Step 1: Write the Lewis picture
Take $\text{H}_2\text{O}$ as the model. Oxygen has the configuration $2s^2 2p^4$, so 6 valence electrons.
Step 2: Distribute electrons
Two of the unpaired electrons of oxygen pair up with the electrons of the two hydrogens. The other four electrons stay on oxygen as two filled pairs.
Step 3: Result
Bond pairs = number of H atoms = 2. Lone pairs = $(6 - 2)/2 = 2$. The same count holds for the S, Se and Te hydrides.
Step 4: Rule out the rest
"3 bond pairs" belongs to group 15 hydrides such as $\text{NH}_3$, which has 1 lone pair. The option with 2 bond pairs and 1 lone pair would account for only 4 valence electrons on X (2 shared plus 2 unshared), which does not fit group 16.
Final Answer:
Group 16 hydrides carry 2 bond pairs and 2 lone pairs. This is option (B).
\[ \boxed{\text{(B) }2 \text{ bond pairs and } 2 \text{ lone pairs}} \]