The reaction of white phosphorus with aqueous $N a O H$ gives phosphine along with another phosphorus containing compound. The reaction type, the oxidation states of phosphorus in phosphine and the other product are respectively:

Question

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Answer 1

Hint: Disproportionation reaction is a reaction in which the same element gets oxidized and gets reduced from one initial oxidation state.

Step 1: Write down the reaction:

$P_{4 (w h i t e)} + N a O H \to P H_{3} + N a H_{2} P O_{2}$

Step 2: Calculate the oxidation state of phosphorous in the various phosphorous compounds:

Let the oxidation number of $P$ be $x$ in each case

$P_{4} \to 4 x = 0 ⟹ x = 0$
$P H_{3} \to x + 3 = 0 ⟹ x = - 3$
$N a H_{2} P O_{2} \to + 1 + 2 + x - 4 = 0 ⟹ x = + 1$

Final Answer: Hence the correct option is C.

Answer 2

Hint: Disproportionation reaction is a reaction in which the same element gets oxidized and gets reduced from one initial oxidation state.

Step 1: Write down the reaction:

P_{4 (w h i t e)} + N a O H \to P H_{3} + N a H_{2} P O_{2}

Step 2: Calculate the oxidation state of phosphorous in the various phosphorous compounds:

Let the oxidation number of

P

be

x

in each case

P_{4} \to 4 x = 0 ⟹ x = 0

P H_{3} \to x + 3 = 0 ⟹ x = - 3

N a H_{2} P O_{2} \to + 1 + 2 + x - 4 = 0 ⟹ x = + 1

Final Answer: Hence the correct option is C.