In which of the following , elements has least oxidation number ?

Options

A. `Ni(CN)_(4)`
B. `Ni(CO)_(4)`
C. `Fe_(2)O_(3)`
D. `SF_(6)`

Explanation

To determine which of the given elements has the least oxidation number, we will analyze each compound step by step.

### Step 1: Analyze Nickel Tetra Cyanide (Ni(CN)₄)
1. Let the oxidation state of nickel be \( X \).
2. The charge on each cyanide ion (CN) is -1.
3. Since there are 4 cyanide ions, the total negative charge contributed by cyanides is \( 4 \times (-1) = -4 \).
4. The overall charge of the complex is 0.
5. Therefore, we can set up the equation:
\[ X - 4 = 0 \]
Solving for \( X \):
\[ X = +4 \]

### Step 2: Analyze Nickel Tetra Carbonyl (Ni(CO)₄)
1. Let the oxidation state of nickel be \( Y \).
2. Carbonyl (CO) is a neutral ligand, so it contributes 0 charge.
3. The overall charge of the complex is also 0.
4. Therefore, we can set up the equation:
\[ Y = 0 \]
Thus, the oxidation state of nickel in this case is 0.

### Step 3: Analyze Iron(III) Oxide (Fe₂O₃)
1. Let the oxidation state of iron be \( Z \).
2. There are 2 iron atoms, so the total contribution from iron is \( 2Z \).
3. The charge on oxygen is -2, and there are 3 oxygen atoms, contributing a total charge of \( 3 \times (-2) = -6 \).
4. The overall charge of the compound is 0.
5. Therefore, we can set up the equation:
\[ 2Z - 6 = 0 \]
Solving for \( Z \):
\[ 2Z = 6 \implies Z = +3 \]

### Step 4: Analyze Sulfur Hexafluoride (SF₆)
1. Let the oxidation state of sulfur be \( W \).
2. The charge on each fluoride ion (F) is -1, and there are 6 fluorine atoms, contributing a total charge of \( 6 \times (-1) = -6 \).
3. The overall charge of the compound is 0.
4. Therefore, we can set up the equation:
\[ W - 6 = 0 \]
Solving for \( W \):
\[ W = +6 \]

### Summary of Oxidation States
- Nickel in Ni(CN)₄: +4
- Nickel in Ni(CO)₄: 0
- Iron in Fe₂O₃: +3
- Sulfur in SF₆: +6

### Conclusion
The element with the least oxidation number among the given compounds is **Nickel in Nickel Tetra Carbonyl (Ni(CO)₄)**, which has an oxidation state of **0**.