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.

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.

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 = $\frac{\text{Valence electrons} - \text{Electrons used in bonding}}{2} = \frac{3 - 3}{2} = 0$.

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 = $\frac{\text{Valence electrons} - \text{Electrons used in bonding}}{2} = \frac{7 - 1}{2} = \frac{6}{2} = 3$.

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 $\text{গ) 9}$.