Chemistry - CHEMICAL BONDING Question with Solution | TestHub

Ans. (I-R) ; (II-T); (III-R); (IV-T)

(P) $\mathrm{C}\Rightarrow \sigma 1{\mathrm{s}}^2<{\sigma }_{\ast }1{\mathrm{s}}^2<\sigma 2{\mathrm{s}}^2<{\pi }_{\ast }2{\mathrm{p}}_{\mathrm{x}}{}^2\approx \pi 2{\mathrm{p}}_{\mathrm{y}}{}^2$

B.O. $=2$, (B.O.) ${)}_{\sigma }=0$, (B.O.) ${}_{\pi }=2$

(Q) ${\mathrm{B}}_2\Rightarrow \sigma 1{\mathrm{s}}^2<{\sigma }_{\ast }1{\mathrm{s}}^2<\sigma 2{\mathrm{s}}^2<{\sigma }_{\ast }2{\mathrm{p}}^2<\pi 2{\mathrm{p}}_{\mathrm{x}}{}^1\approx \pi 2{\mathrm{p}}_{\mathrm{y}}{}^1$

B.O. $=1$, (B.O.) ${)}_{\sigma }=0$, (B.O.) $\pi =1$

(R) ${\mathrm{O}}_2\Rightarrow \sigma 1{\mathrm{s}}^2<{\sigma }_{\ast }1{\mathrm{s}}^2<\sigma 2{\mathrm{s}}^2<{\sigma }_{\ast }2{\mathrm{s}}^2<{\sigma }^{\ast }2{\mathrm{p}}_{\mathrm{z}}^2<\pi 2{\mathrm{p}}_{\mathrm{x}}^2\approx \pi 2{\mathrm{p}}_{\mathrm{y}}^2<{\pi }^{\ast }2{\mathrm{p}}_{\mathrm{x}}^1\approx {\pi }^{\ast }2{\mathrm{p}}_{\mathrm{y}}^1$

B.O. $=2$, (B.O.) ${)}_{\sigma }=1$, (B.O.) $\pi =1$

(S) ${\mathrm{N}}_2\Rightarrow \sigma 1{\mathrm{s}}^2<{\sigma }_{\ast }1{\mathrm{s}}^2<\sigma 2{\mathrm{s}}^2<{\sigma }_{\ast }2{\mathrm{s}}^2<\pi 2{\mathrm{p}}_{\mathrm{x}}{}^2\approx \pi 2{\mathrm{p}}_{\mathrm{y}}{}^2<\sigma 2{\mathrm{p}}_{\mathrm{z}}{}^2$

B.O. $=3$, (B.O.) ${)}_{\sigma }=1$, (B.O.) $\pi =2$

If in molecule electron is removed from B.M.O. than ionisation energy of molecule is more than atom. If in molecule electron is removed ABMO from than ionisation energy of molecule is more then atom.