Chemical Bonding Solution

CHEMICAL BONDING HINTS & SOLUTIONS EXERCISE - 1 Single Choice 4. The conditions required for the formation of an ionic bond. (i) Ionization enthalpy [M(g)  M+(g) + e–] of electropositive element must be low. (ii) Negative value of electron gain enthalpy [X (g) + e–  X–(g)] of electronegative element should be high. 6. Here xz = 3 – 1 = 2 > 1.7 yz = 2 – 1 = 1 < 1.7 7. Cs has lowest IE1 amongst the metals and F has higher electron affinity. So Cs and F form most ionic compound. 8. In SF6, PCl5 and IF7 the valence shell has 12, 10 and 14 electrons. As all contain more than 8 electrons in their valence shell they are example of super octet molecules.   XZ is ionic YZ is covalent. 9.  Bond order = 1.5. 11. (D) Position of atoms are different. (E) Has unpaired electrons and this is not possible since the molecule is diamagnetic. 14. N3– , (CNO)– and (NCN)2– all have same number of electrons i.e., 22 ; so all are isoelectronic with CO2 which also has 22 electrons. 15. 1 Bond length  Bond order Bond order of CO = 3 (as isoelectronic with N2) Bond order = No. of bonds in all possible sides No. of resonating structures Bond order of CO2 = Bond order of CO32– = 4 =2 2 4 = 1.33 3 So, order of bond length of C – O is CO < CO2 < CO32– 16. 17. x>y=z (A)  bond is formed by axial over lapping. (B) p-orbital have both axial and side ways over lapping 21. 19 and 4. (A) (B) Z – + + 20. – 18. PZ Py + + + Z + (D) Z – – – Px Zero overlap S Py 24. 25. PZ Py (C) Px covalency = 7 (a) (b) (c) (d) 26.   29. – N3 [N  N  N ] ; NOCl  30. (A) both are sp3d (C) [ClF2O]+ is sp3 but [ClF4O]– is sp3d2 + .. F F O = N – Cl .. F & F F O F –  N2O  N  N  O (B) both are sp3d – Cl Cl O Z (D) both are sp3d2 CHEMICAL BONDING 32. StNO = 4 36. (A) : : hybridization = sp3 O H H : : (B) Number of bond pairs around I = 2. Number of lone pairs around I = 3. : O (C) 38. O : : F (D) F S Cl (A) XeO3 is trigonal pyramid. Cl (B) IOF4+ is see-saw. F (C) PCl5 is trigonal bipyramidal. (D) F F F O  Xe (ii) [N  N  N] 2 sp (iii) PCl5 (s) exists as [PCl4]+ and [PCl6]– , (iv) I2Cl6() [ICl2]+ + [ICl4]– + self ionisation Cl : F sp3d : (i) I sp Cl 3 F Xe F 39. F Xe : : + F F square pyramidal. F O I : 40. F 3 F sp d but see-saw shape F F – O F | S | F : see-saw 41. As the electronegativity of central atom increases the bond angle increases due to repulsion between bond pair and bond pair as bond pairs are more close to the central atom. 43. S1 : SnCl2 ; bonding takes place in ground state. : Due to lp-bp repulsion bond angle slightly decrease from 120º. Sn Cl Cl S2 : F 89º F 104º due to bp-lp repulsions. S 177º : S3 : F 102º 45. 141 12 1 H O (A) F O N 130º O 114º (B) Bond angle is 180º because of sp hybridisation of nitrogen. 102º H 143 N 8 11 O (C) O 111º 119 (D) O N 134º O CHEMICAL BONDING 48. Species C2H2 C2H4 C2H6 C2H2Br2 Bond order 3 2 1 2 49. Presence of two LP.s on Cl, distorts the bond angle. Statement 3 is against VSEPR. 50. PH3 – No Hybridisation NH3 and H2O bond angle is less than 109º28' due to LP–BP and LP–LP repulsion. CH4 bond angle is 109º28' 52. P : CH3 – C  N Q : H -- N = C = O O R : CH 3 C NH 2 . 54. CaC2 exists as Ca2+ and C22– . 56. Hybridization does not take place without s-orbital 57. 58. 60. N atom in NO2+ is sp hybridised while in NO2– and NO3–, it is sp2 hybridised. Bond order = 2   1 Bond order  Bond length So, bond length order is NO3  NO2  NO2 . Bond order =  2 1 = 1.5 2 Bond order = 2  1 1 = 1.33 3 62. Cl Cl | I 3 Cl — I — Br sp d Cl (Linear) (T-shaped) H H 66. H N+ + H+  N 65. H H H H H2S– No hybridisation bond angle  93º NH3 – Pyramidal 104.5º CH4 – Tetrahedral 109º28' BF3 – Trigonalplanar 120º 68. 70. (a) Electronic configuration of boron in ground state is 1s22s22p1. BCl3 : (b) Electronic configuration of nitrogen in ground state is 1s22s22p3. NH3 : (c) Electronic configuration of phosphorus in ground state is 1s22s22p63s23p3. PCl3 : (d) Electronic configuration of boron in ground state is 1s22s2. BeF2 : CHEMICAL BONDING 71. 75.  ONO  O– N = O Bond order = 2 Bond order = 1.5 80. The hybridization & shape of ClO3 is sp3 & pyramidal but hybridization & shape of NO 3 is sp2 & trigonal planar.. 82 (A) BrF5 contains 10 electrons inplace of eight. (B) SF6 contains 12 electrons inplace of eight. (C) IF7 contains 14 electrons inplace of eight. 83 The maximum covalency of an element is equal to the number of s & p electrons in valence shell. 85 Electronegativity difference between two combining elements must be larger for ionic compound and it is the essential condition for the formation of ionic compounds. It is ionic because electronegativity difference between two combining elements is 1.8. y y x x 86. 87. dx2–y2 dx2–y2 sp2 sp sp sp2 sp2 88. (A) H2 C  C  C  C  C  CH2 | H | H (B) (C) In diamond each carbon atom is in sp3 hybridisation. (D) O = C = C = C = O sp sp sp 89. 90. Bond energy  Bond order Species Bond order CO 3 CO2 2 CO32– 1.33 ; there is positive charge on two adjacent nitrogen atoms. This leads to repulsion and thus increases the energy of the molecule. 92. steric No. = 2  Hybridisation sp. 93. A fact, to be remembered. 94. (a) According to VSEPR as electronegativity of central atom decreases, bond angle decreases. So bond angle of H2O > H2S > H2 Se > H2Te (b) C2H2 > C2H4 > CH4 > NH3 with bond pair- lone pair repulsion sp2 sp sp3 sp3 (c) SF6 < NH3 < H2O < OF2 in this case bond angle of NH3 is highest because lp - lp repulsion is absent in. it. (d) ClO2 > H2O > H2S > SF6 ClO2 bond angle is highest due to its sp2 hybridisation, rest all are sp3 or sp3d2 hybridised more repulsion in double bond electrons. : : Cl. : : : O O > O > H O S H > SF6 (Octahedral) S F H 95. C=S H F F F Hydrogen atoms are in a vertical plane with axial fluorine atoms, -bond involving a p-orbital of carbon atom must lie in equatorial plane of the molecule. Six atoms, i.e. 2 H-atoms, C,S. and both axial F-atoms lie in one plane. 96. Due to more polarising power of Li against other members of the family. 97. Oxidation no. of N in N2O5 is + 5 Anhydride of HOCl is Cl2O. The bond length decreases with increase in difference of electronegativity. 98. The cation should be [H – C  N – Xe – F]+ Hybridisations F 99. : F sp sp O S F sp 3 d F Xe F F : : I O F F O O CHEMICAL BONDING 100. Generally lone pair causes more distortion than a double bond so equitorial FSF in SOF4 > equitorial FSF in SF4 F 108° OCF2 : C O F due to low bond pair bond pair repulsion because of fluorine & large repulsion due to double bond. 12 6° F F S SF4 axial FSF < 180° due to lone pair. F F Similar orbitals on hybridisation can not decrease their energy due to energy conservation law 101. (A) H H (B) S F 102. F .. .. B F H B H .. Cl F Kr F .. (C) F H H Cl .. .. Cl Cl 103. 104. 109. In SOCl2 , sulphur atom is sp3 hybridised. Due to the presence of one lone pair of electrons on ‘S’ atom, the molecule has distorted tetrahedral shape. i.e., trigonal pyramidal shape. This shape contains two chlorine and one oxygen atom in a triangle. The S – O – pi bond is d – p pi bond. 110. F F B Vacant 2p-orbital F Filled 2p-orbital Decrease in B – F bond length is due to delocalised p–p bonding between filled p-orbital of F atom and vacant porbital of B atom. 114. N22– : 1s2 *1s2 2s2 *2s2 2px2  2py2 2pz2 2px1 *2py1. B.O. N22– = 10 – 6 = 2. 2 NO– isoelectronic with O2 so B.O. = ; B.O. O2 = 10 – 6 = 2. 2 10 – 6 = 2. 2 All have same number of electrons (i.e. 16) so isoelectronic. 116. The electron density is zero in the nodal plane during the formation of a molecular orbital from atomic orbitals of the same atom. 118. (A) Bond order of N2+ = 2.5 ; The bond order of O2+ 1/2(10 – 5) = 2.5. (B) Bond order of F2 = 1 ; The bond order of Ne2 = 0. (C) Bond order of O2 = 2 ; The bond order of B2 = 1. (D) Bond order of C2 = 2 ; The bond order of N2 = 3. 120. B2 bond order = 1 ; C2 bond order = 2 ; F2 bond order = 1 ; O2– bond order = 1.5 bond order  1/bond length. 121. (A) NO– is derivative of O2 and isoelectronic with O2. So (1s)2 (*1s)2 (2s)2 (*2s)2 2pz)2 (2p2x = 2p2y ) (*2px1 = *2p1y) and 2 unpaired electrons. (B) O22– : (1s)2 (*1s)2 (2s)2 (*2s)2 2pz)2 (2p2x = 2p2y ) (*2px2 = *2p2y) and no unpaired electrons. (C) CN– is derivative of and isoelectronic with N2 : (1s)2 (*1s)2 (2s)2 (*2s)2 (2p2x = 2p2y ) 2pz)2 and no unpaired electron. (D) CO is derivative of and isoelectronic with N2 : (1s)2 (*1s)2 (2s)2 (*2s)2 (2p2x = 2p2y ) 2pz)2 and no unpaired electron. 123. (A) Stability Bond order O 2+ > O 2 > O 2– 2.5 2 1.5 (C) In all these molecules all electrons are paired in molecular orbitals. 126. (C) O2 : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y ) (*2px1 = *2p1y) O2– : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y ) (*2px2 = *2p1y) 128. (I) O2+ : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y ) (*2px1 = *2p0y) Bond order = 1/2(10 – 5) = 2.5. (II) NO is derivative of O2 and isoelectronic with O2+ : so (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y ) (*2px1 = *2p0y) Bond order = 1/2(10 – 5) = 2.5. (III) N2+ : (1s)2 (*1s)2 (2s)2 (*2s)2 (2p2x = 2p2y ) 2pz)1 Bond order = 1/2(9 – 4) = 2.5. O2+ , NO and N2+ have same bond order i.e. 2.5 and have same magnetic property having one unpaired electron. CHEMICAL BONDING N 186 N Hyponitrite ion (C) N2O5 Dinitrogen pentoxide 130º O 117º (B) N2O22– O 122 (A) N2O3 Dinitrogen trioxide 4 11 129. 120 105º O O Dinitrogen tetroxide 12 (D) N2O4 1 O O N 175 135º N O 132. 6– [Si3O9] [Si4O12]8– [Si5O15] 10– [Si6O18] 12– general formula of cyclic silicates is [Sin O3n]2n– 135. P4O10 136. Partial double bond character is developed in B–F bond of BF3 (due to formation of dative  bond) In other case B is sp3 hybridised (tetrahedral molecule or ion) and the possibility for  bonding no longer exists. 138. S1 : as it does not have d–orbitals. S2 : ClF3 S3 : In B2 mixing of the g(2s) orbital with the g(2p) orbital lowers the energy of the g(2s) orbital and increases the energy of the g(2p) orbital to a higher level than the  orbitals. As a result, the last two electrons are unpaired in the degenerate (having the same energy)  orbitals, and the molecule is paramagnetic. In N2 the g(2s) and g(2p) levels of N2 interact (mix) less than the B2 and C2 levels, and the g (2p) and u(2p) are very close in energy. 139. Boron trimethyl is a weaker Lewis acid than the boron trihalides or monoborane. The electron donating effect of the methyl groups hinders the complex formation with trimethyl amine. Hence the bond N  B is weakest in [(CH3)3 N  B(CH3)3]. Me3N as donor (capacity). BBr3 > BCl3 > BF3 ~ BH3 > BMe3. 141. There is p–d delocalization of lone pair of electron on nitrogen atom and empty d-orbital of silicon but not p– p, one of the 2s-orbital electrons jumps to the last Pz orbital and does not participate in sp2 hybridisation. 144. 2 p 2 z   2 px M.O for C2 = 1s2 < 1*s2 < 2s2 < 2*s2 < 1 2 4p4y 2 4 243 { HOMO LUMO It is important to note that double bond in C2 consists of both pi bonds because of the presence of four electrons in two pi molecular orbitals C22– 147. 148. . NO– > NO > Bond order 2.0 2.5 H2 > H2+ > Bond order 1 0.5 (In He2+ more electron in antibonding MO's) NO2+ > NO2 > Bond angle 180º 133º 2– O 2 < O2 + < – No. of unpaired e 0 1 He2+ bond order = C2 bond order = NO+ 3 He2+ 0.5 (bond length) NO2– 115º O2 2 (bond angle) (bond energy) (paramagnetic moment) 2 1 1 10  7  ; O – bond order =  1 .5 2 2 2 2 84 10  5  2 ; NO bond order =  2 .5 2 2 Bond order  bond dissociation energy.. CHEMICAL BONDING 149. diamond (1.54 Å) graphite (1.42 Å) . (r to the sheets there is no covalent bonding) C60 (1.45 Å and 1.38 Å) benzene (1.36 Å). 151. Hybridisation is sp3 . : : : : 3 sp : : P S : : 152. P : P : : P : : S : P 154. (I) [PCl4]+  sp3 (II) has  102° bond angle due to lp–lp repulsion, as bond pairs are closer to F-atoms. O F F (III) All have sp3 hybridisation and one lone pair. .. P (IV) :P P: .P. 156 CH3 CH3 CH3 . . sp F (A) F – N2 = N+ sp F H 2 157. (B) 112° .. N = N = N 2 sp Si CH3 (C) N Si CH3 Si CH3 CH3 sp2 CH3 CH3 158 In BF3 the B - F bond length is shorter than normal bond character on account of p – p back bonding (1.30 Å). In 1 : 1 complex there is no p–p back bonding and hence the B–F bond length is longer,(1.35 Å). CH3 | CH3 — N | CH3 Steric number 4 F | B—F | F 4 Hence both N and B have tetrahedrdal geometry with sp 3 hybridisation. 161 N2 : (1s)2 (*1s)2 (2s)2 (*2s)2 (2p2x = 2p2y ) 2pz)2 The bond order of N2 is 1/2(10 – 4) = 3. N2+ : (1s)2 (*1s)2 (2s)2 (*2s)2 (2p2x = 2p2y ) 2pz)1 The bond order of N2+ is 1/2(9 – 4) = 2.5. O2 : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y ) (*2px1 = *2p1y) The bond order of O2 1/2(10 – 6) = 2. O2– : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y ) (*2px2 = *2p1y) The bond order of O2 1/2(10 – 7) = 1.5. NO+ derivative of O2 and isoelectronic with O22+ ; so (1s)2 (*1s)2 (2s)2 (*2s)2 2pz)2 (2p2x = 2p2y ) The bond order of NO+ 1/2(10 – 4) = 3. NO derivative of O2 and isoelectronic with O2+ ; (1s)2 (*1s)2 (2s)2 (*2s)2 2pz)2 (2p2x = 2p2y ), *2px)1 The bond order of NO is 1/2(10 – 5) = 2.5. bond order  1/bond length  bond dissociation energy.. 163 OF is derivative of O2 and isoelectronic with O2–. So (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y ) (*2px2 = *2p1y) The bond order of OF 1/2(10 – 7) = 1.5. 164 (A) H2O2 = 1.48 Å due to repulsions between non-bonded pairs of electron on O-atoms and O2F2 = 1.217 Å. (B) In O 22 very slightly increases due to charge (–ve) on two O atoms. + O2 (dioxygenyl) (C) O2 (dioxygen) – O2 (superoxide) O2 165. 2– (peroxide) Bond Order Internuclear Distance (pm) Number of unpaired(s) Electrons 2.5 112.3 1 2.0 120.07 2 1.5 128 1 1.0 149 0 Bond order of N2 = 3 bond order of N2+ = 2.5 B.O. of O2 = 2. B.O. of O2+ = 2.5 CHEMICAL BONDING 168. (A) O  C  C  C  O | | | sp sp sp O (B) Resonance takes place O Cr O O Cr No resonance O Resonance takes place O ; Therefore, six Cr – O bond lengths are not equal. O (C) This is an example of 3-centre 2-e– bond which is also known as Banana bond. But Al2Cl6 have covalent bond only and there is no electron deficient bonding as depicted in the given structure. Cl Cl Al Cl Al Cl Cl Cl (D) AgI is bright yellow coloured compound due to the polarisation of anion, I– .The bigger anions are more polarised and hence their electrons get excited by partial absorption of visible light. s 169. (A) NO B.O = 3 triple bond. O = N – Cl there is a double bond between N and ‘O’ atom (B) CaC2  Ca2+ + C22– (–C  C–) H H (C) H C=C C2H4  H KO2 O2– Na2  O22– B.O = 1.5 B.O = 1 H H py (D) H py pz pz C = C = C (S + Pz + Px) (Spx) xz - plane H (S + Py + Px) (xy – plane) EXERCISE - 2 Part # I : Multiple Choice 1. (A) is incorrect, as ionic compounds have higher melting points as well as higher boiling points due to strong electrostatic force of attraction between the ions. (B), (C) and (D) are properties of ionic compounds.  3. H    |  (A) H — N — H Cl–   |   H (B) K+C  N– (C) 4. (D) Na+O–H H–CC–CH3 The two  bonds are in mutually perpendicular plane. The terminal C-atom is sp3 hybridised. O O¯ 7. F O¯ Xe Xe (A) (B) O¯ O¯ F XeF2 (Linear) O Perxenate ion [XeO6]4- O F F F Xe (C) (D) F F O F XeO2F2(one l.p.) XeOF4 (one l.p.) 8.  ONO F I Xe I F 10. – I + I I (A) Due to the presence of lone pair bond angle decreases. CH3 | B (B) H C 3 Xe O CH3 sp hyb. 2 (C) NH4Cl is an ionic compound and ‘N’ is in sp3 hybridisation. (D) S8 molecule has 16 electron parirs left behind after the bonding. I CHEMICAL BONDING Cl 11. B and C  Drago's rule, I ICl3 Cl Cl sp3d T. shaped 12. The incorrect Lewis diagram are (A) and (C) because octet of O and S are incomplete respectively . 15. (b) If X-axis is internuclear axis, then it will result in  bond. (c) It will result in  bond. (d) & (e) It shows zero overlap with no bond formation. 16. C*  1s2 2s1 2p3 4 unpaired electron  4 bonds B*  1s2 2s1 2p2 3 unpaired electron  3 bonds I*  5s2 5p4 5d1 3 unpaired electron  3 bonds P*  3s2 3p3 3 unpaired electron  3 bonds * represent excited state 17. It is the orbital that undergo hybridisation and not the electrons. For example , for orbitals of nitrogen atom ( 2s 2 2p1x 2p1y 2p1z ) belonging to valency shell when hybridise to form four hybrid orbitals , one of which has two electrons (as before) and other three have one electron each. It is not necessary that only half filled orbitals participate in hybridisation. In some cases, even filled orbitals of valence shell take part in hybridisation. 20. (A) Nitrogen is more electronegative than phosphorus. N 2.5 CH3 CH 3 P 3.0 2.5 CH3 CH3 CH 3 2.1 CH 3 So, dipole moment of trimethylamine is greater than trimethy phosphine. .. .. (B) H Si — O — SiH 3 3 H3 Si + .O. — SiH3 In trisilyl ether the lone pair of electron on oxygen atom is less easily available for donation because of p-d delocalisation due to presence of the vacant d-orbital with Si. This however is not possible with carbon in CH 3– O – CH3 due to the absence of d-orbital making it more basic. (C) Bond order of C2 and O2 are same i.e., 2. In C2 molecules both bonds are -bonds whereas, there is one and one -bond in O2 molecule C2 = 131 pm ; O2 = 121 pm. Si(Me)3 N (D) (Me)3Si . F 2 sp (trigonal planar) Si (Me)3 F (B) || N O sp2(trigonal planar) B F (C) || Cl O O O || N — N O O 21. (A) : N  O 22. (CH3 )2 O : can act as lewis base but (SiH3)2O & (SiH3)3N can not, as one pairs on O & N are deloecalised in to empty orbitals of Si. 24. In FO+ total no. of electrons = 16, so bond order will be 2. In FO– total number of electrons = 18, so bond order will be 1. 25. P4O10 O (D) || O .. 3 sp : : : : The P – O bond lengths shows that the bridging bonds on the edges are 1.60 Å but the P = O bonds on the corners are 1.43 Å and this P = O is formed by p – d back bonding. A full p-orbital on the oxygen atom overlaps sideway with an empty d-orbital on the phosphorus atom. The bond angle POP is 127º and there is no P – P bonds. S S P : : P : : S : 26. : : : : P : P 28. Fullerene are cage like molecules. C60 molecule has a shape like soccer ball and called Buckminsterfullerene. It contains twenty 'six -membered rings' and twelve 'five membered rings'. A six membered ring is fused with six or five membered rings but a five membered ring can only fuse with six membered rings. All the carbon atoms are equal and they undergo sp2 hybridisation. Each carbon atom forms three sigma bonds with other three carbon atoms. The remaining electron at each carbon atom is delocalised in molecular orbitals, which in turn give aromatic character to molecule. This ball shaped molecule has 60 vertices and each one is occupied by one carbon atom and it also contains both single and double bonds with C – C distance of 143.5 pm and 138.3 pm respectively. CHEMICAL BONDING 30. Cl2O > F2O and F2O < H2O 111° 102° 102° 104° NO2 > NO + sp – 2 sp2 AsI3 > AsBr3 > AsCl3 (due to size of halogen) 31. (A) Structure is similar to that of ethane. Each N atom is tetrahedrally surrounded by one N, two H and a lone pair. The two halves of the molecules are rotated 95º about N–N bond and occupy a gauche (non-eclipsed) conformation. The bond length is 1.45 Å. H N—N—H H H (B) Has partial double bond character due to p-d delocalisation. (C) OF2=103º (approximate) and OCl2 = 112º (approximate). (D) Exist in polymeric structure as 33. in solid state. (A) O2+ : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y ) (*2px1 = *2py) (B) NO is derivative of O2: NO(O2+) (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y ) (*2px1 = *2py) (C) O2– : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y ) (*2px2 = *2p1y) (D) B2 : (1s)2 (*1s)2 (2s)2 (*2s)2 (2p1x = 2p1y) (pz)0 37. (A) It results from the electrical attractions among positively charged metal ions and mobile, delocalised electrons belonging to the crystal as a whole. (B) In general, greater the number of (n–1)d and ns electrons, stronger is the resultant bonding. (C) Strength of metallic bond depend on the type of hybrid orbitals participating in metallic bonding. (D) As the size of atom increases the attraction between the positive part of the atom and delocalised electrons decreases and thus the strength of the metallic bond decreases. 39 (A) I – Cl, linear and polar because of the difference in the electronegativities of iodine and chlorine. F Cl — F (B)  0. F F F F (C) F Br F F F F  0. I (D) F F = 0. F 40 (A) net dipole moment H H H (B) Dipole moment is a vector quantity as it depends on the magnitude and the direction. I I (C) O  C  O  = 0. (D) As a result of polarisation in covalent bond which arises due to the difference in the electronegativities of combining atoms, the molecule possesses the dipole moment. 42 maleic acid fumaric acid O – C O H C C H O H C O O2[AsF4] = O2+ B.O = 2.5 so O2 + has smaller bond length than O2– – KO2 = O2 B.O = 1.5 Greater the size of the halogen atom greater will be bond angle. 43 (A) Boiling point of ICl > Br2 as ICI is polar and Br2 is non-polar in nature. (B) (C) 44 (A) (B) (C) (D) pyramidal about each N-atom. (D) SbH3 > NH3 > AsH3 > PH3 ( order of B.Pt) As D2O has higher molecular mass than H2O so its density is more. Mn  4s23d5 Ca  [Ar] 4s2 Sr  [Kr]5s2 Rb  [Kr]5s1 Refer notes H2 weak VDW forces CO2  weak VDW force but stronger than H2. H2O  H-bonding. identical due to resonance. CHEMICAL BONDING 45. 46. S2O32– ion : Individual O.N. of S atoms = +6 and –2. Average O.N. = +2 SbH3 has higher bp than NH3, due to vander waal forces. In hydracids, the acidity increases down the group while for oxyacids, the acidity decreases down the group. O sp3 hybridized || P H–O | O–H O H The P = O contain one p-d bond. The PO43– ion show resonance. Due to the presence of H-bonded to O, it show intermolecular H-bonding due to which it is syrupy. 47. The critical temperature depends on the magnitude (of strength) of intermolecular force of attraction between the molecules. If a molecule has dipole moment it means there is dipole dipole attraction between the molecules and thus it will affect the critical temperature. 48. Polarity depend on net dipole moment. If diple moment  = 0 it is nonpolar. F F (A) Xe F F XeF4 =0 non polar XeF6  0 polar XeOF4   0 polar F (B) F Xe F F F F F (C) O F Xe F F F F (D) Xe F XeF5–  0 non polar F F 49. (A) Ethyne being non-polar is more soluble in acetone. (B) CD3F is more polar than CH3F. In CD3F, C–D bond is more polar as deuterium is more electropositive than hydrogen. (C) Silyl isocyanate (SiH3NCO) is linear in shape because of p-d delocalisation of lone pair of electrons on nitrogen. While in methyl isocyanate (CH3NCO) there is no p-d delocalisation of electron as carbon does not have empty d-orbital and thus is bent in shape. In CH2=C=C=C=CH2, the 2H–atoms on one C–atom lie in a plane perpendicular to the plane in which 2H–atoms on other C–atom lie. 50. The correct formula of borax is Na2[B4O5(OH)4]. 8H2O. The structure of anion is : OH |O- sp3 B O O 2 sp OH– B O O O 53. O Bsp3 | OH 2 sp B–OH As charge on cations increases, their polarising power increases and thus covalent character increases.  2 3 4 Li Cl  Be Cl2  BC l3  C Cl4 54. Increase in oxidation state (Ni4+) increases the polarising power of cation and thus increases the polarisation of Br– ion. 55. Because of high charge density on Sn4+ it has high polarising power and thus leads to greater polarisation of anion i.e., greater distortion of electron clouds of the Cl– ions. So SnCl4 is most covalent. 57. The strength of metallic bonds depends upon the number of mobile electron(s) per atom. Sodium has only one mobile electron per atom where as iron has 8 mobile electrons per atom. 59. Cu, Ag, Zn are solids at room temperature where as mercury is liquid. 61. As size of cations increase, their polarising power decrease and thus ionic character increase. 63. Sn4+ has highest polarising power amongst Na+, Pb2+ , Sn4+ and Al3+ because of smaller size and higher charge. So SnCl4 is most covalent and thus has least melting point. 64. Metallic bond results from the electrical attractions among positively charged metal ions and mobile, delocalised electrons belonging to the crystal as a whole. 65. AgI is less soluble than AgF due to covalent nature. LiI is more soluble then LiF due to high size difference of ions more ionic nature, more melting point. Higher the charge density , larger will be the hydrated radii. 66. Polarisation of the I– by the Cu2+ results in the transference of an electron towards Cu2+ makes it as oxidising agent and I– as a reducing agent, reduces Cu2+ to Cu+ and itself oxidised to I2 . 67. (A) N2 contains one  and two  bonds but C2 contains only two  bonds as four electrons are present in bonding  molecular orbitals. (B) Cu2+ has higher hydration energy then Cu+ so Cu2+ is more stable than Cu+ . Moreover, Cu+ undergoes disproportionation. 2Cu+  2Cu2+ + Cu. (a fact) (C) Can be explained by electron sea model as it is due to the presence of mobile electrons. 68. Larger anion has higher polarisability. 70. 1 Dipole moment  bond angle 71. Symmetrical molecules have zero dipole moment. planar,  = 120º,  = 0. CHEMICAL BONDING 72.  0 ; SiF4 , BF3 and PF5 are symmetrical molecules thus = 0. P F F F 73. O = C = O . The values of C=O bond dipoles are same but acting in opposite direction; so cancel out. Thus dipole moment of CO2 is zero. 74. Number of electron pairs = 4 Number of bond pairs = 2 Number of lone pairs = 2 So, according to VSEPR theory to have minimum repulsions it acquires bent shape as shown below. S H H It is unsymmetrical molecule thus it will have some dipole moment. 75. 76. Dipole moment of compound if it would have been completely ionic = (4.8 × 10–10 esu) (2.67 × 10–8 cm) = 12.8 D so % ionic character = 4.0 × 100% = 31.25 % 12.8 77. Dipole moment depends on the electronegativity of the elements as it is the product of charge on one of the ions and the distance between them. Dipole moment of CH3Cl is greater than CH3F due to more charge separation. (a fact) 79. Graphite has layered structure. Layers are held by van der Waal’s forces and distance between two layers is 340 pm. Each layer is composed of planar hexagonal rings of carbon atoms. C – C bond length within the layer is 141.5 pm Each carbon atom in hexagonal ring undergoes sp2 hybridisation and make three sigma bonds with three neighbouring carbon atoms. Fourth electron forms a  bond. The electrons are delocalised over the whole sheet. 80. Nitrogen is less electronegative than oxygen so electron pair is easily available for bonding, thus P has stronger Hbonding than Q. As oxygen is more electronegative so H-bond strength of Q is greater than S. The same explanation is given for R and S as that of P and Q. F H 81. F F H H F CH3 H CH3 CH3 ... O — H ... O — H ... O — H 82. All molecules have London dispersion force because of number of polarizable electrons. Non polar molecules have only London dispersion force because of polarisable electron. Benzene is non polar molecule. Polar molecules have dipole-dipole attraction in addition to London dispersion force. and so on Molecules with, O–H, N–H, or F – H bond have hydrogen bonding in addition to London dispersion force. Molecule IC is polar, so in addition to London dispersion force it has dipole-dipole attraction also. 84. London forces are extremely short range in action and the weakest of all attractive forces. The order of strength of bonds/ forces is ionic bond > covalent bond > hydrogen bond > London force. 86. (A) A fact (B) In the solid state and in liquid HF , the HF2 – ion is held together by hydrogen bonding. In aqueous solutions there is hydrogen bonding but each HF molecule forms hydrogen bonds with the much more prevalent H2O present, instead of with other HF molecules and H3O+ and F – are much more likely to be formed. (C) H3BO3 (solid) has inter molecular hydrogen bonding. H H B H H H H H B B H H H H B H H B B H H H H H H 88. As molecular weight increases, magnitude of Van der Waal's force of attraction increases. Therefore, boiling point increases. Though NH3 has strong H-bonding but boiling point of SbH3 is highest due to highest molecular weight. Boiling point NH3 = 238.5 K and SbH3 = 254.6 K. (A) intermolecular hydrogen bonded. (true) (B) structure of anions are different CO32– – trigonal planar (sp2) 2– : 89. SO32– – trigonal pyramidal (sp3) O S || O O (false) CHEMICAL BONDING (C) no lone pair of electrons (false) (D) True. 90. D2O = 374.4 K ; H2O = 373.0 K (i) H2O has highest boiling point because of H-bonding. (ii) Boiling point also depends on the magnitude of van der Waal's force of attraction, which in turns depends on molecular weight of the compounds. Thus the correct order is H2O > H2Te > H2Se > H2S. 93. (A) (B) (C) (D) F 94. F Xe XeF2 S SF2Cl2 F F F Cl Xe =O XeOF2 O Cl O Cl I IOCl – 4 Cl Cl Cl I ICl2– F Cl Cl F2ClO+ Cl F F 99. Fluorine is more electronegative but dipole moment is the product of charge and distance between combining atoms (covalently bonded) ; due to more charge separation in CH3Cl, it has higher dipole moment. 100. S1 : O O O Due to resonance , both bonds are equal. S2 : Hydration of ions due to force of attraction between ion and water molecule. Which is a dipole. S3 : Due to intermolecular H-bonding, boiling point increases. S4 : A symmetrical molecule with identical bonds have zero dipole moment. 102. S1 : Due to intermolecular H-bonding in HF it boils at higher temperature than HCl S2 : Mol. wt. of HBr < Mol. wt. of HI S3 : Bond order of N2 is more than N2+ . S4 : Molecular mass of F2is less than that of Cl2 103. Strength of H-bond depends on following factors. (i) Electronegativity of element covalently bonded to hydrogen atom. (ii) Size of electronegative element. (iii) Ease of donation of lone pair of electrons by electronegative element. 104. It has intramolecular H-bonding 106. (A) As + charge on central atom increases, the attraction between + and – increases and thus Xe–F bond length decreases. The correct order is XeF2 > XeF4 > XeF6 (B) PH5 can not undergo sp3d hybridisation as there is much large difference in size of s, p and d orbitals. PH5 does not exist as no partial positive charge develops on P atom. (C) Dipole moment of CH3Cl is greater than CH3F due to greater charge sepration on carbon and chlorine atoms in CH3Cl. (D) it is a correct order. The strength of hydrogen bond depends upon : (i) size (ii) electronegativity and (iii) ease of donation of electron pair by electronegative element. Higher the value of electronegativity and smaller the size of the covalently bonded atom to H atom stronger is the hydrogen bonding. 107. (A) in S8 hybridisation of each S-atom is sp3 , in P4 hybridisation of each P-atom is sp3 F 89º F 104º due to bp-lp repulsions S (B) : F F (C) Inter molecular H-bonding (PO43– groups are bonded by many H-bonds) 177º (D) 108. (I) O– - - - H – O HO S2+ HO O– - - S2+ O– - - - H – O O– - - - As a result of H-bonding, number of sulphuric acid molecules are associated to form cluster. Hence it has high boiling point and viscosity. CHEMICAL BONDING (II) All S — O bond lengths are equal due to resonance. filled sp3 orbital H (III) H B (IV) H H sp3 B H H vacant sp3 orbital 110. S1 : Polarising power of a cation is inversely proportional to its size. S2 : H2+ : 1s1 He2+ : 1s2 1s1 B.O. = 1 2 B.O. = 1 2 But He2+ has electron in antibonding molecular orbital so its is unstable. S3 : H-bond formation requires the availability of lone pair of electron. S4 : In Cl2O, the bond angle is more than 109º28’ due to steric repulsion. 112. F N F H N H  ; So dipole moment of NH3 is greater than NF3. H F 113. This is based on the decreasing electronegativity difference between H (2.1) and Sb (1.9) , As (2.0) , P (2.1) and N (3.0). 114. van der Waal's forces  molecular weight. So AgBr will have maximum van der Waals forces. 115. (C) Both compounds have intermolecular H-bonding amongst themselves and with water. 116. ; Hence hybridisation of Cl is sp3 and molecule is polar.. 117. Polarity depend on net dipole moment. If dipole moment  = 0 it is nonpolar. In XeF4 =0 XeOF4   0 118. non polar ; SO3  0 non polar polar ; Cl  0 non polar – 4 In CH4, H2Se and H2S the central atoms are not more electronegative ; hence do not form hydrogen bonds with itself and other molecule. In N2H4 the nitrogen is more electronegative and thus is able to form hydrogen bond with itself and other molecules like water. 122. In pure phosphoric acid the PO43– groups are bonded together through many hydrogen bonds. 123. HF is least volatile (i.e. high boiling point) due to strong intermolecular hydrogen bonding. 124. Boiling point of SbH3 is greater than NH3. The higher boiling point of SbH3 is attributed to higher van der Waal forces because of its higher molecular weight. (a fact) 125. (A) Like hydrated copper sulphate (CuSO4 .5H2O) in which one water molecule is hydrogen bonded to coordinated water molecule, one water molecule is coordinated to lone pair of electrons on SnCl2 and the other is hydrogen bonded to coordinated water molecule. Part # II : Assertion & Reason 2 3. Statement-1 : It is correct statement because all three p-orbitals are involved in sp3 hybridisation and thus no porbital is left for side wise overlapping. 6 :N C H 8. : Statement-2 : XeO3 is sp3 hybridised and Xe form three double bonds through p-d overlapping. H 2 sp C O: sp H Statement-1 : To have minimum repulsion between the I = O and the lone pair of electrons (has greater repulsive effect), the lone pair is opposite to the I = O bond. Statement-2 : correct statement. 9. Statement-1 and Statement-2 both are correct statement and Statement-2 is the correct statement of Statement-1, e.g., NO2+ and I3– have different hybridisation but on account of stability they have linear shape as given below.  O  N  O (linear) sp and I | I 3 | sp d I linear – CHEMICAL BONDING 10. S1 : SO2, NO3– and CO32– have same number of electrons (i.e. 32) but have different molecular geometry. — O O— sp2 S O O O (bent) + N sp2 C O— (trigonal planar) O 2 sp O— (trigonal planar) S2 : The d and f–orbitals do not shield the nuclear charge very effectively. Therefore, there is significant reduction in the size of the ions , just after d or f orbitals have been completely filled. This is called lanthanide contraction. 12. Both Statement-1 and Statement-2 are true but Statement-2 is not true explanation (directly not but indirectly true) of Statement-1. Atomic size of carbon is smaller and thus it has effective overlapping with itself and with other atoms of small size and high electronegativity. Few examples of multiple bonding are C = C, C C, C = O, C = S and C N. Heavier elements do not form p-p bonds because their atomic orbital are too large and diffuse to have effective overlapping. 13 Elemental nitrogen exists as a diatomic molecule because nitrogen can form p–p multiple bonds which is not possible in case of phosphorus due to repulsion between non–bonded electrons of the inner core. There is no such repulsion in case of smaller nitrogen atoms as they have only 1s2 electrons in their inner core. F 0 87 40’ P 16 F 1.698Å Q 1.598 Å 0 87 40’ F 19 Both are correct and Statement-2 is the correct explanation of Statement-1. In [Al(H2O)6]3+ , aluminium is in +3 oxidation state. So, [Al(H2O)6]3+ ; 21 Both statements are true and the Statement-2 is the true explanation of Statement-1. Fullerenes : It contains twenty 'six -membered rings' and twelve 'five membered rings'. A six membered ring is fused with six or five membered rings but a five membered ring can only fuse with six membered rings. All the carbon atoms are equal and they undergo sp2 hybridisation. Each carbon atom forms three sigma bonds with other three carbon atoms. The remaining electron at each carbon atom is delocalised in molecular orbitals, which in turn give aromatic character to molecule. This ball shaped molecule has 60 vertices and each one is occupied by one carbon atom and it also contains both single and double bonds with C – C distance of 143.5 pm and 138.3 pm respectively. 22 Order of acidic character is HCIO > HBrO > H IO as the order of the stability of their conjugate base is CIO– > BrO – > IO – ; because Cl being small size atom can form effective -bond with oxygen. 24 Both are true statements and the Statement-2 in the correct explanation of Statement-1. C2 : (1s)2 (*1s)2 (2s)2 (*2s)2 (2p2x = 2p2y) or KK (2s)2 (*2s)2 (2p2x = 2p2y ) The bond order of C2 is 1/2 (8 – 4) = 2 and C2 should be diamagnetic. It is important to note that double bond in C2 consists of both pi bonds because of the presence of four electrons in two pi molecular orbitals. In most of the other molecules a double bond is made up of a sigma bond and a pi bond. 27. Both the statements are true and the Statement-2 is the correct answer for the Statement-1. S O p – d overlap 28. (CH3)2 + B(CH3)3 (CH3)2 B(CH3)3 In H3Si – O – H3Si, due to p – d delocalisation, availability of lone pairs on O – atom is decreased and thus disilyl ether does not the react with B(CH3)3. 29. LiBr has more solubility being more ionic than LiI. Lattice energy as well as hydration energy of LiBr are more than for LiI due to small size of Br–. 30. Due to poor shielding by d-electrons in Ga3+, its Zeff is more. So its polarising power is more and hence it forms compound with higher covalent character. H 31 In the solid state, the B(OH)3 units are hydrogen bonded H together in to two dimensional sheets with almost hexagonal B H H B H H B H symmetry. The layered are quite a large distance apart (3.18 H Å) and thus the crystal breaks quite easily into very fine particles. H O 32. NaHCO3 ; KHCO3 ; H O–C O C || O O O || C H O H 2– O–H---O O---H O C–O H B H H B O C || O O H H H B H H H H CHEMICAL BONDING 33. London dispersion force exists among the non-polar molecules like F2, O2, 2, Cl2 etc. in solid or liquid states. Even in atoms in molecules which have no permanent dipole, instantaneous dipoles will arise as a result of momentary inbalances in electron distribution. London forces are extremely short range in action and the weakest of all attractive forces. The London forces increase rapidly with molecular weight, or more properly, with the molecular volume and the number of polarizable electrons. 35. F N Bond dipoles of N—F bonds are counter balanced to some extent by the dipole moment of lone F F pair of electron acting in opposite direction. This reduces both the dipole moment and its donor powers. 38. Within the Ca2+ / SO42– layers the ions are held together by strong electrovalent bonds. But these separated Ca2+ / SO42– layers are linked by relative weak H – bond. The weaker H – bonds link SO42– ions in the intermediate region. Consequently , the gypsum can be readily cleaved and stretched along the layer of water molecules. Anhydride has a completely ionic structure involving only Ca2+ and SO42– ions. EXERCISE - 3 Part # I : Matrix Match Type : : : : (A) (B) S F 3. F | Cl | F linear (A) : < 109 28’ < 109 28 i.e. approx. 90 + – sp2 Cl sp3d F bent (B) sp3d F see - saw F 3 sp3d2 ; sp d F F O (square pyramidal) Xe O F T-shaped F F (D) O 3 Xe F F F F F 2 sp d ; (square pyramidal) sp3d2 Br F I O F F I (C) F O ; (C) 0 (D) F – F ’ F F 8' 0 Kr F N 0 9°2 F F <10 2. F F (square pyramidal) .. – ; 3 sp Se || F F O (trigonal pyramid) F O O 5. (A) HO O P P O O OH P O OH O O || || (C) H  O  S  (S)2  S  O  H || || O O 6. O O || || (B) H  O  S  O  S  O  H || || O O (D) (A) N2+ : BO = 2.5 (4 electron in antibonding MO) 1 unpaired electron (Paramagnetic) N2– : BO = 2.5 (5 electron in antibonding MO) 1 unpaired electron (Paramagnetic) (B) NO : BO = 2.5 (5 electron in antibonding MO) 1 unpaired electron (Paramagnetic).Last electron in antibonding MO.So, easily removed. N2 : BO = 3 (4 electron in antibonding MO) 0 unpaired electron (diamagnetic).Last electron in bonding MO.So, not easily removed. (C) NO+ : BO = 3 (0 unpaired electron) diamagnetic [BO   BL ]. (D) He2+ : BO = 0.5 (1 unpaired electron) paramagnetic [1 electron in antibonding MO]. H2+ : BO = 0.5 (1 unpaired electron) paramagnetic [0 electron in antibonding MO]. 9. (A) O O || S 120º  non-polar, p-d bonds present. O All S–O bonds are identical. No lone pair. (B) O F | S | F F F polar, p-d bonds present. No lone pair. (C) O O || F S | O–  polar  p-d bonds present. Due to resonance identical S-O bond length O || (D) F–Cl  polar F | F  p-d bonds present.  lone pair present. CHEMICAL BONDING : (A) I : : 10. – F sp3d ;  F – I – F = 1800 ;  = 0 F sp3d; polar;  F – Cl – F  180º (B) F (C) O sp3d, non–polar,  F – Xe – F = 1800 O Xe O F F (D) : S F sp3d , polar; F – S – F  1800 ; one lone pair.. F F 11. (A) CuSO4. 5H2O SO42– is resonance stablized. (B) Ca2+ / SO42– are held together by electrostatic force of attraction i.e., ionic bond but alternate layers by Hbonds with water molecules. CaSO4 . (C) H3PO4  12. 1 HO 2 2 (D) (A) Br2 is non-polar in nature. All molecules experience London dispersion forces which results from the motion of electrons. (B) HF has hydrogen bonding in liquid as well as in solid state. HF molecules also experience dispersion forces. (C) Ion-dipole forces are found in the solutions of ionic compounds in polar solvent where solvated species such as Na(OH2)x+ and F(H2O)y– exist. Hydrogen bond exist between water molecules. In addition they also experience dispersion London forces. (D) CH3NH2 is a polar molecule that can form hydrogen bonds. In addition, it has dispersion forces. (E) Induced dipole-dipole interaction exists between non-polar molecules and polar molecules such as clathrate compound of Xe gas with water, when water is frozen under a high pressure of the gas. They have formulae approximating to 6H2O : 1 gas atom (i.e. Xe.6H2O). In addition, Xe.6H2O has H-bonding and dispersion forces. Part # II : Comprehension Comprehension # 1 : || O : 1. F | Cl – F | F I I I Cl-atom is in sp3d hybridisation state. Hence geometry is trigonal bi-pyramidal which is similar to 3– 2. Number of electrons pairs = 6 ; number of bond pairs = 5 ; number of lone pairs =1. According to VSEPR theory geometry of the molecule is square bipyramidal. As all positions are equivalent the lone pair of electrons can occupy any position in octahedral geometry as given below.  sp 3d 2 3. (A)  sp3d (B)  sp 3d 2 (C)  sp3 4. Valence shell electron configuration of Cl, 3s2 3p5 To obtain effective p – d overlap the size of the d orbital must be similar to size of the p orbital. Hence in chlorine, p – d bonding is so strong that no polymerization of oxoanions occurs. CHEMICAL BONDING 5. (A) With hydrogen sulphur does not undergo sp3d2 hybridisation because of larger difference in energies between s, p and d-orbitals. Sulphur show +6 oxidation state with highly electronegative elements like O and F. (B) As fluorine is smaller and more electronegative than oxygen. (C) I being large in size, cannot get accomodated around S. Comprehension # 2 : 1. (A) O2+ (dioxygenyl) O2 (dioxygen) O2– (superoxide) 2– O2 (peroxide) Bond Order Internuclear Distance (pm) 2.5 2.0 1.5 1.0 112.3 120.07 128 149 Number of unpaired(s) Electrons /Magnetic property 1 (paramagnetic) 2 (paramagnetic) 1(paramagnetic) 0 (diamagnetic) Bond order  stability (i.e., bond strength) (B) Helium molecule (He2) : He2 : (1s)2 (*1s)2 Bond order of He2 is ½(2 – 2) = 0 The molecular orbital description of He2 predicts two electrons in a bonding orbital and two electrons in an antibonding orbital, with a bond order of zero - in other words, no bond. The noble gas He has not significant tendency to form diatomic molecules and, like the other noble gases, exists in the form of free atoms. (C) Carbon molecule (C2) : (1s)2 (*1s)2 (2s)2 (*2s)2 (2p2x = 2p2y) or KK (2s)2 (*2s)2 (2p2x = 2p2y ) Lithium molecule (Li2) : (1s)2 (*1s)2 (2s)2 Peroxide (O22–) : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y ) (*2px2 = *2p2y) As all electrons are paired so C2, Li2 and O22– are diamagnetic. (D) Fluorine molecule (F2) : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y ) (*2px2 = *2p2y) 3. (B) Oxygen molecule (O2) : O2 : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y ) (*2px1 = *2p1y) Bond order = 1/2(10 – 6) = 2.0, O2+ : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y ) (*2px1 = *2p0y) Bond order = 1/2(10 – 5) = 2.5. Nitrogen molecule (N2) : (1s)2 (*1s)2 (2s)2 (*2s)2 (2p2x = 2p2y ) 2pz)2 The bond order of N2 is 1/2(10 – 4) = 3. N2+ : (1s)2 (*1s)2 (2s)2 (*2s)2 (2p2x = 2p2y ) 2pz)1 Bond order = 1/2(9 – 4) = 2.5. Comprehension # 3 : 1. 2. 3. (D) For more covalent character, small cation and large anion are favourable factors. In MgS, Mg2+ will have higher polarising power and S2– will have higher polarisability. Hence there will be higher polarisation of anion resulting in higher covalent character. Due to smaller size of Be2+ and largest size of – amongst all anions i.e. F– , Cl– , Br– and –, there will be greater polarisation of anion. Thus Be2 will be most covalent i.e. least ionic. As the size of the cations increases in the order Si4+ < Sn4+ < Sn2+ and for size of anions F– < Cl– so the order of increasing ionic character is SiCl4 < SnCl4 < SnF4 < SnCl2 < SnF2. 4. As polarizability of anion increases covalent character increases. Comprehension # 4 : 2. Vector addition of dipole moment cancels the dipole moment due to three C–Cl bonds. Only one remains. sp 3. 3 P | Cl Cl (A) Cl  polar  non polar (C) (B)  polar (D)  polar Comprehension # 5 : A a a 1. 120° X X= a 2  a 2  2a.a cos 120 Net dipole moment = 2a 2  a 2 = a Comprehension # 6 : 1. 2. o-nitrophenol has intramolecular H-bonding so it will be vaporised first. 3. D2O has different physical properties than H2O max. density of D2O is at approximately 10°C. EXERCISE - 4 Subjective Type 1. (a) Ionic (b) covalent. 2. O–S–O >  O–O–O. – O + 3. O N – O 4. 5. O– O S O sp3 –s 6 4 1 5 O O O O– O S – O O O– O – + – O O – (NO3 ) S O N – O – O + N – – O S – O O (SO42–) O S – – O O O O O – O O– S O – O CHEMICAL BONDING 7. Two terminal carbons can be assumed to be at A and B, while the central carbon at O. Then, AB = 2AP But AP = sin AO    2  109 o 28    = sin   2    (in tetrahedral structure  = 109o28) = sin (54o44’)  AP = AO sin (54o44’) = 1.54 × 0.82 = 1.26Å  AB = 2AP = 2.52Å Ans. 2.52 Å 8. d 1 = 2 × 134 × sin 60° pm = 227.8 pm = 228 pm d 2 = 134 × 3 + 2 × 134 cos 60° pm = 536 pm z y 9. (a) (b) (c) s pz No bond No bond (d) (e) (f) 10. OF4 and OF6 are not possible because oxygen does not have d-orbital for sp3d and sp3d2 hybridisation. 11. Because ionic bonds are non directional in nature  12. Cyanate ion has canonical structures in which negative charge resides on electronegative O or N.  The contributing canonical structures in fulminates are less stable as both contain  as well as  charge. 13. In BF3, B is electron defficient whereas in graphite, 'C' has one free electron , which can conduct electricity. 14. (BeH2)n contains 2e – 3c bonds whereas (BeCl2)n contain the usual 2e – 2c bonds. 15. (a) 6 16. 634 17. (b) 6 CHEMICAL BONDING 18 6 shared ‘O’ atoms. (a) (c) (d) (e) 3 S—O—S bond (b) 4 equivalent N – O bonds due to resonance. a b c d e 6 3 3 4 3 20 V O2[AsF6]   O 2  [AsF6] O22– + peroxide ion B.O = H O O 1 (10  8) = 1 2 H O–O bond is non polar so 2 polar bond. carbon suboxide is C3O2 O=C=C=C=O SiMe 3 Me Si 3 N SiMe 3 Number of l.p is present in N(SiMe3)2 is zero. 21 O p -d P ( -bond) O O O O=P O P=O O P O P O sp 3 The P–O bond lengths shows that the bridging bonds on the edges are 1.60 Å but the P=O bonds on the corners are 1.43 Å and this P=O is formed by p–d back bonding. A full p-orbital on the oxygen atom overlaps sideway with an empty d-orbital on the phosphorus atom. The bond angle POP is 127º and there is no P–P bonds. (i) P = O have bond length of 1.43 Å (8 covalent bonds) (ii) P – O have bond length of 1.60Å (12 covalent bonds) (iii) Number of P – O – P linkage = 6. (iv) Number of lone pair on each phosphorus atom is zero. 4 12 6 0 (i) (ii) (iii) (iv) 22. 2n 23. The lone pair of electrons on N atom in trisilyl amine undergoes p-d delocalisation. Thus this lone pair of electron is not easily available for the donation. On the other hand, in trimethyl amine it is easily available for donation as carbon does not have d-orbital for p-d delocalisation. In O2 and N2 energy gap between HOMO and LUMO is large so electronic excitation is not possible with visible light. But for halogens, the electronic excitation can be done with visible light because energy difference between HOMO and LUMO is small. 24. – 25. + y + – + + – – z +  – + + – – + Bonding (3 nodal planes) – – + + – – + z  z + – – – + y + – + + – Anti bonding (4 nodal planes) 26. sheet like planar structure. CHEMICAL BONDING 27. AP, CaS, KI, MgO 28. AlCl3 shows significant covalent character due to high polarising power of Al3+. (Fajan's rule) while NaCl is predominantly ionic. Hence NaCl has higher M.P. and B.P. against AlCl3. 29. Alkali metals have one loosly bound electron which can easily be excited by heat. During de- excitation the electron imparts colour to the flame. 30. (i) Mg2+< Fe2+ < Ni2+ < Zn2+ 31. (i) XeF2 > XeF4 > XeF6 (ii) Al3+ < Fe3+ < Ga3+ (ii) TiCl2 > TiCl3 > TiCl4 (iii) NaCl> MgCl2 >AlCl3 32. Al3+, As per Fajan's rule, higher the charge on cation, higher will be its polarising power. 33. H–C C– H sp Hybridisation. Due to sp hybridisation of C atom, the electronegativity of C is more and hence can form H-bond with HF. 34. The structure of chloral hydrate is a follows H Cl Cl C Cl H O C H O Its stability is due to intermolecular H-bonding. 35. Thio-ethers have higher boiling points due to their higher molecular weights than that of ethers. 36. EN = 1.4 % ionic character = 29.26 obser = 1.4 D  Th = 4.8 D  6  10 18 × 100 = 50% 4.8  10 10  2.5  10 8 37. Percentage of IC = 38. Dipole moment of compound would have been completely ionic = (4.8 × 10–10 esu) (2.25 × 10–8 cm) = 4.8 x 2.25 D so % ionic character = 9 × 100% = 83.33 % 4.8  2.25 d=1Å S 39 H 97° H R R = 2S2H  2S2H  cos97 R = 2 × S –H × 1 R = 2 × S –H × S–H = 0.88 1.5 D 2  0.94 0.15 109  1.6  10 19 10 1030 3 (S–H)Cal = 3 × 0.15 × 1.6 × 10 D (S–H)Cal = 1.5 1    ×100 = 16% 2  0.94  0.15  16.3  So % Ionic character = 41 CO2 and 42. C–F bond is most polar due to maximum difference in electronegativity. 43 (P) O= C = O BF3 No of p-d bonds = 6. (Q) NOHSO4 exists as NO+ and HSO4– ; NO+ is derivative of oxygen and isoelectronic with O22+ . Bond order is 1/2 (10 – 4) = 3. filled sp3 orbital H H H B (R) sp3 B H H H 3 vacant sp orbital No of vacant sp3 hybrid orbitals participating in the formation of banana bonds are 2. H Cl (S) Cl O C H C Cl O H CHEMICAL BONDING 44. Theoretical value of dipole moment of a 100% ionic character = e × d = (1.60 × 10–19 C) (1.41 × 10–10 m) = 2.26 × 10–29 Cm Observed value of dipole moment = 2.60 × 10–30 Cm Percent ionic character = 45. observed value 2.60  10 30 × 100 = × 100 = 11.5% Ans. theoretical value 2.26  10 29 For a diatomic molecule dipole moment µ = e × d  Electronic charge = 18  1.2  10 esu cm = 1.0  10 8 cm d Actual value of electronic charge = 4.8 × 10–10 esu  46. Fraction of the electronic charge = 1.2  10 10 = 0.25 4.8  10 10 (i) Bond order of N2 = 1/2 (10 – 4) = 3 Bond order of O2 = 1/2 (10 – 6) = 2 Bond order of F2 = 1/2 (10 – 8) = 1 Bond order of O2+ = 1/2 (10 – 5) = 2.5 Bond order is directly proportional to the bond strength and so, the increasing order of their bond dissociation energies is : F2 < O2 < O2+ < N2. (ii) The strength of H – bond mainly depends on the electronegativity of the atom attached to the H atom and it increases with increasing electronegativity of the atom attached to polar H atom. So the increasing order of strength of hydrogen bonding (X – H – – – X) is : S < CI < N < O < F F F + N sp3 B sp2 , (iii) N = N = N , , 180º F 120° F H H107.5°H F sp F Xe sp3d2 90° F So, the increasing order of bond angles is XeF4 < NH3 < BF3 > N3– 47. (i) H3C C Cl Cl Cl I H Cl C C Cl Cl Cl H H H II III I = 3 C–CI × cos 70.5 + CH = 1.5 + 0.4 = 1.9 D 3 II = 3 C–CI × cos 70.5 – C–H = 1.5 – 0.4 = 1.1 D III = 3 C–H × cos 70.5 + C–CI = 1.9 D CH3 (ii) 2CH cos 60 3 CH3 C O 2.5D EXERCISE - 5 Part # I : MAIN 1. The correct order of bond angle (smallest first) is H2S < NH3 < SiH4 < BF3 92.6o < 107o < 109o28’ < 120o 4. (A) 9. According to VSEPR theory (B) F 17. 20. 22. 24. 29. 30. 33. F F S C (D) F F Xe F F F F lp = 1 lp = 0 lp = 2 sp3 d-hybridisation sp3-hybridisation sp3d2-hybridisation see-saw shape tetrahedral shape square planar shape The elements of group 14 show an oxidation state of +4 and +2. The compounds showing an oxidation state of +4 forms covalent compound and have tetrahedral structures. e.g. SnCl4, PbCl4, SiCl4, etc. whereas those which show +2 oxidation state are ionic in nature and behave as reducing agent. e.g. SnCI2, PbCI2 etc. Further as we move down the group the tendency of the element to form covalent compound decreases but the tendency to form ionic compound increases. Because of highest electronegativity of F, hydrogen bonding in F – H - - - - F is strongest. O22– (8 + 8 + 2 = 18) : 1s2 *1s2 2s2 *2s2 2pz2 2px2 2py2 *2px2 *2py2 ; all electrons are paired. So diamagnetic. Structure of IF7 is pentagonal bipyramidal XeF2 has maximum lone pairs out of the given options. Number of lone pairs = 3 F 13. (C) F F PF5 trigonal bipyramidal BrF5 square pyramidal (distorted) NO is paramagnetic in gaseous state. Xe has highest boiling point because boiling point increases with increase in molecular mass. CHEMICAL BONDING 39. 40. NO  One unpaired electron is present in * molecular orbital. CO  No unpaired electron is present B2  Two unpaired electrons are present in  bonding molecular orbitals. O2  Two unpaired electrons are present in * molecular orbitals. ions O2– F– Na+ Mg2+ Atomic number = 8 9 11 12 No. of e– = 10 10 10 10 therefore O2–, F–, Na+, Mg2+ are isoelectronic I I 41. I Total number of lone pair in I 3 is 9. The electonic configuration of H 2 2  is  (1s2), (*1s)2 H- H22 - H- (*1s)2 Energy level 42. 1s 1s (1s)2 Bond order of H 2 2  = Nb  Na 22 = = 0. 2 2 Hence H 2 2  does not exist, due to zero bond order.. Part # II : ADVANCED 1. Molecular orbital electronic configuration is 1s2  *1s2  2s2  * 2s2  2p z 2  2p x 2  2py 2  * 2p x1  * 2py 0 As it contains one unpaired electron it is paramagnetic and bond order = (10 – 5) / 2 = 2.5 (O2 = 2.0). 17. Structure of XeO2F2 molecule is see saw. .. F O 19. F Xe O (B, D) (A) H3BO3 is a weak monobasic Lewis acid. H3BO3 + H – OH ƒ B(OH)4– + H+ ......(i) (B) Equilibrium (i) is shifted in forward direction by the addition of syn-diols like ethylene glycol which forms a stable complex with B(OH)4–. O H H O O H H O B B O H H O O O O H O H O + 4H2O O (stable complex) (C) It has a planar sheet like structure due to hydrogen bonding. (D) H3BO3 is a weak electrolyte in water. 34. 6 Br Br 2 – Br Te Br Br Br lone pair in C.A. is 1 F F F S N lone pair of electron = 0 lone pair of electron = 3 Total sum is 0 + 1 + 2 + 3 = 6 Br F + F lone pair = 2 Xe F F F CHEMICAL BONDING 35. 5 or 6 6. Diamagnetic species are (D) O C2, H2, Li2, N2, F2, Be2 (does not exist) (A) MOCK TEST 1. 2. 3. (B) As anionis same, the covalent character in ionic compound will depend upon the size and charge on the cation. According to Fajan’s rule, PbO (most ionic) = Pb2+, Mn2O7 (most covalent) = Mn+7. (D) Hybridisation is calculated by steric number rule. Steric no. = no. of atoms attached to central atom + no. of lone pairs present on central atom. Steric no. for IO4– = 4 + 0 = 4 sp3 ; – Steric no. for ICl4 = 4 + 2 = 6 sp 3d 2 + Steric no. for IF4 = 4 + 1 = 5 sp3d ; Steric no. for SO3 = 3 + 0 = 3 sp 2 2– Steric no. for SO4 = 4 + 0 = 4 sp3 ; 2– Steric no. for SO3 = 3 + 1 = 4 sp 3 + Steric no. for PCl4 = 4 + 0 = 4 sp3 ; Steric no. for BF4– = 4 + 0 = 4 sp 3 Steric no. for XeO3 = 3 + 1 = 4 sp3 ; Steric no. for XeO4 = 4 + 0 = 4 sp 3 Steric no. for XeF4 = 4 + 2 = 6 sp3d2 ; – Steric no. for ClO4 = 4 + 0 = 4 sp 3 3 O 3 7. % charge = 8. (D) Se F O pyra- F q 1.2  1010  100  100  25% e 4.8  1010 (D) (A) In (CH3)3SiOH, because of p-d deocalisation between silicon and oxygen. (CH3)3SiO– becomes stable. (B) Si having larger size is unable to form multiple like C in CO and thus SiO3 is unstable. (C) Due to resonance. Cl Cl 9. Cl C–O Cl Cl C–O Cl (A) BF3 (sp2) + F–   BF4– (sp3). 10. (D) (A), (B) M.O. for C2 = 2 2 1s2 < 1•s2 < 2s2 < 2•s2 < 2 p y  2 p z  2 p x { 1 44 2 4 43 (A) In P4O10, POP angle is not 180° HOMO (D) F O I O F3 sp d see-saw 3 sp 1.2  1018 = 1.2 × 10–10 108 q= C=O O 5. sp (B) O Cl O– O sp (C) –O S O– O midal. (D) =q×d 1.2 × 10–18 = q × 10–8 cm (A) O 1.43Å P O 127° O O 1.60Å O P P O O O P O 4. O sp2 S two  molecular orbitals are involved in bonding . F F F S F3 sp d see-saw LUMO (C) It is isoelectronic with N2 and has one sigma and two pie-bonds. (D) In both, all electrons are paired so diamagnetic. 11. (A, B, C) (C)  NH2OH + 2H2SO4 HNO2 + 2H2SO3 + H2O  S P P S P S P So, number of P — S bonds = 6; numberof P—P bonds = 3; total number of lone pairs = 10 and all atoms i.e. P and S have sp3 hybridisation. 12. (B, C) Plarity depend on net diple momen. If dipole moment  = 0 it is nonpolar. (A) F F sp 180° so2 sp3 sp3 120° 109.5° 107° (C) SF6 < NH3 < H2O < OF2 in this case bond angle of NH3 is highest because lp-lp reuplsion is absent in it. (D) ClO2 > H2O > H2S > SF6 F Xe so bond alngle decreases to 107° from 109.5° F XeF4 =0 non polar XeF6 0 polar F F (B) F Xe F F F F (C) F Note : It is supposed that in H2S the hybrid orbitals donot participate in bonding but pure p-atomic orbitals participate in bonding. 15. (A, C) For sp3d2 steric no. is 6 O F Xe XeOF4 F 0 non polar Steric no, of IF5 = 5 + 1 = 6 ; Steric no. of XeF4 = 4 + 2 = 6 ; Steric no. of XeOF4 = 5 + 1 = 6 ; F (D) F Xe Steric no. of ICl4– = 4 + 2 = 6. F F XeF5– =0 non polar F 16. (B) 13. (A, C, D) Hydrogen bonding is possible in CH3NH2 , CH3COOH, CCl3 CH(OH)2 H CH3 N H H CH3 C Cl CH3 C Cl CH3 ; N O H O O H O Within the Ca2+ / SO42– layers the ions are held together by strong electrovalent bonds. But these separated Ca2+ / SO42– layers are linked by relative weak H-bond. The weaker H-bonds link SO42– ions in the intermediate region. Consequenctly, the gypsum can be readily cleaved and stretched along the layer of water molecules. Anhydride has a completely ionic structure involving only Ca2+ and SO42– ions. H H But in XeF4 & ICl4– dipole moment is zero because of symmetrical molecule and therfore, these are non-polar. C CH3 ; O C H H O 17. (B) Both S-1 and S- 2 are true statements but are different and reason is not true explanation of assertion. High electronegativity of atoms attached to carbon atom indirectly results into the contraction of the size of p-orbitals. As carbon atomic size is smaller and thus it has effective overlapping. 14. (A, B, D) (A) According to VSEPR theory as electronegativity of central atom decreases, bond angle decreases. So bond angle of H2O > H2S > H2 Se > H2Te (B) C2H2 > C2H4 > CH4 > NH3. In NH3 there is bp-lp repulsion 18. (A) CHEMICAL BONDING F 24. (D) 870- 40’ (A) Bond order  P F 1.698Å Q 1.598 Å Bond order O2+ = 2.5 , O2 = 2 O2– = 1.5. so correct order of bond length is O2– > O2 > O2+. 870- 40’ F 19. (D) 1 bondlength (B) Bond order for He2 = 22  0 . So He2 does not 2 exist. (C) In all these molecules all electrons are paired in molecules orbitals. So they are all diamagnetic. (D) F2 MOT configuration. 1s2 *1s2 2s2 *2s2 2pz2 2py2 2pz2 *2py2 *2pz2 *2pz2 25. (A) O2 O2– O22– O2+ Bond Order 2 1.5 1 2.5 20. (A) Transition metal ions have pseudo inert gas configuration which have higher polarising power due to poor shielding of d-electrons. Alkali metal and alkaline earth metal ions have less polarising power due to inert gas configuration (more shielding effect of s- and p-electrons as compared to d-electrons). 21. (A) As size of cation decreases and charge on cation increases, the polarisation increases according to Fajan’s rule. Hence the covalent character increases. 22. (C) As size of cation increases and charge on cation decreases, the polarisation decreases according to Fajan’s rule. Thus the ionic character increases. So SnO with +2 oxidation state will be most ionic. 23. (D) Agl shows highest polarisation due to following two factors (A) Ag+ has pseudo inert gas configuration (B) I– is biggest anion. 26. (B) Species N2 N2+ O2 O2+ Bond Order 3 2.5 2 2.5 27. (C) (A), (B) and (D) have intermolecular H-bonding (Y) while anion of Caro’s acid (i.e. HSO5–) has intramolecular H-bonding (X) as shown below. 28. (C) (A) Cl  HOCl N Cl HO H  H N H H Cl (B) Association of molecules leads to a high melting solid because of inter molecular H-bonding (Y). O– (C) CH3 – C = N H O +2 Ni H3C – C = N H O N = C – CH3 N = C – CH3 O– rosy red ppt (D) Ionic peroxides form well crystallised hydrates such as Na 2O2.8H2O and M''O2.8H2O. These compounds contain discrete O22– ions to which water molecules are hydrogen bonded, giving chains of the type shown in 2– 2– structure O2 (H2O)8 O2 (H2O)8 . The formation of such stable hydrates, accounts for the extreme hygroscopic nature of crystalline peroxide. 29. (B) In the solid state, the B(OH)3 units are intermolecular Hbonded together into two dimensional sheets with almost hexagonal symmetry the layers are quite a large distance apart (3.18Å) and thus crystal break quite easily into very fine particles. 386