Oxidation number of S in `H_(2)SO_(5)` is 6. This is observed, because -

Options

A. There are five oxygen atoms in the molecule
B. The hydrogen atom is directly linked with non-metal
C. There is peroxide linkage in the molecule
D. The sulphur atom shows co-ordinate linkage

Answer

To determine the oxidation number of sulfur (S) in the compound \( H_2SO_5 \), we can follow these steps:

### Step 1: Write the oxidation states of known elements
In \( H_2SO_5 \):
- Hydrogen (H) has an oxidation state of +1.
- Oxygen (O) typically has an oxidation state of -2, but in peroxides, it can be -1. In \( H_2SO_5 \), there are two oxygen atoms that are part of a peroxide linkage, which means they have an oxidation state of -1. The other three oxygen atoms are in the usual -2 state.

### Step 2: Set up the equation for the total oxidation state
Let the oxidation state of sulfur be \( x \). The total oxidation state of the compound must equal 0. Therefore, we can set up the equation based on the contributions from each element:

\[ 2 \times (+1) + x + 3 \times (-2) + 2 \times (-1) = 0 \]

### Step 3: Simplify the equation
Calculating the contributions:
- From hydrogen: \( 2 \times (+1) = +2 \)
- From three oxygen atoms with -2 oxidation state: \( 3 \times (-2) = -6 \)
- From two oxygen atoms with -1 oxidation state: \( 2 \times (-1) = -2 \)

Now we can substitute these values into the equation:

\[ 2 + x - 6 - 2 = 0 \]

### Step 4: Solve for \( x \)
Combine the constants:

\[ x - 6 = 0 \]

Now, isolate \( x \):

\[ x = 6 \]

### Conclusion
The oxidation number of sulfur in \( H_2SO_5 \) is 6.

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