Match the following: Pair of species | vedclass

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(A) To solve this,we analyze the molecular orbital configuration and properties of each species:$A$. $B_2$ ($10$ electrons) is paramagnetic with bond

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

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(A) To solve this,we analyze the molecular orbital configuration and properties of each species:$A$. $B_2$ ($10$ electrons) is paramagnetic with bond order $1$. $O_2$ ($16$ electrons) is paramagnetic with bond order $2$. Both are paramagnetic $(Q)$.$B$. $Be_2$ ($8$ electrons) has bond order $0$ (does not exist). $H_2^{2-}$ ($4$ electrons) has bond order $0$ (does not exist). Both do not exist $(S)$.$C$. $N_2^{+}$ ($13$ electrons) has bond order $2.5$ and is paramagnetic. $N_2^{-}$ ($15$ electrons) has bond order $2.5$ and is paramagnetic. Both have bond order $2.5$ $(P)$ and are paramagnetic $(Q)$.$D$. $O_2^{+}$ ($15$ electrons) has bond order $2.5$ and is paramagnetic. $O_2^{-}$ ($17$ electrons) has bond order $1.5$ and is paramagnetic. Both are paramagnetic $(Q)$.Correct matching: $A-Q, B-S, C-P, Q, D-Q$.

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