To determine the number of bond pairs and lone pairs around the central atom, we need to draw the Lewis structure for each species and apply VSEPR theory.

Step 1: Analyze option A) ${\mathrm{H}}_2\mathrm{O}$

The central atom is Oxygen (O). Oxygen has 6 valence electrons. It forms two single bonds with two Hydrogen atoms. The remaining $6-2=4$ electrons form two lone pairs.

• Bond pairs = 2
• Lone pairs = 2

Step 2: Analyze option B) ${\mathrm{BF}}_3$

The central atom is Boron (B). Boron has 3 valence electrons. It forms three single bonds with three Fluorine atoms. There are no remaining valence electrons on Boron.

• Bond pairs = 3
• Lone pairs = 0

Step 3: Analyze option C) ${\mathrm{NH}}_2^{-}$

The central atom is Nitrogen (N). Nitrogen has 5 valence electrons. The $-1$ charge adds one more electron, so Nitrogen effectively has $5+1=6$ valence electrons. It forms two single bonds with two Hydrogen atoms. The remaining $6-2=4$ electrons form two lone pairs.

• Bond pairs = 2
• Lone pairs = 2

Step 4: Analyze option D) ${\mathrm{PCl}}_3$

The central atom is Phosphorus (P). Phosphorus has 5 valence electrons. It forms three single bonds with three Chlorine atoms. The remaining $5-3=2$ electrons form one lone pair.

• Bond pairs = 3
• Lone pairs = 1

Therefore, ${\mathrm{PCl}}_3$ contains three bond pairs and one lone pair around the central atom.

Correct Answer: (D)