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Among the following species \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 _______________.

See full solution. Explanation: Those species which have unpaired electrons are called paramagnetic species. And those species which have no unpaired electrons are called diamagnetic species. (1) N_2 has 14 electrons. Moleculer orbital configuration of N_2 = {\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 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}} Here in N_2^{ + } , 1 unpaired electron present, so it is paramagnetic. (3) \mathrm{N}_{2}^{2-} has 16 electrons. Moleculer orbital configuration of \mathrm{N}_{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^1}^ * = \,\,\pi _{2p_y^1}^ * Here 2 unpaired electron present, so it is paramagnetic. (4) \mathrm{N}_{2}^{-} has 15 electrons. Moleculer orbital configuration of \mathrm{N}_{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. (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) O_2^{−} has 17 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^1}^ * Here 1 unpaired electron present, 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.

JEE Main 2022ChemistryChemical Bonding And Molecular StructureMolecular Orbital TheorymediumNumerical

JEE Main 2022 — Chemical Bonding And Molecular Structure Question with Solution

From: JEE Main 2022 (Online) 25th July Morning Shift

Question

Among the following species

$N_{2}, N_{2}^{+}, N_{2}^{-}, N_{2}^{2 -}, O_{2}, O_{2}^{+}, O_{2}^{-}, O_{2}^{2 -}$

the number of species showing diamagnesim is _______________.

Enter your answer

Numerical answer

Show full solutionCorrect answer: —

Correct answer

See solution below.

Step-by-step explanation

Those species which have unpaired electrons are called paramagnetic species.

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

(1)

N_{2}

has 14 electrons.

Moleculer orbital configuration of

N_{2}

σ_{1 s^{2}} σ_{1 s^{2}}^{*} σ_{2 s^{2}} σ_{2 s^{2}}^{*} π_{2 p_{x}^{2}} = π_{2 p_{y}^{2}} σ_{2 p_{z}^{2}}

Here no unpaired electron present, so it is diamagnetic.

(2) Moleculer orbital configuration of

N_{2}^{+}

(13 electrons)

=

σ_{1 s^{2}} σ_{1 s^{2}}^{*} σ_{2 s^{2}} σ_{2 s^{2}}^{*} π_{2 p_{x}^{2}} = π_{2 p_{y}^{2}} σ_{2 p_{z}^{1}}

Here in

N_{2}^{+}

, 1 unpaired electron present, so it is paramagnetic.

(3)

N_{2}^{2 -}

has 16 electrons.

Moleculer orbital configuration of

N_{2}^{2 -}

σ_{1 s^{2}} σ_{1 s^{2}}^{*} σ_{2 s^{2}} σ_{2 s^{2}}^{*} σ_{2 p_{z}^{2}} π_{2 p_{x}^{2}} = π_{2 p_{y}^{2}} π_{2 p_{x}^{1}}^{*} = π_{2 p_{y}^{1}}^{*}

Here 2 unpaired electron present, so it is paramagnetic.

(4)

N_{2}^{-}

has 15 electrons.

Moleculer orbital configuration of

N_{2}^{-}

σ_{1 s^{2}} σ_{1 s^{2}}^{*} σ_{2 s^{2}} σ_{2 s^{2}}^{*} σ_{2 p_{z}^{2}} π_{2 p_{x}^{2}} = π_{2 p_{y}^{2}} π_{2 p_{x}^{1}}^{*} = π_{2 p_{y}^{0}}^{*}

Here 1 unpaired electron present, so it is paramagnetic.

(a)

O_{2}^{2 -}

has 18 electrons.

Moleculer orbital configuration of

O_{2}^{2 -}

σ_{1 s^{2}} σ_{1 s^{2}}^{*} σ_{2 s^{2}} σ_{2 s^{2}}^{*} σ_{2 p_{z}^{2}} π_{2 p_{x}^{2}} = π_{2 p_{y}^{2}} π_{2 p_{x}^{2}}^{*} = π_{2 p_{y}^{2}}^{*}

Here is no unpaired electron so it is diamagnetic.

(b)

O_{2}^{-}

has 17 electrons.

Moleculer orbital configuration of

O_{2}^{2 -}

σ_{1 s^{2}} σ_{1 s^{2}}^{*} σ_{2 s^{2}} σ_{2 s^{2}}^{*} σ_{2 p_{z}^{2}} π_{2 p_{x}^{2}} = π_{2 p_{y}^{2}} π_{2 p_{x}^{2}}^{*} = π_{2 p_{y}^{1}}^{*}

Here 1 unpaired electron present, so it is paramagnetic.

(c)

O_{2}

has 16 electrons.

Moleculer orbital configuration of

O_{2}

σ_{1 s^{2}} σ_{1 s^{2}}^{*} σ_{2 s^{2}} σ_{2 s^{2}}^{*} σ_{2 p_{z}^{2}} π_{2 p_{x}^{2}} = π_{2 p_{y}^{2}} π_{2 p_{x}^{1}}^{*} = π_{2 p_{y}^{1}}^{*}

Here 2 unpaired electron present, so it is paramagnetic.

(d)

O_{2}^{+}

has 15 electrons.

Moleculer orbital configuration of

O_{2}^{+}

σ_{1 s^{2}} σ_{1 s^{2}}^{*} σ_{2 s^{2}} σ_{2 s^{2}}^{*} σ_{2 p_{z}^{2}} π_{2 p_{x}^{2}} = π_{2 p_{y}^{2}} π_{2 p_{x}^{1}}^{*} = π_{2 p_{y}^{0}}^{*}

Here 1 unpaired electron present, so it is paramagnetic.

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About this question

This is a previous-year question from JEE Main 2022, covering the Chemical Bonding And Molecular Structure chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.