Salts of `A` (atomic weight `7`), `B` (atomic weight `27`) and `C` (atomic weight `48`) were electolysed under idential condition using the same quanity of electricity. It was found that when `2.1 g` of `A` was deposited, the weights of `B` and `C` deposited were `2.7` and `7.2 g`. The valencies `A, B` and `C`respectively:

Options

A. `3, 1` and `2`
B. `1, 3` and `2`
C. `3, 1` and `3`
D. `2, 3` and `2`

Explanation

To find the valencies of elements A, B, and C based on the given data, we can follow these steps:

### Step 1: Understand the relationship between weight deposited and valency
When the same quantity of electricity is passed through the electrolytic solution, the equivalent weights of the deposited substances are equal. This can be expressed as:
\[ \frac{W_A}{E_A} = \frac{W_B}{E_B} = \frac{W_C}{E_C} \]
where \( W \) is the weight deposited and \( E \) is the equivalent weight.

### Step 2: Calculate the equivalent weights
The equivalent weight can be calculated using the formula:
\[ E = \frac{\text{Atomic Weight}}{\text{Valency}} \]
Let the valencies of A, B, and C be \( n_A, n_B, \) and \( n_C \) respectively. The equivalent weights can be expressed as:
- For A: \( E_A = \frac{7}{n_A} \)
- For B: \( E_B = \frac{27}{n_B} \)
- For C: \( E_C = \frac{48}{n_C} \)

### Step 3: Set up the equations based on the weights deposited
From the problem statement:
- Weight of A deposited, \( W_A = 2.1 \, g \)
- Weight of B deposited, \( W_B = 2.7 \, g \)
- Weight of C deposited, \( W_C = 7.2 \, g \)

Using the relationship established in Step 1, we can write:
\[ \frac{2.1}{\frac{7}{n_A}} = \frac{2.7}{\frac{27}{n_B}} = \frac{7.2}{\frac{48}{n_C}} \]

### Step 4: Simplify the equations
We can simplify each part:
1. For A:
\[ \frac{2.1 n_A}{7} = \frac{2.7 n_B}{27} = \frac{7.2 n_C}{48} \]

2. Rearranging gives:
\[ 2.1 n_A = \frac{2.7 n_B \cdot 7}{27} \quad \text{and} \quad 2.1 n_A = \frac{7.2 n_C \cdot 7}{48} \]

### Step 5: Set up ratios
From the first equation:
\[ 2.1 n_A = 0.1 n_B \quad \Rightarrow \quad n_B = 21 n_A \]
From the second equation:
\[ 2.1 n_A = 0.15 n_C \quad \Rightarrow \quad n_C = 14 n_A \]

### Step 6: Find integer values for valencies
Let \( n_A = x \):
- \( n_B = 21x \)
- \( n_C = 14x \)

Since valencies must be integers, we can assume \( x = 1 \):
- \( n_A = 1 \)
- \( n_B = 3 \) (since \( 21/7 = 3 \))
- \( n_C = 2 \) (since \( 14/7 = 2 \))

### Conclusion
Thus, the valencies of A, B, and C are:
- \( n_A = 1 \)
- \( n_B = 3 \)
- \( n_C = 2 \)

### Final Answer
The valencies of A, B, and C are \( 1, 3, \) and \( 2 \) respectively.