Question

Among ${\mathrm{KO}}_2$, ${{\mathrm{AlO}}_2}^{-}$, ${\mathrm{BaO}}_2$ and ${{\mathrm{NO}}_2}^{+}$ unpaired electron are present in:

• A. ${\mathrm{KO}}_2$
• B. ${\mathrm{BaO}}_2$
• C. ${\mathrm{BaO}}_2$ and ${{\mathrm{NO}}_2}^{+}$
• D. ${\mathrm{KO}}_2$ and ${{\mathrm{AlO}}_2}^{-}$

Answer 1

The correct option is A

${\mathrm{KO}}_2$

The explanation for correct option:

A. ${\mathrm{KO}}_2$

Paramagnetic materials:

• "Paramagnetic properties are due to the presence of some unpaired electrons."
• Potassium has its configuration as $2,8,8,1$. The valence electron is $1$.
• Oxygen has its configuration as $2,6$. The valence electrons will be $6$.
• The total number of valence electrons in ${\mathrm{KO}}_2$ are$13$ electrons.
• The configuration of ${\mathrm{KO}}_2$ will be ${{\sigma }_{1\text{s}}}^2\sigma {{}_{\text{1s}}^{\ast }}^2{{\sigma }_{2\text{s}}}^2\sigma {{}_{2\text{s}}^{\ast }}^2{{\pi }_{2\text{px}}}^2{{\pi }_{2\text{py}}}^2\pi {{}_{2\text{px}}^{\ast }}^1$. ${\pi }_{2\text{px}}^{\ast }$ orbital has one unpaired electron.
• Thus, ${\mathrm{KO}}_2$ is paramagnetic in nature.

The explanation for incorrect options:

B. ${\mathrm{BaO}}_2$

• ${\mathrm{BaO}}_2$ has 14 valence electrons.
• The configuration of will be 2${{\sigma }_{1\text{s}}}^2\sigma {{}_{\text{1s}}^{\ast }}^2{{\sigma }_{2\text{s}}}^2\sigma {{}_{2\text{s}}^{\ast }}^2{{\pi }_{2\text{px}}}^2{{\pi }_{2\text{py}}}^2\pi {{}_{2\text{px}}^{\ast }}^2$. All the orbitals are completely filled with electrons.
• Thus, it is diamagnetic in nature.

C. ${\mathrm{BaO}}_2$ and ${{\mathrm{NO}}_2}^{+}$

• In the case of ${\mathrm{BaO}}_2$, all electrons are completely filled.
• ${{\mathrm{NO}}_2}^{+}$ has 14 valence electrons.
• The configuration of ${{\mathrm{NO}}_2}^{+}$ will be .
• Thus, these both are diamagnetic paramagnetic in nature.

D. ${\mathrm{KO}}_2$ and ${{\mathrm{AlO}}_2}^{-}$

• ${\mathrm{KO}}_2$ is paramagnetic in nature.
• In the case of ${{\mathrm{AlO}}_2}^{-}$, there are total 16 valence electrons.
• The configuration of will be ${{\sigma }_{1\text{s}}}^2\sigma {{}_{\text{1s}}^{\ast }}^2{{\sigma }_{2\text{s}}}^2\sigma {{}_{2\text{s}}^{\ast }}^2{{\sigma }_{2\text{pz}}}^2{{\pi }_{2\text{px}}}^2{{\pi }_{2\text{py}}}^2\pi {{}_{2\text{px}}^{\ast }}^2$.
• Thus, ${{\mathrm{AlO}}_2}^{-}$ is diamagnetic in nature.

Hence, the correct option is A i.e. ${\mathrm{KO}}_2$ has one unpaired electron.