Total number of electron present in left(pi^* | vedclass

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

Question

$ \mathrm{O}_2(16 \mathrm{e}):\left(\sigma_{1 \mathrm{~s}}\right)^2\left(\sigma_{1 \mathrm{~s}}^*\right)^2\left(\sigma_{2 \mathrm{~s}}\right)^2\left(\

Vedclass · Accepted Answer

$6$

Vedclass · Answer

$ \mathrm{O}_2(16 \mathrm{e}):\left(\sigma_{1 \mathrm{~s}}\right)^2\left(\sigma_{1 \mathrm{~s}}^*\right)^2\left(\sigma_{2 \mathrm{~s}}\right)^2\left(\sigma_{2 \mathrm{~s}}^*\right)^2 $

$ \left(\sigma_{2 \mathrm{p}}\right)^2\left[\left(\pi_{2 \mathrm{p}}\right)^2=\left(\pi_{2 \mathrm{p}}\right)^2\right],\left[\left(\pi_{2 \mathrm{p}}^*\right)^1=\left(\pi_{2 \mathrm{p}}^*\right)^1\right]$

Number of $\mathrm{e}^{-}$present in $\left(\pi^*\right)$ of $\mathrm{O}_2=2$

Number of $\mathrm{e}^{-}$present in $\left(\pi^*\right)$ of $\mathrm{O}_2^{+}=1$

Number of $\mathrm{e}^{-}$present in $\left(\pi^*\right)$ of $\mathrm{O}_2^{-}=3$

So total $\mathrm{e}^{-}$in $\left(\pi^*\right)=2+1+3=6$

Total number of electron present in $\left(\pi^*\right)$ molecular orbitals of $\mathrm{O}_2, \mathrm{O}_2^{+}$and $\mathrm{O}_2^{-}$is ............

Similar Questions

Among the following, the molecule expected to be stabilized by anion formation is

According to molecular orbital theory, the number of unpaired electron(s) in $\mathrm{O}_{2}^{2-}$ is :

"Isoelectronic molecules and ions have identical bond orders." Explain by examples.

According to molecular orbital theory, the paramagnetism of ${O_2}$ molecule is due to presence of

${N_2}$ and ${O_2}$ are converted into monoanions $N_2^ - $ and $O_2^ - $ respectively, which of the following statements is wrong