MCQ
What is the number of unpaired electron(s) in the highest occupied molecular orbital of the following species : $N _2: N _2^{+} ; O _2 ; O _2^{+}$?
- ✓$0,1,2,1$
- B$2,1,2,1$
- C$0,1,0,1$
- D$2,1,0,1$
$\sigma 1 s ^2 \sigma^* 1 s ^2 \sigma 2 s ^2 \sigma^* 2 s ^2 \pi 2 p _{ x }^2=\pi 2 p _y^2 \frac{\sigma 2 p _z^2}{\text { HOMO }}$
$N _2^{+}-\sigma 1 s ^2 \sigma^* 1 s ^2 \sigma^* 2 s ^2 \sigma^* 2 s ^2 \pi 2 p _{ x }^2=\pi 2 p _{ y }^2 \frac{\sigma 2 p _z^1}{\text { HOMO }}$
$O _2-\sigma 1 s ^2 \sigma^* 1 s ^2 \sigma 2 s ^2 \sigma^* 2 s ^2 \sigma 2 p _z^2$
$\pi 2 p _{ x }^2=\pi 2 p _{ y }^2$
$\pi^* 2 p _{ x }^1=\pi^* 2 p _{ y }^1( HOMO )$
$O _2^{+}-\sigma 1 s ^2 \sigma * 1 s ^2 \sigma 2 s ^2 \sigma * 2 s ^2 \sigma 2 p _z^2 \pi 2 p _{ x }^2=\pi 2 p _y^2$
$\pi^* 2 p _{ x }^1=\pi^* 2 p _{ y }^0( HOMO )$
$N _2 \Rightarrow 0 \text { unpaired } e ^{-} \text {in HOMO }$
$N _2^{+} \Rightarrow 1 \text { unpaired } e ^{-} \text {in HOMO }$
$O _2 \Rightarrow 2 \text { unpaired } e ^{-} \text {in HOMO }$
$O _2^{+} \Rightarrow 1 \text { unpaired } e ^{-} \text {in HOMO }$
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