 JEE Mains Previous Years Questions with Solutions

4.5
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1

AIEEE 2002

Which of the following are arranged in an increasing order of their bond strengths?
A
$O_2^-$ < O2 < $O_2^+$ < $O_2^{2-}$
B
$O_2^{2-}$ < $O_2^-$ < $O_2$ < $O_2^{+}$
C
$O_2^-$ < $O_2^{2-}$ < $O_2$ < $O_2^{+}$
D
$O_2^{+}$ < $O_2$ < $O_2^-$ < $O_2^{2-}$

Explanation

Note :

(1) $\,\,\,\,$ Bond strength $\propto$ Bond order

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

(3) $\,$ 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 $O_2^ -$ (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 strength $\propto$ Bond order so, correct order is

$O_2^{2 - } < O_2^ - < {O_2} < O_2^ +$
2

AIEEE 2002

Hybridisation of underline atom changes in
A
$\underline {Al} {H_3}$ changes to $AlH_4^-$
B
${H_2}\underline O$ changes to $H_3O^+$
C
$\underline N {H_3}$ changes to $NH_4^+$
D
in all cases

Explanation

(a) $\,\,\,$ AlH3 + H$-$ $\to$ AlH$_4^ -$

Steric number of AlH3 is = ${1 \over 2}$ [3 + 3] = 3

$\therefore\,\,\,$ AlH3 is sp2 hybridized.

Steric number of AlH$_4^ -$ = ${1 \over 2}$ [3 + 4 +1] = 4

$\therefore\,\,\,$ AlH$_4^ -$ is sp3 hybridized.

(b) $\,\,\,$ H2O + H+ $\to$ H3O+

Steric no of H2O = ${1 \over 2}$ (6+ 2) = 4

$\therefore\,\,\,\,$ H2O s sp3 hybridized.

Steric no of H3O+ = ${1 \over 2}$ [ 8 + 3 $-$1] = 4

$\therefore\,\,\,$ H3O+ s also sp3 hybridized.

(c) $\,\,\,$ NH3 + H+ $\to$ NH4+

Steric no of NH3 = ${1 \over 2}$ [5 + 3] = 4

$\therefore\,\,\,$ hybridization of NH3 is sp3

Steric number of NH4+ = ${1 \over 2}$ [5 + 4 $-$ ] = 4

$\therefore\,\,\,$ Hybridization of NH4+ is sp3
3

AIEEE 2002

In which of the following species the interatomic bond angle is 109o28' ?
A
NH3, BF4-
B
NH4+, BF4-
C
NH3, BF3
D
NH2-1, BF3

Explanation

Bond angle 109o 28' means Regular Tetrahedral geometry and hybridization is sp3.

Steric Number (SN) for sp3 hybridization is 4.

In sp3 molecules can have different shapes which is decided by SN number

(1) $\,\,\,\,$ 4 bond pair in a molecule then angle between bonds 109o 28'

(2) $\,\,\,\,$ 3 bond pair and 1 lone pair in a molecule then angle between bonds 107o.

(3) $\,\,\,\,$ 2 bond pair and 2 lone pair in a molecule then angle between bonds 104.5o

(4) $\,\,\,\,$ 1 bond pair and 3 lone pair in a molecule then bond angle is undefined as there is only one bond.

(1) $\,\,\,\,$ In NH3, 3 Bond pair(BP) +1 lone pair (LP) present so angle between bond 107o. (2) $\,\,\,\,$ BF$_4^ -$, 4 bond pair present so angle is 109o 28'. (3) In NH$_4^ +$, 4 bond pair present so angle between bond is 109o 28' (4) $\,\,\,\,$ BF3 has sp2 hybridization. So bond angle is 120o. (5) $\,\,\,\,$ In NH$_2^ -$, 2 bond pair and 2 lone pair present, so bond angle is 104.5o So, $NH{}^ +$ and BF$_4^ -$ has bond angle 109o 28'.