Considering the above reactions, the compound 'A' and compound 'B' respectively are

The cyanide ion is an:

$$\mathrm{Ambidentate\ Nucleophile}$$

It can attack through either:

$$\mathrm{Carbon}$$

or

$$\mathrm{Nitrogen}$$

When ethyl chloride reacts with:

$$\mathrm{AgCN}$$

the product formed is different from that obtained with:

$$\mathrm{NaCN}$$

Silver cyanide is predominantly covalent.

Hence, the carbon atom remains strongly bonded to silver, and the nitrogen atom attacks the alkyl halide.

$$\mathrm{CH_3CH_2Cl + AgCN \longrightarrow CH_3CH_2NC + AgCl}$$

Thus, compound $$\mathrm{A}$$ is:

$$\mathrm{Ethyl\ Isocyanide\ (CH_3CH_2NC)}$$

Sodium cyanide is ionic and dissociates completely to give free:

$$\mathrm{CN^-}$$

ions.

Attack occurs mainly through carbon because formation of a:

$$\mathrm{C-C}$$

bond is more stable.

$$\mathrm{CH_3CH_2Cl + NaCN \longrightarrow CH_3CH_2CN + NaCl}$$

Thus, compound $$\mathrm{B}$$ is:

$$\mathrm{Ethyl\ Cyanide\ (CH_3CH_2CN)}$$

also called:

$$\mathrm{Propanenitrile}$$

Thus, the right option is C.