To determine the oxidation state of magnesium (Mg) in \( \text{Mg(OH)}_2 \), we need to follow a systematic approach considering the known oxidation states of oxygen and hydrogen atoms:
1. Identify the usual oxidation states:
- Oxygen (O) generally has an oxidation state of -2.
- Hydrogen (H) usually has an oxidation state of +1.
2. Analyze the compound \( \text{Mg(OH)}_2 \):
- The compound contains one magnesium (Mg) atom and two hydroxide (\(\text{OH}^-\)) groups.
- The hydroxide ion \(\text{OH}^-\) overall has a charge of -1.
3. Write the oxidation state equation:
- Let \( x \) be the oxidation state of magnesium (Mg).
- The two hydroxide ions will have a combined oxidation state of \((2 \times -1) = -2\).
- The compound \( \text{Mg(OH)}_2 \) is neutral, so the sum of the oxidation states should equal zero.
4. Set up the equation based on oxidation states:
[tex]\[
\text{Oxidation state of Mg} + 2 \times (\text{Oxidation state of OH}) = 0
\][/tex]
Which translates to:
[tex]\[
x + 2(-1) = 0
\][/tex]
5. Solve for \( x \):
[tex]\[
x - 2 = 0
\][/tex]
[tex]\[
x = 2
\][/tex]
Therefore, the oxidation state of magnesium (Mg) in \( \text{Mg(OH)}_2 \) is \( \text{+2} \).
Hence, the correct answer is:
A. +2
