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Question: The reaction of white phosphorus with aqueous \(NaOH\) gives phosphine along with another phosphorus...

The reaction of white phosphorus with aqueous NaOHNaOH gives phosphine along with another phosphorus containing compound. The reaction type, the oxidation states of phosphorus in phosphine and the other product are respectively:
A.Redox reaction; -3 and -5
B.Redox reaction; +3 and +5
C.Disproportionation reaction; -3 and +5
D.Disproportionation reaction; -3 and +3

Explanation

Solution

Redox reactions are also known as oxidation-reduction reactions, where the oxidation states of the elements, which are involved in the reaction, will change by gaining or losing electrons. And Disproportionation reactions are redox reactions in which one of the reactants undergoes both oxidation and reduction.

Step by step answer: Phosphorus is a tetra-atomic molecule which exists in two forms, red phosphorus and white phosphorus, because of its highly reactive nature and the chemical formula is P4{P_4}.
Phosphine’s chemical formula is PH3P{H_3}, which is also called as Pnictogen hydride.
The reaction of white phosphorus with aqueous NaOHNaOH gives phosphine along with another phosphorus containing compound
P4+3NaOH+3H2O→PH3+3NaH2PO2{P_4} + 3NaOH + 3{H_2}O \to P{H_3} + 3Na{H_2}P{O_2}
And the other product is sodium hypophosphite.
Sodium hypophosphite further produces Tetrasodium pyrophosphate and excessive phosphine.
4NaH2PO2→Na4P2O7+2PH3+H2O4Na{H_2}P{O_2} \to N{a_4}{P_2}{O_7} + 2P{H_3} + {H_2}O
The oxidation state of white phosphorus in P4{P_4} is 0.
Oxidation state of Sodium (Na) is +1, oxygen is -2 and hydrogen is +1.
In the first and second reactions, the oxidation state of phosphorus in phosphine is x.
x+3(+1)=0 →x+3=0 ∴x=−3  x + 3\left( { + 1} \right) = 0 \\\ \to x + 3 = 0 \\\ \therefore x = - 3 \\\
The value of x is -3.
Oxidation state of phosphorus in Sodium hypophosphite is y.
1(+1)+2(+1)+y+2(−2)=0 →1+2+y−4=0 →y−1=0 ∴y=+1  1\left( { + 1} \right) + 2\left( { + 1} \right) + y + 2\left( { - 2} \right) = 0 \\\ \to 1 + 2 + y - 4 = 0 \\\ \to y - 1 = 0 \\\ \therefore y = + 1 \\\
So the neutral phosphorus P4{P_4} with oxidation state zero is changed to phosphine with ox. state -3 and Sodium hypophosphite with ox. state +1. So only the phosphorus is reduced as well as oxidized.
Therefore, the reaction is a disproportionation reaction.
Sodium hypophosphite is not the final product. It further produces Tetrasodium pyrophosphate. So find the ox. state of phosphorus in Tetrasodium pyrophosphate, say z.
4(+1)+2(y)+7(−2)=0 →4+2y−14=0 →2y=10 ∴y=102=+5  4\left( { + 1} \right) + 2\left( y \right) + 7\left( { - 2} \right) = 0 \\\ \to 4 + 2y - 14 = 0 \\\ \to 2y = 10 \\\ \therefore y = \dfrac{{10}}{2} = + 5 \\\
The oxidation states of phosphorus in phosphine and the other product are -3 and +5 respectively, and the reaction is a disproportionation reaction.

Therefore, the correct option is Option C.

Note: Oxidation is the loss of electrons and reduction is the gain of electrons. When the atom loses electrons, the no. of electrons reduces and the atom develops a positive charge increasing its oxidation number, that is why the atom is said to be oxidized; whereas when the atom gains electrons, the no. of electrons increases and the atom develops a negative charge decreasing its oxidation number, that is why the atom is said to be reduced. Be careful with these two terms.