The carbon compound “A” forms “B” with sodium metal and again forms “C” with $\mathrm{PC}{l}_5$ but “B” reacts with “C” to form diethyl ether. Therefore A, B and C are respectively.

"A" reacts with sodium metal and forms "B". 

$\underset{\mathrm{A}}{{\mathrm{C}}_2}{\mathrm{H}}_5\mathrm{OH}\stackrel{\mathrm{Na} }{\to }\underset{\mathrm{B}}{{\mathrm{C}}_2}{\mathrm{H}}_5\mathrm{ONa}$

"A" reacts with ${\mathrm{PCl}}_5$ to form "C".

$\underset{\mathrm{A}}{{\mathrm{C}}_2}{\mathrm{H}}_5\mathrm{OH}\stackrel{{\mathrm{PCl}}_5}{\to }\underset{\mathrm{C}}{{\mathrm{C}}_2}{\mathrm{H}}_5\mathrm{Cl}$

"B" reacts with "C" to form diethyl ether.

$\underset{\mathrm{B}}{{\mathrm{C}}_2}{\mathrm{H}}_5\mathrm{ONa}+\underset{\mathrm{C}}{{\mathrm{C}}_2}{\mathrm{H}}_5\mathrm{Cl} \to {\mathrm{C}}_2{\mathrm{H}}_5-\mathrm{O}-{\mathrm{C}}_2{\mathrm{H}}_5$

Hence, A, B and C are ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}, {\mathrm{C}}_2{\mathrm{H}}_5\mathrm{ONa} and {\mathrm{C}}_2{\mathrm{H}}_5\mathrm{Cl}$ respectively.