Assertion : Fluorine molecule has bond order | vedclass

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Question

(A) The $MO$ electronic configuration of the $F_2$ molecule ($18$ electrons) is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^

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If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

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(A) The $MO$ electronic configuration of the $F_2$ molecule ($18$ electrons) is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$. Number of bonding electrons $(N_b)$ $= 10$. Number of antibonding electrons $(N_a)$ $= 8$. Bond order $= \frac{N_b - N_a}{2} = \frac{10 - 8}{2} = 1$. Since $N_b - N_a = 2$,the number of electrons in antibonding orbitals is indeed two less than in bonding orbitals. Thus,both Assertion and Reason are correct,and the Reason is the correct explanation for the Assertion.

Assertion : Fluorine molecule has bond order one.
Reason : The number of electrons in the antibonding molecular orbitals is two less than that in bonding molecular orbitals.

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Assertion : Fluorine molecule has bond order one. Reason : The number of electrons in the antibonding molecular orbitals is two less than that in bonding molecular orbitals.