Question

Which of the following is paramagnetic ?

• A. NO⁺
• B. CO
• C. O₂^-
• D. CN^-

Hint:

CO, CN^- and NO⁺ are isoelectronic with 14 electrons. Hence, there are no unpaired electrons present within the MO configuration of those species. So these are diamagnetic. 

Answer 1

The Correct Option is C

Paramagnetic species contain unpaired electrons in their molecular orbital electronic configuration. A molecule with one or more unpaired electrons is paramagnetic or odd electron system is paramagnetic.

The molecular orbital configuration of the given species is as 

CO (6+8 = 14 ) = \(\sigma 1s^2,{\sigma}^{*} 1s^2,\)\(\sigma 2s^2 ,{\sigma}^{*} 2s^2, \pi 2P_X^2 \approx \pi 2 P_Y^2 ,\sigma 2 P_Z^2\) 

All the electrons are paired hence, the molecule is diamagnetic.

O₂^- (8+8+1=17)
\(= \sigma 1s^2 . {\sigma} 1s^{*2}, \sigma 2s^2, {\sigma} 2s^{*2} , \sigma 2 P_Z^2 , \pi 2P_X^2 \pi 2P_Y^2,\)\({\pi}^{*} 2 P_Y^{*2}{\pi} 2 P_X^{*1}\) 

The molecule contains one unpaired electron so, it is paramagnetic.

CN^- (6 + 7+ 1 = 14) = same as CO 

NO⁺ (7 + 8 - 1 = 14) = same as CO 

Hence, among the given species only O₂^- is paramagnetic.
\(​O2^{​−} ⇒ σ1 s^2σ^∗1 s^2σ2 s^2σ^∗2 s^2σ2p_z^2​(π2p_x^2​ = π2p_y^2​)\)
\((π^∗2p_x^2​=π^∗2p_y^1​)​\)

One unpaired electron is present in π^∗ ABMO (Anti-Bonding Molecular Orbitals).

Answer 2

A molecule having one or more unpaired electrons in its valence shell or having an odd electron system is paramagnetic. 

Complete step-by-step answer:- There are 3 types of magnetic substances: paramagnetic, diamagnetic, and ferromagnetic substances. 

• Paramagnetic substances are weakly attracted by magnets.
• Diamagnetic substances are repelled by magnets.
• Ferromagnetic substances are strongly attracted by magnets.

Molecular orbitals are obtained by combining the atomic orbitals on the atoms within the molecule. Molecular orbital (MO) theory describes the behaviour of electrons in an exceedingly large molecule in terms of combinations of the atomic wave functions. Materials with unpaired electrons or odd-numbered electrons are paramagnetic and drawn to a magnetic field, while those with all-paired electrons (even number of electrons) are diamagnetic and repelled by a magnetic force field. 

Now let's come to the answer part: CO, CN^- and NO⁺ are isoelectronic with 14 electrons each and there are no unpaired electrons present within the MO configuration of those species. So these are diamagnetic. 

Hence, O₂^- is paramagnetic.

Note: A simple rule of thumb is employed in chemistry to work out whether a particle (atom, ion, or molecule) is paramagnetic or diamagnetic in nature. If all the electrons within the particle are paired, then the substance is diamagnetic, If there are unpaired electrons, then the substance is paramagnetic in nature.