Oxidation state of boron in Magnesium boride

Options

A. `-3`
B. `+3`
C. `+1`
D. `-1`

Answer

To find the oxidation state of boron in magnesium boride (MgB2), we can follow these steps:

### Step 1: Identify the components of the compound
Magnesium boride consists of magnesium (Mg) and boron (B). The formula can be written as MgB2, indicating that there are two boron atoms for every magnesium atom.

### Step 2: Determine the oxidation state of magnesium
Magnesium is an alkaline earth metal, which typically has a fixed oxidation state of +2. Therefore, in MgB2, the oxidation state of magnesium (Mg) is +2.

### Step 3: Set up the equation for the total oxidation states
Let the oxidation state of boron be represented as \( x \). Since there are two boron atoms in the compound, their total contribution to the oxidation state will be \( 2x \).

### Step 4: Write the equation based on the overall charge
The compound is neutral, so the sum of the oxidation states must equal zero. Therefore, we can set up the equation:
\[ \text{Oxidation state of Mg} + \text{Oxidation state of B} = 0 \]
Substituting the known values:
\[ +2 + 2x = 0 \]

### Step 5: Solve for \( x \)
Now, we can solve for \( x \):
\[ 2x = -2 \]
\[ x = -1 \]

### Conclusion
The oxidation state of boron in magnesium boride (MgB2) is -1.