`10800 C` of electricity passed through the electrolyte deposited `2.977g` of metal with atomic mass `106.4 g mol^(-1)`. The charge on the metal cation is

Options

A. `+4`
B. `+3`
C. `+2`
D. `+1`

Explanation

To find the charge on the metal cation, we can follow these steps:

### Step 1: Understand the relationship between charge, mass, and atomic mass
We know that the amount of substance deposited during electrolysis can be calculated using the formula:
\[ \text{Weight deposited} = \frac{\text{Atomic weight} \times Q}{n \times F} \]
where:
- \( Q \) is the total charge passed (in coulombs),
- \( n \) is the number of electrons transferred (n-factor),
- \( F \) is Faraday's constant (approximately \( 96500 \, C/mol \)).

### Step 2: Rearrange the formula to find the n-factor
The formula can be rearranged to solve for the n-factor:
\[ n = \frac{\text{Atomic weight} \times Q}{\text{Weight deposited} \times F} \]

### Step 3: Substitute the known values
Given:
- Weight deposited = \( 2.977 \, g \)
- Atomic weight = \( 106.4 \, g/mol \)
- Charge \( Q = 10800 \, C \)
- Faraday's constant \( F = 96500 \, C/mol \)

Substituting these values into the rearranged formula:
\[ n = \frac{106.4 \, g/mol \times 10800 \, C}{2.977 \, g \times 96500 \, C/mol} \]

### Step 4: Calculate the n-factor
Now, we can perform the calculation:
1. Calculate the numerator:
\[ 106.4 \times 10800 = 1153920 \]
2. Calculate the denominator:
\[ 2.977 \times 96500 = 287,205.5 \]
3. Now divide the numerator by the denominator:
\[ n = \frac{1153920}{287205.5} \approx 4.01 \]

### Step 5: Round the n-factor
Since the n-factor must be a whole number, we round \( 4.01 \) to \( 4 \).

### Step 6: Determine the charge on the metal cation
The charge on the metal cation is equal to the n-factor, which we found to be \( 4 \). Therefore, the charge on the metal cation is \( +4 \).

### Final Answer
The charge on the metal cation is \( +4 \).

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