Problem on oxidation number

Question: The reaction of white phosphorous with aq. NaOH gives phosphine along with another phosphorous containing compound. What are the oxidation states of P in phosphine and other product? Also tell what kind of reaction is it.

Options:

a    a)      Redox reaction; -3 and -5

b    b)   Redox reaction; +3 and +5 

c    c)  Disproprotionation reaction; -3 and +5

d    d)  Disproprotionation reaction; -3 and +3

Solution:

Step: 1 first let us see what happened when white phosphorous reacts with aq. Sodium hydroxide:

Thus oxidations states of phosphorous in products are -3 and +5.

Step: 2 Now lets find out type of reaction. Here phosphorous in P₄ (0) oxidized in Na₃PO₄ (+5) and reduced in PH₃ ( -3)in the same reaction. In other words P in reactant side appears in two different oxidation states in product side. Such kind of reaction is known as ‘disproprotionation reaction’’.

Basically disproprotionation reaction begins with an element in one oxidation state but produce products in which that element appears in two different oxidation states. The starting element has been both oxidized and reduced.

Disproprotionation reaction is a kind of redox reaction. If you are getting confused between them, than make it clear that:

In redox reactions, A + B ---> C  +  D,   either A can undergo oxidation and B can undergo Reduction or B can undergo oxidation and A can undergo reduction. Two different substances undergo reduction and oxidation respectively.

Eg:    Fe + CuSO₄ → FeSO₄ + Cu

 But in disproportionation, same chemical substance undergoes oxidation as well as reduction. 

Eg:      2 Sn²⁺ → Sn + Sn⁴⁺

Thus, option C s correct