1
MCQ (Single Correct Answer)

AIPMT 2010 Prelims

In which one of the following species the central atom has the type of hybridization which is not the same as that present in the other three ?
A
SF4
B
I3$$-$$
C
SbCl52$$-$$
D
PCl5

Explanation

The hybridisation of the central atom can be calculated as

H = $${1 \over 2}\left[ \matrix{ \left( \matrix{ No.\,of\,electrons \hfill \cr in\,valence\,shell \hfill \cr of\,atom \hfill \cr} \right) + \left( \matrix{ No.\,of\,monovalent \hfill \cr atoms\,around \hfill \cr central\,atom \hfill \cr} \right) \hfill \cr - \left( \matrix{ Ch{\mathop{\rm arge}\nolimits} \,on \hfill \cr cation \hfill \cr} \right) + \left( \matrix{ Ch{\mathop{\rm arge}\nolimits} \,on \hfill \cr anion \hfill \cr} \right) \hfill \cr} \right]$$

Applying that formula we find that all the given species except [SbCl5]2- have central atom with sp3d ( corresponding to H = 5) hybridisation. In [SbCl5]2- Sb is sp3d2 hybridized.
2
MCQ (Single Correct Answer)

AIPMT 2010 Prelims

Which one of the following species does not exist under normal conditions?
A
Be2+
B
Be2
C
B2
D
Li2

Explanation

Be2 does not exists

Be2 has an electronic configuration of :

$$\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}$$

$$ \therefore $$ Bond order = $${{4 - 4} \over 2} = 0$$

Thus, Be2 does not exists
3
MCQ (Single Correct Answer)

AIPMT 2010 Prelims

In which of the following pairs of molecules/ ions, the central atoms have sp2 hybridisation ?
A
NO$$_2^{ - }$$ and NH3
B
BF3 and NO$$_2^{ - }$$
C
NH$$_2^{ - }$$ and H2O
D
BF3 and NH$$_2^{ - }$$

Explanation

The hybridisation of the central atom can be calculated as

H = $${1 \over 2}\left[ \matrix{ \left( \matrix{ No.\,of\,electrons \hfill \cr in\,valence\,shell \hfill \cr of\,atom \hfill \cr} \right) + \left( \matrix{ No.\,of\,monovalent \hfill \cr atoms\,around \hfill \cr central\,atom \hfill \cr} \right) \hfill \cr - \left( \matrix{ Ch{\mathop{\rm arge}\nolimits} \,on \hfill \cr cation \hfill \cr} \right) + \left( \matrix{ Ch{\mathop{\rm arge}\nolimits} \,on \hfill \cr anion \hfill \cr} \right) \hfill \cr} \right]$$

$$ \therefore $$ For BF3, H = $${1 \over 2}\left[ {(3) + (3) - (0) + (0)} \right] \Rightarrow 3 \Rightarrow $$ sp2 hybridisation

$$ \therefore $$ For NO2-, H = $${1 \over 2}\left[ {(5) + (0) - (0) + (1)} \right] \Rightarrow 3 \Rightarrow $$ sp2 hybridisation
4
MCQ (Single Correct Answer)

AIPMT 2009

According to MO theory which of the lists ranks the nitrogen species in terms of increasing bond order ?
A
$$N_2^{2 - } < N_2^ - < {N_2}$$
B
$${N_2} < N_2^{2 - } < N_2^ - $$
C
$$N_2^ - < N_2^{2 - } < {N_2}$$
D
$$N_2^ - < {N_2} < N_2^{2 - }$$

Explanation

According to MOT, the molecular orbital electronic configuration of

$${N_2}:{\left( {\sigma 1s} \right)^2}{\left( {{\sigma ^*}1s} \right)^2}{\left( {\sigma 2s} \right)^2}{\left( {{\sigma ^*}2s} \right)^2}{\left( {\pi 2{p_x}} \right)^2}{\left( {\pi 2{p_y}} \right)^2}{\left( {\pi 2{p_z}} \right)^2}$$

$$ \therefore $$ B.O = $${{10 - 4} \over 2} = 3$$

$${N_2}^ - :{\left( {\sigma 1s} \right)^2}{\left( {{\sigma ^*}1s} \right)^2}{\left( {\sigma 2s} \right)^2}{\left( {{\sigma ^*}2s} \right)^2}{\left( {\pi 2{p_x}} \right)^2}{\left( {\pi 2{p_y}} \right)^2}{\left( {\pi 2{p_z}} \right)^2}{\left( {{\pi ^*}2{p_x}} \right)^1}$$

$$ \therefore $$ B.O = $${{10 - 5} \over 2} = 2.5$$

$${N_2}^{2 - }:{\left( {\sigma 1s} \right)^2}{\left( {{\sigma ^*}1s} \right)^2}{\left( {\sigma 2s} \right)^2}{\left( {{\sigma ^*}2s} \right)^2}{\left( {\pi 2{p_x}} \right)^2}{\left( {\pi 2{p_y}} \right)^2}{\left( {\pi 2{p_z}} \right)^2}{\left( {{\pi ^*}2{p_x}} \right)^1}{\left( {{\pi ^*}2{p_y}} \right)^1}$$

$$ \therefore $$ B.O = $${{10 - 6} \over 2} = 2$$

Hence the order : $$N_2^{2-}$$ < $$N_2^-$$ < $$N_2$$

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