Question

Consider following species:Cn+, Cn-, NO,CN...... Which have highest bond order??? ​

Answer 1

Answer:

I think cn have to highest bond

Answer 2

Answer:

CN- has highest bond order.

Explanation:

According to the molecular orbital theory, the electronic configurations of the given elements are as follows:

Bond order of NO:

Electronic configuration of “NO”$=(\sigma 1 s)^{2},\left(\sigma^{*} 1 s\right)^{2},(\sigma 2 s)^{2},\left(\sigma^{*} 2 s\right)^{2},\left(\sigma 2 p_{z}\right)^{2},\left(\pi 2 p_{x}\right)^{2}$

$=\left(\pi^{*} 2 p_{y}\right)^{2},\left(\pi^{*} 2 p_{x}\right)^{2}=,\left(\pi^{*} 2 p_{y}\right)^{0}$

$\text {Bond order}=\frac{\text {Bonding electrons -Antibonding electrons}}{2}$

$=\frac{10-5}{2}=2.5$

Bond order of CN¬:

Electronic configuration of “CN-”$=(\sigma 1 s)^{2},\left(\sigma^{*} 1 s\right)^{2},(\sigma 2 s)^{2},\left(\sigma^{*} 2 s\right)^{2},\left(\pi 2 p_{x}\right)^{2}$

$=\left(\pi^{*} 2 p_{y}\right)^{2},\left(\sigma 2 p_{z}\right)^{2}$

$\text {Bond order}=\frac{\text {Bonding electrons - Antibonding electrons}}{2}$

$=\frac{10-4}{2}=3$

Bond order of CN:

Electronic configuration of “CN” $= (\sigma 1 s)^{2},\left(\sigma^{*} 1 s\right)^{2},(\sigma 2 s)^{2},\left(\sigma^{*} 2 s\right)^{2},\left(\pi 2 p_{x}\right)^{2}$

$=\left(\pi^{*} 2 p_{y}\right)^{2},\left(\sigma 2 p_{z}\right)^{1}$

$\text {Bond order}=\frac{\text {Bonding electrons -Antibonding electrons}}{2}$

$=\frac{9-4}{2}=2.5$

Bond order of CN+:

Electronic configuration of “CN+” =$(\sigma 1 s)^{2},\left(\sigma^{*} 1 s\right)^{2},(\sigma 2 s)^{2},\left(\sigma^{*} 2 s\right)^{2},\left(\pi 2 p_{x}\right)^{2}=\left(\pi^{*} 2 p_{y}\right)^{2}$

$\text {Bond order}=\frac{\text {Bonding electrons - Antibonding electrons}}{2}$

$=\frac{8-4}{2}=2$