JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2003

An ethar is more volatile than an alcohol having the same molecular formula. This is due to
A
alcohols having resonance structures
B
inter-molecular hydrogen bonding in ethers
C
inter-molecular hydrogen bonding in alcohols
D
dipole characters of ethers

Explanation

Alcohol and ether are isomer with each other. So, with same molecular formula we can make ether as well as alcohol.

For ex,

With molecular formula C2H6O

(1) $\,\,\,$ alcohol will be CH3CH2 OH

(2) $\,\,\,$ ether will be CH3 $-$ O $-$ CH3

In Alcohol there is hydrogen bond and in Ether there is Van der walls force of attraction.

We know that H bond is stronger bond than van der walls force of attraction as the atoms of alcohol are strongly attached with each other by hydrogen bonding so tendency of vaporization of alcohol is less compared to ether.

In alcohol inter-molecular hydrogen bonding look like this -

2

AIEEE 2002

A square planar complex is formed by hybridisation of which atomic orbitals ?
A
s, px, py, dxy
B
s, px, py, dx2 - y2
C
s, px, py, dz2
D
s, px, py, dyz

Explanation

Hybridization of square planar complex is dsp2 .

In square planar complex all 4 surrounding atoms and central atom are in the same plane and let this plane is x $-$ y plane. As it is dsp2 hybridized, so it has 1 d orbital, 1 s orbital and 2 p orbital.

s orbital is non-directional, so we just write it as s.

In px the two lobes are along x-axis and in py two lobes are along y-axis

One d orbital is dx2 $-$ y2, it means out of four lobes of d orbital two along x-axis and two along y-axis.

3

AIEEE 2002

In which of the following species is the underlined carbon having sp3 hybridisation?
A
$C{H_3}\underline C OOH$
B
$C{H_3}\underline C H_2OH$
C
$C{H_3}\underline C OCH_3$
D
$C{H_2} = \underline C H - C{H_3}$

Explanation

(a) $\,\,\,\,$

(b) $\,\,\,\,$

(c) $\,\,\,\,$

(d) $\,\,\,\,$

So, option (B) is correct

Note :
If in carbon(C) all 4 bonds are sigma bond then it is sp3 hybridization and if there is 3 sigma bond then sp2 and if there is only 2 sigma bond there sp hybridization.
4

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

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