Regarding the molecular orbital (MO) energy l | vedclass

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Question

$(i) Ne _2 \Rightarrow\left(\sigma 1 s^2\right)\left(\sigma^* 1 s^2\right)\left(\sigma 2 s^2\right)\left(\sigma^* 2 s^2\right)\left(\sigma 2 p_z^2\rig

Vedclass · Accepted Answer

$(B,C)$

Vedclass · Answer

$(i) Ne _2 \Rightarrow\left(\sigma 1 s^2ight)\left(\sigma^* 1 s^2ight)\left(\sigma 2 s^2ight)\left(\sigma^* 2 s^2ight)\left(\sigma 2 p_z^2ight)\left(\pi 2 p_{ x }^2=\pi 2 p_{ y }^2ight)\left(\pi^* 2 p_{ x }^2=\pi^* 2 p_{ y }^2ight)\left(\sigma^* 2 p_z^2ight)$
$B.O. =\frac{6-6}{2}=0$
$(ii) F _2 \Rightarrow\left(\sigma 1 s^2ight)\left(\sigma^* 1 s^2ight)\left(\sigma 2 s^2ight)\left(\sigma^* 2 s^2ight)\left(\sigma 2 p _{ z }^2ight)\left(\pi 2 p _{ x }^2=\pi 2 p _{ y }^2ight)\left(\pi^* 2 p _{ x }^2=\pi^* 2 p _{ y }^2ight)$
$	ext{HOMO}$
$(iii) O _2^{\oplus} \Rightarrow\left(\sigma 1 s^2ight)\left(\sigma^* 1 s^2ight)\left(\sigma 2 s^2ight)\left(\sigma^* 2 s^2ight)\left(\sigma 2 p_z^2ight)\left(\pi 2 p_x^2=\pi 2 p_y^2ight)\left(\pi^* 2 p_x^1=\pi^* 2 p_yight)$
$	ext { B.O. }=\frac{6-1}{2}=2.5$
$O_2 \Rightarrow\left(\sigma 1 s^2ight)\left(\sigma^* 1 s^2ight)\left(\sigma 2 s^2ight)\left(\sigma^* 2 s^2ight)\left(\sigma 2 p_z^2ight)\left(\pi 2 p_x^2=\pi 2 p_y^2ight)\left(\pi^* 2 p_x^1=\pi^* 2 p_y^1ight)$
$B.O. \frac{6-2}{2}=2 ($Bond order increases, Bond strength increases$)$
$(iv)$ Size of atom increases, Bond length increases Size of $Li > B$
So, Bond length of $Li _2 >  B _2$

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