According to Molecular Orbital Theory,which o | vedclass

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(B) The correct statements are $(A)$ and $(C)$.$(A)$ $C_2^{2-}$ has $14$ electrons. Its configuration is $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2

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$A, C$

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(B) The correct statements are $(A)$ and $(C)$.$(A)$ $C_2^{2-}$ has $14$ electrons. Its configuration is $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^2$. Since all electrons are paired,it is diamagnetic.$(B)$ Bond order of $O_2^{2+}$ is $3$ and $O_2$ is $2$. Higher bond order implies shorter bond length. Thus,$O_2$ has a longer bond length than $O_2^{2+}$.$(C)$ Bond order of $N_2^{+}$ ($13$ electrons) is $(9-4)/2 = 2.5$. Bond order of $N_2^{-}$ ($15$ electrons) is $(10-5)/2 = 2.5$. They have the same bond order.$(D)$ $He_2^{+}$ has a bond order of $0.5$. Since the bond order is non-zero,it is more stable than two isolated $He$ atoms,meaning it has lower energy.

Column $I$	Column $II$
$(A)$ $B_2$	$(p)$ Paramagnetic
$(B)$ $N_2$	$(q)$ Undergoes oxidation
$(C)$ $O_2^{-}$	$(r)$ Undergoes reduction
$(D)$ $O_2$	$(s)$ Bond order $\geq 2$
	$(t)$ Mixing of '$s$' and '$p$' orbitals

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