Which of the following statements is correct | vedclass

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(C) The electronic configuration of the peroxide ion $(O_2^{2-})$ is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x

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$A, B, C$

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(C) The electronic configuration of the peroxide ion $(O_2^{2-})$ is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$.$1.$ The $\pi^*$ antibonding molecular orbitals are completely filled with $4$ electrons,so statement $A$ is correct.$2.$ Since all electrons are paired,the ion is diamagnetic,so statement $B$ is correct.$3.$ Bond order = $\frac{1}{2} (N_b - N_a) = \frac{1}{2} (10 - 8) = 1$. Thus,statement $C$ is correct.$4.$ The total number of electrons in $O_2^{2-}$ is $8 + 8 + 2 = 18$,while Neon has $10$ electrons. Thus,statement $D$ is incorrect.Therefore,statements $A, B,$ and $C$ are correct.

Which of the following statements is correct for the peroxide ion $(O_2^{2-})$?
$A.$ It has completely filled $\pi^*$ antibonding molecular orbitals.
$B.$ It is diamagnetic.
$C.$ It has a bond order of one.
$D.$ It is isoelectronic with Neon.

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Which of the following statements is correct for the peroxide ion $(O_2^{2-})$?$A.$ It has completely filled $\pi^*$ antibonding molecular orbitals.$B.$ It is diamagnetic.$C.$ It has a bond order of one.$D.$ It is isoelectronic with Neon.