 JEE Mains Previous Years Questions with Solutions

4.5
star star star star star
1

AIEEE 2008

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.
2

AIEEE 2008

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 2007

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.
4

AIEEE 2007

In which of the following ionization processes, the bond order has increased and the magnetic behaviour has changed?
A
$C_2 \to C_2^+$
B
$N_2 \to N_2^+$
C
$NO \to NO^+$
D
$O_2 \to O_2^+$

Explanation

(A) Moleculer orbital configuration of $C_2$ (12 electrons)

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

$\therefore\,\,\,\,$Na = 4

Nb = 8

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {8 - 4} \right] = 2$

Here no unpaired electron present, so it is diamagnetic.

Moleculer orbital configuration of $C_2^+$ (11 electrons)

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

$\therefore\,\,\,\,$Na = 4

Nb = 7

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {7 - 4} \right] = 1.5$

Here 1 unpaired electron present, so it is paramagnetic.

(B) Moleculer orbital configuration of $N_2$ (14 electrons)

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

$\therefore\,\,\,\,$Na = 4

Nb = 10

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

Here no unpaired electron present, so it is diamagnetic.

Moleculer orbital configuration of $N_2^+$ (13 electrons)

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

$\therefore\,\,\,\,$Na = 4

Nb = 9

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {9 - 4} \right] = 2.5$

Here 1 unpaired electron present, so it is paramagnetic.

(C) Moleculer orbital configuration of NO (15 electrons) is

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

$\therefore\,\,\,\,$Na = 5

Nb = 10

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

Here is 1 unpaired electron, So it is paramagnetic.

Moleculer orbital configuration of NO+ (14 electrons) is

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

$\therefore\,\,\,\,$Na = 4

Nb = 10

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

Here is no unpaired electron, So it is diamagnetic.

(D) 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 = ${1 \over 2}\left[ {10 - 6} \right] = 2$

Here 2 unpaired electrons present, so it is paramagnetic.

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

Here 1 unpaired electrons present, so it is also paramagnetic.