Concept: Paramagnetism arises from the presence of unpaired electrons in a species. To determine if a species is paramagnetic, we need to analyze its electronic configuration or the nature of its bonding and the presence of odd electrons or unpaired electrons in molecular orbitals.

Why (A) is correct:

In ${\mathrm{KO}}_2$, potassium (K) is an alkali metal and forms ${\mathrm{K}}^{+}$ ion. Therefore, the species responsible for paramagnetism would be the superoxide ion, ${\mathrm{O}}_2^{-}$.

Let's analyze the molecular orbital configuration of ${\mathrm{O}}_2^{-}$:

• Total number of electrons in ${\mathrm{O}}_2$ is $8+8=16$.
• For ${\mathrm{O}}_2^{-}$, there is one additional electron, so the total number of electrons is $16+1=17$.
• The molecular orbital configuration for ${\mathrm{O}}_2^{-}$ is: ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\sigma }_{2p_z}^2{\pi }_{2p_x}^2{\pi }_{2p_y}^2{\pi }_{2p_x}^{\ast 2}{\pi }_{2p_y}^{\ast 1}$.

As seen from the configuration, there is one unpaired electron in the ${\pi }_{2p_y}^{\ast }$ antibonding molecular orbital. Therefore, ${\mathrm{O}}_2^{-}$ (and thus ${\mathrm{KO}}_2$) is paramagnetic.

Option Analysis:

• A) ${\mathrm{KO}}_2$: Contains the superoxide ion ${\mathrm{O}}_2^{-}$, which has one unpaired electron, making it paramagnetic.
• B) ${\mathrm{SiO}}_2$: Silicon dioxide is a covalent network solid. All electrons are paired in covalent bonds. It is diamagnetic.
• C) ${\mathrm{TiO}}_2$: Titanium dioxide. Titanium is in the +4 oxidation state (Ti has electronic configuration $[Ar]3d^{2}4s^2$, so ${\mathrm{Ti}}^{4+}$ has $[Ar]$ configuration, i.e., no unpaired electrons). Oxygen is in the -2 oxidation state. All electrons are paired. It is diamagnetic.
• D) ${\mathrm{BaO}}_2$: Barium peroxide. Barium is in the +2 oxidation state (${\mathrm{Ba}}^{2+}$). The peroxide ion is ${\mathrm{O}}_2^{2-}$. The total number of electrons in ${\mathrm{O}}_2^{2-}$ is $16+2=18$. Its molecular orbital configuration is ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\sigma }_{2p_z}^2{\pi }_{2p_x}^2{\pi }_{2p_y}^2{\pi }_{2p_x}^{\ast 2}{\pi }_{2p_y}^{\ast 2}$. All electrons are paired. It is diamagnetic.

Correct Answer: (A)