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QJEEM22Cha5f0: JEE Main 2022 Chemistry - Class 11 JEEM22Cha5f0 Solution: Among the following species the number of species showing…

Among the following species

N2, N2+,N2−,N22−,O2,O2+,O2−,O22−\mathrm{N}_{2}, \mathrm{~N}_{2}^{+}, \mathrm{N}_{2}^{-}, \mathrm{N}_{2}^{2-}, \mathrm{O}_{2}, \mathrm{O}_{2}^{+}, \mathrm{O}_{2}^{-}, \mathrm{O}_{2}^{2-}

the number of species showing diamagnesim is _______________.

Step-by-Step Verified Solution

Those species which have unpaired electrons are called paramagnetic species.

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

(1)   N2N_2 has 14 electrons.

Moleculer orbital configuration of N2N_2

= σ1s2 σ1s2∗ σ2s2 σ2s2∗ π2px2 = π2py2 σ2pz2{\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}}

Here no unpaired electron present, so it is diamagnetic.

(2) Moleculer orbital configuration of N2+N_2^{ + } (13 electrons)

= σ1s2 σ1s2∗ σ2s2 σ2s2∗ π2px2 = π2py2 σ2pz1{\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}}

Here in N2+N_2^{ + } , 1 unpaired electron present, so it is paramagnetic.

(3)   N22−\mathrm{N}_{2}^{2-} has 16 electrons.

Moleculer orbital configuration of N22−\mathrm{N}_{2}^{2-} is

σ1s2 σ1s2∗  σ2s2  σ2s2∗  σ2pz2  π2px2 = π2py2 π2px1∗=  π2py1∗{\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.

(4)   N2−\mathrm{N}_{2}^{-} has 15 electrons.

Moleculer orbital configuration of N2−\mathrm{N}_{2}^{-} is

σ1s2 σ1s2∗  σ2s2  σ2s2∗  σ2pz2  π2px2 = π2py2 π2px1∗=  π2py0∗{\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.

(a)    O22−O_2^{2−} has 18 electrons.

Moleculer orbital configuration of O22−O_2^{2−} is

σ1s2 σ1s2∗ σ2s2 σ2s2∗ σ2pz2 π2px2 = π2py2 π2px2∗ = π2py2∗{\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)    O2−O_2^{−} has 17 electrons.

Moleculer orbital configuration of O22−O_2^{2−} is

σ1s2 σ1s2∗ σ2s2 σ2s2∗ σ2pz2 π2px2 = π2py2 π2px2∗ = π2py1∗{\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}^ *

Here 1 unpaired electron present, so it is paramagnetic.

(c)   O2O_2 has 16 electrons.

Moleculer orbital configuration of O2O_2 is

σ1s2 σ1s2∗  σ2s2  σ2s2∗  σ2pz2  π2px2 = π2py2 π2px1∗=  π2py1∗{\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)   O2+O_2^{+} has 15 electrons.

Moleculer orbital configuration of O2+O_2^{+} is

σ1s2 σ1s2∗  σ2s2  σ2s2∗  σ2pz2  π2px2 = π2py2 π2px1∗=  π2py0∗{\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.

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