH4, , H-C-H, , 109⁰ 28', , 2, , NH3, , H-N-H, , 107⁰ 18', , 3, , H2O, , H-O-H, , 104⁰ 35', , Table 10.4 Bond lengths and bond, enthalpies of some common bonds:, , 10.5.4 Bond enthalpy, The bond enthalpy is defined as, the minimum amount of energy required, to break one mole of a particular bond in, molecules in their gaseous state. The unit, of bond enthalpy is kJ mol-1. Larger the, bond enthalpy, stronger will be the bond., The bond energy value depends on the, size of the atoms and the number of bonds, between the bonded atoms. Larger the size, of the atom involved in the bond, lesser is, the bond enthalpy., In case of polyatomic molecules with,, two or more same bond types, in the term, average bond enthalpy is used. For such, , UpBright Classes, Haldia WhatsApp 9734878427, Unit 10 layout.indd 77, , ΔH2 = 427 kJ mol-1, , 77, , S., No., , Bond, type, , Bond, Enthalpy, (kJ mol-1), , Bond, Length, (Å), , 1, , H-H, , 432, , 0.74, , 2, , H-F, , 565, , 0.92, , 3, , H-Cl, , 427, , 1.27, , 4, , H-Br, , 363, , 1.41, , 5, , H-I, , 295, , 1.61, , 6, , C-H, , 413, , 1.09, , 7, , C-C, , 347, , 1.54, , 8, , C-Si, , 301, , 1.86, , 9, , C-N, , 305, , 1.47, , 10, , C-O, , 358, , 1.43, , 11, , C-P, , 264, , 1.87, , 12, , C-S, , 259, , 1.81, , 13, , C-F, , 453, , 1.33, , 14, , C-Cl, , 339, , 1.77, , 15, , C-Br, , 276, , 1.94, , 16, , C-I, , 216, , 2.13, , Page 11 of 43, 27-08-2018 17:53:03

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10.5.5 Resonance, , (OB and OC) thus creating three similar, structures as shown below in which the, relative position of the atoms are same., They only differ in the position of bonding, and lone pair of electrons. Such structures, are called resonance structures (canonical, structures) and this phenomenon is called, resonance., , When we write Lewis structures, for a molecule, more than one valid Lewis, structures are possible in certain cases. For, example let us consider the Lewis structure, of carbonate ion [CO3]2-., The skeletal structure of carbonate ion (The, oxygen atoms are denoted as OA, OB & OC, , O, , B, , O, , B, , C, , OA, , A, , OC, , Total number of valence electrons = [1 x, 4(carbon)] + [3 x 6 (oxygen)] + [2 (charge)], = 24 electrons., , C, 1, , O, , A, , B, , O, , O, , A, , C, , C, , Fig 10. 14 (a) Lewis Structure of CO32-, , In this case, we can draw two, additional Lewis structures by moving, the lone pairs from the other two oxygens, UpBright Classes, Haldia WhatsApp 9734878427, Unit 10 layout.indd 78, , O, , 3, , It is evident from the experimental, results that all carbon-oxygen bonds in, carbonate ion are equivalent. The actual, structure of the molecules is said to be the, resonance hybrid, an average of these three, resonance forms. It is important to note that, carbonate ion does not change from one, structure to another and vice versa. It is not, possible to picturise the resonance hybrid by, drawing a single Lewis structure. However,, the following structure gives a qualitative, idea about the correct structure., , 2–, , O, , C, , Fig 10. 14 (b) Resonance structures of CO32-, , B, , C, , C, , 2–, , O, , O, A, , O, , 2, , Complete the octet for carbon by, moving a lone pair from one of the oxygens, (OA) and write the charge of the ion (2-) on, the upper right side as shown in the figure., , O, , 2–, , C, , B, , C, , C, , B, , O, A, , O, , O, , Distribution of these valence electrons gives, us the following structure., , O, , 2–, , O, , 78, , Page 12 of 43, 27-08-2018 17:53:03

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fluoride. The electronegativities of hydrogen, and fluorine on Pauling's scale are 2.1 and 4, respectively. It means that fluorine attracts, the shared pair of electrons approximately, twice as much as the hydrogen which leads, to partial negative charge on fluorine and, partial positive charge on hydrogen. Hence,, the H-F bond is said to be polar covalent, bond., , –2, 3, , o, –2, 3, , o, , C, , o–23, , 4, , Fig 10. 14 (c) Resonance Hybrid structures of, CO32-, , It is found that the energy of the, resonance hybrid (structure 4) is lower, than that of all possible canonical structures, (Structure 1, 2 & 3). The difference in, energy between structure 1 or 2 or 3, (most, stable canonical structure) and structure 4, (resonance hybrid) is called resonance energy., , Evaluate Yourself, , Here, a very small, equal and opposite, charges are separated by a small distance (91, pm) and is referred to as a dipole., Dipole moment:, The polarity of a covalent bond can, be measured in terms of dipole moment, which is defined as, , ?, , 4) Write the resonance structures for, i), , μ = q × 2d, , Ozone molecule ii) N2O, , Where μ is the dipole moment, q is, the charge and 2d is the distance between the, two charges. The dipole moment is a vector, and the direction of the dipole moment, vector points from the negative charge to, positive charge., , 10.5.6 Polarity of Bonds, Partial ionic character in covalent, bond:, When a covalent bond is formed, between two identical atoms (as in the, case of H2, O2, Cl2 etc...) both atoms have, equal tendency to attract the shared pair, of electrons and hence the shared pair of, electrons lies exactly in the middle of the, nuclei of two atoms. However, in the case, of covalent bond formed between atoms, having different electronegativities, the, atom with higher electronegativity will have, greater tendency to attract the shared pair, of electrons more towards itself than the, other atom. As a result the cloud of shared, electron pair gets distorted., , 2d, +q, , Unit 10 layout.indd 79, , -q, , Fig 10. 15 Representation of Dipole, , The unit for dipole moment is, columb meter (C m). It is usually expressed, in Debye unit (D). The conversion factor is, 1 Debye = 3.336 x 10 -30 C m, Diatomic molecules such as H2, O2,, F2 etc... have zero dipole moment and are, called non polar molecules and molecules, such as HF, HCl, CO, NO etc... have non, zero dipole moments and are called polar, molecules., , Let us consider the covalent bond, between hydrogen and fluorine in hydrogen, , UpBright Classes, Haldia WhatsApp 9734878427, , μ, , 79, , Page 13 of 43, 27-08-2018 17:53:03

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The extent of ionic character in a, covalent bond can be related to the electro, negativity difference to the bonded atoms., In a typical polar molecule, Aδ--Bδ+, the, electronegativity difference (χA- χB) can, be used to predict the percentage of ionic, character as follows., , Molecules having polar bonds will, not necessarily have a dipole moment. For, example, the linear form of carbon dioxide, has zero dipole moment, even though it, has two polar bonds. In CO2, the dipole, moments of two polar bonds (CO) are equal, in magnitude but have opposite direction., Hence, the net dipole moment of the CO2 is,, μ = μ1 + μ2 = μ1 + (-μ1) = 0, , If the electronegativity difference, (χA- χB ), is, equal to 1.7, then the bond A-B has 50%, ionic character, , O== C == O, μ1, , μ2, , if it is greater than 1.7, then the bond A-B, has more than 50% ionic character,, , → →, , In this case μ= μ1 + μ2, , and if it is lesser than 1.7, then the bond A-B, has less than 50% ionic character., , → →, , = μ1 + (-μ1) = 0, , Evaluate Yourself, , Incase of water net dipole moment is, the vector sum of μ1+ μ2 as shown., –, μ1 2δ μ2, O, δ+, δ+, H → H, μ, , μ, , 5) Of the two molecules OCS and, CS2 which one has higher dipole, moment value? why?, , →, μ2, , Partial covalent character in ionic, bonds:, , →, μ1, , Fig 10. 16 Dipole moment in water, , Like the partial ionic character in, covalent compounds, ionic compounds, show partial covalent character. For example,, the ionic compound, lithium chloride shows, covalent character and is soluble in organic, solvents such as ethanol., , Dipole moment in water is found to be, 1.85D, Table 10. 5 Dipole moments of common, molecules, S., No., , Molecule, , Dipole moment, (in D), , 1, , HF, , 1.91, , 2, , HCl, , 1.03, , 3, , H 2O, , 1.85, , 4, , NH3, , 1.47, , 5, , CHCl3, , 1.04, , UpBright Classes, Haldia WhatsApp 9734878427, Unit 10 layout.indd 80, , ?, , The partial covalent character in, ionic compounds can be explained on the, basis of a phenomenon called polarisation., We know that in an ionic compound, there, is an electrostatic attractive force between, the cation and anion. The positively charged, cation attracts the valence electrons of anion, while repelling the nucleus. This causes a, distortion in the electron cloud of the anion, and its electron density drifts towards the, cation, which results in some sharing of the, , 80, , Page 14 of 43, 27-08-2018 17:53:04

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(iii) Cations having ns2 np6 nd10, configuration show greater polarising, power than the cations with ns2 np6, configuration. Hence, they show greater, covalent character., , valence electrons between these ions. Thus,, a partial covalent character is developed, between them. This phenomenon is called, polarisation., The ability of a cation to polarise, an anion is called its polarising ability and, the tendency of an anion to get polarised, is called its polarisability. The extent of, polarisation in an ionic compound is given, by the Fajans rules, , CuCl is more covalent than NaCl., Compared to Na+ (1.13 Å) . Cu+ (0.6, Å) is small and have 3s2 3p6 3d10, configuration., Electronic configuration of Cu+ , , [Ar] 3s2, 3p6, 3d10, , Fajans Rules, (i) To show greater covalent character, both, the cation and anion should have high, charge on them. Higher the positive, charge on the cation, greater will be the, attraction on the electron cloud of the, anion. Similarly higher the magnitude, of negative charge on the anion, greater, is its polarisability. Hence, the increase, in charge on cation or in anion increases, the covalent character, , Electronic Configuration of Na+, , [He] 2s2, 2p6, , 10.6 Valence Shell Electron Pair, Repulsion (VSEPR) theory, Lewis concept of structure of, molecules deals with the relative position, of atoms in the molecules and sharing of, electron pairs between them. However, we, cannot predict the shape of the molecule, using Lewis concept. Lewis theory in, combination with VSEPR theory will be, useful in predicting the shape of molecules., , Let us consider three ionic compounds, aluminum, chloride,, magnesium, chloride and sodium chloride. Since, the charge of the cation increase in the, order Na+ < Mg2+ < Al3+, the covalent, character also follows the same order, NaCl < MgCl2 < AlCl3., , Important principles of VSEPR Theory, are as follows:, 1. The shape of the molecules depends on, the number of valence shell electron pair, around the central atom., , (ii) The smaller cation and larger anion, show greater covalent character due to, the greater extent of polarisation., , 2. There are two types of electron pairs, namely bond pairs and lone pairs. The, bond pair of electrons are those shared, between two atoms, while the lone pairs, are the valence electron pairs that are not, involved in bonding., , Lithium chloride is more covalent, than sodium chloride. The size of Li+, is smaller than Na+ and hence the, polarising power of Li+ is more. Lithium, iodide is more covalent than lithium, chloride as the size of I- is larger than, the Cl-­. Hence I- will be more polarised, than Cl- by the cation, Li+ ., , UpBright Classes, Haldia WhatsApp 9734878427, Unit 10 layout.indd 81, , 3. Each pair of valence electrons around, the central atom repels each other and, hence, they are located as far away as, possible in three dimensional space to, minimize the repulsion between them., 81, , Page 15 of 43, 27-08-2018 17:53:04

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4. The repulsive interaction between the different types of electron pairs is in the following, order., lp - lp > lp - bp> bp-bp, lp- lone pair ; bp- bond pair, The lone pair of electrons are localised only on the central atom and interacts with only, one nucleus whereas the bond pairs are shared between two atoms and they interact with two, nuclei. Because of this the lone pairs occupy more space and have greater repulsive power than, the bond pairs in a molecule., The following Table illustrates the shapes of molecules predicted by VSEPR theory., Consider a molecule ABx where A is the central atom and x represents the number of atoms of, B covalently bonded to the central atom A. The lone pairs present in the atoms are denoted as, L., , 2, , AB2, , 2, , -, , A, , B, , B, , B, , AB3, , 3, , -, , 3, , AB2L, , 2, , A, B, , Trigonal planar, , 1, , 1200, , B, B, Trigonal planar, A, , A, B, , B, , A, Linear, , B, A, , B, , Examples, , 1800, , B, , Linear, , geometry, , Molecular, , Shape, , No.of. lone pairs, , No.of bond pairs, , Molecule, , Number of, electron Pairs, , Table 10. 6 Shapes of molecules predicted by VSEPR theory., , B, , B, Bent or V - Shape, , B, , Bent or V - Shape, , B, , B, 109.50, , 4, , AB4, , 4, , -, , B, , B, Tetrahedral, , UpBright Classes, Haldia WhatsApp 9734878427, Unit 10 layout.indd 82, , A, , A, B, , 82, , B, B, Tetrahedral, , B, , BeCl2,, HgCl2, CO2,, CS2, HCN,, BeF2, , BF3, BCl3,, NO–3, BF3,, CO32–, HCHO,, SO2, O3,, PbCl2, SnBr2, CH4, CCl4,, CCl2F2,, SO42–,ClO–4 ,, NH+4, , Page 16 of 43, 27-08-2018 17:53:04

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Examples, , geometry, , Molecular, , Shape, , No.of. lone pairs, , No.of bond pairs, , Molecule, , Number of, electron Pairs, , B, , B, A, , AB2L3, , 2, , A, , 3, , XeF2, I–3, IF2–, , B, , Linear, , B, Linear, , B, , B, , AB6, , 6, , -, , B, , A, , B, , B, , B, , B, , B, , B, , B, , B, Octathedral, , B, , B, , 6, , AB5 L, , 5, , 1, , B, , B, A, , B, , 4, , 2, , B, A, , B, , Square pyramidal, , AB4L2, , B, , B, B, , B, B, , Square pyramidal, , B, , A, , B, , Square Planar, , B, , AB7, , 7, , -, , B, , B, B, , B, B, , B, , B, A, , B, , B, , IF7, , B, pentagonal bi-pyramidal, , B, , Unit 10 layout.indd 84, , B, , XeF4, ICl–4, , B, , A, , UpBright Classes, Haldia WhatsApp 9734878427, , BrF5,, IF5, TeF–5,, XeOF4,, , B, , B, , Square planar, , 7, , B, , B, , A, , B, , SF6, IOF5,, , A, , 84, , Page 18 of 43, 27-08-2018 17:53:04

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Evaluate Yourself, , ?, , 6) Arrange the following in the decreasing order of Bond angle, i), , CH4, H2O, NH3, , ii) C2H2, BF3, CCl4, , 10.7 Valence Bond Theory, Heitler and London gave a theoretical treatment to explain the formation of covalent, bond in hydrogen molecule on the basis of wave mechanics of electrons. It was further, developed by Pauling and Slater. The wave mechanical treatment of VB theory is beyond, the scope of this textbook. A simple qualitative treatment of VB theory for the formation of, hydrogen molecule is discussed below., Consider a situation wherein two hydrogen atoms (Ha and Hb) are separated by infinite, distance. At this stage there is no interaction between these two atoms and the potential energy, of this system is arbitrarily taken as zero. As these two atoms approach each other, in addition, to the electrostatic attractive force between the nucleus and its own electron (purple arrows),, the following new forces begins to operate., , +, Ha, , –, , –, , +, Hb, , Fig 10. 17 (a) VB theory for the formation of hydrogen molecule, , The new attractive forces (green arrows) arise between, (i), , nucleus of Ha and valence electron of Hb, , (ii), , nucleus of Hb and the valence electron of Ha., , The new repulsive forces (red arrows) arise between, (i), , the nucleus of Ha and Hb, , (ii), , valence electrons of Ha and Hb., , The attractive forces tend to bring Ha and Hb together whereas the repulsive forces, tends to push them apart. At the initial stage, as the two hydrogen atoms approach each other,, the attractive forces are stronger than the repulsive forces and the potential energy decreases., A stage is reached where the net attractive forces are exactly balanced by repulsive forces and, the potential energy of the system acquires a minimum energy., , UpBright Classes, Haldia WhatsApp 9734878427, Unit 10 layout.indd 85, , 85, , Page 19 of 43, 27-08-2018 17:53:04

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4, , 3, , 2, , 1, , 0, , 1, , Potential energy (KJ mol-1), , -100, -200, 2, , 4, , -300, -400, -432, -500, , 3, , (H2 bond length), 74, , 100, , 200, , Internuclear distance (pm), Fig 10. 17 (b) VB theory for the formation of hydrogen molecule, , At this stage, there is a maximum overlap between the atomic orbitals of Ha and Hb, and, the atoms Ha and Hb are now said to be bonded together by a covalent bond. The internuclear, distance at this stage gives the H-H bond length and is equal to 74 pm. The liberated energy, is 436 kJ mol-1 and is known as bond energy. Since the energy is released during the bond, formation, the resultant molecule is more stable. If the distance between the two atoms is, decreased further, the repulsive forces dominate the attractive forces and the potential energy, of the system sharply increases, 10.7.1 Salient features of VB Theory:, (i) When half filled orbitals of two atoms overlap, a covalent bond will be formed between, them., (ii) The resultant overlapping orbital is occupied by the two electrons with opposite spins. For, example, when H2 is formed, the two 1s electrons of two hydrogen atoms get paired up, and occupy the overlapped orbital., (iii) The strength of a covalent bond depends upon the extent of overlap of atomic orbitals., Greater the overlap, larger is the energy released and stronger will be the bond formed., UpBright Classes, Haldia WhatsApp 9734878427, Unit 10 layout.indd 86, , 86, , Page 20 of 43, 27-08-2018 17:53:04

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(iv) Each atomic orbital has a specific direction (except s-orbital which is spherical) and hence, orbital overlap takes place in the direction that maximizes overlap, Let us explain the covalent bond formation in hydrogen, fluorine and hydrogen fluoride, using VB theory., , 10.8 Orbital Overlap, When atoms combines to form a covalent molecule, the atomic orbitals of the combining, atoms overlap to form a covalent bond. The bond pair of electrons will occupy the overlapped, region of the orbitals. Depending upon the nature of overlap we can classify the covalent, bonding between the two atoms as sigma (σ) and pi (π) bonds., 10.8.1 Sigma and Pi bonds, When two atomic orbitals overlap linearly along the axis, the resultant bond is called, a sigma (σ) bond. This overlap is also called 'head-on overlap' or 'axial overlap'. Overlap, involves an s orbital (s-s and s-p overlaps) will always result in a sigma bond as the s orbital, is spherical. Overlap between two p orbitals along the molecular axis will also result in sigma, bond formation. When we consider x-axis as molecular axis, the px-px overlap will result in, σ-bond., When two atomic orbitals overlaps sideways, the resultant covalent bond is called a pi, (π)bond. When we consider x-axis as molecular axis, the py-py and pz-pz overlaps will result in, the formation of a π-bond., Following examples will be useful to understand the overlap:, 10.8.2 Formation of hydrogen (H2) Molecule, Electronic configuration of hydrogen atom is 1s1, During the formation of H2 molecule, the 1s orbitals of two hydrogen atoms containing, one unpaired electron with opposite spin overlap with each other along the internuclear axis., This overlap is called s-s overlap. Such axial overlap results in the formation of a σ-covalent, bond., , H, ↿, , H2, , H, , +, , 1s, , ↿, , →, , 1s, , ↿⇂, , ss overlapping, , Fig 10. 18 Formation of hydrogen molecule, UpBright Classes, Haldia WhatsApp 9734878427, Unit 10 layout.indd 87, , 87, , Page 21 of 43, 27-08-2018 17:53:04

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the oxygen atoms. Thus, in oxygen molecule, two oxygen atoms are connected by two covalent, bonds (double bond). The other two pair of electrons present in the 2s and 2px orbital do not, involve in bonding and remains as lone pairs on the respective oxygen., , y, py, , z, pz, , π, , ↿, , y, py, , ↿, ↿⇂, , o, , z, pz, , o, , σ, , π, y, , y, , Fig 10. 21 Formation of π bond in O2 Molecule, , Evaluate Yourself, , ?, , 7) Bond angle in PH4+ is higher than in PH3 why?, , 10.9 Hybridisation, Bonding in simple molecules such as hydrogen and fluorine can easily be explained, on the basis of overlap of the respective atomic orbitals of the combining atoms. But, the observed properties of polyatomic molecules such as methane, ammonia, beryllium, chloride etc... cannot be explained on the basis of simple overlap of atomic orbitals. For, example, it was experimentally proved that methane has a tetrahedral structure and the, four C-H bonds are equivalent. This fact cannot be explained on the basis of overlap of, atomic orbitals of hydrogen (1s) and the atomic orbitals of carbon with different energies, (2s2 2px2 2py 2pz)., In order to explain these observed facts, Linus Pauling proposed that the valence, atomic orbitals in the molecules are different from those in isolated atom and he introduced, the concept of hybridisation. Hybridisation is the process of mixing of atomic orbitals of, the same atom with comparable energy to form equal number of new equivalent orbitals, with same energy. The resultant orbitals are called hybridised orbitals and they posses, maximum symmetry and definite orientation in space so as to minimize the force of, repulsion between their electrons ., UpBright Classes, Haldia WhatsApp 9734878427, Unit 10 layout.indd 89, , 89, , Page 23 of 43, 27-08-2018 17:53:05

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10.9.1 Types of hybridisation, geometry of molecules, , and, , In BeCl2 both the Be-Cl bonds are, equivalent and it was observed that the, molecule is linear. VB theory explain this, observed behaviour by sp hybridisation., One of the paired electrons in the 2s orbital, gets excited to 2p orbital and the electronic, configuration at the excited state is shown., , sp Hybridisation:, Let us consider the bond formation, in beryllium chloride. The valence shell, electronic configuration of beryllium in the, ground state is shown in the figure., , Now, the 2s and 2p orbitals hybridise, and produce two equivalent sp hybridised, orbitals which have 50 % s-character and, 50 % p-character. These sp hybridised, orbitals are oriented in opposite direction as, shown in the figure., , Ground State, E, , 2px1, , 2py0, , 2pz0, , ↿⇂, 2s2, , Excited state, E, , ↿, , 2p, , 1, x, , 2py, , sp Hybridisation, , 2pz, , Hybridised State, ↿, , sp, , ↿, , sp, , ↿, , 2s1, Overlap with orbital of chlorine, Each of the sp hybridized orbitals linearly overlap with pz orbital of the chlorine to form, a covalent bond between Be and Cl as shown in the Figure., , Cl, 3pz, , ↿⇂ sp, , sp, , ↿⇂, , Fig 10.22 sp Hybridisation : BeCl2, , UpBright Classes, Haldia WhatsApp 9734878427, Unit 10 layout.indd 90, , Be, , 90, , Cl, , 3pz, , Page 24 of 43, 27-08-2018 17:53:05

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sp2 Hybridisation:, , In the ground state boron has only, one unpaired electron in the valence shell., In order to form three covalent bonds with, fluorine atoms, three unpaired electrons are, required. To achieve this, one of the paired, electrons in the 2s orbital is promoted to the, 2py orbital in the excite state., , Consider boron trifluoride molecule., The valence shell electronic configuration of, boron atom is [He]2s2 2p1., Ground State, E, , ↿, 2px1, , 2py0, , 2pz0, , In boron, the s orbital and two p, orbitals (px and py) in the valence shell, hybridses, to generate three equivalent sp2, orbitals as shown in the Figure. These three, orbitals lie in the same xy plane and the, angle between any two orbitals is equal to, 1200, , ↿⇂, 2s2, , Excited state, E, , ↿, , 2p, , 1, x, , ↿, , 2p, , sp2 Hybridisation, , ↿, , sp, , 2p, , 1, y, , Hybridised State, 2, , 0, z, , ↿, , sp, , 2, , ↿, , sp2, , ↿, , 2s1, Overlap with 2pz orbitals of fluorine:, The three sp2 hybridised orbitals of boron now overlap with the 2pz orbitals of fluorine, (3 atoms). This overlap takes place along the axis as shown below., , 2pz, , F, ↿⇂, , sp2, , B, , sp, , 2, , F, , ↿⇂, , sp2 ↿⇂, , F, , 2pz, , Fig 10.23 sp2 Hybridation BF3, , 2pz, UpBright Classes, Haldia WhatsApp 9734878427, Unit 10 layout.indd 91, , 91, , Page 25 of 43, 27-08-2018 17:53:05

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Evaluate Yourself, , ?, , 3., , The number of molecular orbitals, formed is the same as the number of, combining atomic orbitals. Half the, number of molecular orbitals formed, will have lower energy than the, corresponding atomic orbital, while the, remaining molecular orbitals will have, higher energy. The molecular orbital, with lower energy is called bonding, molecular orbital and the one with, higher energy is called anti-bonding, molecular orbital. The bonding, molecular orbitals are represented, as σ (Sigma), π (pi), δ (delta) and the, corresponding antibonding orbitals are, denoted as σ*, π* and δ*., , 4., , The electrons in a molecule are, accommodated in the newly formed, molecular orbitals. The filling of, electrons in these orbitals follows, Aufbau's principle, Pauli's exclusion, principle and Hund's rule as in the case, of filling of electrons in atomic orbitals., , 5., , Bond order gives the number of covalent, bonds between the two combining, atoms. The bond order of a molecule, can be calculated using the following, equation, , 8) Explain the bond formation in SF4 and, CCl4 using hybridisation concept., , Evaluate Yourself, , ?, , 9) The observed bond length of N2+ is larger, than N2 while the bond length in NO+, is less than in NO. Why?, 10.10 Molecular orbital theory, Lewis concept and valence bond, theory qualitatively explains the chemical, bonding and molecular structure. Both, approaches are inadequate to describe some, of the observed properties of molecules. For, example, these theories predict that oxygen is, diamagnetic. However, it was observed that, oxygen in liquid form was attracted towards, the poles of strong magnet, indicating that, oxygen is paramagnetic. As both these, theories treated the bond formation in, terms of electron pairs and hence they fail to, explain the bonding nature of paramagnetic, molecules. F. Hund and Robert. S. Mulliken, developed a bonding theory called molecular, orbital theory which explains the magnetic, behaviour of molecules., The salient features of this theory are as, follows., 1., , 2., , When atoms combines to form, molecules, their individual atomic, orbitals lose their identity and forms, new orbitals called molecular orbitals., , Nb − Νa, 2, , Where, Nb = Total number of electrons, present in the bonding molecular orbitals, Na = Total number of electrons present in, the antibonding molecular orbitals and, , The shapes of molecular orbitals depend, upon the shapes of combining atomic, orbitals., , UpBright Classes, Haldia WhatsApp 9734878427, Unit 10 layout.indd 97, , Bond order =, , A bond order of zero value indicates that the, molecule doesn't exist., , 97, , Page 31 of 43, 27-08-2018 17:53:07

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10.10.1 Linear combination of atomic, orbitals, , Constructive interaction: The two 1s orbitals, are in phase and have the same sign., , The wave functions for the molecular, orbitals can be obtained by solving, Schrödinger wave equation for the molecule., Since solving the Schrödinger equation, is too complex, approximation methods, are used to obtain the wave function for, molecular orbitals. The most common, method is the linear combination of atomic, orbitals (LCAO)., , +, , 1s, , add, , +, , 1s, , 1s, , represented by :, , +, , +, , 1s, , bonding molecular orbital, , +, , +, , σ bonding MO, , We know that the atomic orbitals are, represented by the wave function Ψ. Let us, consider two atomic orbitals represented, by the wave function ψA and ψB with, comparable energy, combines to form, two molecular orbitals. One is bonding, molecular orbital(ψbonding) and the other is, antibonding molecular orbital(ψantibonding)., The wave functions for these two molecular, orbitals can be obtained by the linear, combination of the atomic orbitals ψA and, ψB as below., , Destructive interaction The two 1s, Orbitals are out phase, , add, +, , +, , +, –, , node, , anti-bonding molecular, orbital, , represented by, +, +, , ψbonding = ψA + ψB, ψantibonding = ψA - ψB, , +, , –, , node, , The formation of bonding molecular, orbital can be considered as the result of, constructive interference of the atomic, orbitals and the formation of anti-bonding, molecular orbital can be the result of the, destructive interference of the atomic, orbitals. The formation of the two molecular, orbitals from two 1s orbitals is shown below., , σ*, energy, , 1s, , 1s, , atomic orbital, , antibonding, , ↿⇂, , σ, , atomic orbital, , bonding molecular orbital, , Fig 10.29 Linear Combination of atomic, orbitals, , UpBright Classes, Haldia WhatsApp 9734878427, Unit 10 layout.indd 98, , 98, , Page 32 of 43, 27-08-2018 17:53:08

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Metallic bonding, , ductility. Pure metals can undergo 40 to, 60% elongation prior to rupturing under, mechanical loading. As each metal ion, is surrounded by electron cloud in all, directions, the metallic bonding has no, directional properties., , Metals have some special properties, of lustre, high density, high electrical and, thermal conductivity, malleability and, ductility, and high melting and boiling, points. The forces that keep the atoms of, the metal so closely in a metallic crystal, constitute what is generally known as the, metallic bond. The metallic bond is not just, an electrovalent bond(ionic bond), as the, latter is formed between atoms of different, electro negativities. Similarly, the metallic, bond is not a covalent bond,as the metal, atoms do not have sufficient number of, valence electrons for mutual sharing with 8, or 12 neighboring metal atoms in a crystal., So, we have to search for a new theory to, explain metallic bond. The first successful, theory is due to Drude and Lorentz, which, regards metallic crystal as an assemblage, of positive ions immersed in a gas of free, electrons. The free electrons are due to, ionization of the valence electrons of the, atoms of the metal. As the valence electrons, of the atoms are freely shared by all the, ions in the crystal, the metallic bonding is, also referred to as electronic bonding. As, the free electrons repel each other, they are, uniformly distributed around the metal, ions. Many physical properties of the metals, can be explained by this theory, nevertheless, there are exceptions. , , As the electrons are free to move, around the positive ions, the metals exhibit, high electrical and thermal conductivity., The metallic luster is due to reflection of, light by the electron cloud. As the metallic, bond is strong enough, the metal atoms are, reluctant to break apart into a liquid or gas,, so the metals have high melting and boiling, points., The bonding in metal is better treated, by Molecular orbital theory. As per this, theory, the atomic orbitals of large number of, atoms in a crystal overlap to form numerous, bonding and antibonding molecular, orbitals without any band gap. The bonding, molecular orbitals are completely filled with, an electron pair in each, and the antibonding, molecular orbitals are empty. Absence, of band gap accounts for high electrical, conductivity of metals. High thermal, conductivity is due to thermal excitation, of many electrons from the valence band, to the conductance band. With an increase, in temperature, the electrical conductivity, decreases due to vigorous thermal motion of, lattice ions that disrupts the uniform lattice, structure, that is required for free motion, of electrons within the crystal. Most metals, are black except copper, silver and gold. It is, due to absorption of light of all wavelengths., Absorption of light of all wavelengths is due, to absence of bandgap in metals., , The electrostatic attraction between, the metal ions and the free electrons yields, a three-dimensional close packed crystal, with a large number of nearest metal ions., So, metals have high density. As the close, packed structure contains many slip planes, along which movement can occur during, mechanical loading, the metal acquires, UpBright Classes, Haldia WhatsApp 9734878427, Unit 10 layout.indd 102, , 102, , Page 36 of 43, 27-08-2018 17:53:11

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Lewis theory in combination with VSEPR, theory will be useful in predicting the shape, of molecules. According to this theory,, the shape of the molecules depends on the, number of valance shell electron pair (lone, pairs and bond pairs) around the central, atom. Each pair of valance electrons around, the central atom repels each other and hence,, they are located as far away as possible in, three-dimensional space to minimise the, repulsion between them., , SUMMARY, In molecules, atoms are held together by, attractive forces, called chemical bonds., Kossel and Lewis are the first people to, provide a logical explanation for chemical, bonding. They proposed that atoms try, to attain the nearest noble gas electronic, configuration by losing, gaining or sharing, one or more electrons during the bond, formation. The noble gases contain eight, electrons in their valance shell which, is considered to be stable electronic, configuration. The idea of Kossel – Lewis, approach to chemical bond lead to the octet, rule, which states that “the atoms transfer or, share electrons so that all atoms involved in, chemical bonding obtain 8 electrons in their, outer shell (valance shell)”., , Heitler and London gave a theoretical, treatment to explain the formation of, covalent bond in hydrogen molecule on the, basis of wave mechanics of electrons. It was, further developed by Pauling and Slater., According to this theory when half-filled, orbitals of two atoms overlap, a covalent, bond will be formed between them., Linus Pauling introduced the concept of, hybridisation. Hybridisation is the process, of mixing of atomic orbitals of the same, atom with comparable energy to form equal, number of new equivalent orbitals with, same energy. There are different types of, hybridization such as sp, sp2, sp3, sp3d2, etc.., , There are different types of chemical bonds., In compounds such as sodium chloride,, the sodium atom loses an electron which, is accepted by the chlorine atom resulting, in the formation of Na+ and Cl- ions., These two ions are held together by the, electrostatic attractive forces. This type of, chemical bond is known as ionic bonds or, electrovalent bonds. In certain compounds,, instead of the complete transfer of electrons,, the electrons are shared by both the bonding, atoms. The two combining atoms are held, together by their mutual attraction towards, the shared electrons. This type of bond is, called covalent bonding. In addition, there, also another bond type known as coordinate, covalent bonds, where the shared electrons, of a covalent bond are provided by only one, of the combining atoms. Metallic bonding is, another type of bonding which is observed, in metals., , UpBright Classes, Haldia WhatsApp 9734878427, Unit 10 layout.indd 103, , F. Hund and Robert. S. Mulliken developed, a bonding theory called molecular orbital, theory. According to this theory, when, atoms combines to form molecules, their, individual atomic orbitals lose their identity, and forms new orbitals called molecular, orbitals. The filling of electrons in these, orbitals follows Aufbau's principle, Pauli's, exclusion principle and Hund's rule as in the, case of filling of electrons in atomic orbitals., , 103, , Page 37 of 43, 27-08-2018 17:53:11

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Reason :, , Evaluation, 1., , 2., , 3., , 4., , 5., , 6., , 7., , In which of the following Compounds, does the central atom obey the octet, rule?, a) XeF4, , b) AlCl3 , , c) SF6 , , d) SCl2, , In the molecule OA, C, OB, the, formal charge on OA, C and OB are, respectively., a) -1, 0, + 1, , b) +1, 0,-1, , c) -2,0,+2, , d) 0,0,0, , a) PH3 , , b) (CH3)2, , c) BH3 , , d) NH3, , a), , both assertion and reason are, true and reason is the correct, explanation of assertion, , b), , both assertion and reason are, true but reason is not the correct, explanation of assertion, , c), , assertion is true but reason is, false, , d), , Both assertion and reason are, false, , According to Valence bond theory, a, bond between two atoms is formed, when, a), , fully filled atomic orbitals overlap, , b), , half filled atomic orbitals overlap, , Which of the following molecule, contain no л bond?, , c), , non- bonding atomic orbitals, overlap, , a) SO2 , , b) NO2, , d), , empty atomic orbitals overlap, , c) CO2 , , d) H2 O, , 9., , The ratio of number of sigma (σ) and, pi (л) bonds in 2- butynal is, a) 8/3 , , b) 5/3, , c) 8/2 , , d) 9/2, , Which one of the following is the likely, bond angles of sulphur tetrafluoride, molecule?, a) 1200,800, , b) 1090.28, , c) 900 , , d) 890,1170, , Assertion:, , Oxygen molecule, paramagnetic., , UpBright Classes, Haldia WhatsApp 9734878427, Unit 10 layout.indd 104, , 8., , Which of the following is electron, deficient?, , It has two unpaired, electron in its bonding, molecular orbital, , In ClF3 ,NF3 and BF3 molecules the, chlorine, nitrogen and boron atoms, are, a), , sp3 hybridised, , b), , sp3 ,sp3 and sp2 respectively, , c), , sp2 hybridised, , d), , sp3d, sp3 and sp, respectively, , hybridised, , 10. When one s and three p orbitals, hybridise,, is, , 104, , a), , four equvivalent orbitals at 900 to, each other will be formed, , b), , four equvivalent orbitals at 1090, , Page 38 of 43, 27-08-2018 17:53:12

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c) BF3 and NO2-, , a) SnCl2, , b) NO2, , d) BF3 and NH2-, , c) C2 H2 , , d) All of these., , 25. According to VSEPR theory, the, repulsion between different parts of, electrons obey the order., , 20. Some of the following properties of two, species, NO3- and H3O+ are described, below. which one of them is correct?, , a) l.p – l.p > b.p–b.p> l.p–b.p, , a) dissimilar in hybridisation for, the central atom with different, structure., , b) b.p–b.p> b.p–l.p> l.p–b.p, c) l.p–l.p> b.p–l.p > b.p–b.p, , b) isostructural, with, same, hybridisation for the Central atom., , d) b.p–b.p> l.p–l.p> b.p–l.p, 26. Shape of ClF3 is, , c) different hybridiration for the, central atom with same structure, , a) Planar triangular, , d) none of these, , b) Pyramidal, c) 'T' Shaped, , 21. The types of hybridiration on the five, carbon atom from right to left in the,, 2,3 pentadiene., , d) none of these, 27. Non- Zero dipole moment is shown by, , a) sp3, sp2, sp, sp2, sp3, , a) CO2, , b) sp3, sp, sp, sp, sp3, , b) p-dichlorobenzene, , c) sp2, sp, sp2,sp2, sp3, , c) carbontetrachloride, , d) sp3, sp3, sp2, sp3, sp3, , d) water., , 22. Xe F2 is isostructural with, , 28. Which of the following conditions, is not correct for resonating, structures?, , a) SbCl2 b) BaCl2, c) TeF2 d) ICl2–, 23. The percentage of s-character of the, hybrid orbitals in methane, ethane,, ethene and ethyne are respectively, a) 25, 25,33.3,50, b) 50,50,33.3,25, c) 50,25,33.3,50, , a), , the, contributing, structure, must have the same number of, unpaired electrons, , b), , the contributing structures, should have similar energies, , c), , the resonance hybrid should, have higher energy than any of, the contributing structure., , d), , none of these, , d) 50,25,25,50, 24. Of the following molecules, which, have shape similar to carbondixide?, , UpBright Classes, Haldia WhatsApp 9734878427, Unit 10 layout.indd 106, , 106, , Page 40 of 43, 27-08-2018 17:53:12

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29. Among the following, the compound, that contains, ionic, covalent and Coordinate linkage is, a) NH4Cl, , i) NO3–, iv) O3, , 43. Which bond is stronger σ or π? Why?, , c) NaCl d) none of these, 30. CaO and NaCl have the same crystal, structure and approximately the same, radii. It U is the lattice energy of NaCl,, the approximate lattice energy of CaO, is, U , , b) 2U, , c), , U/2, , d) 4U, , 44. Define bond energy., 45. Hydrogen gas is diatomic where as inert, gases are monoatomic – explain on the, basis of MO theory., 46. What is Polar Covalent bond? explain, with example., 47. Considering x- axis as molecular axis,, which out of the following will form a, sigma bond., , 31. Define the following, i) Bond order, , i) 1s and 2py, ii) 2P x and 2P x, iii) 2px and 2pz iv) 1s and 2pz, , ii) Hybridisation, iii) σ- bond, , 48. Explain resonance with reference to, carbonate ion?, , 32. What is a pi bond?, 33. In CH4, NH3 and H2O, the central atom, undergoes sp3 hybridisation - yet their, bond angles are different. why?, , 49. Explain the bond formation in ethylene, and acetylene., 50. What type of hybridisations are possible, in the following geometeries?, , 34. Explain Sp2 hybridisation in BF3, 35. Draw the M.O diagram for oxygen, molecule calculate its bond order and, show that O2 is paramagnetic., , a) octahedral, b) tetrahedral , c) square planer., , 36. Draw MO diagram of CO and calculate, its bond order., , 51. Explain VSEPR theory. Applying this, theory to predict the shapes of IF7, and, SF6, , 37. What do you understand by Linear, combination of atomic orbitals in MO, theory., , 52. CO2 and H2O both are triatomic, molecule but their dipole moment, values are different. Why?, , 38. Discuss the formation of N2 molecule, using MO Theory, , 53. Which one of the following has highest, bond order?, , 39. What is dipolment?, 40. Linear form of carbondioxide molecule, has two polar bonds. yet the molecule, has Zero dipolement why?, , N2, N+2 or N2–, 54. Explain the covalent character in ionic, bond., , 41. Draw the Lewis structures for the, following species., UpBright Classes, Haldia WhatsApp 9734878427, Unit 10 layout.indd 107, , i i i ) H NO 3, , 42. Explain the bond formation in BeCl2, and MgCl2., , b) NH3, , a), , ii) SO42–, , 55. Describe fajan's rule., 107, , Page 41 of 43, 27-08-2018 17:53:12

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ICT Corner, , Shapes of Molecule (VSEPR Theory), By using this tool, you will be able, to create and visualise molecules, with different number of bonds, and lone pairs., , Please go to the URL, https://phet.colorado.edu/sims/html/, molecule-shapes/latest/moleculeshapes_en.html, (or), Scan the QR code on the right side, , Steps, •, , Open the Browser and type the URL given (or) Scan the QR Code., , •, , Double click the ‘models’ icon given in this page. Now you will see the web page as shown in the, figure., , •, , Now, by selecting the number of bond pairs and lone pairs using the options in the box 3 and 4 effectively, you visualize the corresponding molecular geometry., , •, , You can see the angles between the bonds selecting the “Show bond angles” indicated in box 5., , •, , You can also visualize the molecular geometry and the electron geometry by selecting corresponding option in box 2., , Structure of simple molecules:, By selecting the real molecules icon in indicated in the box 6, you will be able to see a screen as shown, below. In this screen, use the dropdown menu indicated in box 7 you can select a molecule. Use the, option in the box 8 you can visualise the lone pairs and bond pairs in the molecules., , UpBright Classes, Haldia WhatsApp 9734878427, Unit 10 layout.indd 109, , 109, , Page 43 of 43, 27-08-2018 17:53:13