GATE 2025 CY – Question 30

Inorganic Chemistry · Main Group Elements: Shapes and structures of molecules using VSEPR postulates, Hydrides, halides, oxides, oxoacids, nitrides, sulfides - shapes and reactivity. Structure and bonding of boranes, carboranes, silicones, silicates, boron nitride, borazine and phosphazenes: electron counting in polyhedral boranes, isolobal analogy. Allotropes of carbon, phosphorus and sulfur. Industrial synthesis of compounds of main group elements such as NH3, H2SO4 and HNO3. Chemistry of noble gases and interhalogen compounds. Acid-base concepts and principles (Lewis, Brønsted, HSAB and acid-base catalysis). · 1 mark · Numerical answer

Xe and $\mathrm{F_2}$ in 1:1 molar ratio when mixed in a closed flask and kept in the sunlight for a day, gave white crystals of a compound Q. Two equivalents of Q on reaction with one equivalent of $\mathrm{AsF_5}$ gave an ionic compound $X^+Y^-$ with the cation having two Xe atoms. The total number of lone pairs present on the cation $X^+$ is ______ (in integer).

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Correct answer: 14

Explanation

**Step 1: compound Q.** Xe and $F_2$ in a 1:1 ratio, in sunlight (or heat), give **xenon difluoride**, $\text{XeF}_2$ (white crystals): Q $=\text{XeF}_2$.

**Step 2: reaction with $\text{AsF}_5$.** $\text{AsF}_5$ is a strong fluoride-ion acceptor. Two equivalents of $\text{XeF}_2$ with one of $\text{AsF}_5$ give
$$2\,\text{XeF}_2+\text{AsF}_5\longrightarrow[\text{Xe}_2\text{F}_3]^+[\text{AsF}_6]^-.$$
The cation is $[\text{F–Xe–F–Xe–F}]^+$, with a bridging fluorine.

**Step 3: count the lone pairs on $\text{Xe}_2\text{F}_3^+$.**
- Valence electrons: $2(8)+3(7)-1=36$.
- Bonding electrons: 4 Xe–F bonds, which is $4\times2=8$ electrons.
- Non-bonding electrons: $36-8=28$, i.e. **14 lone pairs**.

(Check by atom: each of the two terminal F atoms has 3 lone pairs, giving 6; the bridging F has 2; each Xe has 3, giving 6. The total is $6+2+6=14$.)

Answer **14**.