Select the incorrect statement.

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(C) $1$. $O_3$ is diamagnetic (all electrons paired). $O_2^{2-}$ (peroxide ion) has $18$ electrons,all paired,so it is diamagnetic. Statement $A$ is c

Vedclass · Accepted Answer

Out of $O_2$,$O_2^+$,and $O_2^-$,only $O_2$ is paramagnetic.

Vedclass · Answer

(C) $1$. $O_3$ is diamagnetic (all electrons paired). $O_2^{2-}$ (peroxide ion) has $18$ electrons,all paired,so it is diamagnetic. Statement $A$ is correct.
$2$. Bond orders: $O_2$ $(2.0)$,$O_2^+$ $(2.5)$,$O_3$ (approx $1.5$). Higher bond order means shorter bond length. $O_2^+$ has the highest bond order,so it has the least $O-O$ bond length. Statement $B$ is correct.
$3$. Molecular orbital configurations: $O_2$ has $2$ unpaired electrons (paramagnetic). $O_2^+$ has $1$ unpaired electron (paramagnetic). $O_2^-$ has $1$ unpaired electron (paramagnetic). Statement $C$ is incorrect because $O_2^+$,$O_2$,and $O_2^-$ are all paramagnetic.
$4$. Spin magnetic moment $(\mu = \sqrt{n(n+2)} \ BM)$: $O_2$ has $n=2$ $(\mu = 2.83 \ BM)$,$O_2^+$ has $n=1$ $(\mu = 1.73 \ BM)$,$O_2^-$ has $n=1$ $(\mu = 1.73 \ BM)$. $O_2$ has the maximum spin magnetic moment. Statement $D$ is correct.

Pair of species	Identical property
$A$. $B_2 \& O_2$	$P$. Bond order $= 2.5$
$B$. $Be_2 \& H_2^{2-}$	$Q$. Paramagnetic nature
$C$. $N_2^{+} \& N_2^{-}$	$R$. Diamagnetic nature
$D$. $O_2^{+} \& O_2^{-}$	$S$. Doesn't exist

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Match the following:

Pair of species Identical property

$A$. $B_2 \& O_2$ $P$. Bond order $= 2.5$

$B$. $Be_2 \& H_2^{2-}$ $Q$. Paramagnetic nature

$C$. $N_2^{+} \& N_2^{-}$ $R$. Diamagnetic nature

$D$. $O_2^{+} \& O_2^{-}$ $S$. Doesn't exist

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