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 solve the problem, we need to find the valencies of the elements A, B, and C based on the weights deposited during electrolysis. Let's break down the solution step by step.

### Step 1: Understand the relationship between weight deposited and equivalent weight
When salts are electrolyzed under identical conditions using the same quantity of electricity, 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 of a substance 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. Then, we can express the equivalent weights 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
Given the weights deposited:
- \( W_A = 2.1 \, \text{g} \)
- \( W_B = 2.7 \, \text{g} \)
- \( W_C = 7.2 \, \text{g} \)

Using the relationship from 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
From the first part:
\[ \frac{2.1 n_A}{7} = \frac{2.7 n_B}{27} \]
Cross-multiplying gives:
\[ 2.1 n_B = \frac{2.7 \times 7}{27} n_A \]
\[ 2.1 n_B = 0.7 n_A \]
Thus,
\[ n_B = \frac{0.7 n_A}{2.1} = \frac{n_A}{3} \]

From the second part:
\[ \frac{2.7 n_B}{27} = \frac{7.2 n_C}{48} \]
Cross-multiplying gives:
\[ 2.7 \times 48 n_C = 7.2 \times 27 n_B \]
\[ 129.6 n_C = 194.4 n_B \]
Thus,
\[ n_C = \frac{194.4 n_B}{129.6} = \frac{3 n_B}{2} \]

### Step 5: Substitute \( n_B \) into the equation for \( n_C \)
Substituting \( n_B = \frac{n_A}{3} \) into the equation for \( n_C \):
\[ n_C = \frac{3 \times \frac{n_A}{3}}{2} = \frac{n_A}{2} \]

### Step 6: Solve for the integer values of valencies
We have:
1. \( n_B = \frac{n_A}{3} \)
2. \( n_C = \frac{n_A}{2} \)

Let \( n_A = 6 \) (the smallest integer that satisfies both equations):
- Then \( n_B = \frac{6}{3} = 2 \)
- And \( n_C = \frac{6}{2} = 3 \)

Thus, the valencies are:
- \( n_A = 6 \)
- \( n_B = 2 \)
- \( n_C = 3 \)

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