Properties and Reactions of Chlorine Nitrogen and Fluorine Compounds

To determine the correct statements about compounds $\mathbf{X}$, $\mathbf{Y}$, and $\mathbf{Z}$, let us first identify each compound from the given chemical reactions:

1. Identification of $\mathbf{X}$: When manganese dioxide ($\text{MnO}_2$) reacts with concentrated hydrochloric acid ($\text{HCl}$), it undergoes a redox reaction to produce manganese(II) chloride ($\text{MnCl}_2$), water ($\text{H}_2\text{O}$), and chlorine gas ($\text{Cl}_2$), which is a characteristic greenish-yellow gas: $\text{MnO}_2 + 4\text{HCl (conc.)} \longrightarrow \text{MnCl}_2 + \mathbf{Cl}_2\uparrow + 2\text{H}_2\text{O}$ Therefore, compound $\mathbf{X}$ is Chlorine gas ($\text{Cl}_2$).

2. Identification of $\mathbf{Y}$: When ammonia ($\text{NH}_3$) reacts with excess chlorine gas ($\mathbf{X}$), nitrogen trichloride ($\text{NCl}_3$) and hydrogen chloride ($\text{HCl}$) are formed: $\text{NH}_3 + 3\text{Cl}_2\text{ (excess)} \longrightarrow \mathbf{NCl}_3 + 3\text{HCl}$ Therefore, compound $\mathbf{Y}$ is Nitrogen trichloride ($\text{NCl}_3$).

3. Identification of $\mathbf{Z}$: The reaction between chlorine gas ($\mathbf{X}$) and fluorine ($\text{F}_2$) at $573\text{ K}$ yields chlorine trifluoride ($\text{ClF}_3$): $\text{Cl}_2 + 3\text{F}_2 \xrightarrow{573\text{ K}} 2\mathbf{ClF}_3$ Therefore, compound $\mathbf{Z}$ is Chlorine trifluoride ($\text{ClF}_3$).

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Evaluation of Options:

• Option A: $\mathbf{X}$ ($\text{Cl}_2$) is widely used as a disinfectant and sterilizing agent for purification of drinking water. $\text{This statement is \textbf{CORRECT}.}$

• Option B: The central nitrogen atom in $\text{NCl}_3$ ($\mathbf{Y}$) is $sp^3$-hybridized with three bond pairs and one lone pair. Its molecular geometry is trigonal pyramidal, not planar. $\text{This statement is \textbf{INCORRECT}.}$

• Option C: Chlorine trifluoride ($\text{ClF}_3$, compound $\mathbf{Z}$) is used to produce uranium hexafluoride ($\text{UF}_6$), which is crucial for the enrichment of $^{235}\text{U}$ nuclear fuel: $\text{U}(s) + 3\text{ClF}_3(l) \longrightarrow \text{UF}_6(g) + 3\text{ClF}(g)$ $\text{This statement is \textbf{CORRECT}.}$

• Option D: In $\text{NCl}_3$, the three highly electronegative chlorine atoms exert a strong electron-withdrawing inductive effect ($-I$), reducing the electron density on the central nitrogen atom. Consequently, $\text{NCl}_3$ is a much weaker Lewis base than $\text{NH}_3$. $\text{This statement is \textbf{INCORRECT}.}$

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Conclusion:

The correct statements are A and C.