Explanation:

${\mathrm{O}}_2^{2-}$ (Number of electrons $=18$ does not contain unpaired electrons)
${\mathrm{O}}_2^{2-}$
$=\sigma 1{\mathrm{s}}^2,\stackrel{\ast }{\sigma }1{\mathrm{s}}^2,\sigma 2{\mathrm{s}}^2,\stackrel{\ast }{\sigma }2{\mathrm{s}}^2,\sigma 2\mathrm{p}{\mathrm{z}}^2,\pi 2{\mathrm{p}}_z^2$
$\approx {\pi 2\mathrm{p}}_{\mathrm{y}}^2,\stackrel{\ast }{\pi }{2\mathrm{p}}_{\mathrm{x}}^2\approx \stackrel{\ast }{\pi }{2\mathrm{p}}_{\mathrm{y}}^2$

Explanation:

According to molecular orbital theory, Number of unpaired electrons in ${\mathrm{O}}_2$ molecule is two, whereas in ${\mathrm{O}}_2^{+}$ molecule is one, so paramagnetic character decreases in the conversion of ${\mathrm{O}}_2\mathrm{t}\mathrm{o}{\mathrm{O}}_2^{+}$.
$\mathrm{B}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{o}\mathrm{r}\mathrm{d}\mathrm{e}\mathrm{r}=\left(\frac{{\mathrm{N}}_{\mathrm{b}}-{\mathrm{N}}_{\mathrm{a}}}{2}\right)$
$\therefore \mathrm{B}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{o}\mathrm{r}\mathrm{d}\mathrm{e}\mathrm{r}\mathrm{i}\mathrm{n}$ ${\mathrm{O}}_2=\left(\frac{8-4}{2}\right)=2$
Bond order in ${\mathrm{O}}_2^{+}=\left(\frac{8-3}{2}\right)=2.5$
$\therefore$ Bond order increases in the conversion of ${\mathrm{O}}_2\mathrm{t}\mathrm{o}{\mathrm{O}}_2^{+}$.

Explanation:

The bond order of ${\mathrm{B}}_2$ molecule is 1. The ground state electron configuration of B is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^1$. So, ${\mathrm{B}}_2$ molecule has a total of ten electrons, which are arranged in MOs.
${\left(\sigma 1\mathrm{s}\right)}^2{\left({\sigma }^{\ast }1\mathrm{s}\right)}^2{\left({\sigma }^{\ast }2\mathrm{s}\right)}^2{\left({\sigma }^{\ast }2\mathrm{s}\right)}^2\left({\pi }_{{2\mathrm{p}}_{\mathrm{y}}^1}={\pi }_{{2\mathrm{p}}_{\mathrm{y}}^1}\right)$
Bond order $=\frac{1}{2}$ (Number of bonding electrons - number of anti-bonding electrons)
$=\frac{1}{2}\left(6-4\right)=1$

Explanation:

${\mathrm{O}}_2^{2-}$ (Number of electrons $=18$ does not contain unpaired electrons)
${\mathrm{O}}_2^{2-}$
$=\sigma 1{\mathrm{s}}^2,\stackrel{\ast }{\sigma }1{\mathrm{s}}^2,\sigma 2{\mathrm{s}}^2,\stackrel{\ast }{\sigma }2{\mathrm{s}}^2,\sigma 2\mathrm{p}{\mathrm{z}}^2,\pi 2{\mathrm{p}}_z^2$
$\approx {\pi 2\mathrm{p}}_{\mathrm{y}}^2,\stackrel{\ast }{\pi }{2\mathrm{p}}_{\mathrm{x}}^2\approx \stackrel{\ast }{\pi }{2\mathrm{p}}_{\mathrm{y}}^2$

Explanation:

Explanation:

According to molecular orbital theory, Number of unpaired electrons in ${\mathrm{O}}_2$ molecule is two, whereas in ${\mathrm{O}}_2^{+}$ molecule is one, so paramagnetic character decreases in the conversion of ${\mathrm{O}}_2\mathrm{t}\mathrm{o}{\mathrm{O}}_2^{+}$.
$\mathrm{B}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{o}\mathrm{r}\mathrm{d}\mathrm{e}\mathrm{r}=\left(\frac{{\mathrm{N}}_{\mathrm{b}}-{\mathrm{N}}_{\mathrm{a}}}{2}\right)$
$\therefore \mathrm{B}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{o}\mathrm{r}\mathrm{d}\mathrm{e}\mathrm{r}\mathrm{i}\mathrm{n}$ ${\mathrm{O}}_2=\left(\frac{8-4}{2}\right)=2$
Bond order in ${\mathrm{O}}_2^{+}=\left(\frac{8-3}{2}\right)=2.5$
$\therefore$ Bond order increases in the conversion of ${\mathrm{O}}_2\mathrm{t}\mathrm{o}{\mathrm{O}}_2^{+}$.

Explanation:

The bond order of ${\mathrm{B}}_2$ molecule is 1. The ground state electron configuration of B is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^1$. So, ${\mathrm{B}}_2$ molecule has a total of ten electrons, which are arranged in MOs.
${\left(\sigma 1\mathrm{s}\right)}^2{\left({\sigma }^{\ast }1\mathrm{s}\right)}^2{\left({\sigma }^{\ast }2\mathrm{s}\right)}^2{\left({\sigma }^{\ast }2\mathrm{s}\right)}^2\left({\pi }_{{2\mathrm{p}}_{\mathrm{y}}^1}={\pi }_{{2\mathrm{p}}_{\mathrm{y}}^1}\right)$
Bond order $=\frac{1}{2}$ (Number of bonding electrons - number of anti-bonding electrons)
$=\frac{1}{2}\left(6-4\right)=1$

Explanation:

${\mathrm{O}}_2^{2-}$ (Number of electrons $=18$ does not contain unpaired electrons)
${\mathrm{O}}_2^{2-}$
$=\sigma 1{\mathrm{s}}^2,\stackrel{\ast }{\sigma }1{\mathrm{s}}^2,\sigma 2{\mathrm{s}}^2,\stackrel{\ast }{\sigma }2{\mathrm{s}}^2,\sigma 2\mathrm{p}{\mathrm{z}}^2,\pi 2{\mathrm{p}}_z^2$
$\approx {\pi 2\mathrm{p}}_{\mathrm{y}}^2,\stackrel{\ast }{\pi }{2\mathrm{p}}_{\mathrm{x}}^2\approx \stackrel{\ast }{\pi }{2\mathrm{p}}_{\mathrm{y}}^2$

Explanation:

Explanation:

According to molecular orbital theory, Number of unpaired electrons in ${\mathrm{O}}_2$ molecule is two, whereas in ${\mathrm{O}}_2^{+}$ molecule is one, so paramagnetic character decreases in the conversion of ${\mathrm{O}}_2\mathrm{t}\mathrm{o}{\mathrm{O}}_2^{+}$.
$\mathrm{B}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{o}\mathrm{r}\mathrm{d}\mathrm{e}\mathrm{r}=\left(\frac{{\mathrm{N}}_{\mathrm{b}}-{\mathrm{N}}_{\mathrm{a}}}{2}\right)$
$\therefore \mathrm{B}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{o}\mathrm{r}\mathrm{d}\mathrm{e}\mathrm{r}\mathrm{i}\mathrm{n}$ ${\mathrm{O}}_2=\left(\frac{8-4}{2}\right)=2$
Bond order in ${\mathrm{O}}_2^{+}=\left(\frac{8-3}{2}\right)=2.5$
$\therefore$ Bond order increases in the conversion of ${\mathrm{O}}_2\mathrm{t}\mathrm{o}{\mathrm{O}}_2^{+}$.

Explanation:

The bond order of ${\mathrm{B}}_2$ molecule is 1. The ground state electron configuration of B is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^1$. So, ${\mathrm{B}}_2$ molecule has a total of ten electrons, which are arranged in MOs.
${\left(\sigma 1\mathrm{s}\right)}^2{\left({\sigma }^{\ast }1\mathrm{s}\right)}^2{\left({\sigma }^{\ast }2\mathrm{s}\right)}^2{\left({\sigma }^{\ast }2\mathrm{s}\right)}^2\left({\pi }_{{2\mathrm{p}}_{\mathrm{y}}^1}={\pi }_{{2\mathrm{p}}_{\mathrm{y}}^1}\right)$
Bond order $=\frac{1}{2}$ (Number of bonding electrons - number of anti-bonding electrons)
$=\frac{1}{2}\left(6-4\right)=1$

Explanation:

${\mathrm{O}}_2^{2-}$ (Number of electrons $=18$ does not contain unpaired electrons)
${\mathrm{O}}_2^{2-}$
$=\sigma 1{\mathrm{s}}^2,\stackrel{\ast }{\sigma }1{\mathrm{s}}^2,\sigma 2{\mathrm{s}}^2,\stackrel{\ast }{\sigma }2{\mathrm{s}}^2,\sigma 2\mathrm{p}{\mathrm{z}}^2,\pi 2{\mathrm{p}}_z^2$
$\approx {\pi 2\mathrm{p}}_{\mathrm{y}}^2,\stackrel{\ast }{\pi }{2\mathrm{p}}_{\mathrm{x}}^2\approx \stackrel{\ast }{\pi }{2\mathrm{p}}_{\mathrm{y}}^2$

Explanation:

Explanation:

According to molecular orbital theory, Number of unpaired electrons in ${\mathrm{O}}_2$ molecule is two, whereas in ${\mathrm{O}}_2^{+}$ molecule is one, so paramagnetic character decreases in the conversion of ${\mathrm{O}}_2\mathrm{t}\mathrm{o}{\mathrm{O}}_2^{+}$.
$\mathrm{B}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{o}\mathrm{r}\mathrm{d}\mathrm{e}\mathrm{r}=\left(\frac{{\mathrm{N}}_{\mathrm{b}}-{\mathrm{N}}_{\mathrm{a}}}{2}\right)$
$\therefore \mathrm{B}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{o}\mathrm{r}\mathrm{d}\mathrm{e}\mathrm{r}\mathrm{i}\mathrm{n}$ ${\mathrm{O}}_2=\left(\frac{8-4}{2}\right)=2$
Bond order in ${\mathrm{O}}_2^{+}=\left(\frac{8-3}{2}\right)=2.5$
$\therefore$ Bond order increases in the conversion of ${\mathrm{O}}_2\mathrm{t}\mathrm{o}{\mathrm{O}}_2^{+}$.

Explanation:

The bond order of ${\mathrm{B}}_2$ molecule is 1. The ground state electron configuration of B is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^1$. So, ${\mathrm{B}}_2$ molecule has a total of ten electrons, which are arranged in MOs.
${\left(\sigma 1\mathrm{s}\right)}^2{\left({\sigma }^{\ast }1\mathrm{s}\right)}^2{\left({\sigma }^{\ast }2\mathrm{s}\right)}^2{\left({\sigma }^{\ast }2\mathrm{s}\right)}^2\left({\pi }_{{2\mathrm{p}}_{\mathrm{y}}^1}={\pi }_{{2\mathrm{p}}_{\mathrm{y}}^1}\right)$
Bond order $=\frac{1}{2}$ (Number of bonding electrons - number of anti-bonding electrons)
$=\frac{1}{2}\left(6-4\right)=1$