Which of the following statements is correct?

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Question

(A) According to Molecular Orbital Theory $(MOT)$: $1$. The electronic configuration of $O_2$ ($16$ electrons) is $\sigma 1s^2, \sigma^* 1s^2, \sigma

Vedclass · Accepted Answer

Oxygen and nitric oxide molecules are both paramagnetic because both contain unpaired electrons.

Vedclass · Answer

(A) According to Molecular Orbital Theory $(MOT)$: $1$. The electronic configuration of $O_2$ ($16$ electrons) is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. It has two unpaired electrons in the antibonding $\pi^*$ orbitals,making it paramagnetic. $2$. The electronic configuration of $NO$ ($15$ electrons) is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. It has one unpaired electron in the antibonding $\pi^*$ orbital,making it paramagnetic. Therefore,both $O_2$ and $NO$ are paramagnetic due to the presence of unpaired electrons.

Similar Questions

Give the electron configuration,bond order,magnetic property,and energy diagram for the beryllium $(Be_{2})$ molecule and discuss its existence.

Between $1s-1s$ and $1s-2s$ combinations,which one does not form a molecular orbital? Why?

What is the difference in the $MO$ energy level diagram for diatomic molecules from $Li_2$ to $N_2$ compared to those from $O_2$ to $Ne_2$?

How are molecular orbitals formed? Explain.

Bonding in which of the following diatomic molecule$(s)$ becomes stronger,on the basis of $M.O.$ Theory,by removal of an electron?
$(A)$ $NO$
$(B)$ $N_2$
$(C)$ $O_2$
$(D)$ $C_2$
$(E)$ $B_2$
Choose the most appropriate answer from the options given below.

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