#### Answer

The number of Hydrogen atoms in the compounds are:
(a) $C_{8}H_{?}O_{2}$ has 12 H atoms
(b) $C_{7}H_{?}N$ has 13 H atoms
(c) $C_{9}H_{?}NO$ has 13 H atoms

#### Work Step by Step

(a) In $C_{8}H_{?}O_{2}$, there are two rings and one double bond present, i.e. the degree of unsaturation in this compound is 3.
Further, the equivalent hydrocarbon of this compound will have the formula $C_{8}H_{18}$. Subtracting the number of H atoms present in the original compound (let it be $x$) from the number of H atoms in the equivalent formula gives the degree of unsaturation, or,
$H_{18}-H_{x} = H_{6}= 3H_{2}$
i.e. $H_{18}-H_{6}=H_{x}$
i.e. $H_{12}=H_{x}$
Thus $C_{8}H_{?}O_{2}$ has 12 H atoms.
(b) $C_{7}H_{?}N$ has two double bonds or two degrees of unsaturation. The formula of the equivalent saturated hydrocarbon of this compound is $C_{7}H_{16}$. (let the number of H atoms be $x$ in the compound)
Following the same method as above,
$H_{16}-H_{x-1}=H_{4}=2H_{2}$
(we have subtracted 1 from the number of H atoms since the number of nitrogen atoms have to be subtracted from the number of hydrogen atoms to reach the equivalent formula for the unsaturated hydrocarbon)
i.e. $H_{16}-H_{4}=H_{x-1}$
i.e $H_{12}=H_{x-1}$
or $H_{13}=H_{x}$
Thus $C_{7}H_{?}N$ has 13 H atoms.
(c) $C_{9}H_{?}NO$ has one ring and three double bonds, i.e. 4 degrees of unsaturation. The formula of the equivalent hydrocarbon of this compound is $C_{9}H_{20}$.
And,
$H_{20}-H_{x-1}=H_{8}=4H_{2}$
i.e. $H_{20}-H_{x-1}=H_{8}$
i.e. $H_{20}-H_{8}=H_{x-1}$
$H_{12}=H_{x-1}$
or
$H_{13}=H_{x}$
Thus $C_{9}H_{?}NO$ has 13 H atoms.