Question

Among, $\text{K}{{\text{O}}_{2}}$, $\text{AlO}_{2}^{-}$, $\text{Ba}{{\text{O}}_{2}}$ and $\text{NO}_{2}^{+}$ unpaired electron is present in ___________.
A. $\text{NO}_{2}^{+}\text{ and Ba}{{\text{O}}_{2}}$
B. $\text{K}{{\text{O}}_{2}}\text{ and AlO}_{2}^{-}$
C. $\text{K}{{\text{O}}_{2}}\text{ only}$
D. $\text{Ba}{{\text{O}}_{2}}\text{ only}$

Answer 1

Hint: Find the total number of valence electrons in the molecules and then write configuration of these molecules. If the electrons are not coupled, the electrons are said to be unpaired. The configuration will be written according to molecular orbital theory.

Complete step by step answer:
(1) $\text{K}{{\text{O}}_{2}}$: 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 valence electrons are 6+6+1 are 13 electrons. The configuration of will be ${{\sigma }_{1\text{s}}}^{2}\sigma {{_{\text{1s}}^{*}}^{2}}{{\sigma }_{2\text{s}}}^{2}\sigma {{_{2\text{s}}^{*}}^{2}}{{\pi }_{2\text{px}}}^{2}{{\pi }_{2\text{py}}}^{2}\pi {{_{2\text{px}}^{*}}^{1}}$. The $\pi _{2\text{px}}^{*}$ orbital has one unpaired electron.

(2) $\text{Ba}{{\text{O}}_{2}}$: Barium has its configuration as $\left[ \text{Xe} \right]6{{\text{s}}^{2}}$. The valence electrons are 2. Oxygen has its configuration as $2,6$. The valence electrons will be 6. The total valence electrons are 6+6+2 are 14 electrons. The configuration of will be ${{\sigma }_{1\text{s}}}^{2}\sigma {{_{\text{1s}}^{*}}^{2}}{{\sigma }_{2\text{s}}}^{2}\sigma {{_{2\text{s}}^{*}}^{2}}{{\pi }_{2\text{px}}}^{2}{{\pi }_{2\text{py}}}^{2}\pi {{_{2\text{px}}^{*}}^{2}}$. All the orbitals are completely filled with electrons.

(3) $\text{AlO}_{2}^{-}$ : Aluminium has its configuration as $2,8,3$. The valence electrons are 3. Oxygen has its configuration as $2,6$. The valence electrons will be 6. The total valence electrons are 6+6+3+1 are 16 electrons. The configuration of will be ${{\sigma }_{1\text{s}}}^{2}\sigma {{_{\text{1s}}^{*}}^{2}}{{\sigma }_{2\text{s}}}^{2}\sigma {{_{2\text{s}}^{*}}^{2}}{{\sigma }_{2\text{pz}}}^{2}{{\pi }_{2\text{px}}}^{2}{{\pi }_{2\text{py}}}^{2}\pi {{_{2\text{px}}^{*}}^{2}}$. All the orbitals are completely filled with electrons.

(4) $\text{NO}_{2}^{+}$ : Nitrogen has its configuration as $2,5$. The valence electrons are 5. Oxygen has its configuration as $2,6$. The valence electrons will be 6. The total valence electrons are 5+6+6-1 are 16 electrons. The configuration of will be ${{\sigma }_{1\text{s}}}^{2}\sigma {{_{\text{1s}}^{*}}^{2}}{{\sigma }_{2\text{s}}}^{2}\sigma {{_{2\text{s}}^{*}}^{2}}{{\sigma }_{2\text{pz}}}^{2}{{\pi }_{2\text{px}}}^{2}{{\pi }_{2\text{py}}}^{2}\pi {{_{2\text{px}}^{*}}^{2}}$. All the orbitals are completely filled with electrons.
The correct answer of this question is option ‘c’. Among, $\text{K}{{\text{O}}_{2}}$, $\text{AlO}_{2}^{-}$, $\text{Ba}{{\text{O}}_{2}}$ and $\text{NO}_{2}^{+}$ unpaired electron is present in $\text{K}{{\text{O}}_{2}}$.
So, the correct answer is “Option C”.

Note: The electronic configuration for elements having electrons greater than 14 electrons is different with the configuration of elements having electrons less than equal to 14. Let us see the electronic configuration:
(1) Greater than 14 electrons: ${{\sigma }_{\text{1s}}}{{\sigma }^{*}}_{1\text{s}}{{\sigma }_{2\text{s}}}{{\sigma }^{*}}_{2\text{s}}{{\sigma }_{2\text{pz}}}{{\pi }_{2\text{px}}}{{\pi }_{2\text{py}}}{{\pi }^{*}}_{2\text{px}}{{\pi }^{*}}_{2\text{py}}$
(2) Electrons less than equal to 14 electrons: ${{\sigma }_{\text{1s}}}{{\sigma }^{*}}_{1\text{s}}{{\sigma }_{2\text{s}}}{{\sigma }^{*}}_{2\text{s}}{{\pi }_{2\text{px}}}{{\pi }_{2\text{py}}}{{\pi }^{*}}_{2\text{px}}{{\sigma }_{2\text{pz}}}{{\pi }^{*}}_{2\text{py}}\sigma _{2\text{pz}}^{*}$