Animated Solution for Chemistry - s and p-Block Elements: Correct statement(s) about the compounds $\mathbf{X}$, $\mathbf{Y}$ and $\mathbf{Z}$ is (are) ${\text{MnO}}_2+\text{Conc. HCl}⟶{\text{MnCl}}_2+\underset{\text{(greenish yellow gas)}}{\mathbf{X}}+{\text{H}}_2\text{O}$ ${\text{NH}}_3+\underset{\text{(excess)}}{\mathbf{X}}⟶\mathbf{Y}+\dots$ $\mathbf{X}+\underset{\text{(excess)}}{{\text{F}}_2}\stackrel{573\text{K}}{\to }\mathbf{Z}$

Let's start by identifying the mysterious greenish-yellow gas, $\mathbf{X}$. When manganese dioxide (${\text{MnO}}_2$) reacts with concentrated hydrochloric acid ($\text{HCl}$), a classic redox reaction occurs. The manganese dioxide acts as an oxidizing agent, oxidizing the chloride ions to produce chlorine gas.

The balanced chemical equation is:

So, we can confidently say that $\mathbf{X}$ is chlorine, ${\text{Cl}}_2$. This greenish-yellow gas is a fundamental building block in inorganic chemistry.

Now look at the second equation. We react ammonia (${\text{NH}}_3$) with an excess of our newly identified chlorine gas. The keyword here is excess. Because chlorine is present in abundance, it has the oxidizing power to replace all the hydrogen atoms in the ammonia molecule.

This exhaustive substitution gives us nitrogen trichloride, ${\text{NCl}}_3$, along with hydrogen chloride as a byproduct:

Therefore, compound $\mathbf{Y}$ is ${\text{NCl}}_3$.

For the third reaction, chlorine reacts with an excess of fluorine gas at $573\text{K}$. Fluorine is the most electronegative and reactive element on the periodic table. When it encounters chlorine, it forces chlorine to abandon its usual $-1$ state and adopt a positive oxidation state.

Because fluorine is in excess, chlorine is pushed to the $+3$ oxidation state, forming chlorine trifluoride, ${\text{ClF}}_3$:

Thus, compound $\mathbf{Z}$ is ${\text{ClF}}_3$.

Now let's check the options one by one. Option A says $\mathbf{X}$ is used for sterilizing drinking water. We know $\mathbf{X}$ is chlorine, and chlorine is indeed a standard disinfectant used worldwide to kill bacteria in municipal water supplies. So, Option A is absolutely correct.

Moving to Option B, it claims $\mathbf{Y}$ has a planar structure. Let's look at the Lewis structure of ${\text{NCl}}_3$. Nitrogen has five valence electrons, forms three single bonds with chlorine, and has one lone pair left over. This gives it a steric number of 4, meaning an $s{p}^3$ hybridization. Due to the lone pair-bond pair repulsion, its molecular geometry is pyramidal, not planar. So, Option B is incorrect.

Option C states that $\mathbf{Z}$ is used in the enrichment of Uranium-235. $\mathbf{Z}$ is ${\text{ClF}}_3$, which is an incredibly powerful fluorinating agent. It reacts with solid uranium metal to produce volatile uranium hexafluoride (${\text{UF}}_6$), which is essential for separating isotopes via gas centrifugation. This makes Option C correct.

Finally, Option D claims $\mathbf{Y}$ is a stronger Lewis base than ammonia. A Lewis base is an electron pair donor. In ${\text{NCl}}_3$, the highly electronegative chlorine atoms pull the electron density away from nitrogen through a strong $-I$ (inductive) effect. This makes the lone pair on nitrogen much less available for donation compared to the lone pair in ammonia. So, ${\text{NCl}}_3$ is actually a weaker base, making Option D incorrect.

To wrap it up, we've successfully identified the compounds and evaluated their chemical properties. The correct statements are (A) and (C). Understanding these fundamental inorganic reactions and molecular structures is key to mastering the chemistry of p-block elements.