Choose the disproportionation reaction among the following redox reactions.

Options

A. `3Mg (s) + N_(2)(g) to Mg_(3) N_(2)(s)`
B. `P_(4)(s) + 3NaOH(aq) + 3H_(2)O (l) to PH_(3)(g) + 3NaH_(2)PO_(2)(aq)`
C. `Cl_(2)(g) + 2KI(aq) to 2KCl (aq) + I_(2)(s)`
D. `Cr_(2)O_(3)(s) + 2AI(s) to Al_(2)O_(3)(s) + 2Cr(s)`

Answer

B

Explanation

Oxidation number of P (in `P_(4)`) is 0 which is the intermmediate of `-3` (in `PH_(3)`) and `+1` ( in `NaH_(2)PO_(2)`). So, that `P_(4)` is reduced to `PH_(3)` and oxidised to `NaH_(2)PO_(2)`.