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### JEE Mains Previous Years Questions with Solutions

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### 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.
2

### AIEEE 2008

MCQ (Single Correct Answer)
Which of the following pair of species have the same bond order?
A
CN- and NO+
B
CN- and CN+
C
$O_2^-$ and CN-
D
NO+ and CN+

## Explanation

Number of electron in NO+ = number of electron in CN = 14 electrons.

As both have same number of electrons so their bond order is equal.

Moleculer orbital configuration of NO+ (14 electrons) is

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

$\therefore$ B.O = ${1 \over 2}\left[ {10 - 4} \right]$ = 3

Moleculer orbital configuration of CN- (14 electrons) is

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

$\therefore$ B.O = ${1 \over 2}\left[ {10 - 4} \right]$ = 3
3

### AIEEE 2008

MCQ (Single Correct Answer)
The bond dissociation energy of B - F in BF3 is 646 jJ mol-1 whereas that of C - F in CF4 is 515 kj mol-1. The correct reason for higher B - F bond dissociation energy as compared to that of C - F is
A
stronger $\sigma$ bond between B and F in BF3 as compared to that between C and F in CF4
B
significant $p\pi - p\pi$ interaction between B and F in BF3 whereas there is no possibility of such interaction between C and F in CF4
C
lower degree of $p\pi - p\pi$ interaction between B and F in BF3 than that between C and F in CF4
D
smaller size of B - atom as compared to that of C- atom.

## Explanation

In BF3, B is sp2 hybridised and by Back Bonding method strong p$\pi$-p$\pi$ bond is created between filled p-orbital of F and vacant p-orbital of B which leads to shortening of B–F bond length which results in higher bond dissociation energy of the B–F bond. However in CF4, C does not have any vacant p-orbitals to undergo $\pi$-bonding.
4

### AIEEE 2007

MCQ (Single Correct Answer)
Which of the following hydrogen bonds is the strongest?
A
O−H…….N
B
F−H…….F
C
O−H…….O
D
O−H…….F

## Explanation

Among F, O and N, F is most electronegative so F pulls bond pair of electron in F - H towards itself and develops highly positive charge on H atom.

This highly positive charged H atom creates stongest hydrogen bonding by taking lone pair of electron form electronegative atom F/N/O. Hence among the given options F – H ...... F is the strongest bond.

Note : F – H ...... N has strongest hydogen bonding among F – H ...... F, F – H ...... O and F – H ...... N because N is least electronegative among F, O and N and can easily donate lone pair of electron.

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