Q6 / 54 · JEE Advanced 2026 (Paper 2) · Chemistry
Correct statement(s) about the compounds $\textbf{P}$, $\textbf{Q}$ and $\textbf{R}$ is(are)


- Correct answerA$\textbf{P}$ has two lone pairs of electrons on the central atom.
- B$\textbf{Q}$ has a perfect octahedral geometry.
- Correct answerC$\textbf{Q}$ can act as a fluorinating agent.
- Correct answerDThe molecular structure of $\textbf{R}$ is trigonal pyramidal.
Solution
From the given reaction conditions:
$\text{Xe(g)} + 2\text{F}_2\text{(g)} \xrightarrow[\text{1 : 5 ratio}]{\text{873 K, 7 bar}} \text{XeF}_4$
Thus, $\textbf{P}$ is $\text{XeF}_4$.
$\text{XeF}_4 + \text{O}_2\text{F}_2 \xrightarrow{\text{143 K}} \text{XeF}_6 + \text{O}_2$
Thus, $\textbf{Q}$ is $\text{XeF}_6$.
$\text{XeF}_6 + 3\text{H}_2\text{O} \xrightarrow{\text{complete hydrolysis}} \text{XeO}_3 + 6\text{HF}$
Thus, $\textbf{R}$ is $\text{XeO}_3$.
Evaluating the given options:
(A) $\textbf{P}$ ($\text{XeF}_4$) has $8$ valence electrons on the central Xe atom. It forms $4$ single bonds with F, leaving $4$ electrons which form $2$ lone pairs. This statement is correct.
(B) $\textbf{Q}$ ($\text{XeF}_6$) has $6$ bond pairs and $1$ lone pair on the central Xe atom. Its geometry is distorted octahedral, not perfect octahedral. This statement is incorrect.
(C) $\textbf{Q}$ ($\text{XeF}_6$) is a strong fluorinating agent. This statement is correct.
(D) $\textbf{R}$ ($\text{XeO}_3$) has $3$ double bonds and $1$ lone pair on the central Xe atom. Its molecular structure is trigonal pyramidal. This statement is correct.
Answer: $\textbf{P}$ has two lone pairs of electrons on the central atom.; $\textbf{Q}$ can act as a fluorinating agent.; The molecular structure of $\textbf{R}$ is trigonal pyramidal.
$\text{Xe(g)} + 2\text{F}_2\text{(g)} \xrightarrow[\text{1 : 5 ratio}]{\text{873 K, 7 bar}} \text{XeF}_4$
Thus, $\textbf{P}$ is $\text{XeF}_4$.
$\text{XeF}_4 + \text{O}_2\text{F}_2 \xrightarrow{\text{143 K}} \text{XeF}_6 + \text{O}_2$
Thus, $\textbf{Q}$ is $\text{XeF}_6$.
$\text{XeF}_6 + 3\text{H}_2\text{O} \xrightarrow{\text{complete hydrolysis}} \text{XeO}_3 + 6\text{HF}$
Thus, $\textbf{R}$ is $\text{XeO}_3$.
Evaluating the given options:
(A) $\textbf{P}$ ($\text{XeF}_4$) has $8$ valence electrons on the central Xe atom. It forms $4$ single bonds with F, leaving $4$ electrons which form $2$ lone pairs. This statement is correct.
(B) $\textbf{Q}$ ($\text{XeF}_6$) has $6$ bond pairs and $1$ lone pair on the central Xe atom. Its geometry is distorted octahedral, not perfect octahedral. This statement is incorrect.
(C) $\textbf{Q}$ ($\text{XeF}_6$) is a strong fluorinating agent. This statement is correct.
(D) $\textbf{R}$ ($\text{XeO}_3$) has $3$ double bonds and $1$ lone pair on the central Xe atom. Its molecular structure is trigonal pyramidal. This statement is correct.
Answer: $\textbf{P}$ has two lone pairs of electrons on the central atom.; $\textbf{Q}$ can act as a fluorinating agent.; The molecular structure of $\textbf{R}$ is trigonal pyramidal.
Hence, the correct answer is options: 'A', 'C', 'D'.