Question

The oxidation number of phosphorus in \( {Ba(H}_2{PO}_2)_2 \) is:

• A. +3
• B. +2
• C. +1
• D. -1

Hint:

In molecules with multiple elements, balance the oxidation numbers to ensure that the total charge of the molecule is zero or the charge of the ion.

Answer 1

The Correct Option is C

Step 1: The chemical formula \( {Ba(H}_2{PO}_2)_2 \) indicates that it contains barium (\( {Ba} \)), hydrogen (\( {H} \)), oxygen (\( {O} \)), and phosphorus (\( {P} \)). 
Step 2: In this compound, the barium ion (\( {Ba}^{2+} \)) has an oxidation state of +2. The hydrogen ion (\( {H}^+ \)) has an oxidation state of +1, and oxygen in peroxides (\( {O}_2^{2-} \)) typically has an oxidation state of -1. 
Step 3: Now, to find the oxidation number of phosphorus (\( {P} \)), we assume the oxidation states of the ions are balanced. The total oxidation states for the two \( {PO}_2^{2-} \) groups are: \[ 2 \times ({oxidation number of P} + 2 \times (-1)) = 2 \times ({oxidation number of P} - 2) \] The total oxidation state of the \( {Ba(H}_2{PO}_2)_2 \) molecule must be zero since the compound is neutral. We have: \[ +2 \, ({from Ba}) + 2 \times (+2) \, ({from H}) + 2 \times ({oxidation number of P} - 2) + 2 \times (-1) = 0 \] Simplifying: \[ +2 + 4 + 2 \times ({oxidation number of P} - 2) - 2 = 0 \] \[ 4 + 2 \times ({oxidation number of P} - 2) = 0 \] \[ 2 \times ({oxidation number of P} - 2) = -4 \] \[ {oxidation number of P} - 2 = -2 \] \[ {oxidation number of P} = +1 \] Therefore, the oxidation number of phosphorus in \( {Ba(H}_2{PO}_2)_2 \) is +1.