What is the oxidation state of Phosphorus in \(H_3PO_4\)?
Show Hint
To quickly find oxidation states in oxyacids, remember that hydrogen contributes \(+1\) and oxygen contributes \(-2\). Set the sum equal to zero for neutral molecules and solve for the unknown element.
Concept:
The oxidation state is the formal charge an atom would carry if all its bonds were ionic. It is calculated based on the sum of oxidation numbers of all atoms in a neutral molecule being zero. Step 1: Understanding the Question:
We are given Phosphoric acid (\(H_3PO_4\)) and need to find the oxidation state of the Phosphorus (\(P\)) atom. Step 2: Key Formula or Approach:
Use the standard oxidation numbers:
- Hydrogen (\(H\)) is usually \(+1\).
- Oxygen (\(O\)) is usually \(-2\).
- Let the oxidation state of Phosphorus be \(x\).
Sum of all charges = \(0\). Step 3: Detailed Solution:
Equation: \(3(\text{Oxidation state of } H) + 1(x) + 4(\text{Oxidation state of } O) = 0\)
\[ 3(+1) + x + 4(-2) = 0 \]
\[ 3 + x - 8 = 0 \]
\[ x - 5 = 0 \]
\[ x = +5 \] Step 4: Final Answer:
The oxidation state of Phosphorus in \(H_3PO_4\) is +5.
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