For F_2 and OF molecules,consider the followi | vedclass

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(A) $1$. For $F_2$ ($18$ electrons): Electronic configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 =

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(A) $1$. For $F_2$ ($18$ electrons): Electronic configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$. Bond order = $(10-8)/2 = 1$. It is diamagnetic.$2$. For $OF$ ($17$ electrons): Electronic configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. Bond order = $(10-7)/2 = 1.5$. It is paramagnetic.$3$. Statement $(a)$: False,as bond order of $F_2$ is $1$ and $OF$ is $1.5$.$4$. Statement $(b)$: True,$OF$ is paramagnetic and $F_2$ is diamagnetic.$5$. Statement $(c)$: True,$F_2$ has a lower bond order $(1)$ compared to $OF$ $(1.5)$,making it easier to dissociate.$6$. Statement $(d)$: True,for both,$BMO$ electrons $(10)$ > $ABMO$ electrons ($8$ for $F_2$,$7$ for $OF$).

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