Question

In the unbalanced reactions given below, the oxidation numbers of oxygen in X, Z and Y are respectively: \[ Li+O_2 \rightarrow X \] \[ Na+O_2(\text{excess}) \rightarrow Z \] \[ K+O_2(\text{excess}) \rightarrow Y \]

• A. \(-1,-2,-\frac12\)
• B. \(-2,-1,-\frac12\)
• C. \(-2,-\frac12,-1\)
• D. \(-1,-\frac12,-2\)

Hint:

Memorize: \[ Li_2O \;(\text{oxide}) \Rightarrow -2 \] \[ Na_2O_2 \;(\text{peroxide}) \Rightarrow -1 \] \[ KO_2 \;(\text{superoxide}) \Rightarrow -\frac12 \]

Answer 1

The Correct Option is B

Concept: Alkali metals form different oxygen compounds depending upon their size.
• Lithium forms oxide.
• Sodium forms peroxide.
• Potassium forms superoxide.

Step 1: Identify compound X. Lithium reacts with oxygen to form lithium oxide. \[ 4Li+O_2\rightarrow 2Li_2O \] In \(Li_2O\), \[ \text{Oxidation number of oxygen}=-2 \]

Step 2: Identify compound Z. Sodium in excess oxygen forms sodium peroxide. \[ 2Na+O_2\rightarrow Na_2O_2 \] In peroxide ion, \[ O_2^{2-} \] each oxygen has oxidation number \[ -1 \]

Step 3: Identify compound Y. Potassium in excess oxygen forms potassium superoxide. \[ K+O_2\rightarrow KO_2 \] In superoxide ion, \[ O_2^{-} \] oxidation number of each oxygen atom is \[ -\frac12 \] Hence, \[ X=-2,\qquad Z=-1,\qquad Y=-\frac12 \] Therefore, \[ \boxed{(-2,-1,-\frac12)} \]