JEE Challenger
More from p-Block Elements
ChemMedAdv 2010 • Paper 2

Match Reactions of Water with Inorganic Compounds

All the compounds listed in Column I react with water. Match the result of the respective reactions with the appropriate options listed in Column II.

Column IColumn IIA) (CH3)2SiCl2p) Hydrogen halide formationB) XeF4q) Redox reactionC) Cl2r) Reacts with glassD) VCl5s) Polymerizationt) O2 formation\begin{array}{ll} \textbf{Column I} & \textbf{Column II} \\ A)\ (CH_3)_2SiCl_2 & p)\ \text{Hydrogen halide formation} \\ B)\ XeF_4 & q)\ \text{Redox reaction} \\ C)\ Cl_2 & r)\ \text{Reacts with glass} \\ D)\ VCl_5 & s)\ \text{Polymerization} \\ & t)\ O_2\ \text{formation} \end{array}

Options

A

(A)→(P,S),  (B)→(P,Q,R,T),  (C)→(P,Q),  (D)→(P)(\text{A}) \rightarrow (\text{P}, \text{S}),\; (\text{B}) \rightarrow (\text{P}, \text{Q}, \text{R}, \text{T}),\; (\text{C}) \rightarrow (\text{P}, \text{Q}),\; (\text{D}) \rightarrow (\text{P})

Correct
B

(A)→(P),  (B)→(P,Q,R,T),  (C)→(P,Q),  (D)→(P)(\text{A}) \rightarrow (\text{P}),\; (\text{B}) \rightarrow (\text{P}, \text{Q}, \text{R}, \text{T}),\; (\text{C}) \rightarrow (\text{P}, \text{Q}),\; (\text{D}) \rightarrow (\text{P})

C

(A)→(P,S),  (B)→(P,R,T),  (C)→(Q),  (D)→(P)(\text{A}) \rightarrow (\text{P}, \text{S}),\; (\text{B}) \rightarrow (\text{P}, \text{R}, \text{T}),\; (\text{C}) \rightarrow (\text{Q}),\; (\text{D}) \rightarrow (\text{P})

D

(A)→(S),  (B)→(Q,R,T),  (C)→(P,Q),  (D)→(P)(\text{A}) \rightarrow (\text{S}),\; (\text{B}) \rightarrow (\text{Q}, \text{R}, \text{T}),\; (\text{C}) \rightarrow (\text{P}, \text{Q}),\; (\text{D}) \rightarrow (\text{P})

Topics & Concepts

p-Block Elementsp-Block ElementsRedox Reactions

Step-by-Step Solution

To solve the given matching problem, we analyze the chemical reaction of each compound in Column I with water (H2O\text{H}_2\text{O}) and compare the products/characteristics with the options in Column II:

1. (CH3)2SiCl2(\text{CH}_3)_2\text{SiCl}_2 (Dichlorodimethylsilane)

  • Hydrolysis Reaction:
    (CH3)2SiCl2+2H2O⟶(CH3)2Si(OH)2+2HCl(\text{CH}_3)_2\text{SiCl}_2 + 2\text{H}_2\text{O} \longrightarrow (\text{CH}_3)_2\text{Si(OH)}_2 + 2\text{HCl}
    This produces hydrogen chloride (HCl\text{HCl}), which is a hydrogen halide (p).

  • Polymerization:
    The formed dialkylsilanediol derivative, (CH3)2Si(OH)2(\text{CH}_3)_2\text{Si(OH)}_2, undergoes condensation polymerization to yield a linear silicone polymer:
    n(CH3)2Si(OH)2⟶[−O−Si(CH3)2−]n+nH2On(\text{CH}_3)_2\text{Si(OH)}_2 \longrightarrow \left[-\text{O}-\text{Si}(\text{CH}_3)_2-\right]_n + n\text{H}_2\text{O}
    This represents polymerization (s).

Thus, A matches with p, s.

2. XeF4\text{XeF}_4 (Xenon Tetrafluoride)

  • Hydrolysis Reaction:
    6XeF4+12H2O⟶2XeO3+4Xe+3O2+24HF6\text{XeF}_4 + 12\text{H}_2\text{O} \longrightarrow 2\text{XeO}_3 + 4\text{Xe} + 3\text{O}_2 + 24\text{HF}

  • Analysis of Products and Properties:

    1. Hydrogen Halide Formation (p): Hydrofluoric acid (HF\text{HF}) is produced.
    2. Redox Reaction (q): Xenon undergoes disproportionation where its oxidation state changes from +4+4 in XeF4\text{XeF}_4 to +6+6 in XeO3\text{XeO}_3 and 00 in Xe\text{Xe}. Additionally, oxygen is oxidized from −2-2 in H2O\text{H}_2\text{O} to 00 in O2\text{O}_2. Thus, this is a redox reaction.
    3. Reacts with Glass (r): The generated HF\text{HF} reacts with silica (SiO2\text{SiO}_2) present in glass:
      SiO2+4HF⟶SiF4+2H2O\text{SiO}_2 + 4\text{HF} \longrightarrow \text{SiF}_4 + 2\text{H}_2\text{O}
    4. O2\text{O}_2 Formation (t): Dioxygen gas (O2\text{O}_2) is released as a product.

Thus, B matches with p, q, r, t.

3. Cl2\text{Cl}_2 (Chlorine Gas)

  • Hydrolysis Reaction:
    Cl2+H2O⇌HCl+HOCl\text{Cl}_2 + \text{H}_2\text{O} \rightleftharpoons \text{HCl} + \text{HOCl}

  • Analysis of Products and Properties:

    1. Hydrogen Halide Formation (p): HCl\text{HCl} is formed.
    2. Redox Reaction (q): Chlorine undergoes disproportionation, changing its oxidation state from 00 in Cl2\text{Cl}_2 to −1-1 in HCl\text{HCl} and +1+1 in HOCl\text{HOCl}.

Thus, C matches with p, q.

4. VCl5\text{VCl}_5 (Vanadium Pentachloride)

  • Hydrolysis Reaction:
    VCl5+H2O⟶VOCl3+2HCl\text{VCl}_5 + \text{H}_2\text{O} \longrightarrow \text{VOCl}_3 + 2\text{HCl}
    (On complete hydrolysis: 2VCl5+5H2O⟶V2O5+10HCl2\text{VCl}_5 + 5\text{H}_2\text{O} \longrightarrow \text{V}_2\text{O}_5 + 10\text{HCl})

  • Analysis of Products and Properties:

    1. Hydrogen Halide Formation (p): Hydrogen chloride (HCl\text{HCl}) is formed.
    2. No Oxidation State Change: The oxidation state of Vanadium remains +5+5 throughout the hydrolysis, so it is not a redox reaction.

Thus, D matches with p.

Final Matching:

  • A →\rightarrow p, s
  • B →\rightarrow p, q, r, t
  • C →\rightarrow p, q
  • D →\rightarrow p
PreviousNext