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Question

(A) $1$. $B_2$ ($10$ electrons) is paramagnetic ($HOMO$ $\pi_{2p_y}^1 \pi_{2p_z}^1$) and $O_2$ ($16$ electrons) is paramagnetic ($HOMO$ $\pi_{2p_y}^{*

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$A-Q, B-S, C-P, D-Q$

Vedclass · Answer

(A) $1$. $B_2$ ($10$ electrons) is paramagnetic ($HOMO$ $\pi_{2p_y}^1 \pi_{2p_z}^1$) and $O_2$ ($16$ electrons) is paramagnetic ($HOMO$ $\pi_{2p_y}^{*1} \pi_{2p_z}^{*1}$). Thus,$A-Q$.$2$. $Be_2$ ($8$ electrons) and $H_2^{2-}$ ($4$ electrons) both have bond order $0$ and do not exist. Thus,$B-S$.$3$. $N_2^{+}$ ($13$ electrons) and $N_2^{-}$ ($15$ electrons) both have a bond order of $2.5$. Thus,$C-P$.$4$. $O_2^{+}$ ($15$ electrons) and $O_2^{-}$ ($17$ electrons) are both paramagnetic. Thus,$D-Q$.Therefore,the correct match is $A-Q, B-S, C-P, D-Q$.

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

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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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