Consider the following compounds. (I) NO [ BF 4 ] (II) Xe [ PtF 6 ] (III) O 2 [ BF 4 ] (IV) KO 2 (V) KO 3 (VI) BaO 2

Question

Consider the following compounds. (I) NO [ BF 4 ​ ] (II) Xe [ PtF 6 ​ ] (III) O 2 ​ [ BF 4 ​ ] (IV) KO 2 ​ (V) KO 3 ​ (VI) BaO 2 ​

TestHub · Accepted Answer

The question asks to identify compounds with paramagnetic cations. Paramagnetism arises from unpaired electrons.

(I) $NO [B F_{4}]$ : $NO$ forms $N O^{+}$ , which is diamagnetic.

(II) $Xe [Pt F_{6}]$ : Xenon is oxidized to $X e^{+}$ , which is paramagnetic.

(III) $O_{2} [B F_{4}]$ : $O_{2}$ forms $O_{2}^{+}$ , which is paramagnetic.

(IV) $K O_{2}$ : Contains the superoxide anion, $O_{2}^{-}$ , which is paramagnetic, but the cation $K^{+}$ is diamagnetic.

(V) $K O_{3}$ : Contains the ozonide anion, $O_{3}^{-}$ , which is paramagnetic, but the cation $K^{+}$ is diamagnetic.

(VI) $Ba O_{2}$ : Contains the peroxide anion, $O_{2}^{2 -}$ , which is diamagnetic, and the cation $B a^{2 +}$ is diamagnetic.

Therefore, compounds (II) and (III) contain paramagnetic cations. Option C is the correct answer.

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