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Chemical Bonding & Molecular Structure

NEET Chemistry · 95 questions · Page 3 of 10 · Click an option or "Show Solution" to reveal answer

Q21
Identify a molecule which does not exist.
A Li 2
B C 2
C O 2
D He 2
Correct Answer
Option D
Solution

For He 2 molecule Total number of electron = 4 Electronic configuration is σ\sigma1s 2 , σ\sigma*1s 2 Bond order =

[Nb−Na]2{{[{N_b} - {N_a}]} \over 2}

=

[2−2]2{{[2 - 2]} \over 2}

= 0 So bond order = 0 So bond order = 0 (Remove this line, not needed.)

Since, bond order is zero, so He 2 molecule does not exist.

Q22
Which of the following diatomic molecular species has only π\pi bonds according to Molecular Orbital Theory?
A C 2
B Be 2
C O 2
D N 2
Correct Answer
Option A
Solution
C2C_2

has 12 electrons. Moleculer orbital configuration of

C2C_2

=

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

Here no unpaired electron present, so it is diamagnetic. Here, Bond order =

8−42{{8 - 4} \over 2}

= 2. So here two bonds are present. For this two bond formation 4 electrons are required which is present in

π2px2 {\pi _{2p_x^2}}\,

and

 π2py2\,{\pi _{2p_y^2}}

So, in C 2 molecule, only π\pi electron occupied the bonding molecular orbital.

Q23
The number of sigma (σ\sigma ) and pi (π\pi ) bonds in pent-2-en-4-yne is -
A 11 σ\sigma bonds and 2 π\pi bonds
B 13 σ\sigma bonds and no π\pi bonds
C 10 σ\sigma bonds and 3 π\pi bonds
D 8 σ\sigma bonds and 5 π\pi bonds
Correct Answer
Option C
Solution

H–C ≡\equiv C – CH = CH – CH 3 Here, 10 σ\sigma bonds and 3 π\pi bonds are presents.

Q24
Consider the following species : CN + , CN – , NO and CN. Which one of these will have the highest bond order?
A NO
B CN –
C CN +
D NO
Correct Answer
Option B
Solution

Molecular orbital configuration of NO (15 electrons) is =

σ1s2 σ1s2∗ σ2s2 σ2s2∗ σ2pz2 π2px2 = π2py2 π2px1∗ = π2pyo∗{\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ *
∴    \therefore\,\,\,\,

N b = 10 N a = 5

∴    \therefore\,\,\,\,

BO =

12[10−5]{1 \over 2}\left[ {10 - 5} \right]

= 2.5 Moleculer orbital configuration of CN – (14 electrons) is =

σ1s2 σ1s2∗ σ2s2 σ2s2∗ π2px2 = π2py2 σ2pz2{\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^2}}
∴    \therefore\,\,\,\,

N b = 10 N a = 4

∴    \therefore\,\,\,\,

BO =

12[10−4]{1 \over 2}\left[ {10 - 4} \right]

= 3 Moleculer orbital configuration of CN (13 electrons) is =

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\,\,\,\,

N b = 9 N a = 4

∴    \therefore\,\,\,\,

BO =

12[9−4]{1 \over 2}\left[ {9 - 4} \right]

= 2.5 Moleculer orbital configuration of CN + (12 electrons) is =

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

N b = 8 N a = 4

∴    \therefore\,\,\,\,

BO =

12[8−4]{1 \over 2}\left[ {8 - 4} \right]

= 2 Hence, CN – has highest bond order.

Q25
In the structure of ClF 3 , the number of lone pairs of electrons on central atom ‘Cl’ is
A one
B two
C four
D three
Correct Answer
Option B
Solution

Hence, Cl has 2 lone pairs of electrons.

Q26
Which one of the following pairs of species have the same bond order ?
A O 2 , NO +
B CN −- , CO
C N 2 , O 2 −-
D CO, NO
Correct Answer
Option B
Solution

CN - and CO have same no. of electrons and have same bond order equal to 3.

Q27
The species, having bond angles of 120 o is
A ClF 3
B NCl 3
C BCl 3
D PH 3
Correct Answer
Option C
Solution

BCl 3 -Trigonal planar, sp 2 -hybridised, 120 o angle

Q28
The hybridizations of atomic orbitals of nitrogen in NO2+_2^ + , NO3−_3^ - and NH4+_4^ + respectively are
A sp, sp 3 and sp 2
B sp 2 , sp 3 and sp 3
C sp, sp 2 and sp 3
D sp 2 , sp and sp 3
Correct Answer
Option C
Solution

X =

121 \over 2

(VE + MA - c + a) For

NO2+NO^+_2

, X =

121 \over 2

(5 + 0 - 1) = 2, so sp hybridisation For

NO3−NO^-_3

, X =

121 \over 2

(5 + 0 + 1) = 3, so sp 2 hybridisation For

NH4+NH^+_4

, X =

121 \over 2

(5 + 4 - 1) = 4, so sp 3 hybridisation

Q29
The correct geometry and hybridization for XeF 4 are
A octahedral, sp 3 d 2
B trigonal bipyramidal, sp 3 d
C planar triangle, sp 3 d 3
D square planar, sp 3 d 2
Correct Answer
Option A
Solution

sp 3 d 2 hybridisation (octahedral geometry, square planar shape)

Q30
Which one of the following compounds shows the presence of intramolecular hydrogen bond?
A H 2 O 2
B HCN
C Cellulose
D Concentrated acetic acid
Correct Answer
Option C
Solution

H 2 O 2 , HCN and conc.

CH 3 COOH form intermolecular hydrogen bonding while cellulose has intramolecular hydrogen bonding.

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