Calculate the bond order of ${{\rm{N}}_2},{{\rm{O}}_2}{\rm{,O}}_2^ + $ and ${\rm{O}}_2^ - $
Calculate the bond order of ${{\rm{N}}_2},{{\rm{O}}_2}{\rm{,O}}_2^ + $ and ${\rm{O}}_2^ - $
In $\mathrm{N}_{2} \mathrm{Z}=7$, so total electron $=14$
Electron configuration of in $MO$ for $\mathrm{N}_{2}:(\sigma 1 s)^{2}\left(\sigma^{*} 1 s\right)^{2}(\sigma 2 s)^{2}\left(\sigma^{*} 2 s\right)^{2}\left(\pi 2 p_{x}\right)^{2}\left(\pi 2 p_{y}\right)^{2}\left(\sigma 2 p_{z}\right)^{2}$ $\mathrm{BO}=\frac{1}{2}\left(\mathrm{~N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}\right)=\frac{1}{2}(10-4)=3$ (Triple bond)
Calculate the bond order of $\mathrm{O}_{2}$
In $\mathrm{O}_{2}, \mathrm{Z}=8 \mathrm{So}$, Total electron $=16$
Electron configuration in $MO$ for $\mathrm{O}_{2}:(\sigma 1 s)^{2}\left(\sigma^{*} 1 s\right)^{2}(\sigma 2 s)^{2}\left(\sigma^{*} 2 s\right)^{2}\left(\sigma 2 p_{z}\right)^{2}\left(\pi 2 p_{x}\right)^{2}\left(\pi 2 p_{y}\right)^{2}$
$\left(\pi^{*} 2 p_{x}\right)^{1}$ $\left(\pi^{*} 2 p_{y}\right)^{1}$
$\mathrm{BO}=\frac{1}{2}\left(\mathrm{~N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}\right)=\frac{1}{2}(10-6)=($ Double bond $)$
Total electron in $\mathrm{O}_{2}=16$ and total electron in
$\mathrm{O}_{2}^{+}=15$
Electron configuration in $MO$ for $\mathrm{O}_{2}:\left(\sigma_{1 s}\right)^{2}\left(\sigma_{1 s}^{*}\right)^{2}\left(\sigma_{2 s}\right)^{2}\left(\sigma_{2 s}^{*}\right)^{2}\left(\sigma_{2 p_{z}}\right)^{2}\left(\pi_{2 p_{x}}\right)^{2}\left(\pi_{2 p_{y}}\right)^{2}\left(\pi_{2 p_{x}}^{*}\right)^{1}$
$\left(\pi_{2 p_{y}}^{*}\right)^{0}$ BO $=\frac{1}{2}\left(N_{b}-N_{a}\right)=\frac{1}{2}(10-5)=2.5$
Calculate the bond order of $\mathrm{O}_{2}^{-}$
Total electron in $\mathrm{O}_{2}^{-}=16+1=17$
Electron configuration in $MO$ for $\mathrm{O}_{2}^{-}:\left(\sigma_{1 s}\right)^{2}\left(\sigma_{1 s}^{*}\right)^{2}\left(\sigma_{2 s}\right)^{2}\left(\sigma_{2 s}^{*}\right)^{2}\left(\sigma_{2 p_{z}}\right)^{2}\left(\pi_{2 p_{x}}\right)^{2}\left(\pi_{2 p_{y}}\right)^{2}\left(\pi_{2 p_{x}}^{*}\right)^{2}$
$\left(\pi_{2 p_{y}}^{*}\right)^{1}$
$\begin{aligned} \mathrm{BO} =\frac{1}{2}\left(\mathrm{~N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}\right) \\ =\frac{1}{2}(10-7)=\frac{3}{2}=1.5 \end{aligned}$
Similar Questions
Choose correct order
Choose correct order
Which of the following statement is correct for peroxide ion
$A.$ has completely filled antibonding molecular orbitals
$B.$ is diamagnetic
$C.$ has bond order one
$D.$ is isoelectronic with Neon
Which of the following statement is correct for peroxide ion
$A.$ has completely filled antibonding molecular orbitals
$B.$ is diamagnetic
$C.$ has bond order one
$D.$ is isoelectronic with Neon
In which of the following ionization processes, the bond order has increased and the magnetic behaviour has changed ?
In which of the following ionization processes, the bond order has increased and the magnetic behaviour has changed ?
The correct order of the $O-O$ bond length in $O_2$ , $H_2O_2$ and $O_3$ is
The correct order of the $O-O$ bond length in $O_2$ , $H_2O_2$ and $O_3$ is
Match $List-I$ with $List-II$.
$List-I$
$List-II$
$(A)$ $\Psi_{ MO }=\Psi_{ A }-\Psi_{ B }$
$(I)$ Dipole moment
$(B)$ $\mu=Q \times I$
$(II)$ Bonding molecular orbital
$(C)$ $\frac{N_{b}-N_{a}}{2}$
$(III)$ Anti-bonding molecualr orbital
$(D)$ $\Psi_{ MO }=\Psi_{ A }+\Psi_{ B }$
$(IV)$ Bond order
Match $List-I$ with $List-II$.
| $List-I$ | $List-II$ |
| $(A)$ $\Psi_{ MO }=\Psi_{ A }-\Psi_{ B }$ | $(I)$ Dipole moment |
| $(B)$ $\mu=Q \times I$ | $(II)$ Bonding molecular orbital |
| $(C)$ $\frac{N_{b}-N_{a}}{2}$ | $(III)$ Anti-bonding molecualr orbital |
| $(D)$ $\Psi_{ MO }=\Psi_{ A }+\Psi_{ B }$ | $(IV)$ Bond order |
- [JEE MAIN 2022]