Question 6

MCQMEDIUM

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

MnO2+Conc. HCl⟶MnCl2+X+H2OMnO_2 + \text{Conc. } HCl \longrightarrow MnCl_2 + \mathbf{X} + H_2O (greenish yellow gas)

NH3+X (excess)⟶Y+HClNH_3 + \mathbf{X} \text{ (excess)} \longrightarrow \mathbf{Y} + HCl

X+F2 (excess)→573 KZ\mathbf{X} + F_2 \text{ (excess)} \xrightarrow{573\text{ K}} \mathbf{Z}

(A)

X\mathbf{X} is used for sterilizing drinking water.

(B)

Y\mathbf{Y} has a planar structure.

(C)

Z\mathbf{Z} is used in the enrichment of 235U^{235}U.

(D)

Y\mathbf{Y} is a stronger Lewis base than ammonia.

Detailed Solution

  1. Identification of X\mathbf{X}: The reaction of MnO2MnO_2 with concentrated HClHCl produces chlorine gas, which is greenish-yellow.

MnO2+4HCl→MnCl2+Cl2+2H2OMnO_2 + 4HCl \rightarrow MnCl_2 + Cl_2 + 2H_2O. Thus, X=Cl2\mathbf{X} = Cl_2.

  1. Identification of Y\mathbf{Y}: Reaction of excess chlorine with ammonia yields nitrogen trichloride. NH3+3Cl2(excess)→NCl3+3HClNH_3 + 3Cl_2 (excess) \rightarrow NCl_3 + 3HCl. Thus, Y=NCl3\mathbf{Y} = NCl_3.
  2. Identification of Z\mathbf{Z}: Reaction of chlorine with excess fluorine at 573 K573\text{ K} yields chlorine trifluoride. Cl2+3F2(excess)→2ClF3Cl_2 + 3F_2 (excess) \rightarrow 2ClF_3. Thus, Z=ClF3\mathbf{Z} = ClF_3.

Analysis of options:

(A) Cl2Cl_2 is well known for its disinfectant properties and is used for sterilizing drinking water. Correct.

(B) NCl3NCl_3 has sp3sp^3 hybridization with one lone pair on nitrogen, resulting in a pyramidal structure, not planar. Incorrect.

(C) ClF3ClF_3 is used as a fluorinating agent to produce UF6UF_6 from UU for the gaseous diffusion process of 235U^{235}U enrichment. Correct.

(D) In NCl3NCl_3, the electronegative chlorine atoms exert a strong inductive effect (−I-I), reducing the electron density on Nitrogen compared to NH3NH_3, making it a much weaker Lewis base. Incorrect.

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