$HCHO$ with conc. alkali forms two compounds. The change in oxidation number would be 

Oxidation number of any species change when they either gain or loose their fixed number of electrons in the stable compounds. When $HCHO$ & $NaOH$ react, it is called disproportionation reaction in which one compound is alcohol and other on hydrolysis gives carboxylic acid. 

Formaldehyde has no alpha-hydrogen atom when reacts with $NaOH$ it is called Cannizzaro’s reaction. 

 $2\mathrm{H}\stackrel{0}{\mathrm{C}}\mathrm{HO}+\mathrm{NaOH}\to \mathrm{H}\stackrel{+2}{\mathrm{C}}\mathrm{OONa}+\stackrel{-2}{\mathrm{C}}{\mathrm{H}}_3\mathrm{OH}$ 

Hence, the correct answer is option (C) (0 to + 2) in one compound and (0 to -2) in the second compound.