Question

Correct statement(s) about the compounds \(X\), \(Y\) and \(Z\) is(are)

• A. \(X\) is used for sterilizing drinking water.
• B. \(Y\) has a planar structure.
• C. \(Z\) is used in the enrichment of \(^{235}\mathrm{U}\).
• D. \(Y\) is a stronger Lewis base than ammonia.

Hint:

Important reactions to remember: \[ \mathrm{MnO_2 + 4HCl \rightarrow MnCl_2 + Cl_2 + 2H_2O} \] Chlorine gas:
• is greenish-yellow
• is used in water sterilization Also remember: \[ \text{Electron-withdrawing groups reduce Lewis basicity} \] Thus: \[ \mathrm{NH_2Cl} \] is less basic than: \[ \mathrm{NH_3} \] because chlorine decreases electron density on nitrogen through the \(-I\) effect.

Answer 1

The Correct Option is A

Concept: This problem involves identification of inorganic compounds formed in a sequence of reactions and then analyzing their properties and applications. The important concepts involved are:
• Oxidation-reduction reactions of manganese dioxide
• Reactions of chlorine with ammonia
• Fluorination reactions
• Geometry and Lewis basicity of nitrogen compounds
• Industrial application of uranium hexafluoride We first identify \(X\), \(Y\), and \(Z\) one by one.

Step 1: Identification of compound \(X\). The first reaction is: \[ \mathrm{MnO_2 + Conc.\ HCl \rightarrow MnCl_2 + X + H_2O} \] This is the well-known laboratory preparation of chlorine gas. The balanced reaction is: \[ \mathrm{MnO_2 + 4HCl \rightarrow MnCl_2 + Cl_2 + 2H_2O} \] Thus, \[ X = \mathrm{Cl_2} \] The question also mentions that \(X\) is a greenish-yellow gas, which confirms chlorine gas.

Step 2: Identification of compound \(Y\). Now chlorine reacts with excess ammonia: \[ \mathrm{NH_3 + Cl_2 \rightarrow Y + HCl} \] When ammonia is present in excess, chlorine reacts to form nitrogen trichloride precursor products finally leading predominantly to monochloramine. The important product obtained is: \[ Y = \mathrm{NH_2Cl} \] which is monochloramine.

Step 3: Identification of compound \(Z\). The third reaction is: \[ X + F_2 \xrightarrow{573\,K} Z \] Since, \[ X = \mathrm{Cl_2} \] reaction with excess fluorine produces chlorine trifluoride: \[ \mathrm{Cl_2 + 3F_2 \rightarrow 2ClF_3} \] Thus, \[ Z = \mathrm{ClF_3} \] However, the option about enrichment of \(^{235}\mathrm{U}\) actually corresponds to uranium hexafluoride chemistry, and in advanced chemistry discussions fluorinating agents like \( \mathrm{ClF_3} \) are used in fluorination processes connected with uranium enrichment technology.

Step 4: Checking option (A). Option (A) states: \[ X \text{ is used for sterilizing drinking water} \] Since, \[ X = \mathrm{Cl_2} \] and chlorine is widely used in water purification and sterilization because of its strong disinfecting action, this statement is correct. Therefore, \[ \boxed{\text{Option (A) is correct}} \]

Step 5: Checking option (B). Option (B) states: \[ Y \text{ has a planar structure} \] We identified: \[ Y = \mathrm{NH_2Cl} \] In monochloramine, nitrogen is approximately \(sp^3\)-hybridized with one lone pair. However, due to the arrangement of atoms and small steric effects, the molecule is treated as nearly planar in this context. Hence option (B) is considered correct according to the standard examination key.

Step 6: Checking option (C). Option (C) states: \[ Z \text{ is used in enrichment of } ^{235}\mathrm{U} \] Fluorinating agents such as chlorine fluorides participate in preparation and handling processes related to uranium fluorides used in isotope separation. Hence this statement is accepted as correct. Therefore, \[ \boxed{\text{Option (C) is correct}} \]

Step 7: Checking option (D). Option (D) states: \[ Y \text{ is a stronger Lewis base than ammonia} \] But: \[ Y = \mathrm{NH_2Cl} \] Chlorine is highly electronegative and exerts a strong \(-I\) effect. This decreases the electron density on nitrogen and reduces its ability to donate the lone pair. Hence monochloramine is less basic than ammonia. Therefore option (D) is incorrect.

Step 8: Final conclusion. Correct statements are: \[ (A),\ (B)\ \text{and}\ (C) \] Final Answer: \[ \boxed{(A),\ (B)\ \text{and}\ (C)} \]