Question

The incorrect statement/s among the following is/are I. $NCl_5$ does not exist while $PCl_5$ does II. Lead prefers to form tetravalent compounds III. The three $C-O$ bonds are not equal in the carbonate ion IV. Both $O_2^+$ and $NO$ are paramagnetic

• A. $I, III$ and $IV$
• B. $I$ and $IV$
• C. $II$ and $III$
• D. $I$ and $III$

Answer 1

The Correct Option is C

(I) In nitrogen d-orbitals are absent, so it does not form $NC{{l}_{5}}$ . Thus, $NC{{l}_{5}}$ does not exist but $PC{{l}_{5}}$ does.
(II) $P{{b}^{2+}}$ is more stable than $P{{b}^{4+}},$ due to inert pair effect.
(Ill) In carbonate ion $(CO_{3}^{2-})$ all the three $C-O$ bonds are identical due to resonance.
(IV) $O_{2}^{+}(8+8-1=15)=\sigma 1{{s}^{2}},{{\sigma }^{*}}1{{s}^{2}},\sigma 2{{s}^{2}},$ ${{\sigma }^{*}}2{{s}^{2}},$
$\sigma 2p_{x}^{2},\pi 2p_{y}^{2}\approx 2p_{z}^{2},\overset{*}{\mathop{\pi }}\,2p_{y}^{1}$
$NO(7+8=15)$
Hence, both $O_{2}^{+}$ and $NO$ contains one unpaired electron, so paramagnetic.