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 I Column II (A) (CH 3 ) 2 SiCl 2 (P) Hydrogen halide formation (B) XeF 4 (Q) Redox reaction (C) Cl 2 (R) Reacts with glass (D) VCl 5 (S) Polymerisation (T) O 2 formation

(A)→(P), (S); (B)→(P), (Q), (R), (T); (C)→(P), (Q); (D)→(P)

(A)→(P); (B)→(P), (Q), (R), (T); (C)→(P); (D)→(P)

(A)→(P), (S); (B)→(P), (R), (T); (C)→(P), (Q); (D)→(P)

(A)→(P), (S); (B)→(P), (Q), (R); (C)→(P), (Q); (D)→(P)

P Block Elements PYQ — JEE Advanced 2010 · Shift 2 · Q19 with Solution

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 I	Column II
(A) (CH₃)₂SiCl₂	(P) Hydrogen halide formation
(B) XeF₄	(Q) Redox reaction
(C) Cl₂	(R) Reacts with glass
(D) VCl₅	(S) Polymerisation
	(T) O₂ formation

A
(A)→(P); (B)→(P), (Q), (R), (T); (C)→(P); (D)→(P)
B
(A)→(P), (S); (B)→(P), (Q), (R), (T); (C)→(P), (Q); (D)→(P)
C
(A)→(P), (S); (B)→(P), (R), (T); (C)→(P), (Q); (D)→(P)
D
(A)→(P), (S); (B)→(P), (Q), (R); (C)→(P), (Q); (D)→(P)

View written solutionFree

Correct answer: B

Analyze each compound with water

(A) $(CH_3)_2SiCl_2$

Hydrolysis of dialkyl dichlorosilane:

(CH_3)_2SiCl_2 + 2H_2O \rightarrow (CH_3)_2Si(OH)_2 + 2HCl

So, hydrogen halide formation occurs $$ $(P)$ .

Now the silanediol formed condenses:

n(CH_3)_2Si(OH)_2 \rightarrow [-(CH_3)_2Si-O-]_n + nH_2O

This is polymerisation $$ $(S)$ .

Hence:

(A) \rightarrow (P), (S)

(B) $XeF_4$

Hydrolysis of xenon tetrafluoride:

6XeF_4 + 12H_2O \rightarrow 2XeO_3 + 24HF + 4Xe + 3O_2

From this reaction:

$HF$ is formed $$ hydrogen halide formation $(P)$
Xe changes oxidation state, so it is a redox reaction $(Q)$
$HF$ reacts with glass ( $SiO_2$ ), so effectively the products react with glass $(R)$
$O_2$ is formed $$ $(T)$

Hence:

(B) \rightarrow (P), (Q), (R), (T)

(C) $Cl_2$

Chlorine reacts with water as:

Cl_2 + H_2O \rightleftharpoons HCl + HOCl

So hydrogen halide formation occurs $$ $(P)$ .

Also, chlorine undergoes disproportionation:

$Cl_2$ : oxidation state $0$
In $HCl$ : $-1$
In $HOCl$ : $+1$

So this is a redox reaction $$ $(Q)$ .

Hence:

(C) \rightarrow (P), (Q)

(D) $VCl_5$

Hydrolysis:

VCl_5 + H_2O \rightarrow  oxychlorides/oxide + HCl

The key result is formation of hydrogen halide $(P)$ .

No polymerisation, no oxygen formation, and no essential redox assignment here.

Hence:

(D) \rightarrow (P)

Now compare with options

We found:

$(A) \rightarrow (P), (S)$
$(B) \rightarrow (P), (Q), (R), (T)$
$(C) \rightarrow (P), (Q)$
$(D) \rightarrow (P)$

This exactly matches Option B.

Comparison with stored correct answer

Stored correct answer: B

Derived answer: B

So, the derived answer agrees with the stored correct answer.

More from P Block Elements

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 I	Column II
(A) (CH₃)₂SiCl₂	(P) Hydrogen halide formation
(B) XeF₄	(Q) Redox reaction
(C) Cl₂	(R) Reacts with glass
(D) VCl₅	(S) Polymerisation
	(T) O₂ formation

A
(A)→(P); (B)→(P), (Q), (R), (T); (C)→(P); (D)→(P)
B
(A)→(P), (S); (B)→(P), (Q), (R), (T); (C)→(P), (Q); (D)→(P)
C
(A)→(P), (S); (B)→(P), (R), (T); (C)→(P), (Q); (D)→(P)
D
(A)→(P), (S); (B)→(P), (Q), (R); (C)→(P), (Q); (D)→(P)

View written solutionFree

Correct answer: B

Analyze each compound with water

(A) $(CH_3)_2SiCl_2$

Hydrolysis of dialkyl dichlorosilane:

(CH_3)_2SiCl_2 + 2H_2O \rightarrow (CH_3)_2Si(OH)_2 + 2HCl

So, hydrogen halide formation occurs $$ $(P)$ .

Now the silanediol formed condenses:

n(CH_3)_2Si(OH)_2 \rightarrow [-(CH_3)_2Si-O-]_n + nH_2O

This is polymerisation $$ $(S)$ .

Hence:

(A) \rightarrow (P), (S)

(B) $XeF_4$

Hydrolysis of xenon tetrafluoride:

6XeF_4 + 12H_2O \rightarrow 2XeO_3 + 24HF + 4Xe + 3O_2

From this reaction:

$HF$ is formed $$ hydrogen halide formation $(P)$
Xe changes oxidation state, so it is a redox reaction $(Q)$
$HF$ reacts with glass ( $SiO_2$ ), so effectively the products react with glass $(R)$
$O_2$ is formed $$ $(T)$

Hence:

(B) \rightarrow (P), (Q), (R), (T)

(C) $Cl_2$

Chlorine reacts with water as:

Cl_2 + H_2O \rightleftharpoons HCl + HOCl

So hydrogen halide formation occurs $$ $(P)$ .

Also, chlorine undergoes disproportionation:

$Cl_2$ : oxidation state $0$
In $HCl$ : $-1$
In $HOCl$ : $+1$

So this is a redox reaction $$ $(Q)$ .

Hence:

(C) \rightarrow (P), (Q)

(D) $VCl_5$

Hydrolysis:

VCl_5 + H_2O \rightarrow  oxychlorides/oxide + HCl

The key result is formation of hydrogen halide $(P)$ .

No polymerisation, no oxygen formation, and no essential redox assignment here.

Hence:

(D) \rightarrow (P)

Now compare with options

We found:

$(A) \rightarrow (P), (S)$
$(B) \rightarrow (P), (Q), (R), (T)$
$(C) \rightarrow (P), (Q)$
$(D) \rightarrow (P)$

This exactly matches Option B.

Comparison with stored correct answer

Stored correct answer: B

Derived answer: B

So, the derived answer agrees with the stored correct answer.

P Block Elements question

P Block Elements question

(A) $(CH_3)_2SiCl_2$

(B) $XeF_4$

(C) $Cl_2$

(D) $VCl_5$

More from P Block Elements

P Block Elements question

P Block Elements question

(A) $(CH_3)_2SiCl_2$

(B) $XeF_4$

(C) $Cl_2$

(D) $VCl_5$

More from P Block Elements

(A) (CH3)2SiCl2(CH_3)_2SiCl_2(CH3​)2​SiCl2​

(B) XeF4XeF_4XeF4​

(C) Cl2Cl_2Cl2​

(D) VCl5VCl_5VCl5​

More from P Block Elements

(A) (CH3)2SiCl2(CH_3)_2SiCl_2(CH3​)2​SiCl2​

(B) XeF4XeF_4XeF4​

(C) Cl2Cl_2Cl2​

(D) VCl5VCl_5VCl5​

More from P Block Elements

(A) $(CH_3)_2SiCl_2$

(B) $XeF_4$

(C) $Cl_2$

(D) $VCl_5$

(A) $(CH_3)_2SiCl_2$

(B) $XeF_4$

(C) $Cl_2$

(D) $VCl_5$