1
MCQ (Single Correct Answer)

AIPMT 2008

The correct order of increasing bond angles in the following triatomic species is
A
NO$$_2^ + $$ < NO2 < NO$$_2^ - $$
B
NO$$_2^ + $$ < NO$$_2^ - $$ < NO2
C
NO$$_2^ - $$ < NO$$_2^ + $$ < NO2
D
NO$$_2^ - $$ < NO2 < NO$$_2^ + $$

Explanation


We know that higher the number of lone pair of electron on central atom, greater is the lp – lp repulsion between Nitrogen and oxygen atoms. Thus smaller is bond angle.

The correct order of bond angle is

NO$$_2^ - $$ < NO2 < NO$$_2^ + $$
2
MCQ (Single Correct Answer)

AIPMT 2008

Four diatomic species are listed below. Identify the correct order in which the bond order is increasing in them
A
$$NO < {O_2}^ - < {C_2}^{2 - } < He_2^ + $$
B
$${O_2}^ - < NO < {C_2}^{2 - } < He_2^ + $$
C
$${C_2}^{2 - } < He_2^ + < {O_2}^ - < NO$$
D
$$He_2^ + < {O_2}^ - < NO < {C_2}^{2 - }$$

Explanation

Molecular orbital configuration of NO (15 electrons) is

$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ * $$

$$\therefore\,\,\,\,$$ Nb = 10

Na = 5

$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 5} \right]$$ = 2.5

Molecular orbital configuration of $$O_2^ - $$ (17 electrons) is

$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^1}^ * $$

$$\therefore\,\,\,\,$$ Nb = 10

Na = 7

$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 7} \right]$$ = 1.5

Molecular orbital configuration of $$C_2^ {2-} $$ (14 electrons) is

$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$$

$$\therefore\,\,\,\,$$ Nb = 10

Na = 4

$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 4} \right]$$ = 3

$$\,\,\,$$ Configuration of $$He_2^ + $$ (3 electrons) is = $${\sigma _{1{s^2}}}$$ $$\sigma _{1{s^1}}^ * $$

$$\therefore\,\,\,$$ Bond order = $${1 \over 2}$$ (2 $$-$$1) = 0.5

$$ \therefore $$ Correct order is :

$$He_2^ + < {O_2}^ - < NO < {C_2}^{2 - }$$
3
MCQ (Single Correct Answer)

AIPMT 2007

The correct order of C $$-$$ O bond length among CO, CO$${_3^{2 - }}$$, CO2 is
A
CO < CO$${_3^{2 - }}$$ < CO2
B
CO$${_3^{2 - }}$$ < CO2 < CO
C
CO < CO2 < CO$${_3^{2 - }}$$
D
CO2 < CO$${_3^{2 - }}$$ < CO

Explanation

All these structures exhibits resonance and can be represented by the following resonating structures.

More single bond character in resonance hybrid, more is the bond length. Hence the increasing bond length is

CO $$<$$ CO2 $$<$$ CO$${_3^{2 - }}$$
4
MCQ (Single Correct Answer)

AIPMT 2007

In which of the following pairs, the two species are isostructural?
A
SO$${_3^{2 - }}$$ and NO$${_3^{ - }}$$
B
BF3 and NF3
C
BrO$${_3^{ - }}$$ and XeO3
D
SF4 and XeF4

Explanation

Hybridisation of Br in BrO3 :

= $${1 \over 2}\left( {7 + 0 - 0 + 1} \right)$$ = 4

Four hybrid orbital means sp3 hybridisation.

Hybridisation of Xe in XeO3 :

= $${1 \over 2}\left( {8 + 0 - 0 + 0} \right)$$ = 4

Four hybrid orbital means sp3 hybridisation.

Thus, both BrO3 and XeO3 are sp3 hybridised with three bond pairs of electrons and one lone pair of electrons and results in trigonal pyramidal shape.

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