Answer:
Bond order is
defined as half of the difference between the number of electrons present in
the bonding and antibonding orbitals, i.e.,
Bond order = \[\frac{{{N}_{b}}-{{N}_{a}}}{2}\]
\[{{N}_{2}}:KK\sigma {{(2s)}^{2}}\overset{*}{\mathop{\sigma
}}\,{{(2s)}^{2}}\pi {{(2{{p}_{x}})}^{2}}\pi {{(2{{p}_{y}})}^{2}}\pi
{{(2{{p}_{z}})}^{2}}\]
Bond order = \[\frac{8-2}{2}=2\]
\[{{O}_{2}}:KK\sigma {{(2s)}^{2}}\overset{*}{\mathop{\sigma
}}\,{{(2s)}^{2}}\sigma {{(2{{p}_{z}})}^{2}}\pi {{(2{{p}_{x}})}^{2}}\pi
{{(2{{p}_{y}})}^{2}}\]
\[\overset{*}{\mathop{\pi
}}\,{{(2{{p}_{x}})}^{1}}\overset{*}{\mathop{\pi }}\,{{(2{{p}_{y}})}^{1}}\]
Bond order = \[\frac{8-4}{2}=2\]
\[O_{2}^{+}:KK\sigma
{{(2s)}^{2}}\overset{*}{\mathop{\sigma }}\,{{(2s)}^{2}}\sigma
{{(2{{p}_{z}})}^{2}}\]
\[\pi {{(2{{p}_{x}})}^{2}}\pi
{{(2{{p}_{y}})}^{2}}\overset{*}{\mathop{\pi }}\,{{(2{{p}_{x}})}^{1}}\]
Bond order \[=\frac{8-3}{2}=2.5\]
\[O_{2}^{-}:KK\sigma {{(2s)}^{2}}\overset{*}{\mathop{\sigma
}}\,{{(2s)}^{2}}\sigma {{(2{{p}_{z}})}^{2}}\pi {{(2{{p}_{x}})}^{2}}\pi
{{(2{{p}_{y}})}^{2}}\] \[\overset{*}{\mathop{\pi
}}\,{{(2{{p}_{x}})}^{1}}\overset{*}{\mathop{\pi }}\,{{(2{{p}_{y}})}^{1}}\]
Bond
order \[=\frac{8-5}{2}=1.5\]
You need to login to perform this action.
You will be redirected in
3 sec