In the Cannizzaro reaction given : $2C_6H_5-CHO\quad \underrightarrow {OH^-}\quad C_6H_5-CH_2OH +C_6H_5COO^-$ the slowest step is

Question

$\begin{array}{1 1}(a)\;\text{The abstraction of proton from the caboxylic acid}\\(b)\;\text{the deprotonation of }C_6H_5-CH_2OH\\(c)\;\text{the transfer of hydride to the carbonyl group}\\(d)\;\text{the attack of }OH^-\text{at the carbonyl groups}\end{array}$