Question

Column-I contains four entries and Column-II contains five entries. Entries of Column-I are to be matched with some entries of Column-II. One or more than one entries of Column-I may have the matching with the same entries of Column-II

Answer 1

(A) $CN\left(13\right);\sigma 1s^2{\sigma }^{\ast }1x^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p{x}^2\pi 2p{y}^2\sigma 2p{z}^1$
Bond order $=\frac{9-4}{2}=2.5$; paramagnetic, ${\mu }_{BM}=1.73$
$HOMO=\sigma 2pzLUMO={\pi }^{\ast }2px$ or ${\pi }^{\ast }2py$
$C{N}^{+}$ bond order $=\frac{8-4}{2}=2;C{N}^{-}$ bond order $=\frac{10-4}{2}=3$
(B) $OF17);\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p{z}^2\pi 2p_x^2=\pi 2p_u^2{\pi }^{\ast }2p_x={\pi }^{\ast }2p_y^1$
Bond order $=\frac{10-7}{2}=1.5$; paragmanetic, ${\mu }_{BM}=1.73$
$HOMO={\pi }^{\ast }2p$ orbital; $O{F}^{+}$ bond order $=\frac{10-6}{2}=2;O{F}^{-}$ bond order $=\frac{10-8}{2}=1$
(C) $BN\left(12\right)=\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p_x^2=\pi 2p_y^2$
Bond order $=\frac{8-4}{2}=2;diamagnetic,{\mu }_{BM}=0$
$B{N}^{+}$ bond order $=\frac{7-4}{2}=1.5;B{N}^{-}=\frac{9-4}{2}=2.5$
$HOMO=\pi 2p$
(D) $NO\left(15\right)=\sigma 1s^2{\sigma }^{\ast }2s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^2=\pi 2p_y^2{\pi }^{\ast }2p_x^1$
Bond order $=\frac{10-5}{2}=2.5paramagnetic,{\mu }_{BM}=1.73$
$HOMO={\pi }^{\ast }2p$
$N{O}^{+}$ bond order $\frac{10-4}{2}=3;N{O}^{-}$ bond order $=\frac{10-6}{2}=2$
$O_2^{-}$ has one unpaired electron, so ${\mu }_{BM}=1.73$