The pair of species that has the same bond or | vedclass

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Question

$\sigma 1 \mathrm{s}^{2}, \sigma^{*} 1 \mathrm{s}^{2}, \sigma 2 \mathrm{s}^{2}, \sigma^{*} 2 \mathrm{s}^{2}, \sigma 2 \mathrm{p}_{2}^{2}, \pi 2 \mathr

Vedclass · Accepted Answer

$CO, NO^+$

Vedclass · Answer

$\sigma 1 \mathrm{s}^{2}, \sigma^{*} 1 \mathrm{s}^{2}, \sigma 2 \mathrm{s}^{2}, \sigma^{*} 2 \mathrm{s}^{2}, \sigma 2 \mathrm{p}_{2}^{2}, \pi 2 \mathrm{p}_{\mathrm{x}}^{2} \approx \pi 2 \mathrm{p}_{\mathrm{x}}^{2}, \pi^{*} 2 \mathrm{p}_{\mathrm{x}}^{2} \approx \mathrm{\pi}^{*} 2 \mathrm{p}_{\mathrm{y}}^{2}$

Bond order $=\frac{N_{b}-N_{a}}{2}$ (where $N_{b}=$ number of electrons in bonding molecular orbital

$\mathrm{N}_{\mathrm{a}}=$ number of electrons in anti $-$ bonding molecular orbital)

$O_{2}^{2-}(8+8+2=18)$

$\mathrm{BO}=\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}=\frac{10-8}{2}=1$

$\mathrm{B}_{2}(5+5=10)=\sigma 1 \mathrm{s}^{2}, \sigma^{*} 1 \mathrm{s}^{2}, \sigma 2 \mathrm{s}^{2}, \sigma^{*} 2 \mathrm{s}^{2}, \pi 2 \mathrm{p}_{\mathrm{x}}^{1} \approx \pi 2 \mathrm{p}_{\mathrm{y}}^{1}$

$B O=\frac{6-4}{2}$

similarly, order of;

$O_{2}^{+}=2.5, \mathrm{NO}^{+}=3$

$\mathrm{NO}=2.5, \mathrm{CO}=3$

$\mathrm{N}_{2}=3$ and $\mathrm{O}_{2}=2$

$\sigma 1 \mathrm{s}^{2}, \sigma^{*} 1 \mathrm{s}^{2}, \sigma 2 \mathrm{s}^{2}, \sigma^{*} 2 \mathrm{s}^{2}, \sigma 2 \mathrm{p}_{2}^{2}, \pi 2 \mathrm{p}_{\mathrm{x}}^{2} \approx \pi 2 \mathrm{p}_{\mathrm{y}}^{2}, \pi^{*} 2 \mathrm{p}_{\mathrm{x}}^{2} \approx \pi^{*} 2 \mathrm{p}_{\mathrm{y}}^{2}$

Bond order $=\frac{N_{\mathrm{b}}-\mathrm{N}_{a}}{2}$

(where $N_{b}=$ number of electrons in bonding molecular orbital

$\mathrm{N}_{a}=$ number of electrons in anti - bonding molecular orbital $O_{2}^{2-}(8+8+2=18)$

$\mathrm{BO}=\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}=\frac{10-\mathrm{g}}{2}=1$

$\mathrm{B}_{2}(5+5=10)=\sigma 1 \mathrm{s}^{2}, \sigma^{*} 1 \mathrm{s}^{2}, \sigma 2 \mathrm{s}^{2}, \sigma^{*} 2 \mathrm{s}^{2}, \pi 2 \mathrm{p}_{\mathrm{x}}^{1} \approx \pi 2 \mathrm{p}_{\mathrm{y}}^{1}$

$B O=\frac{6-4}{2}$

similarty, order of:

$O_{2}^{+}=2.5, \mathrm{NO}^{+}=3$

$\mathrm{NO}=2.5, \mathrm{CO}=3$

$\mathrm{N}_{2}=3$ and $\mathrm{O}_{2}=2$

The pair of species that has the same bond order in the following is

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