What is the number of unpaired electron(s) in | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$N _2$

$\sigma 1 s ^2 \sigma^* 1 s ^2 \sigma 2 s ^2 \sigma^* 2 s ^2 \pi 2 p _{ x }^2=\pi 2 p _y^2 \frac{\sigma 2 p _z^2}{	ext { HOMO }}$

$N _2^

Vedclass · Accepted Answer

$0,1,2,1$

Vedclass · Answer

$N _2$

$\sigma 1 s ^2 \sigma^* 1 s ^2 \sigma 2 s ^2 \sigma^* 2 s ^2 \pi 2 p _{ x }^2=\pi 2 p _y^2 \frac{\sigma 2 p _z^2}{	ext { HOMO }}$

$N _2^{+}-\sigma 1 s ^2 \sigma^* 1 s ^2 \sigma^* 2 s ^2 \sigma^* 2 s ^2 \pi 2 p _{ x }^2=\pi 2 p _{ y }^2 \frac{\sigma 2 p _z^1}{	ext { HOMO }}$

$O _2-\sigma 1 s ^2 \sigma^* 1 s ^2 \sigma 2 s ^2 \sigma^* 2 s ^2 \sigma 2 p _z^2$

$\pi 2 p _{ x }^2=\pi 2 p _{ y }^2$

$\pi^* 2 p _{ x }^1=\pi^* 2 p _{ y }^1( HOMO )$

$O _2^{+}-\sigma 1 s ^2 \sigma * 1 s ^2 \sigma 2 s ^2 \sigma * 2 s ^2 \sigma 2 p _z^2 \pi 2 p _{ x }^2=\pi 2 p _y^2$

$\pi^* 2 p _{ x }^1=\pi^* 2 p _{ y }^0( HOMO )$

$N _2 \Rightarrow 0 	ext { unpaired } e ^{-} 	ext {in HOMO }$

$N _2^{+} \Rightarrow 1 	ext { unpaired } e ^{-} 	ext {in HOMO }$

$O _2 \Rightarrow 2 	ext { unpaired } e ^{-} 	ext {in HOMO }$

$O _2^{+} \Rightarrow 1 	ext { unpaired } e ^{-} 	ext {in HOMO }$

What is the number of unpaired electron(s) in the highest occupied molecular orbital of the following species : $N _2: N _2^{+} ; O _2 ; O _2^{+}$?

Similar Questions

According to $MO$ theory which of the lists ranks the nitrogen species in terms of increasing bond order?

Number of paramagnetic oxides among the following given oxides is ..... .

$\mathrm{Li}_{2} \mathrm{O}, \mathrm{CaO}, \mathrm{Na}_{2} \mathrm{O}_{2}, \mathrm{KO}_{2}, \mathrm{MgO}$ and $\mathrm{K}_{2} \mathrm{O}$

The number of species from the following which are paramagnetic and with bond order equal to one is___________

$\mathrm{H}_2, \mathrm{He}_2^{+}, \mathrm{O}_2^{+}, \mathrm{N}_2^{2-}, \mathrm{O}_2^{2-}, \mathrm{F}_2, \mathrm{Ne}_2^{+}, \mathrm{B}_2$

In which of the following processes, the bond order has increased and paramagnetic character has changed to diamagnetic?

The paramagnetic nature of oxygen can be explained by