Question

The common features of the species ${{\mathrm{N}}_2}^{2-}$ ,${\mathrm{O}}_2$ and ${\mathrm{NO}}^{-}$ are

• A. bond order is three and are isoelectronic
• B. bond order is two and are isoelectronic
• C. bond order is three but are not isoelectronic
• D. bond order is two but are not isoelectronic

Answer 1

The correct option is B

bond order is two and are isoelectronic

Explanation for correct answer:-

Option (B) bond order is two and are isoelectronic

• Isoelectronic species: Are those which have equal number of electrons in the outer most shell
• Electronic configuration of ${{\mathrm{N}}_2}^{2-}$ = $\left[\mathrm{\sigma }\right(1\mathrm{s}\left)\right]2[\mathrm{\sigma }\ast (1\mathrm{s}\left)\right]2\left[\mathrm{\sigma }\right(2\mathrm{s}\left)\right]2[\mathrm{\sigma }\ast (2\mathrm{s}\left)\right]2\left[\mathrm{\pi }\right(2\mathrm{px}\left)\right]2\left[\mathrm{\pi }\right(2\mathrm{py}\left)\right]2\left[\mathrm{\sigma }\right(2\mathrm{pz}\left)\right]2[\mathrm{\pi }\ast (2\mathrm{px}\left)\right]1[\mathrm{\pi }\ast (2\mathrm{py}\left)\right]1$
• We know, Bond order = $\frac{(No.ofbondingelectrons–No.ofantibondingelectrons)}{2}$ = $\frac{(10-6)}{2}=2$
• Electronic configuration of ${\mathrm{O}}_2$= $\left[\mathrm{\sigma }\right(1\mathrm{s}\left)\right]2[\mathrm{\sigma }\ast (1\mathrm{s}\left)\right]2\left[\mathrm{\sigma }\right(2\mathrm{s}\left)\right]2[\mathrm{\sigma }\ast (2\mathrm{s}\left)\right]2\left[\mathrm{\sigma }\right(2\mathrm{pz}\left)\right]2\left[\mathrm{\pi }\right(2\mathrm{px}\left)\right]2\left[\mathrm{\pi }\right(2\mathrm{py}\left)\right]2[\mathrm{\pi }\ast (2\mathrm{px}\left)\right]1[\mathrm{\pi }\ast (2\mathrm{py}\left)\right]1$
• Bond order = $\frac{(10-6)}{2}=2$ = 2
• Electronic configuration of ${\mathrm{NO}}^{-}$= $\left[\sigma \right(1s\left)\right]2[\sigma \ast (1s\left)\right]2\left[\sigma \right(2s\left)\right]2[\sigma \ast (2s\left)\right]2\left[\sigma \right(2pz\left)\right]2\left[\pi \right(2px\left)\right]2\left[\pi \right(2py\left)\right]2[\pi \ast (2px\left)\right]1[\pi \ast (2py\left)\right]1$
• Bond order = $\frac{(10-6)}{2}=2$
• Calculating , No.of electrons in ${{\mathrm{N}}_2}^{2-}$ = $7\times 2+2=16$ electrons.
• No.of electrons in ${\mathrm{O}}_2$= $(8\times 2)=16$ electrons.
• No.of electrons in ${\mathrm{NO}}^{-}$ = $7+8+1=16$ electrons.

Hence the three molecules ${{\mathrm{N}}_2}^{-2},{\mathrm{O}}_2\mathrm{and}{\mathrm{NO}}^{–}$are bond order is two and are isoelectronic and option b is correct.