Calculate the total number of valence electrons.

To determine the number of lone pairs of electrons in a molecule of $\text{H}_2\text{O}$, we need to draw its Lewis structure. Step 1: Calculate the total number of valence electrons. • Hydrogen ($\text{H}$) is in Group 1, so it has 1 valence electron. There are 2 hydrogen atoms: $2 \times 1 = 2$ valence electrons. • Oxygen ($\text{O}$) is in Group 16, so it has 6 valence electrons. There is 1 oxygen atom: $1 \times 6 = 6$ valence electrons. Total valence electrons = $2 + 6 = 8$ electrons. Step 2: Determine the central atom and form single bonds. Oxygen is less electronegative than hydrogen (and hydrogen can only form one bond), so oxygen is the central atom. Form single bonds between the oxygen atom and each hydrogen atom. $$\text{H} - \text{O} - \text{H}$$ Each single bond uses 2 electrons. So, $2 \text{ bonds} \times 2 \text{ electrons/bond} = 4$ electrons used for bonding. Step 3: Distribute the remaining electrons as lone pairs on the central atom. Remaining electrons = Total valence electrons - Electrons used for bonding Remaining electrons = $8 - 4 = 4$ electrons. These 4 electrons are placed on the central oxygen atom as lone pairs. Since each lone pair consists of 2 electrons, there will be $4 \text{ electrons} / 2 \text{ electrons/pair} = 2$ lone pairs. Step 4: Verify the octet rule (or duet rule for hydrogen). The oxygen atom now has 2 bonding pairs (4 electrons) and 2 lone pairs (4 electrons), totaling 8 electrons, which satisfies the octet rule. Each hydrogen atom has 1 bonding pair (2 electrons), satisfying the duet rule. The Lewis structure of $\text{H}_2\text{O}$ is: $$\text{H} - \ddot{\text{O}} - \text{H}$$ The two dots above and below the oxygen represent the two lone pairs. Therefore, there are 2 lone pairs of electrons in a molecule of $\text{H}_2\text{O}$. The final answer is $\boxed{\text{b. 2}}$. That's 2 down. 3 left today — send the next one.