Consider the anions present in Na2O2 and KO2. Select the correct statement(s).

KO2 is paramagnetic since the anion has one unpaired electron.

The O-O bond order is 1.5 in the anion of KO2 and 1 in the anion of Na2O2.

The O-O bond length in the peroxide ion is longer than in the superoxide ion.

KO2 is paramagnetic since the anion has one unpaired electron.

The O-O bond order is 1.5 in the anion of KO2 and 1 in the anion of Na2O2.

Na2O2 is paramagnetic since the peroxide ion has two unpaired electrons.

The O-O bond length in the peroxide ion is longer than in the superoxide ion.

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JEE Advanced levelOne or more correct optionsDifficulty 3 of 3sb144

Consider the anions present in $Na X_{2} O X_{2}$ and $KO X_{2}$ . Select the correct statement(s).

A $KO X_{2}$ is paramagnetic since the anion has one unpaired electron.
BThe O-O bond order is 1.5 in the anion of $KO X_{2}$ and 1 in the anion of $Na X_{2} O X_{2}$ .
C $Na X_{2} O X_{2}$ is paramagnetic since the peroxide ion has two unpaired electrons.
DThe O-O bond length in the peroxide ion is longer than in the superoxide ion.

Answer: A, B, D

O X_{2} X^{-}

has configuration

π^{* 3}

, so one unpaired electron and bond order

(8 - 5) /2 = 1.5

O X_{2} X^{2 -}

has

π^{* 4}

, no unpaired electron and bond order

(8 - 6) /2 = 1

. Lower bond order means a longer bond, so the peroxide bond is longer. Thus the three statements other than the paramagnetic-peroxide one are correct.

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Consider the anions present in NaX2​OX2​ and KOX2​. Select the correct statement(s).

Answer: A, B, D

More questions on Oxides, peroxides, superoxides and hydroxides

Consider the anions present in $Na X_{2} O X_{2}$ and $KO X_{2}$ . Select the correct statement(s).