Chemical Bonding (Package Solutions)

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Solutions of Assignment (Set-2) Chemical Bonding and Molecular Structure (Solutions)

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Section-A

Q.No. Solution

1. Answer (4)

Cl – is common hence larger the cation more will be the ionic character (Fajan’s rule). Hence, CsCl willhave highest ionic character.

2. Answer (3)

NH3

)ACMV(2

1VSEP

=2

1 (5 + 3 – 0 + 0) = 4 sp

3

3. Answer (4)

BF3 , = 0

NH3 has greater dipole moment than NF3 hence order is

BF3 < NF3 < NH3

4. Answer (4)

22O has zero unpaired electrons.

5. Answer (3)

Both S and P are sp3 hybridized in 2

4SO and 34PO respectively.

6. Answer (3)

Both 24SO and

4BF have sp3 hybridization.

7. Answer (2)

Number of Hydrogen-bonds formed by water molecule = 4

8. Answer (1)

Difference of electronegativity is maximum between H and F in H.

It will form strongest H-Bond.

4Chapter

Chemical Bonding andMolecular Structure



Chemical Bonding and Molecular Structure (Solutions) Solutions of Assignment (Set-2)


Q.No. Solution

9. Answer (1)

OCO

Linear molecule with two equal and opposite dipoles.

10. Answer (1)

NCK

KCN has ionic and covalent bonds.

11. Answer (1)

C

OH

O Intramolecular Hydrogen bonding

H

12. Answer (2)

4NH = 7 + 4 – 1 = 10e –

– 4BH = 5 + 4 + 1= 10e –

Both have same number of electrons and same total number of atoms, hence, isostructural.

13. Answer (2)

O2 is paramagnetic because it contains unpaired electrons. N2 does not have unpaired electrons henceit is diamagnetic.

14. Answer (1)

Cl F

F

F

sp d3

T-Shaped

I

I

I

sp d3

Linear

Xe

O OO

sp3

Pyramidal

15. Answer (2)

Due to H-bonding, boiling point and solubility in water increases

16. Answer (3)

Xe

F O

sp3d

2

Squareplanar

FO

Xe

F

sp3d

3

Distorted octahedral

F

F F

F

F





Q.No. Solution

17. Answer (1)

4sp

3sp

HNHHN33

18. Answer (1)

CN – = 6 + 7 + 1 = 14

CO = 6 + 8 = 14

Both have same number of electrons and same number of atoms.

19. Answer (1)

CO2 = 6 + 2 × 8 = 22

2NO = 7 + 8 × 2 – 1= 22

Both have same number of atoms and electrons, hence, isostructural.

20. Answer (1)

ClNHHClNH 43

When NH3 converts into 4NH number of lone pair decreases, hence, bond angle increases.

21. Answer (3)

22 OKKO

2232 OAlAlO

22

22 OBaBaO

2NO

N

O O

+

Species Bond order No. of unpaired e –

2O

2

3 1

22O 1 0

only KO2 has unpaired electrons

22. Answer (1)

Cl

Cl

Compound is not planar while

Cl

Cl

is planar.





Q.No. Solution

23. Answer (3)

3I , BP = 2 , LP = 3 Total no. of hybridized orbital = 5

(5, 2, 3) linear

XeF2, BP = 2, LP = 3 Total no. of hybridized orbital = 5(5, 2, 3) linear

SO2 , OSO

, BP = 2 , LP = 1 Total no. of hybridized orbital = 3

(3, 2, 1) sp2 bent or V shaped or angular

CHC–CCH linear

24. Answer (2)

Except BF3, all molecules have complete octet hence only BF3 can act as Lewis acid.

25. Answer (4)

All are sp2 hybridized with 120° bond angle.

26. Answer (1)

BF3 has sp2 hybridization hence 3 dipole moment vectors of equal magnitude are at 120° from each

other. Hence = 0

120°

120° 120°

P

P P

= 0

27. Answer (4)

O

Hare Isoelectronic

H H

N

H H Hand

C

H H H

28. Answer (1)

2ClO

VSEP =2

1 (V + M –C + A)

=2

1 (7 + 0 – 0 + 1)

= 42

8 sp

3





Q.No. Solution

29. Answer (2)

CIF3 is T-shaped.

30. Answer (4)

P

Fis pyramidal with one lone and three bond pairs.

F F

31. Answer (1)

CCl4 ;

Cl |C |Cl

Cl Cl

all atoms in surrounding are same and no lone pairs present on central atom

Hence = 0

32. Answer (1)

Smaller the cation and larger the Anion, higher is polarization.

33. Answer (4)

lengthbondO — OIncreasing

ONaOHKOO 222222

34. Answer (1)

Cl–P–Cl | Cl

ClCl

Since P-has 5e – in its outermost shell and is forming 5 bonds hence all the 5e – are consumed and nolone pair will remain.

35. Answer (3)

Å115.1Å128.1lengthBond COCO

36. Answer (2)

++

– Nonbonding overlap





Q.No. Solution

37. Answer (2)

2

3

NO 2.5

NO 3

3NO24

NO3

Compound Bond order

Bond length1

Bond order

38. Answer (4)

Cl

Cl

I

Cl

Cl

II

Cl

ClIII

II < I < III (increasing order of Bond angle)

1

III < I < II ( Increasing order of dipole moment ).

39. Answer (4)

OO

O O

P P

P

P

O

OO

O

O

O

Each P is attached to 4 oxygen atoms.

40. Answer (1)

P

F F F

O

greatest

More electronegative substituent

41. Answer (1)

Bond order of CO = 3 & NO – = 2.5





Q.No. Solution

42. Answer (4)

43. Answer (4)

S

O FF

Pyramidal

44. Answer (4)

Xe

F

F

O

O

Seesaw.

Section-B

Q.No. Solution

1. Answer (1, 3)

CO2, Hg2Cl2 and C2H2 all have linear geometry.

2. Answer (1, 3)

SF4 BP = 4, LP = 1 VSEP = 5 , (5, 4, 1) seesaw

O ||F–Xe–F || O

BP = 4, LP = 1 VSEP = 5 , (5, 4, 1) seesaw

3. Answer (2, 3, 4)

F do not have d orbital hence it is always monovalent but Cl, Br, I have d orbital hence can showvariable valency.

4. Answer (2, 3)

If z is intermolecular axis than Px – Px and Py – Py will overlap sidewise

5. Answer (3, 4)

)CH,NH,XeO( 444 and )NF,NH,CH( 333

are isostructural.

6. Answer (1, 3)

Hybridization of P in 4PCl is sp3 while in 6PCl is sp

3d2.





Q.No. Solution

7. Answer (2, 3, 4)

SnCl2 has lone pair on central atom hence it is not linear

Dipole moment of CH3Cl is more than CH3F because bond length of C–F is less (charge in both issame).

= q × d3I has a linear shape.

8. Answer (1, 3, 4)

In (SiH3)3N. Lone pair is delocalized hence Lone pair-Bond pair repulsion decreases, hence, bondangle increases.

9. Answer (1, 2)

SF4 and XeO2F2 both have 5 VSEP, 4 bond pair and 1 lone pair

10. Answer (2, 4)NaCl is more soluble and p-hydroxy benzoic acid is more soluble.

11. Answer (1, 3)

When CO CO+, bond order increases from 3 to 3.5 and when O2 2O bond order increases

from 2 to 2.5 because in both case electron is removed from antibonding molecular orbital.

12. Answer (1, 2, 4)

Molecular orbital No. of nodes

2pz 0

* 2pz 1

2py 1

* 2py 2

N |H

H H

P |H

H H

As |H

H H

Size N < P < As

EN N > P > As

Bond angle increases with increasing electronegativity

CH3 is electron donating and CN is electron attracting, hence, order of dipole moment will be o < m < p CH3

, CH3 is donor hence bond length decreases.

CH2=CH–CH3, Here CH3 can show hyper conjugation hence bond length will be between single anddouble bond.

CH3 –CH2 –CH=CH2.





Q.No. Solution

13. Answer (1, 2)

H2 gas molecule can not form H-bonding

HCl gas molecules have dipole interaction.

14. Answer (1, 3, 4)In option 2, N-atom has 5 bonds.

15. Answer (1, 3)

MgF2 I2 Na3Bi

Ionic Covalent Metallic

BeF2 S8 Na3Sb

Ionic Covalent Metallic

Section-C

Q.No. Solution

Comprehension-I

1. Answer (1)

2N 2.5

2N 2.5

N2 3.0

2N &

2N both have same bond order but N2 has higher number of antibonding e – . Hence correct order

is 2N <

2N < N2

2. Answer (4)

NO is paramagnetic

3. Answer (3)

F2 1

Ne2 0

C2 2

N2 3

2N 2.5

2O 2.5

Both 2N &

2O have bond order 2.5





Q.No. Solution

Comprehension-II

1. Answer (1)

Cations are same, hence, smaller the size of anion, lesser is polarization and greater will be ioniccharacter.

2. Answer (1)

Li2CO3 has smaller cation hence more covalent character hence decompose on heating.

3. Answer (3)

LiCl BeCl2

RbCl MgCl2

Anions are same hence greatest ionic character is possessed by species having largest cation andleast ionic character will be possessed by species containing smallest cation.

Comprehension-III

1. Answer (2)

– 4ICl is square planar molecule having zero dipole moment.

2. Answer (2)

CH3 — is e – donar and —CN is e – acceptor groups.

3. Answer (2)

In

OH

OH

, H atoms are not in same plane. In p-xylene both moments are cancelled.

4. Answer (1)

It should be sp hybridized and should have zero lone pair.

Comprehension-IV

1. Answer (1)

Lattice energy 1 22

q q

r

2. Answer (2)

3. Answer (4)

LiF have very high lattice energy.





Q.No. Solution

Comprehension-V

1. Answer (1)

SnCl2 is bent shape molecule.

2. Answer (4)

3. Answer (4)

2zd have largest lobes and can easily takes part in all three hybridization.

Section-D

Q.No. Solution

1. Answer (2)

Factual.

2. Answer (1)

More EN atom shall repel each other strongly, increasing the bond angle. But back-bonding is possiblein PF3 due to vacand d-orbitals in P and availability of lp on F.

3. Answer (3)

AlF3 is ionic, hence, a solid. SiF4 is covalent, hence, a gas.

4. Answer (2)

S=C=S

C, S have almost same electronegativity, hence, CS2 is non polar.

CS2 is linear (BP = 2, LP = 0 , VSEP = 2)

5. Answer (3)

In PCl5 all bond angles and bond lengths are not equal.

6. Answer (4)

O=C=O

CO2 has two polar bonds although it is non polar.

7. Answer (2)

H2O is liquid due to Hydrogen-bonding; O is more electronegative than S but for Hydrogen-bondingone atom should be highly electronegative while S is not highly electronegative.

8. Answer (2)

Boiling point of H2O is higher because it forms four hydrogen bonds per molecule.





Q.No. Solution

9. Answer (1)

Larger will be anoin more will be polarisability.

10. Answer (3)

Bond order of – 2 2O and O are 2.5 and 1.5 respectively.

11. Answer (4)

o-hydroxybenzoic acid has lower boiling point as it has intramolecular hydrogen bonding.

Section-E

Q.No. Solution

1. Answer A(s), B(p), C(q), D(r)

CH3CH3CH3CH3

N |CH3

CH3

xx

CH3CH3CH3

Si(CH )

|N

3

3

(CH ) Si3 3Si(CH )3 3

In NSi(CH3)3 , Si uses its vacant d-orbital for back bonding with lone pair electrons of central N atom,hence, has less tendency to release its electrons, hence, less basic.

Ortho nitro phenol shows intramolecular hydrogen bonding while para nitro phenol showsintermolecular hydrogen bonding hence ortho isomer is more volatile.

Na+ is smaller than Ca+2, hence, Na2CO3 has more lattice energy than CaCO3. Hence, Na2CO3 is morethermally stable.

Water has high density than ice because ice forms cage like structure and large vacancies are createdin ice.

2. Answer A(q), B(r), C(s), D(p)

ClF3 :F

|F–Cl–F

322LP3BP

XeF4 :

F |F–Xe–F |

F

422LP4BP

3I

LP = 3, BP = 2 3 + 2

H2S : H–S–H

LP = 2, BP = 2 2 + 2

I –

xx

xx xx

I

I





Q.No. Solution

3. Answer A(r), B(s), C(p), D(q)

H2Te

VSEP =2

1 (V + M – C + A)

=21 (6 + 2 – 0 + 0)

=2

8 = 4 sp

3

XeF4

VSEP =2

1 (8 + 4 – 0 + 0)

=2

12 = 6 sp

3d

2

IClF2

Cl is centre atom

VSEP =2

1 (7 + 3 – 0 + 0)

=2

10 = 5 sp

3d

O ||H–C–O–H

central atom is C

If C attached with, one double bond then sp2

4. Answer A(p), B(s), C(r), D(q, s)

Section-F

Q.No. Solution

1. Answer (6)

Fact.

2. Answer (6)

Fact.

3. Answer (4)

Hybridization is sp3d

Five hybrid orbitals and one hybrid orbital is unused carrying lone pair.

4. Answer (0)

Its shape is distorted square pyramidal.





Q.No. Solution

5. Answer (1)

Na2O2 contain 22O anion having bond order one.

6. Answer (6)

PCl6 – exist

7. Answer (6)

Due to its trigonal bipyramidal shape.

8. Answer (5)

4 between C & N and one between C —C.

9. Answer (0)

Br

FeqFeq

FeqFeq

Fa

Presence of lone pair will distort all the bond angles and none of the bond angle is exactly equal to 90º

FaBrFeq = 84°

FeqBrFeq = 89°

Section-G

Q.No. Solution

1. Answer (3)

Fact.

2. Answer (4)

Bond order of CO+

is 3.5.

3. Answer (2)

4. Answer (1)

5. Answer (2)





Section-H

Q.No. Solution

1. Answer (3)

Hybridisation compound/ion shape

(i) sp3d XeF3

+

T(ii) sp3d2 XeF5

+ Square pyramid

2. Answer (1)

In PF3 due to presence of vacant d orbital.

3. Answer (2)

PBr 5 exist on PBr 4+Br – in solid.

4. Answer (4)Fact

5. Answer (1)

In hydride down the group bond angle decreases.

6. Answer (1)

Due to network structure in water.

7. Answer (2)

Fact

8. Answer (2)

Opposite sign of wave function gives antibonding.

9. Answer (4)

Due to formation of pseudo dipole.

10. Answer (4)

Based on M.O. theory.

11. Answer (2)

In PCl5, P—Claxial bond length are large.

12. Answer (3)

Fact





Q.No. Solution

13. Answer (3)

Fact

14. Answer (2)

Hybridization of S in orthorhombic S8.

15. Bond order of H2 = 1, H2+ = 0.5

He2+ = 1.5, He2 = 0

H2 is more stable than H2+ but He2

+ is more stable than He2

16. CO > SiO (stability order) because there is no tendency of p-p bonding in Si.

17.

H

S(bond angle)

O

H>

H H

O is more electronegative than S, so it has more tendency to attract bonded electron pairs.

18. More the carbonate ion is polarized by the cation more is the chance of formation of CO2 and thereforehigher is probability of decomposition.

19. AlCl3 (anhydrous) is covalent because of very high ionization energy required for formation of Al+3. Butin hydrated condition, this excessive energy is obtained by hydration energy of smaller Al+3 ion.

20. White P has following tetrahedral structure where each phosphorus is attached to 3 other phosphorusatoms by sigma bonds :

PP

P

P

In red phosphorus, molecule forms a chain structure (polymer like) where 2 inter molecular bonds aremade by each P4 unit.

21. Silicon has lower bond energy than carbon, so less tendency of catenation is present.

22. IF7 has lower covalent character i.e., higher ionic character this strengthens the I– F interaction, hence,stabilizing the molecule.




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Q.No. Solution

23. Calculated µ = 4.8 × 10 –10 esu × 1.62 × 10 –8 cm = 7.776 D

Percentage ionic character 100776.7

39.0 = 5%

Percentage covalent character = 100 – 5 = 95%.

24. To avoid strong repulsion of lone pair and bond pair.

Chemical Bonding (Package Solutions)

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