The molecular formula of a non - stoichiometric tin oxide containing Sn (II) and Sn (IV) ions is `Sn_(4.44) O_8`.
Therefore , the molar ratio of Sn (II) to Sn (IV) is approximately

Options

A. `1:8`
B. `1:6`
C. `1:4`
D. `1:1`

Answer

To solve the problem of finding the molar ratio of Sn(II) to Sn(IV) in the non-stoichiometric tin oxide Sn₄.₄O₈, we can follow these steps:

### Step 1: Define Variables
Let:
- \( x \) = moles of Sn(II) (Sn²⁺)
- \( y \) = moles of Sn(IV) (Sn⁴⁺)

From the molecular formula Sn₄.₄O₈, we know that the total moles of Sn is:
\[ x + y = 4.4 \]
This is our first equation.

### Step 2: Charge Balance
The overall charge of the compound must be neutral. The charge contributed by oxygen is:
- Each oxygen has a charge of -2, and there are 8 oxygens, so the total charge from oxygen is:
\[ 8 \times (-2) = -16 \]

Let the total charge from Sn be \( U \). Since the compound is neutral:
\[ U + (-16) = 0 \]
Thus:
\[ U = +16 \]

### Step 3: Calculate Total Oxidation State
The total oxidation state contributed by Sn can be expressed in terms of \( x \) and \( y \):
- The contribution from Sn(II) is \( 2x \) (since each Sn(II) contributes +2).
- The contribution from Sn(IV) is \( 4y \) (since each Sn(IV) contributes +4).

Setting up the equation for total oxidation state:
\[ 2x + 4y = 16 \]
This is our second equation.

### Step 4: Solve the Equations
Now we have a system of two equations:
1. \( x + y = 4.4 \)
2. \( 2x + 4y = 16 \)

We can solve these equations simultaneously.

From the first equation, we can express \( y \) in terms of \( x \):
\[ y = 4.4 - x \]

Substituting this into the second equation:
\[ 2x + 4(4.4 - x) = 16 \]
Expanding this gives:
\[ 2x + 17.6 - 4x = 16 \]
Combining like terms:
\[ -2x + 17.6 = 16 \]
Rearranging gives:
\[ -2x = 16 - 17.6 \]
\[ -2x = -1.6 \]
\[ x = 0.8 \]

Now substituting \( x \) back into the first equation to find \( y \):
\[ 0.8 + y = 4.4 \]
\[ y = 4.4 - 0.8 \]
\[ y = 3.6 \]

### Step 5: Calculate the Molar Ratio
Now we have:
- \( x = 0.8 \) (moles of Sn(II))
- \( y = 3.6 \) (moles of Sn(IV))

The molar ratio of Sn(II) to Sn(IV) is:
\[ \text{Ratio} = \frac{x}{y} = \frac{0.8}{3.6} \approx \frac{1}{4.5} \]

### Final Answer
The approximate molar ratio of Sn(II) to Sn(IV) is:
\[ \text{Ratio} \approx 1 : 4.5 \]