Alum is the name used for all double salts having the composition `M_(2)^(I)SO_(4).M_(2)^(III)(SO_(4))_(3).24H_(2)O`. Where `M^(III)` stands for `AI^(+3), Cr^(+3), Fe^(+3)`, while `M^(I)` stands for

Options

A. `Li^(+), Cu^(+), Ag^(+)`
B. `Li^(+), NH_(4)^(+), Na^(+)`
C. `Na^(+), K^(+), Rb^(+)`
D. `Ca^(+2), Mg^(+2), Sr^(+2)`

Explanation

To solve the question regarding the composition of alum and the identity of `M^(I)`, we can follow these steps:

### Step 1: Understand the Composition of Alum
Alum is a double salt with the general formula:
\[ M_2^{I}SO_4 \cdot M_2^{III}(SO_4)_3 \cdot 24H_2O \]
Here, `M^(I)` represents a monovalent cation, while `M^(III)` represents a trivalent cation.

### Step 2: Identify the Trivalent Cations
From the question, we know that `M^(III)` can be:
- \( Al^{+3} \) (Aluminum)
- \( Cr^{+3} \) (Chromium)
- \( Fe^{+3} \) (Iron)

### Step 3: Determine the Nature of `M^(I)`
`M^(I)` must be a monovalent cation. Monovalent cations are ions that have a charge of +1. Common examples include:
- \( Na^{+} \) (Sodium)
- \( K^{+} \) (Potassium)
- \( Rb^{+} \) (Rubidium)
- \( Cs^{+} \) (Cesium)

### Step 4: Consider the Size of `M^(I)`
The problem specifies that `M^(I)` should not be an extremely small cation. For example, \( Li^{+} \) (Lithium) is too small and cannot be used. Therefore, we need to choose a monovalent cation that is larger than lithium.

### Step 5: Identify Suitable Cations
Considering the size and the requirement of being monovalent, the suitable candidates for `M^(I)` are:
- \( Na^{+} \)
- \( K^{+} \)
- \( Rb^{+} \)

### Conclusion
Thus, the possible monovalent cations that can represent `M^(I)` in the composition of alum are:
- Sodium (\( Na^{+} \))
- Potassium (\( K^{+} \))
- Rubidium (\( Rb^{+} \))

### Final Answer
The cations that can represent `M^(I)` are Sodium, Potassium, and Rubidium.

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