Question

Oxidation number of hydrogen is (-1) in:

• A. \(\mathrm{NaH}_2\mathrm{PO}_4\)
• B. \(\mathrm{NaHSO}_4\)
• C. \(\mathrm{NaBH}_4\)
• D. \(\mathrm{H}_2\)

Hint:

For oxidation numbers:
- Assign known values (e.g., Na: \(+1\), O: \(-2\)).
- Hydrogen is \(-1\) in metal hydrides, \(+1\) with non-metals.
- Sum to zero for neutral compounds.

Answer 1

The Correct Option is C

The oxidation number indicates the hypothetical charge of an atom in a molecule based on electronegativity rules. Hydrogen typically has an oxidation number of \(+1\) when bonded to non-metals (e.g., in \(\mathrm{H}_2\mathrm{O}\)) but \(-1\) in metal hydrides where it acts as a hydride ion (\(\mathrm{H}^-\)).
Step 1: Oxidation number rules
- Alkali metals (e.g., Na) have \(+1\).
- Oxygen is typically \(-2\) (except in peroxides).
- The sum of oxidation numbers in a neutral compound is zero.
- Hydrogen is \(+1\) with non-metals, \(-1\) with metals.
Step 2: Analyze each compound
- (A) \(\mathrm{NaH}_2\mathrm{PO}_4\): Sodium dihydrogen phosphate. Assign: Na: \(+1\), O: \(-2\), H: \(x\), P: \(y\). Neutral compound: \[ (+1) + 2x + y + 4(-2) = 0 \quad \Rightarrow \quad 1 + 2x + y - 8 = 0 \quad \Rightarrow \quad 2x + y = 7 \] In phosphates, P is typically \(+5\). If H is \(+1\): \[ 2(+1) + y = 7 \quad \Rightarrow \quad 2 + y = 7 \quad \Rightarrow \quad y = 5 \] H is \(+1\), not \(-1\). - (B) \(\mathrm{NaHSO}_4\): Sodium hydrogen sulfate. Na: \(+1\), H: \(x\), S: \(y\), O: \(-2\). Neutral: \[ (+1) + x + y + 4(-2) = 0 \quad \Rightarrow \quad 1 + x + y - 8 = 0 \quad \Rightarrow \quad x + y = 7 \] In sulfates, S is \(+6\). If H is \(+1\): \[ 1 + 6 = 7 \] H is \(+1\), not \(-1\). - (C) \(\mathrm{NaBH}_4\): Sodium borohydride. Na: \(+1\), B: \(y\), H: \(x\). Neutral: \[ (+1) + y + 4x = 0 \quad \Rightarrow \quad 1 + y + 4x = 0 \] In borohydrides, H is \(-1\): \[ 1 + y + 4(-1) = 0 \quad \Rightarrow \quad 1 + y - 4 = 0 \quad \Rightarrow \quad y = 3 \] B is \(+3\), H is \(-1\), which matches. - (D) \(\mathrm{H}_2\): Elemental form, oxidation number is 0.
Step 3: Conclusion Only \(\mathrm{NaBH}_4\) has hydrogen with an oxidation number of \(-1\).