03 Chemical Bonding (solution) Final E 5n5u5

Chemical Bonding 24.

(a)

133

M 3 (PO4 )2 means M is divalent so formula of its sulphate is

MSO 4 . 25.

(b) As the molecular formula of chloride of a metal M is MCl3 , it is trivalent so formula of its carbonate will be M 2 (CO 3 )3 .

26.

Electrovalent bonding

27. 



(d) Sodium chloride is electrovalent compound so it dissolves in water which is a polar solvent. (d) When sodium chloride is dissolved in water, the sodium ion is hydrated.

1.

(b) NaCl is ionic crystal so it is formed by Na

2.

(a) Bond formation is always exothermic. Compounds of sodium are ionic.

3.

(a) According to Fajan’s rule ionic character is less.

32.

4.

(c) Valencies of L, Q, P and R is – 2, – 1, + 1 and + 2 respectively so they will form P2 L, RL, PQ and RQ 2 .

(b) Molten sodium chloride conducts electricity due to the presence of free ions.

33.

(b) The phosphate of a metal has the formula MHPO4 it means

34. 35.

(d) (b) Cs is highly electropositive while F is highly electronegative so they will form ionic bond. (b) Na is highly electropositive while Cl is highly electronegative so they will form ionic bond. (a) Ionic compounds are good conductors of heat and electricity so they are good electrolyte. (a) Metal tends to lose electrons due to low ionization energy. (c) As the formula of calcium pyrophosphate is Ca 2 P2 O7 means valency of pyrophosphate radical is – 4 so formula of ferric pyrophosphate is Fe4 P2 O7 3 .

5.

and Cl

ions.

(c) Electrovalent compounds are good conductor of heat and electricity in molten state or in aqueous solution.

30.

M 3  ion so formula of its phosphate would be MPO4 .

metal is divalent so its chloride would be MCl 2 .

7.

(d) Electrovalent bond formation depends on ionization energy of cation, electron affinity of anion and on lattice energy.

8.

(b) Because CsF is electrovalent compound.

37.

9.

(c)

38.

10.

(d) Valency of metal is + 2 by formula MO so its phosphate would be M 3 (PO4 )2 because valency of [PO4 ] is – 3.

NaCl is formed by electrovalent bonding.

11.

(b) Li, Na and K are alkali metals with low ionization energy and one electron in their outermost shell so they will form cation easily.

12.

(a) Melting point and boiling point of electrovalent compounds are high due to strong electrostatic force of attraction between the ions.

13.

(d) The value of lattice energy depends on the charges present on the two ions and distance between them. It shell be high if charges are high and ionic radii are small.

39. 40.

41. 42. 43.

14.

(a) Cs is more electropositive.

15.

(a) X loses electron, Y gains it.

44.

16.

  (c) Formation of NaCl occurs by Na ion and Cl ion .

45.

17.

(b)

18.

(a) Electrovalent compounds generally have high m.pt and high b.pt due to stronger coulombic forces of attractions.

19.

(d) Water is a polar solvent so it decreases the interionic attraction in the crystal lattice due to solvation.

20.

21.

MgCl2 has electrovalent linkage because magnesium is electropositive metal while chlorine is electronegative.

2

2

5

(c) Element C has electronic structure 1s , 2 s 2 p , it requires only one electron to complete its octet and it will form anion so it will form electrovalent bond. (b) Since the chloride of a metal is MCl 2 therefore metal ‘M’ must be divalent i.e. M 2 . As a result the formula of its phosphate is M 3 (PO4 )2 .

22.

48. 50. 51. 52. 53.

(a) Ion is formed by gaining or losing electrons. To form cation electron are lost from the valency shell, so Zn atoms to Zn  ions there is a decrease in the no. of valency electron.

61.

of nitrate is M ( NO 3 )3 . 23.

47.

54. 55. 57. 58. 59.

(d) In MPO4 the oxidation state of M is +3. Hence, the formula

(d) Yet the formula of sulphate of a metal (M) is M 2 (SO 4 )3 , it is

M  X bond is a strongest bond so between Na  Cl is a strongest bond. (b) The solubility order is : BeF2  MgF2  CaF2  SrF2 so SrF2 is least soluble.

(c)

(d) NaF has maximum melting point, melting point decreases of sodium halide with increase in size of halide their bond energy get lower. (b) Sulphanilic acids have bipolar structure so their melting point is high and insoluble in organic solvents. (c) CaCl 2 will have electrovalent bonding because calcium is electropositive metal while chlorine is electronegative so they will combined with electrovalent bond. (a) Electrovalent bond is formed by losing electrons from one atom and gaining electron by other atom i.e. redox reaction. (b) Electrovalent compound are polar in nature because they are formed by ions. (b) CsCl has ionic bonding. (b) As soon as the electronegativity increases, ionic bond strength increases. (b) This X element is a second group element so its chloride will be XCl 2 . (a) When electronegativity difference is from 1.7 to 3.0. This bond is called as ionic bond. (a) Ethyl chloride is an organic compound so it will be covalent. (a) Lithium oxide and calcium fluoride show ionic characters. (a) Generally cation and anion form ionic bond. (c) Those atoms which contain +ve and –ve sign are known as ion. (a) Generally Br-F contain maximum electronegativity difference compare to other compound. (a) Due to greater electronegativity difference. 3d

4s

4s

134 Chemical Bonding 31. 62.

(b) Valency of phosphorus in H 3 PO4 is supposed ‘x’ then

(b) Co 3   3d 6 4 s 0 ,

Ni 4   3d 6 4 s 0 ,

3  x 8  0, x 5  0 , x  5.

33.

(d) (1)  x  3(2)  0  1  x  6  0  x  6  1  5 .

34. 35.

(a) HCl molecule has covalent bond. (d) Electrovalent compounds have high melting point and high boiling point.

36.

(b)

64.

(d)

BaCl2 contain higher ionic character.

66.

(a) Electrolytes are compound which get dissociated into their ion in water so it contains electrovalent bond.

67.

(abc) CaH 2 , BaH2 , SrH 2 are ionic hydride.

Length of H 

68.

(bcd) Generally MgCl2 , SrCl 2 , BaCl2 are ionic compounds so they conduct electricity in fused state.

Middle length of Cl 2  198 pm

3.

(c) In N 2 molecule each Nitrogen atom contribute 3 e  so total no. of electron’s are 6. (b) Non-metals readily form diatomic molecules by sharing of electrons. Element M (1s 2 2 s 2 2 p 5 ) has seven electrons in its valence shell and thus needs one more electron to complete its octet. Therefore, two atoms share one electron each to form a diatomic molecule (M 2 )

198  99 pm 2 Bond length of HCl = Length of H + Length of Cl = 37 + 99 = 136 pm 37.

(d) Covalent character depend on the size of cation and anion.

6.

(a) In graphite all carbon atoms are sp 2 -hybridised and have covalent bond. (c) Silica has tendency to form long chain covalent structure such as carbon so it has giant covalent structure.

7. 8.

(a) All have linear structure. O = C = O, Cl – Hg – Cl, HC  CH

9. 10.

(d) Similar atoms form covalent bond. (a) Covalent bond forms when electronegativity difference of two atom is equal to 1.7 or less than 1.7 (b) Similar atoms form covalent bond. (b) Water is a polar solvent while covalent compounds are nonpolar so they usually insoluble in water.

11. 12. 13. 14.

BCl 3 is electron deficient compound because it has only ‘6’ electrons after forming bond. (b) Due to its small size and 2 electrons in s-orbital Be forms covalent compound.

(c)

18.

(c)

21.

(a) Two identical atoms are ed with covalent bond so H 2 will be covalent. (c) Element ‘X’ has atomic no. 7 so its electronic configuration will . be 2, 5. So its electron dot symbol would be : X . . (c) C-S will be most covalent. Covalent character depend on the size of cation and anion. (c) HCl has ionic character yet it has covalent compound because electronegativity of chlorine is greater than that of hydrogen.

23.

24. 25. 26.

H 2 O will formed by covalent bonding.

(c) Order of polarising power Be   Li   Na 

Hence order of covalent character BeCl 2  LiCl  NaCl .

(d) Compound has 254 gm of I 2 means 80 gm O 2 means

254  2 mole, while 127

80  5 mole so they will form compound 16

I2 O5 . 38.

.. .. .. .. : M ..M :  : M : M : .. .. .. .. 5.

74  37 pm 2

Length of Cl 

Covalent bonding 2.

Middle length of H 2  74 pm

39. 41.

(c)

NH 4 Cl has covalent as well as ionic bond.

H     |   H  N  H  Cl    |   H   (d) Covalent character increases when we come down a group so CaI2 will have highest covalent character. (b) In water molecule three atom are linked by covalent bond.

O H H .. (b) : N  N   O : or N  N  O. .. (b) The electronic configuration Structure is

42. 44.

of

Na (Z  11) is

1s 2 , 2 s 2 2 p 6 , 3 s1 . The oxide of Na is Na 2 O . 45. 47. 48.

(b) Covalent bond is directional. (d) Bond dissociation energy decreases with increase in size. So D is smallest. (b) Molecule X is nitrogen because nitrogen molecule has triple bond. It’s configuration will be 1s 2 , 2 s 2 2 p 3 .

49.

(a)

50.

(a) The compound will be A2 B3 (By criss cross rule).

51. 52.

(b) Each nitrogen share 3 electrons to form triple bond. (d) Urea solution does not conduct electricity because it is a covalent compound. (d) Due to the small size and higher ionization energy, boron forms covalent compound.

54.

PCl5 does not follow octet rule, it has 10 electrons in its valence shell.

BF 3 contain 6 electron so it is lewis acid.

58.

(a)

59.

(d) Among the given species. The bond dissociation energy of C  O bond is minimum in case of CO 32  by which

Chemical Bonding

135

C  O bond become more weaker in CO 32  or the bond

order of CO 32  (1.33) is minimum so the bond become weaker. 60.

(a) Valency of

Na 2 S 2 O3

is supposed to be

x, then

2  2 x  ( 6)  0 , 2 x  4  0 , x  2 . O

O

||

61.

(c)

||

H  O  S  O  O  S  O  H (Marshall acid) ||

62.

63.

64.

65.

||

O O (a) Among the given choice Al is least electropositive therefore, the bond between Al and Cl will be least ionic or most covalent or the difference in electronegativeity of two atom is less than 1.8.

2.

(b)

O

(b) Electronic configuration of 16 S 32  1s 2 ,2 s 2 ,2 p 6 ,3 s 2 ,3 p 4 . In the last orbit it has only 6 electron. So it require 2 electron to complete its octet, therefore it share 2 electron with two hydrogen atom and forms 2 covalent bond with it. (b) The acidity of hydrides of VI group elements increase from top to bottom as the bond strength X  H decrease from top to bottom H 2 O  H 2 S  H 2 Se  H 2 Te

3.

4.

(d)

(b) We know that Al 3 cation is smaller than Na  (because of greater nuclear change) According to Fajan's rule, small cation

7.

H O  N  O . (a) Structure of N 2 O5 is O  N  O  N  O .

3



H O  SO H 

O (c) NH 3 has lone pair of electron while BF3 is electron deficient compound so they form a co-ordinate bond. NF3  BF3

66.

between Al and Cl atoms. (b) Sulphur has the second highest catenation property after carbon. Its molecule has eight atom bonded together (i.e. S 8 )

67.

(b)

9. 10.

H 2 O 2 has open book structure.

(a)

H O H 7

electronic

configuration

of

nitrogen

72. 73. 74. 76.

(ad) CO has only 6 electrons while PCl5 has 10 electrons after sharing so both don’t follow octet rule. (a) Among these, NaH and CaH 2 are ionic hydrides and B2 H 6

(d) F

– Cl

F F

F Xe

O



(a)

13.

(a)

H 3 PO4 is orthophosphoric acid.

O

F

H O  PO  H |

O |

15.

H (c) Sulphuric acid contain, covalent and co-ordinate bond.

Dipole moment 1. 2. 3. 5. 6.

8.

(b) CO 2 is a symmetrical molecule so its dipole moment is zero. (d) These all have zero dipole moment. (d) HF has largest dipole moment because electronegativity difference of both is high so it is highly polar. (c) Due to its symmetrical structure. (c) Chloroform has 3 chlorine atom and one hydrogen atom attached to the carbon so it is polarised and it will show dipole moment. (a) The dipole moment of two dipoles inclined at an angle  is given by the equation  

Co-ordinate or Dative bonding

O

SO 32  has one coordinate bond.  O  S  O 



.. .. H   N   H or H  N  H | . H  H (c) Multiple bonds have more bond energy so C  N will be the strongest. (c) Diamond, silicon and quartz molecule bounded by covalent bond. (cd) C 2 H 4 and N 2 has multiple bonds.

O



O

12.

N  1s 2 , 2 s 2 , 2 p 3

and NH 3 are covalent hydrides.

1.



O

H 2 SO 4 . Therefore all these contains coordinate bond.  CH 3 N  C contain dative bond.

is

It has 5 electrons in valency shell, hence in ammonia molecule it complete its octet by sharing of three electron with three H atom, therefore it has 8 electrons in its valence shell in ammonia molecule

71.

does not have co-ordinate bond. Structure is

O (d) Co-ordinate bond is a special type of covalent bond which is formed by sharing of electrons between two atoms, where both the electrons of the shared pair are contributed by one atom. Since this type of sharing of electrons exits in O 3 , SO 3 and

O

(b) The

HNO 2



polarise anion upto greater extent. Hence Al polarise Cl ion upto greater extent, therefore AlCl3 has covalent bond

69.

H 2 SO 4 has co-ordinate covalent bond.

F

X 2  Y 2  2 XY cos

cos 90  0 . Since the angle increases from 90  180 , the value of cos becomes more and more – ve and hence resultant decreases. Thus, dipole moment is maximum when   90 .

136 Chemical Bonding 9.

(c) Due to distorted tetrahedral geometry SF4 has permanent

dipole moment



F F :

1.03  100  16.83%  17% 6.12

Polarisation and Fajan's rule

S

F F 10.

(b) CCl 4 has no net dipole moment because of its regular tetrahedral structure.

12.

(d)

14.

(c)

16. 20.

1.

(d)

BF3 is planar while NF3 is pyramidal due to the presence of

lone pair of electron on nitrogen in NF3 . 2.

(c)

3.

(b) When electronegativity difference is more between two ed atoms then covalent bond becomes polar and electron pair forming a bond don’t remain in the centre.

(c) Dipole moment of CH 3 OH is maximum in it.

4.

(d) Hexane has symmetrical structure so does not have polarity.

(b) CH 4 have regular tetrahedron so its dipole moment is zero.

5.

(c) When two identical atoms form a bond, bond is non-polar.

22.

(b) Ammonia have some dipole moment.

6.

23.

(b) Charge of e   1.6  10 19

(a) According to Fajan’s rule, polarisation of anion is influenced by charge and size of cation more is the charge on cation, more is polarisation of anion.

8.

(a) When two atoms shares two electrons it is an example of covalent bond. This covalent bond may be polar or may be non-polar depends on the electronegativity difference. In given example formula is AB. So it is polar.

9.

(c) HCl is most polar due to high electronegativity of Cl.

10.

(b)

11.

(d) p-dichloro benzene have highest melting point.

13.

(b)

H-F is polar due to difference of electronegativity of hydrogen and fluorine so it shows positive dipole moment. BCl 3 has zero dipole moment because of its trigonal planar geometry.

Dipole moment of HBr  1.6  10 30 Inter atomic spacing  1Å  1  10 10 m % of ionic character in

HBr  

dipolemoment of HBr  100 inter spacing distance q

1 .6  10 30  100 1 .6  10 19  10 10

 10 30  10 29  100  10 1  100  0.1  100  10% 25.

(b)

29.

(c) Given ionic charge  4.8  10 10 e.s.u. and ionic distance

BF3 has zero dipole moment.

× ionic distance 30.

 4.8  10 10  10 8

 4.8  10 8 e.s.u. per cm  4.8 debye. (a) Higher is the difference in electronegativity of two covalently bonded atoms, higher is the polarity. In HCl there is high difference in the electronegativity of H and Cl atom so it is a polar compound.

31.

(a) Linear molecular has zero dipole moment CO 2 has linear

32.

(c)

structure so it does not have the dipole moment O  C  O .

33.

34.

35.

SF6 is symmetrical and hence non polar because its net dipole moment is zero. (a) Polarity create due to the difference in electronegativity of both atom in a molecule except H 2 all other molecule have the different atom so they will have the polarity while H 2 will be non polar. (bd) cis isomer shows dipole moment while that of trans is zero or very low value. Trans 1, 2 di-chloro-2-pentene will also show dipole moment due to unsymmetry. (a) % of ionic character Experiment al value of dipole moment = Expected value of dipole moment

NH 4 Cl has both types of bonds polar and non polar



   H  Cl   

14.

(b) Greater the charge of cation more will be its polarising power (according to Fajan’s rule).

15.

(d)

16.

(d) As the size of anion increases, polarity character increases.

20.

(d) Due to the electronegativity difference.

21.

(a) We know that greater the difference in electronegativity of two atoms forming a covalent bond. More is its polar nature. In HF there is a much difference in the electronegatives of hydrogen and flourine. Therefore (HF) is a polar compound.

22.

(c) Silicon tetrafloride has a centre of symmetry.

23.

(d)

25.

(b) According to Fajan’s rule largest cation and smallest anion form ionic bond. (b) Polarity character is due to the difference in electronegativity of two atoms or molecule.

8

 1 A  10 cm we know that dipole moment = ionic charge

NH 3 has sp 3 hybridised central atom so it is non planar.

H  |   H  N|  H 

(a) Carbon tetrachloride has a zero dipole moment because of its regular tetrahedral structure.

27.

H 2 O is a polar molecule due to electronegativity difference of hydrogen and oxygen.

26.

AlI3 Aluminiumtriiodide shows covalent character. According to Fajan’s rule.

BF3 have zero dipole moment.

Overlaping-  and - bonds 1.



(c)

H–C 2.

 C–H

 molecule formation p-p orbitals take part in bond (c) In fluorine formation.

Chemical Bonding 3.

(b) -bond is formed by lateral overlapping of unhybridised p-p orbitals.

5.

(d) No. of e pair  3  –

C 4.

1 [3  3] =0 2

No. of e pair  3 +0

1 and 2

F

C 5.

(c) In a double bond connecting two atom sharing of 4 electrons take place as in H 2 C  CH 2 .

6. 9.

(c) C  C is a multiple bond so it is strongest. (d) As the bond order increases, C  H bond energy also increases so it will be greatest in acetylene because its B.O. is 3. H H

11.

(b)

|

16. 17.

(a)

N

|

8.

(b) In ethylene both Carbon atoms are sp 2 - hybridised so 120 o .

|

9.

(d) Structure of bipyramidal. ||



(c)

2, 2 bond and one lone pair.

.. .. (d) : O  S  O : 5 atoms has 12 electrons in its outermost ||

(d) In H  C  C  O  H the asterisked carbon has a valency of * 5 and hence this formula is not correct.

11.

(d) dsp 3 hybrid orbitals have bond angles 120 o , 90 o .

13.

(a) In BeF3 , Be is not sp 3 –hybridised it is sp 2 hybridised.

17.

(c) In molecule OF2 oxygen is sp 3 hybridised.

18.

(a) In sp 3 hybrid orbitals s-character is 1/4 means 25%.

19.

(d)

20.

th

XeF4 molecule has ‘Xe’ sp 3 d 2 hybridised and its shape is square planar. (b) The bond angle is maximum for sp hybridisation because two

sp hybridised orbitals lies at angle of 180 o .

O: .. shell. One (S  O) bond will be (p-p)  bond while two (S  O) bond will be (p-d)  bond. 20.

hybridized compound is Trigonal

10.

(d) We know that trisilylamine is sp 2 –hybridized therefore

.. :O

sp 3 d

O

S

19.

No. of atom bonded to the central atom  3 In case of 3, 3 geometry is Trigonal planar.

N

.. :O

F

(a) In sp 3 –hybridisation each sp 3 hybridised orbital has 1/4 scharacter.

p  d bonding is possible due to the availability of vacant d-orbitals with silicon. 18.

120°

120°

H



B

6.

H C  C  C  C



F 120°

–

(b) Ca

137

(d) Structure of P4 O10 is

C 2 H 4 Br2 has all single bonds so C  H bond distance is the largest.

21.

(c)

23.

(a) In methane molecule C is sp 3 hybridised so its shape will be tetrahedral.

24.

(c) In compound CH 2  C  CH 2 the second carbon sphybridised. .. .. (a) : Cl   Cl : is the correct electronic formula of Cl 2 molecule .. .. because each chlorine has 7 electrons in its valence shell.

O

3

P O

O

25.

O P

O

P

O O

O

1

XeF4 has sp 3 d 2 hybridisation, its shape is square planar.

26.

(a)

27.

(b) In HCHO, carbon is sp 2 hybridized

O P

2

H O

|

H  C2  O

Each phosphorus is attached to 4 oxygen atoms.

sp

Hybridisation 1.

(d)

H 2 O is not linear because oxygen is sp 3 hybridised in H 2O .

2.

(d)

O 95.7 pm (104.5)

H

4.

(c)

28.

(c) Because of the triple bond, the carbon-carbon bond distance in ethyne is shortest.

29.

(b) The hybridisation of Ag in complex [ Ag (NH 3 )2 ] will be sp because it is a Linear complex.

30.

(a) Structure of CO 2 is linear O  C  O while that of H 2 O

O

o

is

H

CO 2 has sp – hybridization and is linear.

H

H

i.e. bent structure so in CO 2 resultant dipole

moment is zero while that of H 2 O has some value. 31.

(d) CO 2 is not sp 3 hybridised, it is sp hybridised.

138 Chemical Bonding 32.

(a) As compare to pure atomic orbitals, hybrid orbitals have low energy. sp 2

(a) As p-character increases the bond angle decreases.

sp 3

sp 2

sp

70.

33.

(d) CH 2  C  CH – CH 3 1, 2-butadiene.

36.

(b) CCl 4 is sp 3 hybridised so bond angle will be approximately

109 o . 40.

o

2

1 , bond angle - 180 o 2

In sp 2 - p-character

2 , bond angle - 120 o 3

In sp 3 - p-character

3 , bond angle - 109 o 4

(b) Ethene has sp hybridised carbon so bond angles are 120 .

O 44.

In sp - p-character

(a) Acetate ion is CH 3

(a)

72.

(a) S-atom in SF6 has sp 3 d 2 hybridisation. So, the structure of

i.e. one C  O single bond

C O



sp 3 -hybridization called tetrahedral because it provides tetrahedral shape to the molecule.

71.

and one C  O double bond.

SF6 will be octahedral.

46.

(c) Benzene has all carbons sp 2 hybridised and planar in shape.

74.

(b) Structure of H 2 O 2 is non-planar. It has open book structure.

47.

(d) In methane C is sp 3 hybridized and bond angle is 109 o .

75.

(d) Structure of N 2 O is similar to CO 2 both have linear structure.

H H H 56.

(d)

|

|

|

78.

(a)

|

|

|

79.

(d) In NH 4 nitrogen is sp 3 hybridised so 4 hydrogen situated at the corners of a tetrahedron.

81.

(c) Increasing order of bond angle is sp 3  sp 2  sp .

H  C C C H H H H

58.

There are 10 shared pairs of electrons. (a) The diborane molecule has two types of B – H bond : (ii) B  H b – It is a three centred bond.

86.

(b) Bond angle increases with change in hybridisation in following

Ht

B Ht

Hb (b) PF5 involves sp d hybridization and hence has trigonal bipyramidal structure. 3

(c) s-character in sp 

NH 4

3

has sp –hybridized nitrogen so its shape is tetrahedral.

order sp 3  sp 2  sp .

Hb

1  100  50% 2

s-character in sp 2 

88.

(c) In Diborane boron shows sp 3 –hybridization.

89.

(a) Alkene does not show linear structure but it has planar structure due to sp 2 –hybridisation.

90.

(c) Generally SF4 consist of 10 electrons, 4 bonding electron pair

92.

and one lone pair of electron, hence it shows hybridization. (c) Atom/Ion Hybridisation

1  100  33.3% 3

1 s-character in sp   100  25% 4 3

Hence, maximum s-character is found in sp-hybridisation.

64.

180 

(a)

Ht

63.

120 

84.

B

62.

109 

(i) B  H t – It is a normal covalent bond.

Ht

61.

SnCl 2 is V–shaped.

93.

(a)

sp

SF4

sp 3 d with one lone pair of electron

PF6

sp 3 d 2

PF3 consist of three bonding pair electrons and one lone pair of electron hence it shows sp 3 – hybridization.

3

(b) The molecule of PCl5 has sp d hybridisation, structure is trigonal bipyramidal. (b) Merging (mixing) of dissimilar orbitals of different energies to form new orbitals is known as hybridisation and the new orbital formed are known as hybrid oribitals. They have similar energy.

NO 2

sp 3 d

NO 2 shows sp–hybridization. So its shape is linear.

94.

(b)

95.

(c) Generally octahedral compound show sp 3 d 2 – hybridization.

96.

(a) In fifth group hydride bond angle decreases from top to bottom

65.

(b) In SO 3 sulphur is sp 2 hybridized so its shape will be trigonal planar.

66.

(a) These all are triangular with sp 2 hybridization.

97.

(b) Generally NH 4 shows sp 3 hybridization.

67.

(c) Bond length depends upon bond order and in benzene all C  C bonds have same bond order.

98.

(b) We know that single, double and triple bond lengths of carbon

(b) In

99.

(b) It shows sp 2 –hybridization so it is planar.

101.

(a) Bond angle of hydrides decreases down the group.

68.

C2 H 2

H  C  C H sp

sp

each

carbon

has

sp -hybridization

NH 3  PH3  AsH3  SbH 3  BiH3 .

in carbon dioxide are 1.22 Å,1.15 Å and 1.10Å respectively.

Chemical Bonding 102.

(b) Hybridization of N in NH 3 is sp 3 that of Pt in [PtCl4 ]2 2

125.

3

is dsp that P in PCl5 is sp d and that of B in BCl 3 is

126.

sp 2 . NH 4 and SO 42  tetrahedral structure.

(d)

104.

(a) It is shows sp 3 d 3 –hybridization. Hence the bond angle is

107.

both show

sp

33.3

120 o

M

sp3 d 2 hybrid orbital have octahedral shape

127.

(d)

128.

(c) In the formation of d 2 sp3 hybrid orbitals two (n  1)d orbitals of e.g., set [i.e., (n  1) dz 2 and (n  1)dx 2  y 2 orbitals] one

25

109.28

3 1

20

90 o and 120 o 3

X

according to thisX geometry, the number of X  M  X bond X at 180° must be three. X

3

sp d

X X

about 72 o . (a) s-character increases with increase in bond angle. Hybridization s% Angle sp 50 180 o

sp 2

(b) There is sp hybridization in C 2 H 2 so it has the linear structure. (c) In octahedral molecule six hybrid orbitals directed towards the corner of a regular octahedron with a bond angle of 90°.

sp 3 –hybridization and

103.

o

ns and three np [ np x , npy and np z ] 2

together and form six d sp hybrid orbitals.

3

129.

(c) The correct order of bond angle (Smallest first) is

(b)

IF7 molecule show sp d –hybridization.

110.

(a)

PCl3 contain three bonding and one lone pair electron. Hence

H 2 S  NH 3  SiH4  BF3

shows sp 3 –hybridization.

92.6  107  10928'  120

112.

113.

92.6°

2

(b) For square planar geometry hybridization is dsp involving y2

S

117.

(c) (c)

BCl 3 molecule structure.

show

sp 2 –hybridization

119.

(c) Due to multiple bonding in N 2 molecule.

120.

(a) % of s-character in

CH 4  3

(sp )

and

122.

H

130.

(a)

( 120) due to 2 lone pair of electron over S atom. CO 2

F CN

F

Square planar

Regular tetrahedral

F

F

F F

S

131.

F

H

F Square planar

See saw shaped

(b)

F Xe

F

··

O

··

(a) In H 2 CO 3 and BF3 central atom are in sp 2 hybridization but in H 2 CO 3 due to the ionic character of O  H bond it will be polar (High electronegativity of oxygen).

124.

F

B

NC

and N 2 O have the sp hybridization. 123.

–

CN Ni

100 100  25 , C 2 H 4   33 , 2 3 4 (sp )

SO 2 has sp 2 hybridization have the V shape structure

F

2–

NC

(a) Acidic character increases when we come down a group, so HI is the strongest acid. (c)

H

H

F

F

planar

100 C2 H 2   50 2 ( sp ) 121.

Si

B

SO 2 molecule shows sp 2 –hybridization and bent structure.

(b)

109° 28'

120°

BCl 3 Boron trichloride molecule show sp 2 –hybridization and trigonal planar structure.

118.

H

F

(b) All carbon atoms of benzene consist of alternate single and

H

H

H

orbital.

double bond and show sp 2 hybridization. 116.

N H 107°

H

(a) Ammonia and (BF4 )1 shows sp 3 –hybridization.

s, p x , p y and d x 2 

orbitals combine

3

108.

111.

139

H sp

(a) Due to sp 3 hybridization and presence of lone pair of electron on p atom PCl3 are of pyramidal shape like that of NH 3 .

132.

3

·· O ··

B sp2

··

H

O sp3 ··

(b) In the formation of BF3 molecule, one s and 2p orbital hybridise. Therefore it is sp 2 hybridization.

140 Chemical Bonding 133.

(e) In NCl 3 and H 2 S the central atom of both (N and S) are in

I  [ Xe] 5 s 2 ,5 p 5 hence

sp 3 hybridization state

··

··

N

Cl

S

··

107°

Cl

Cl

92.5°

H

H

5d

5s

5p

5d

and I F7 in excited state

sp 3 hybridization respectively. In H 2 S and BeCl 2 central

F

F

F

I

F

F

F

atom are in sp 3 and sp 2 hybridization In BF3 , NCl 3 &

IF shows sp d hybridization. So, sp itsd structure is pentagonal

H 2 S central atom are in sp 2 , sp 3 & sp 3 hybridization

bipyramidal. 141. (a) Compound containing highly electronegative element (F, O, N) attached to an electropositive element (H) show hydrogen bonding. Fluorine (F) is highly electronegative and has smaller size. So hydrogen fluoride shows the strongest hydrogen bonding in the liquid phase.

and in the central atom are in sp 3 and sp hybridization. (c)

5p

I in excited state

while in BF3 and NCl 3 central atoms are in sp 2 and

134.

5s

I in ground state

1 1 1 Cground state  2 s 2 ,2 p x 1 py1 ; Cexcited state  2 s1 ,2 p x py p z

Oground state  2 s 2 ,2 p x p y p z 2

1

1

In the formation of CO 2 molecule, hybridization of orbitals of carbon occur only to a limited extent involving only one s and one p orbitals there is thus sp hybridisation of valence shell orbitals of the carbon atom resulting in the formation of two sp hybrid orbitals.

3

2

3 3

7

142.

(b) In the ammonia molecule N atom is sp 3  hybridized but due to the presence of one lone pair of e  (i.e. due to greater L p  b p repulsion) it has distorted tetrahedral (or pyramidal) geometry.

Oxygen atom in ground state

sp – p  bonded

N

p – p

Carbon atom in excited state

H p – p

sp – p  bonded

143.

(a)

136. 137.

1s 3

(d) In NH 3 , N undergoes sp hybridization. Due to the presence of one lone pair, it is pyramidal in shape. (b) NO 2 SF4 PF6 sp sp 3 d sp 3 d 2 (b) The configuration of

5B

2s

2p

Be in excited state BeCl2 Cl

 1s 2 , 2 s 2 2 p 1

Cl

sp hybridisation

B in ground state

144. 1s

H

Be in ground state

Oxygen atom in ground state

135.

H

2 2 0 4 Be  1s ,2 s ,2 p

2s

2p

(Linear diagonal hybridization)

(a) Except CO 3 other choice CO 2 , CS 2 and BeCl 2 have sp  hybridization and shows the linear structure while

B in excited state

CO 3 have 1s

2s

2p

1s

2s

2p

In BCl3 state

sp3 hybridization and show the non linear

structure because sp3 generate tetrahedral structure. 145.

(a)

dsp3 or sp3 d hybridization exhibit trigonal bipyramidal geometry e.g., PCl 5

Cl

Cl

Cl Cl

sp2hybridisation

138.

(d) In SO 3 molecule, S atom remains sp hybrid, hence it has O trigonal planar structure

Cl

O

O

(a) In PCl3 molecule, phosphorous is sp 3  hybridised but due to presence of lone pair of electron, it has pyramidal structure

P 140.

(a) The electronic configuration of Cl

Cl Cl

sp3d2 (Trigonal bipyramidal)

P Cl

S

139.

Cl

2

Cl

146.

(b) Carbon has only two unpaired electrons by its configuration but hybridization is a concept by which we can explain its valency 4.

147.

(c) Hybridization is due to overlapping of orbitals of same energy content.

Chemical Bonding 148.

(d) MX 3 show the sp 2 hybridization in which 3sp 2 hybridized orbital of M bonded by 3 X from  bond and having the zero dipole moment.

149.

(bcd) SnCl 2 has V–shaped geometry.

150.

(a)

XeF6 is distorted Octahedral. It has sp 3 d 3 hybridisation with lone pair of electron on Xe, so its shape is distorted.

9.

(b)

10. 11.

(a) (c) Xe ground state

Xe double excitation

NF3 is predominantly covalent in nature and has pyramidal

structure (the central atom is sp 3 hybridised) with a lone pair of electrons in the fourth orbital. 151.

(a)

12. 13.

dsp 3 or sp 3 d : one s  three p  one d (d z 2 ) .

2.

(b) In NH 3 nitrogen has one lone pair of electron. 

(b) In CN ion formal negative charge is on nitrogen atom due to lone pair of electrons.  

9.

(b) There are three resonance structure of CO 32  ion.

O

C

11.

O

4.

17. 18.

SO 42  has 42 electrons; CO 32  has 32 electrons; NO 3 has 32 electrons. (c) Molecular oxygen contains unpaired electron so it is paramagnetic (according to MOT). (b) Structure of H 2 O is a bent structure due to repulsion of lone pair of oxygen. (d) Bond angle between two hybrid orbitals is 105 o it means orbitals are sp 3 hybridised but to lone pair repulsion bond angle get changed from 109 o to 105 o . So its % of scharacter is between 22-23%.

(I) (II) (abcd) It has all the characteristics.



O

O (III)

22.

(d) Number of electrons in ClO2– = 7 + 6 + 6 + 1 = 20 Number of electrons in ClF2+ = 7+7+7 – 1=20.

23.

(b) Central atom having four electron pairs will be of tetrahedral shape.

(a) The bond angle in PH 3 would be expected to be close to

F F

24.

6. 7. 8.

..

F

(d)

F C

S. . F

F F

F Xe

F

F

F

F

.. sp 2 –hybridization and show trigonal planar

(b) In BF3 molecule Boron is sp 2 hybridised so its all atoms are co-planar. (c) Due to lp  lp repulsions, bond angle in H 2 O is lower

26.

(c) It shows structure.

(104 o .5 o ) than that in NH 3 (107 o ) and CH 4 (109 o 28 ) .

28.

(b)

BeF2 on the other hand, has sp-hybridization and hence has a

31.

(a) Bond angle of hydrides is decreases top to bottom in the group. NH 3  PH3  AsH3  SbH 3

32.

(c)

o

5.

5d

(c)

C O

90 o . (The bond angle H  P  H in PH 3 is 93 o ) 3.

5p

15.

VSEPR Theory 2.

5s

(a) CO 2 has bond angle 180 o . (a) As the s-character of hybridized orbitals decreases the bond angle also decreases

O

C O

5d

14.

|

(a) CH 3  C  CH 2 has 9, 1 and 2 lone pairs. (c) In resonance structure there should be the same number of electron pairs.

O

5p

In sp hybridisation: s-character 1/2, bond angle 180 o (a) XeF2 molecule is Linear because Xe is sp hybridised.

16.

O H

7. 8.



5s

In sp 2 hybridisation: s-character 1/3, bond angle 120 o

(d) Choice (a), (b), (c) are the resonance structures of CO 2 .

O

5d

In sp 3 hybridisation: s-character 1/4, bond angle 109 o

1.

 

5p

sp 3 d 2 - hybridization

Resonance

5.

5s

XeF4

(ac) PCl3 , NH 3  Pyramidal.

CH 4 , CCl 4  Tetrahedral. 152.

141

bond angle of 180 . (c) Compound is carbontetrachloride because CCl 4 has sp 3 – hybridization 4 orbitals giving regular tetrahedron geometry. In others the geometry is little distorted inspite of sp 3 hybridization due to different atoms on the vertices of tetrahedron. (b) SO 42  ion is tetrahedral since hybridization of S is sp 3 .

N Three bond pair and one lone pair of electron. H 33.

(b) NH 3 molecule has one lone pair of electrons on the central atom i.e. Nitrogen. (c) C 2 H 2 has linear structure because carbons are sp-hybridised o

and lies at 180 .

H 2 S show bond angle nearly 90 o .

H

H

(c) Unpaired electrons are present in KO 2 while others have paired electron

NO 2 = 22 electrons ; BaO2 = 72 electrons

AlO2  30 electrons ; KO 2  35 electrons 34.

(a) Bond angle decreases from H 2 O to H 2 Te .

142 Chemical Bonding 35.

(c)

36.

(b)

BF3 does not contain lone pair of electron. Bent T-shaped geometry in which both lone pairs occupy the equatorial position of the trigonal bipyramidal here F (l p  l p ) repulsion = 0

F

..

Br

(lp  b p ) repulsion = 4 and

.. F

37.

(b p  b p ) repulsion =2 (b) The overall value of the dipole moment of a polar molecule depends on its geometry and shape i.e., vectorial addition of dipole moment of the constituent bonds water has angular structure with bond angle 105° as it has dipole moment. However BeF2 is a linear molecule since dipole moment summation of all the bonds present in the molecule cancel each other.

Molecular orbital theory

3.

No. of bonding e   No. of antibonding e  2 8 3 5    2 .5 . 2 2 (b) One bonding M.O. and one anti-bonding M.O.

4.

(b) O 22  is least stable.

5.

(c) B.O. of O 2 is 2, B.O. of O 21 is 1.5, B.O. of O 21 is 2.5 and of

2.

38.

(d)

Be

F

6.

40.

H

P

107°

(b) Bond order of O 2 is highest so its bond length is smallest. (c) Oxygen is paramagnetic due to the presence of two unpaired electron :

O 2   (1s)2   (1s)2  (2 s)2   (2 s)2

 (2 p x )2  (2 p y )2  (2 p x )2   (2 p y )1   (2 p z )1

107°

Less than

17. H

H

H H

12. 13.

H

. .H As

.. H

11.

NO(7  8  15) has odd number of electrons and hence it is paramagnetic. No. of N b  No. of N a 5 (c) B.O.    2 .5 . 2 2

N

H

18.

H

(d) In CH 3 CN bond order between C and N is 3 so its bond length is minimum. (b)

As the electronegativity of central atom decreases bond angle is decreases  NH 3 has largest bond angle. 41.

 (1s) *

 (1s)

(c) In NH 3 , sp 3 -hybridization is present but bond angle is

B.O.

106 o 45  because Nitrogen has lone pair of electron according to VSEPR theory due to bp-lp repulsion bond angle decreases from 109 o 45' to 106 o 45  . 42.

44.

(b) (c)

NH 3 has pyramidal structure, yet nitrogen is sp 3 hybridised. This is due to the presence of lone pair of electron.

geometry which arise due to sp3 d hybridization of central sulphur atom and due to the presence of lone pair of electron in one of the equatorial hybrid orbital. 45.

O

O

dsp2 hybridization

sp3d hybridization

sp3d2 hybridization

(Four 90° angles between bond pair and bond pair)

(Six 90° angle between bond pair and bond pair)

(Twelve 90° angle between bond pair and bond pair)

H 2 

H 2 

1

1 2

 1 2

(c) Due to unpaired e  ClO2 is paramagnetic.

20.

(c) The Bond order in N 2 molecule is 3, N  N

Here,

N b  2  4  2  8 and N a  2  B.O.  ( 8  2) / 2  3. 21. 22.

(d) O

 1 2

H2 

19.

3

SiF4 has symmetrical tetrahedral shape which is due to sp hybridization of the central sulphur atom in its excited state configuration. SF4 has distorted tetrahedral or Sea- Saw

He 2 

Magnetic P D P P nature (P = Paramagnetic, D = Diamagnetic)

(a) Bond strength decreases as the size of the halogen

increases from F to I. 43.

O 2 (2  8  1  17) has odd number of electrons and hence it is paramagnetic. All the remaining molecules/ions, i.e., CN  (6  7  1  14) diamagnetic

trigonal planar ( sp hybridization) Hence bond angle is same for all of them (i.e., equal to 120°) (d) We know that molecule of ( NH 3 ) has maximum repulsion due to lone pair of electron. Its shape is pyramidal and is sp3 hybridization. :O: (b) .. 105°

(c)

F

BCl H3 , BBr3 and H BF3 , all of these have same structure i.e.

H

(d) Hydride of boron does not exist in BH 3 form. It is stable as its dimer di borane (B2 H 6 ) .

2

39.

B.O. 

O 22  is 1.

10. O

(c)

23.

24.

(d)

H 2 has the bond order

1 , it has only one electron so it will 2

be paramagnetic. (c) When bond forms between two atom then their energy get lower than that of separate atoms because bond formation is an exothermic process. (b) Valency of A is 3 while that of B is 2 so according to Criss Cross rule the formula of the compound between these two will be A2 B3 . (c) Due to resonance bond order of C  C bonds in benzene is between 1 and 2.

Chemical Bonding

26. 27.

(a) Nitrogen does not have vacant ‘d’-orbitals so it can’t have +5 oxidation state i.e. the reason PCl5 exists but NCl 5 does not. (d) Molecules having unpaired electrons show paramagnetism. (b) NO 2 has unpaired electrons so it would be paramagnetic.

30.

(c) Helium molecule does not exist as bond order of He 2  0 .

31.

(c) Structure of P4 O10 is

25.

55.

(a)

56.

(c)

57.

(c)

59.

(a)

60.

(a)

61.

(a)

64.

(c)

O P O

O O O

P

P

O O

O

O P

33. 34. 37. 38.

143

H 2 O 2 contain bond angle between two O  H planes about 90 o . Nitrogen molecule has highest bond energy due to presence of triple bond. Cu 2  [ Ar18 ] 3d 9 4 s 0 it has one unpaired electron so it is paramagnetic. CN   14 electrons ; CO =14 electrons 1 6 B.O. = [10  4 ]   3 . 2 2 1 5 B.O. = [10  5]   2 .5 , paramagnetic 2 2 P P P P The paramagnetic property in oxygen came through unpaired electron which can be explained by molecular orbital theory.

Each phosphorus is attached to 4Ooxygen atoms. N  Na 8  4 (c) B.O. of carbon  b  2. 2 2 N  N a 10  4 (a) B.O.  b  3. 2 2 N  Na 8  3 5 (b) B.O.  b    2.5 . 2 2 2 (a) Electronic configuration of O2 is

Antibonding

2px*

2pz* 2py*

Px

O2   (1s)2   (1s)2  (2 s)2   (2 s)2  (2 p x )2  (2 p y )2

Py

Pz

Px

Py

Pz

 (2 p z )2   (2 p y )1   (2 p z )1 The molecule has two unpaired electrons So, it is paramagnetic

 2 p y has two nodal planes. *

40.

(c)

42. 43.

(a) Element with atomic number 26 is Fe. It is a ferromagnetic. (b) Correct Sequence of bond order is 65.

44.

O 2  O 2  O 22 B.O – 2.5 2 1.5 (a) Due to small bond length.

45. 46.

(a) S 2 have all paired electrons so it is diamagnetic. (c) NO has 15 electrons.

66.

47.

(b) In the conversion of O 2 into O 2 bond order decreases.

49.

(c)

O 22 

50.

(a)

O 22  consist of four antibonding electron pair [1s and 2s have

51.

two antibonding and 2 p x 2 p y have two antibonding electron pair]. (c) The electron’s distribution in molecular orbitals is 1s 2 , 2 s 1

53.

54.

(b)

Total number of bonds between atoms 2Px bonding Total number of resonating structure 5   1 .25 4 (c) We know that carbonate ion has following resonating structures

(a) Bond order 

–

O

2 1 1   0 .5 . 2 2

–

(a)

C – O–

O

O

Total number of bonds between atoms Total number of resonating structure 

67.

O

C–O  –

–

O

Bond order 

ClO2

–

O C=O 

does not have any unpaired electron so it is diamagnetic.

B.O.  52.

So 2 unpaired of electron present in  2 p y* and  2 p *z .

1 1  2 4   1 .33 . 3 3

O 2 (15e  )  K : K * ( 2 s)2 ( * 2 s)2 ( 2 p x )2 ( 2 p y )2 ( 2 p z )2 ( * 2 p y )1 ( * 2 p z )0

has all paired electrons hence it does not show paramagnetism. 1 (a) B.O.  [ N b  N a ] 2 1 6 1 6 N 2  [10  4 ]   3 ; O 22   [10  4 ]   3 . 2 2 2 2 1 1 5  (a) B.O. for N 2 = [ N b  N a ] = [9  4 ]   2 .5 . 2 2 2

Hence, bond order 

1 (10  5)  2 .5 2

N 2 (13e  )  KK * ( 2 s)2 ( * 2 s)2 ( 2 p x )2 ( 2 p y )2 ( 2 p z )1

1 (9  4 )  2 .5 . 2 (a) Electronic configuration of O 2 is Hence, bond order 

68.

144 Chemical Bonding O 2  ( 1s)2 ( * 1s)2 ( 2 s)2 ( * 2 s)2 ( * 2 s)2 ( 2 p z )2 ( 2 p x2

69. 70.



 2 p y2 ) ( * 2 p 1x



*

1 1 Hence bond order  N b  N a   [10  6]  2 . 2 2 (c) Nitrogen form triple bond N  N In which 6 electron take part. (a) As bond order increase bond length decrease the bond order of species are number of bonding electron - Number of a.b. electron  2 10  6 For O 2  2 ; 2 10  5 O 2   2 .5 2 10  7 O 2   1 .5 2

O 2  O 2  O 2

So, bond order

O 2

 O2 

O 2

So only B 2 exist unpaired electron and show the paramagnetism.

2 p 1y )

 2 py 2  * 2 py1 76.

(b) O2   1s 2 ,  * 1s 2 ,  2 s 2 ,  * 2 s 2 ,  2 p x 2

 2 pz 2  * 2 pz 2

77. 78.

So two unpaired electron found in O 2 at ground stage by which it shows paramagnetism. (b) Due to greater electron affinity Cl 2 has the highest bond energy. (a) Molecular orbital electronic configuration of these species are :

O2 (17 e  )   1s 2 * 1s 2 ,  2 s 2 * 2 s 2 ,  2 p x 2 ,  2 p y 2 ,

 2 p z 2 ,  * 2 p y 2 * 2 p z 1 O2 (16 e )   1s 2 * 1s 2 ,  2 s 2 * 2 s 2 ,  2 p x 2 ,  2 p y 2 ,

 2 p z 2 * 2 p y 1 * 2 p z 1

and bond length are

O22  (18 e )   1s 2 * 1s 2 ,  2 s 2 * 2 s 2 ,  2 p x 2 ,  2 p y 2 ,

.

 2 p z 2 * 2 p y 2 * 2 p z 2

 2py 2  * 2py1 71.

Hence number of antibonding electrons are 7,6,and 8 respectively.

(b) O 2 :  1s 2 ,  * 1s 2 ,  2 s 2 ,  * 2 s 2 ,  2 p x 2

 2pz 2  * 2pz1 10  6  2.0 2 unpaired electrons

79.

(c) Species with unpaired electrons is paramagnetic O 2 has 2

Bond order  (Two orbital)

72.

73.

74. 75.

unpaired electrons, O 2 has one unpaired, O 22  has zero in

antibonding

unpaired electrons, O 22  has one unpaired.

molecular

2 * 1 2  2 py  2 py O 2 :  1s 2 ,  * 1s 2 ,  2 s 2 ,  * 2 s 2 ,  2 p x  2 * 0  2 pz  2 pz 10  5 Bond order   2 .5 2 (One unpaired electron in antibonding molecular orbital so it is paramagnetic) (b) Higher the bond order, shorter will be the bond length, thus NO  having the higher bond order that is 3 as compared to NO having bond order 2 so NO  has shorter bond length.

(d) Oxygen molecule (O 2 ) boron molecule (B 2 ) and N 2 ion, all of them have unpaired electron, hence they all are paramagnetic.

80.

(a)

unpaired electrons while O 22  does not have any unpaired electron. 81.

(c)

(c) Bond order of NO , NO and NO are 3, 2 .5 and 2 respectively, bond energy  bond order (a) Paramagnetic property arise through unpaired electron. B 2 molecule have the unpaired electron so it show paramagnetism.

O

C 2   1s  1s ,  2 s  2 s ,  2 p x . 2 p y 2

2

*

2

2

O



(a) From valency bond theory, bond order in CO, i.e. : C  O : is 3, that of O  C  O is 2 while that of CO 32  ion is 1.33. Since the bond length increases as the bond order decreases,

i.e. CO  CO 2  CO 32  . 83.

(c)

N 2 : KK (2 s)2  * (2 s)2  (2 p x )2  (2 py )2  (2 p z )2 (diamagnetic)

C2 : KK (2 s)  * (2 s)  (2 p x )  (2 py ) 2

2

2

2

(diamagnetic)

N 2 : KK (2 s)2  * (2 s)2  (2 p x )2  (2 py )2  (2 p z )2

(No unpaired electron)

(paramagnetic)

N 2   1s 2 * 1s 2 ,  2 s 2 * 2 s 2 ,  2 p x 2 ,  2 p y 2 2 p z 2

O22 

(No unpaired electron)

: KK (2 s)  * (2 s)  (2 p z )  (2 p x )  (2 py )2 2

2

2

2

 * (2 p x )2  * (2 py )2 (diamagnetic)

F2  s 2 ,  *1s 2 ,  2 s 2 ,  * 2 s 2 ,  2 p x 2 ,  2 p y 2 ,  2 p z 2 , 84.

 * 2 py 2 , * 2 pz 2

O



82.

2

(No unpaired electron)

O

O  O and O  O .

(2 unpaired electron) *

O

Due to resonance in O 3 O  O bond length will be in b/w

B2   1s 2 * 1s 2 ,  2 s 2 * 2 s 2 ,  2 p x 1   2 p y 1 2

H OOH ,O  O  O,O  O O





O 2 has 2 unpaired electron while O 2 and O 2 has one each

(d)

NH 3  107, PH3  93, H 2O  104.5

H 2 Se  91, H 2 S  92.5

Chemical Bonding

Hydrogen bonding 1. 2. 3. 6. 7. 8. 9.

10.

(d) Hydrogen bonding will be maximum in F-H bond due to greater electronegativity difference. (b) Ice has hydrogen bonding. (b) H – F has highest boiling point because it has hydrogen bonding. (d) CO 2 is sp-hybridised

48.

(b) sp-hybridization gives two orbitals at 180 o with Linear structure. (d) Hydrogen bonding increases the boiling point of compound. (c) o-Nitrophenol has intramolecular hydrogen bonding but pNitrophenol has intermolecular hydrogen bonding so boiling point of p-Nitrophenol is more than o-Nitrophenol. (c) The strongest hydrogen bond is in hydrogen fluoride because the power of hydrogen bond  electronegativity of atom and

1.

electronegativity 

11.

(d)

1 atomic size

So fluorine has maximum electronegativity and minimum atomic size. H 2 O can form hydrogen bonds rest CH 4 and CHCl 3 are organic compound having no oxygen while NaCl has itself intraionic attraction in the molecule. PH 3 has the lowest boiling point because it does not form Hydrogen bond. Hydrogen bonding increases heat of vaporisation. Only NH 3 forms H-bonds.

12.

(b)

14. 15.

(b) (d)

22.

25.

(a) Water molecule has hydrogen bonding so molecules get dissociated so it is liquid. (d) In case of water, five water molecules are attached together through four hydrogen bonding. (c) Hydrogen bond is strongest in hydrogen fluoride.

28.

(c) Boiling point of H 2 O is more than that of H 2 S because

23.

H 2 O forms hydrogen bonding while H 2 S does not.

30.

(c) O C

31. 34. 35. 36. 37. 38. 39. 40. 41. 43. 45. 46.

47.

H



Interamolecular H-bonding.

 O

(a) HydrogenHbond is formed when hydrogen is attached with the atom which is highly electronegative and having small radius. (a) Water is dense than ice because of hydrogen bonding interaction and structure of ice. (a) Ethanol have hydrogen bonding so its boiling point is higher than its isomer dimethyl ether. (a) A compound having maximum electronegative element will form strong Hydrogen bond. (a) Due to electronegativity difference of N 2 and H 2 , NH 3 form hydrogen bond. (b) Intermolecular hydrogen bonding compound contain more b.p. compare to intramolecular hydrogen bonding compound. (d) Water molecule contain hydrogen bonding. (c) It contain intermolecular hydrogen bonding. (b) Ethyl alcohol has a intermolecular hydrogen bond. (b) HCl contain weak covalent bond. (c) Due to intermolecular hydrogen bonding water molecules come close to each other and exist in liquid state. (b) Due to greater resonance stabilization.

49.

145

(d) C 2 H 5 OH will dissolve in water because it forms hydrogen bond with water molecule. (b) In ice cube all molecules are held by inter molecular hydrogen bond. (d) Hydrogen bonding is developed due to inter atomic attraction so it is the weakest.

Types of bonding and Forces in solid

13. 14.

(b) In electrovalent crystal has cation and anion are attached by electrostatic forces. (d) Mercury has very weak interatomic forces so it remains in liquid state. (c) The melting and boiling points of argon is low hence, in solid argon atoms are held together by weak Vander Waal’s forces. (c) NaF is the strongest ionic crystal so its melting point would be highest. (b) Diamond is the hardest substance it’s melting point would be highest. (c) Bond is formed by attractive and repulsive forces of both the atoms. (a) Generally zero group elements are linked by the Vander Waal’s force. Hence these show weakest intermolecular forces. (d) Glycerol has a three OH group hence it is viscous in nature. (c) Vander waal's forces is the weakest force of attraction.

16.

(b)

2. 3. 4. 9. 10. 12.

NH 4 contain all three types of bond in its structure H   |   H  N  H   |   H  

17. 18. 22.

23.



(d) In NaOH covalent bond is present in O  H bond while ionic bond is formed between OH  and Na  . (a) Bond formation is an exothermic reaction so there is decrease in energy of product. (d) Blue vitriol is CuSO 4 . 5 H 2 O and it has all types of bonds.

 H   |   (a)  H  N  H  Cl  |   H   Ionic bond = 1, Covalent bond = 3 Co-ordinate bond = 1.

Critical Thinking Questions 1.

(d) We know that ionic characters

 16 [EA  EB ]  3.5  [EA  EB ]2 or ionic characters = 72.24% 3.

(c) Configuration of O 2 molecule is

[ (1s)2   (1s)2  (2 s)2  * (2 s)2  (2 p x )2  (2 p y )2

 (2 p z )2   (2 p x )1   (2 p y )1 ] No. of pair are 7 so total no. of paired electrons are 14.

  H  O :  H  H  O  H

6.

(a)

7.

H H (b) The correct order of increasing dipole moment is

|

|

146 Chemical Bonding p-dichlorobenzene < Toluene < m-dichlorobenzene

< o-

27.

(c)

dichlorobenzene. 8.

(a) The

dipole

moment

of

CH 4  0 D,

and NF3  0.2 D, NH 3  1.47 D H 2 O  1.85 D . Therefore the correct order of the dipole moment is CH 4  NF3  NH 3  H 2O . 10.

(d) Ammonia molecule is more basic than nitrogen trifluoride and Boron trifluoride because ammonia molecule easily gives lone pair of electron.

11.

(a) Chlorine atom in ClO2 is sp 3 hybridised but its shape is angular.

12. 13.



(c) [ NF3 and H 3 O ] are pyramidal while [ are planar. Hence answer (c) is correct. (d) CH 2  CH  CH 2  CH 2  C  CH sp 2

and BF3 ]

Cu

1. 2. 5.

(d) B.O. in CO i.e., : C  O : is 3, that of O  C  O is 2 while the bond order decreases i.e. CO  CO 2  CO 32 . Thus option (d) is correct. (b) Dichromate dianion has following structure 2

9.

6, Cr  O bonds are equivalent. (b) ClF3 is a [ AB 3 ] type of molecule because it consist of three bonding pair and two lone pair of electrons hence this

21.

BeF3 does not show compound is not formed. (a) K 3 [Fe(CN )6 ] (a)

sp 3 –hybridization because this

Fe26  4 s 2 3d 6

10.

=

22. 23.

24.

11.

12.

Fe 3   3d 5 4 s 0

 



  

Unpaired electron d 2sp 3 –hybridization  (d) N 2 has one unpaired electron so it would be paramagnetic. (a) Each of the species has 14 electron so isoelectronic and shows bond order 3. 1 1 6 B.O.  [ N b  N a ]  [10  4 ]   3 . 2 2 2 (d) O O S O O

O O

S

Trimer of SO 3 .

O

O



SiF4 have sp 3 hybridization & shape of regular tetrahedral where the bond angle of F F  S  F are found

which

is

o

but

109.5o

S less than 180 o . F Repulsion sequence are F Lp  Lp  Lp  Bp  Bp  Bp F so assertion are true but the reason are false. (c) N 2 molecule is diamagnetic. The diamagnetic character is due

to the presence of paired electron N 2 molecule does not contain any unpaired electron. Thus, assertion is coorect but the reason is false. (a) It is correct that during formation of Ice from water there are vacant spaces between hydrogen bonded molecules of Ice. Ice has a cage like structure. Due to this reason Ice is less dense than liquid water. hence both assertion & reason are true & reason are the correct explanation of assertion. (b) Water is liquid while H 2 S is gas because oxygen is of small size & more electronegative in comparision to sulphur. Hence water molecules exist as associated molecules to form liquid state due to hydrogen bonding H 2 S does not have hydrogen bonding & can’t associated hence it is gas. (d) Iodine is more soluble in CCl 4 than in H 2 O because iodine

13.

(a)

14.

(e)

15.

(c)

16.

(c)

17.

(c)

18.

(b)

O S

   O . 5 H 2O .  

greater than 90

compound shows sp 3 d hybridization. 20.

(c)

109.5 o

O O       O  Cr  O  Cr  O      O O   17.

O     O  S   O 

(a) Solubility in water depends on hydration energy and lattice energy. (a) Polarity in covalent bond developed due to shifting of electrons towards one of the bonded atoms.



that of CO 32  ion is 1.33. Since the bond length increases as

15.

2

Assertion & Reason

sp 3 hybridised 

14.

NO 3

CuSO 4 .5 H 2 O has electrovalent, covalent and coordinate bonds.

is non polar & thus it dissolve in CCl 4 because like dissolves like. o & p -nitrophenols can be separated by steam distillation because o -nitrophenol is steam volatile. Here, both assertion & reason are correct & reason is correct explanation of assertion. Fluorine is highly reactive F  F bond has low bond dissociation energy. Here assertion is false but reason is true. It is true that sigma ( ) bond is stronger than pi ( ) bond but the reason that there is free rotation of atoms is false. Energy is released in the formation of the crystal lattice. It is qualitative measure of the stability of an ionic compound so assertion is true & reason are false. Li, Na & K are alkali metals & not alkaline earth metal so, size of alkali metal increases So. Assertion is true & reason are false. Hess’s law states that the enthalpy of a reaction is the same, whether it takes place in a single step or in more than one step. In born haber cycle the formation of an cycle ionic compound may occur either by direct combination of the element or by a stepwise process involving vaporization of elements, conversion of the gaseous atoms into ions & the combination of the gaseous ions to form the ionic solid.

Chemical Bonding 19.

20.

21.

(a) With increase in bond order, bond length decreases & hence bond energy increases so both assertion & reason are true & reason are the correct explanation of assertion. (c) Electron affinity is experimentally measurable while electronegativity is a relative number so assertion is true but reason are false. (b) Assertion & reason both are correct but reason is not the correct explanation of assertion sulphur has five electrons pairs whose arrangement should be trigonal bipyramidal according to VSEPR theory. Two structure are possible F F F | .. :S FS F F F | F (b) Lone pair in the equatorial position (two L.p – b.p repulsion)

(a) Lone pair in the axial position (three l.p – b.p repulsion at 90o)

22.

(e)

BF3 has zero dipole moment because of its structure. F F  B    0 F

23.

24.

H 2 S has two lone pairs on sulphur atom & hence. It has irregular shape. Thus it possess dipole moment. So assertion is false but reason are true. (d) Both assertion & reason are false because pairs of electron will have different spins. Electrons are equally shared between them. (d) In B2, total number of electrons = 10 B  (1s) *(1s ) (2s) *(2s) (2p ) (2p ) Presence of unpaired electron shows the paramagnetic nature. The highest occupied molecular orbital is of -type. (a) Both assertion & reason are true & reason is the correct explanation of the assertion because. At any given instant, at room temperature each water molecules forms hydrogen bonds with other water molecules. The H 2 O molecules are in continuous motion. So hydrogen bonds are constantly & rapidly broken & formed. In Ice H 2 O molecules are however fixed in the space lattice. (a) Both assertion & reason are true & reason is the correct explanation of assertion, because helium molecule is formed by linking two helium atoms. both have 1s orbitals. These will 2

2

25.

26.

2

2

2

1

x

1

y

combine to form two molecular orbitals  ( 1s ) &  * ( 1s ) four available electrons are accommodated as  (1s)2 &

 * (1s)2 .

147