In Cannizzaro reaction given below:
$2 PhCHO \overset{- OH}{⟶} {PhCH}_{2} OH + {PhCO}_{2}^{-}$
The slowest step is

A
The abstraction of proton from the carboxylic group
B
The deprotonation of ${PhCH}_{2} OH$
C
The attack of $^{-} OH$ at the carboxylic group
D
The transfer of hydride to the carbonyl group

Solution:

The mechanism of the reaction involves the following steps.

The shifting of hydride ion to the carbonyl group is the slowest step.

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