Lithium only forms monoxide when heated in oxygen. Sodium forms monoxide and peroxide in excess of oxygen. Other alkali metals form superoxide metals superoxide with oxygen i.e., `MO_(2)`. The abnormal behaviour of lithium is due to small size. The larger size of higher metals also decides the role in formation of superoxide. All the three anions abstract proton from water. The three anions are raleted to each other as follows:
`underset("oxide ion")(O_(2)^(2-)overset((1)/(2)O_(2))(to)underset("peroxide ion")(O_(2)^(2-))overset(O_(2))(to)underset("superoxide ion")(2O_(2)^(-))`
Q. Which compounds will liberate oxygen when reacts with ice cold water?

Options

A. `Na_(2)CO_(2)`
B. `KO_(2)`
C. `Na_(2)O`
D. `Cs_(2)O_(2)`

Explanation

To determine which compounds will liberate oxygen when they react with ice-cold water, we need to analyze the behavior of the alkali metal oxides, peroxides, and superoxides. Here’s a step-by-step solution:

### Step 1: Understand the Types of Compounds
- **Oxides (MO)**: Formed by lithium, do not liberate oxygen when reacting with water.
- **Peroxides (M2O2)**: Formed by sodium, can liberate oxygen.
- **Superoxides (MO2)**: Formed by other alkali metals, liberate oxygen when reacting with water.

### Step 2: Identify the Compounds
We are given four compounds to analyze:
1. Na2O2 (Sodium peroxide)
2. KO2 (Potassium superoxide)
3. Na2O (Sodium oxide)
4. CS2O2 (Cesium peroxide)

### Step 3: Determine the Oxidation States
We will calculate the oxidation state of oxygen in each compound to identify which ones are superoxides.

1. **Na2O2**:
- Let the oxidation state of oxygen be \( x \).
- Sodium (Na) has an oxidation state of +1.
- Equation: \( 2(+1) + 2x = 0 \)
- Solving gives: \( 2 + 2x = 0 \) → \( 2x = -2 \) → \( x = -1 \) (Peroxide)

2. **KO2**:
- Let the oxidation state of oxygen be \( x \).
- Potassium (K) has an oxidation state of +1.
- Equation: \( 1 + 2x = 0 \)
- Solving gives: \( 1 + 2x = 0 \) → \( 2x = -1 \) → \( x = -\frac{1}{2} \) (Superoxide)

3. **Na2O**:
- Let the oxidation state of oxygen be \( x \).
- Sodium (Na) has an oxidation state of +1.
- Equation: \( 2(+1) + x = 0 \)
- Solving gives: \( 2 + x = 0 \) → \( x = -2 \) (Oxide)

4. **CS2O2**:
- Let the oxidation state of oxygen be \( x \).
- Cesium (Cs) has an oxidation state of +1.
- Equation: \( 2(+1) + 2x = 0 \)
- Solving gives: \( 2 + 2x = 0 \) → \( 2x = -2 \) → \( x = -1 \) (Peroxide)

### Step 4: Identify the Superoxide
From the calculations:
- Na2O2 is a peroxide.
- KO2 is a superoxide.
- Na2O is an oxide.
- CS2O2 is a peroxide.

### Step 5: Reaction with Ice Cold Water
Superoxides liberate oxygen when they react with water:
\[ 2O_2^- + 2H_2O \rightarrow 2OH^- + O_2 \]

### Conclusion
The only compound that is a superoxide and will liberate oxygen when reacting with ice-cold water is **KO2**.

### Final Answer
**KO2 (Potassium superoxide) will liberate oxygen when it reacts with ice-cold water.**

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