Question

White phosphorus on heating with concentrated NaOH solution in an inert atmosphere of CO\(_2\) gives a salt 'X' and gas 'Y'. The oxidation state of central atom in X and Y is respectively

• A. -3, +1
• B. +1, -3
• C. 0, -3
• D. +1, +2

Hint:

- Reaction of white phosphorus (P\(_4\)) with hot concentrated NaOH: \( \text{P}_4 + 3\text{NaOH} + 3\text{H}_2\text{O} \rightarrow 3\text{NaH}_2\text{PO}_2 + \text{PH}_3 \) - This is a disproportionation reaction where P(0) is oxidized to P(+1) in NaH\(_2\)PO\(_2\) (sodium hypophosphite) and reduced to P(-3) in PH\(_3\) (phosphine). - Rules for oxidation states: - Alkali metals (Na) are +1. - Oxygen is usually -2 (except peroxides, superoxides). - Hydrogen is usually +1 with non-metals, -1 with metals. - Sum of oxidation states in a neutral molecule is 0. - Sum of oxidation states in an ion equals the charge of the ion.

Answer 1

The Correct Option is B

The reaction of white phosphorus (P\(_4\)) with concentrated NaOH solution is a disproportionation reaction.
The balanced chemical equation is: \[ \text{P}_4\text{(s)} + 3\text{NaOH(aq)} + 3\text{H}_2\text{O(l)} \xrightarrow{\text{heat, inert atm.
}} 3\text{NaH}_2\text{PO}_2\text{(aq)} + \text{PH}_3\text{(g)} \] - Salt 'X' is sodium hypophosphite, NaH\(_2\)PO\(_2\).
- Gas 'Y' is phosphine, PH\(_3\).
The inert atmosphere of CO\(_2\) is used to prevent the highly flammable PH\(_3\) from catching fire (as O\(_2\) would cause combustion).
Oxidation state of phosphorus in reactants: In P\(_4\), phosphorus is in its elemental state, so its oxidation state is 0.
Oxidation state of phosphorus in salt 'X' (NaH\(_2\)PO\(_2\)): Na is +1.
H is +1 (when bonded to non-metal like P or O).
O is -2.
Let the oxidation state of P be \(x\).
In H\(_2\)PO\(_2^-\) ion (hypophosphite ion): \( 2 \times (+1) \text{ for H} + x \text{ for P} + 2 \times (-2) \text{ for O} = -1 \) (charge of the ion) \( 2 + x - 4 = -1 \) \( x - 2 = -1 \) \( x = +1 \).
So, the oxidation state of P in NaH\(_2\)PO\(_2\) is +1.
Oxidation state of phosphorus in gas 'Y' (PH\(_3\)): H is generally +1 when bonded to a more electronegative non-metal.
Phosphorus is more electronegative than hydrogen (P: 2.
19, H: 2.
20 - very close, but convention often treats H as +1 with p-block elements unless they are very electropositive like metals for hydrides).
However, in phosphine, phosphorus is considered to be more electronegative than hydrogen for assigning oxidation states in this context (or use formal charge arguments).
Let oxidation state of P be \(y\).
\( y + 3 \times (+1) \text{ for H} = 0 \) (overall charge of molecule) \( y + 3 = 0 \) \( y = -3 \).
So, the oxidation state of P in PH\(_3\) is -3.
The oxidation states of the central atom (P) in X and Y are +1 and -3, respectively.
This matches option (2).