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JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2007

MCQ (Single Correct Answer)
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.
2

AIEEE 2007

MCQ (Single Correct Answer)
Which of the following species exhibits the diamagnetic behaviour?
A
$$O_2^{2−}$$
B
NO
C
$$O_2^+$$
D
O2

Explanation

Those species which have unpaired electrons are called paramagnetic species.

And those species which have no unpaired electrons are called diamagnetic species.

(a)    $$O_2^{2−}$$ has 18 electrons.

Moleculer orbital configuration of $$O_2^{2−}$$ is

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

Here is no unpaired electron so it is diamagnetic.

(b)   NO has 15 electrons.

Moleculer orbital configuration of NO is

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

Here is 1 unpaired electron, So it is Paramagnetic.

(c)   $$O_2$$ has 16 electrons.

Moleculer orbital configuration of $$O_2$$ is

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

Here 2 unpaired electron present, so it is paramagnetic.

(d)   $$O_2^{+}$$ has 15 electrons.

Moleculer orbital configuration of $$O_2^{+}$$ is

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

Here 1 unpaired electron present, so it is paramagnetic.
3

AIEEE 2007

MCQ (Single Correct Answer)
The charge/size ratio of a cation determines its polarizing power. Which one of the following sequences represents the increasing order of the polarizinig order of the polarizing power of the cationic species, K+, Ca2+, Mg2+, Be2+?
A
Mg2+ < Be2+ < K+ < Ca2+
B
K+ < Ca2+ < Mg2+ < Be2+
C
Be2+ < K+ < Ca2+ < Mg2+
D
Ca2+ < Mg2+ < Be2+ < K+

Explanation

As charge/size ratio of a cation determines its polarizing power so high charge and small size of the cations increases polarisation.

As the size of the given cations decreases as

K+ > Ca2+ > Mg2+ > Be2+

Hence, polarising power decreases as

K+ < Ca2+ < Mg2+ < Be2+
4

AIEEE 2006

MCQ (Single Correct Answer)
The decreasing values of bond angles from NH3 (106o) to SbH3 (91o ) down group-15 of the periodic table is due to
A
increasing bp-bp repulsion
B
increasing p-orbital character in sp3
C
decreasing lp-bp repulsion
D
decreasing electronegativity

Explanation

Nitrogen(N) atom is smaller in size so it's lone pair is relatively unstable. That is why repulsion between lone pairs present on nitrogen atom and bonded pairs of electrons is relatively high.

But size of Sb is more so its lone pair is relatively stable. That is why it does not want to perticipate in bond pair lone pair repulsion and decreases lp-bp repulsion.

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