Step 1: Identify the elements P and Q. The problem states that P, Q, R, X are four elements with atomic numbers 5, 9, 11, and 12. The compound is given as PQ₃. This formula suggests that P has a valency of 3 and Q has a valency of 1. • Element with atomic number 5 is Boron (B), which has 3 valence electrons and typically forms 3 bonds. • Element with atomic number 9 is Fluorine (F), which has 7 valence electrons and typically forms 1 bond. • Element with atomic number 11 is Sodium (Na), which has 1 valence electron. • Element with atomic number 12 is Magnesium (Mg), which has 2 valence electrons. Based on the formula PQ₃, P is likely Boron (atomic number 5) and Q is Fluorine (atomic number 9). The compound is therefore BF₃. Step 2: Determine the number of lone pairs on the central atom (P). In BF₃, Boron (P) is the central atom. Boron has 3 valence electrons. It forms 3 single bonds with 3 Fluorine atoms. All 3 valence electrons of Boron are used in bonding. Number of lone pairs on Boron = $\frac{\text{Valence electrons} - \text{Electrons used in bonding}}{2} = \frac{3 - 3}{2} = 0$. Step 3: Determine the number of lone pairs on the surrounding atoms (Q). There are 3 Fluorine (Q) atoms. Each Fluorine atom has 7 valence electrons. Each Fluorine atom forms 1 single bond with Boron. Number of electrons used by each Fluorine in bonding = 1. Number of lone pairs on each Fluorine atom = $\frac{\text{Valence electrons} - \text{Electrons used in bonding}}{2} = \frac{7 - 1}{2} = \frac{6}{2} = 3$. Step 4: Calculate the total number of lone pairs in the PQ₃ molecule. Total lone pairs = (Lone pairs on P) + (Number of Q atoms $\times$ Lone pairs on each Q atom) Total lone pairs = $0 + (3 \times 3) = 0 + 9 = 9$. The total number of lone pairs in the PQ₃ molecule is 9. The final answer is $\boxed{\text{গ) 9}}$. That's 2 down. 3 left today — send the next one.