The diamagnetic species among the following i | vedclass

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(D) To be diamagnetic,no unpaired electron must be present.Write the molecular orbital electronic configuration of each species.For all options given,

Vedclass · Accepted Answer

$O_{2}^{2-}$

Vedclass · Answer

(D) To be diamagnetic,no unpaired electron must be present.Write the molecular orbital electronic configuration of each species.For all options given,the configuration $\sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*2} \sigma_{2p_z}^{2} \pi_{2p_x}^{2} = \pi_{2p_y}^{2}$ is common.After that:$O_{2}^{+}: \pi_{2p_x}^{*1}$ (one unpaired electron,paramagnetic)$O_{2}^{-}: \pi_{2p_x}^{*2} = \pi_{2p_y}^{*1}$ (one unpaired electron,paramagnetic)$O_{2}: \pi_{2p_x}^{*1} = \pi_{2p_y}^{*1}$ (two unpaired electrons,paramagnetic)$O_{2}^{2-}: \pi_{2p_x}^{*2} = \pi_{2p_y}^{*2}$ (no unpaired electrons,diamagnetic).Therefore,$O_{2}^{2-}$ is the diamagnetic species.

The diamagnetic species among the following is $.....$

Similar Questions

Which among the following have diamagnetic property?
$(i)$ $B_2$
$(ii)$ $N_2$
$(iii)$ $O_2$
$(iv)$ $C_2$

Which of the following statements is incorrect when $C_2$ changes to $C_2^{2-}$?

Select the incorrect statement about $N_2$ molecule.

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When $\psi_A$ and $\psi_B$ are the wave functions of atomic orbitals,then $\sigma^*$ is represented by :

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