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Chemical Bonding & Molecular Structure

JEE Chemistry · 150 questions · Page 2 of 15 · Click an option or "Show Solution" to reveal answer

Q11
Two pi and half sigma bonds are present in :
A O2
B N2+_2^ +
C O2+_2^ +
D N2
Correct Answer
Option B
Solution

Two pi and half sigma bonds are presents in molecule with bond order 2.5. Moleculer orbital configuration of

N2+N_2^+

(13 electrons) =

σ1s2 σ1s2∗ σ2s2 σ2s2∗ π2px2 = π2py2 σ2pz1{\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^1}}

Bond order =

12(9−4){1 \over 2}\left( {9 - 4} \right)

= 2.5

Q12
The compound that has the largest H–M–H bond angle (M = N, O, S, C) is :
A CH4
B H2S
C NH3
D H2O
Correct Answer
Option A
Solution

CH4 - Bond angle = 109o28' H2S - Bond angle = 92o NH3 - Bond angle = 107o H2O - Bond angle = 104o5'

Q13
The relative strength of interionic/intermolecular forces in decreasing order is :
A dipole-dipole >> ion-dipole >> ion-ion
B ion-dipole >> dipole-dipole >> ion-ion
C ion-dipole >> ion-ion >> dipole-dipole
D ion-ion >> ion-dipole >> dipole-dipole
Correct Answer
Option D
Solution

Ionic interactions are stronger as compared to van der waal interactions. So, correct order is ion-ion

>>

ion-dipole

>>

dipole-dipole

Q14
The correct increasing order for bond angles among BF3,PF3\mathrm{BF}_3, \mathrm{PF}_3 and ClF3\mathrm{ClF}_3 is :
A BF3=PF3<ClF3\mathrm{BF}_3=\mathrm{PF}_3<\mathrm{ClF}_3
B BF3<PF3<ClF3\mathrm{BF}_3<\mathrm{PF}_3<\mathrm{ClF}_3
C ClF3<PF3<BF3\mathrm{ClF}_3<\mathrm{PF}_3<\mathrm{BF}_3
D PF3<BF3<ClF3\mathrm{PF}_3<\mathrm{BF}_3<\mathrm{ClF}_3
Correct Answer
Option C
Solution

Order of bond angle is

ClF3<PF3<BF3\mathrm{ClF}_3<\mathrm{PF}_3<\mathrm{BF}_3
Q15
In which one of the following pairs the central atoms exhibit sp2\mathrm{sp}^2 hybridization ?
A BF3\mathrm{BF}_3 and NO2−\mathrm{NO}_2^{-}
B NH2−\mathrm{NH}_2^{-} and BF3\mathrm{BF}_3
C NH2−\mathrm{NH}_2^{-} and H2O\mathrm{H}_2 \mathrm{O}
D H2O\mathrm{H}_2 \mathrm{O} and NO2\mathrm{NO}_2
Correct Answer
Option A
Solution

have central atom with sp

2^2

hybridisation NH

2−_2^-

and H

2_2

O have sp

3^3

hybridised central atom.

Q16
The correct statement/s about Hydrogen bonding is/are A. Hydrogen bonding exists when H is covalently bonded to the highly electro negative atom. B. Intermolecular H bonding is present in oo-nitro phenol C. Intramolecular H\mathrm{H} bonding is present in HF. D. The magnitude of H\mathrm{H} bonding depends on the physical state of the compound. E. H-bonding has powerful effect on the structure and properties of compounds Choose the correct answer from the options given below:
A A, B, C only
B A only
C A, D, E only
D A, B, D only
Correct Answer
Option C
Solution

In o-nitrophenol intra molecular hydrogen bonding is present.

In HF intermolecular hydrogen bonding is present.

Other statements are correct except B and C.

Q17
The number of species from the following that have pyramidal geometry around the central atom is _________ S2O32−,SO42−,SO32−,S2O72−\mathrm{S}_2 \mathrm{O}_3^{2-}, \mathrm{SO}_4^{2-}, \mathrm{SO}_3^{2-}, \mathrm{S}_2 \mathrm{O}_7^{2-}
A 4
B 3
C 2
D 1
Correct Answer
Option D
Solution
SO32−\mathrm{SO}_3^{2-}

is the only species with pyramidal geometry.

Q18
The shape of carbocation is :
A tetrahedral
B diagonal pyramidal
C diagonal
D trigonal planar
Correct Answer
Option D
Solution

Carbocation is

sp2s p^2

hybridised hence it's shape is trigonal planar

Q19
Aluminium chloride in acidified aqueous solution forms an ion having geometry
A Square planar
B Octahedral
C Trigonal bipyramidal
D Tetrahedral
Correct Answer
Option B
Solution
AlCl3\mathrm{AlCl}_3

in acidified aqueous solution forms octahedral geometry

[Al(H2O)6]3+[\mathrm{Al}(\mathrm{H}_2 \mathrm{O})_6]^{3+}
Q20
In the molecular orbital diagram for the molecular ion, N2+, the number of electrons in the σ\sigma 2pz molecular orbital is :
A 0
B 1
C 2
D 3
Correct Answer
Option B
Solution

Electrons in

N2+N_2^ +

= 7 ×\times 2 −- 1 = 13 Moleculer orbital configuration of

N2+N_2^ +

(13) =

σ1s2 σ1s2∗ σ2s2 σ2s2∗ π2px2 = π2py2 σ2pz1{\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^1}}
∴   \therefore\,\,\,
 σ2pz\,{\sigma _{2{p_z}}}

no. of electrons = 1.

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