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1

AIEEE 2011

MCQ (Single Correct Answer)
The structure of IF7 is :
A
trigonal bipyramid
B
octahedral
C
pentagonal bipyramid
D
square pyramid

Explanation

The structure of IF7 pentagonal bipyramidal having sp3d3 hybridisation.
2

AIEEE 2011

MCQ (Single Correct Answer)
The hybridization of orbitals of N atom in $$NO_3^-$$, $$NO_2^+$$ and $$NH_4^+$$ are respectively :
A
sp , sp2, sp3
B
sp2, sp , sp3
C
sp , sp3, sp2
D
sp2, sp3, sp

Explanation

3

AIEEE 2011

MCQ (Single Correct Answer)
Among the following the maximum covalent character is shown by the compound :
A
SnCl2
B
AlCl3
C
MgCl2
D
FeCl2

Explanation

Charge of cation/Size of cation is called polarising power.

$$ \therefore $$ (i) Polarising power $$ \propto $$ charge of cation

(ii) Polarising power $$ \propto $$
1
size of cation


Here the AlCl3 is satisfying the above two conditions i.e., Al is in +3 oxidation state and also has small size. So it has more covalent character.
4

AIEEE 2009

MCQ (Single Correct Answer)
Using MO theory, predict which of the following species has the shortest bond length?
A
$$O_2^+$$
B
$$O_2^-$$
C
$$O_2^{2-}$$
D
$$O_2^{2+}$$

Explanation

Note :

(1) $$\,\,\,\,$$ Bond length $$ \propto $$ $${1 \over {Bond\,\,order}}$$

(2) $$\,$$ Bond order $$ = {1 \over 2}$$ [Nb $$-$$ Na]

Nb = No of electrons in bonding molecular orbital

Na $$=$$ No of electrons in anti bonding molecular orbital

(4) $$\,\,\,\,$$ upto 14 electrons, molecular orbital configuration is



Here Na = Anti bonding electron $$=$$ 4 and Nb = 10

(5) $$\,\,\,\,$$ After 14 electrons to 20 electrons molecular orbital configuration is - - -



Here Na = 10

and Nb = 10

In O atom 8 electrons present, so in O2, 8 $$ \times $$ 2 = 16 electrons present.

Then in $$O_2^ + $$ no of electrons = 15

in $$O_2^ - $$ no of electrons = 17

in $$O_2^{2 - }$$ no of electrons = 18

$$\therefore\,\,\,\,$$ Molecular orbital configuration of O2 (16 electrons) is

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

$$\therefore\,\,\,\,$$Na = 6

Nb = 10

$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 6} \right] = 2$$

Molecular orbital configuration of O$$_2^ + $$ (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$$_2^ {2+ }$$ (14 electrons) is

$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^0}^ * \, = \,\pi _{2p_y^o}^ * $$

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

Na = 4

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

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 O $$_2^{2 - }$$ (18 electrons) is

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

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

Na = 8

$$\therefore\,\,\,\,$$ BO = $${1 \over 2}$$ [ 10 $$-$$ 8] = 1

As Bond length $$ \propto $$ $${1 \over {Bond\,\,order}}$$

So $$O_2^{2+}$$ has the shortest bond length.

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