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P Block Elements question

2009 · Shift 2 · Q10
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P Block Elements question

2009 · Shift 2 · Q10

JEE AdvancedChemistryP Block ElementsMCQ+3 / −1

Match each of the reactions given in Column I with the corresponding product(s) given in Column II:

Column I Column II
(A) Cu+dil. HNO3\mathrm{Cu+dil.~HNO_3}Cu+dil. HNO3​ (P) NO\mathrm{NO}NO
(B) Cu+conc. HNO3\mathrm{Cu+conc.~HNO_3}Cu+conc. HNO3​ (Q) NO2\mathrm{NO_2}NO2​
(C) Zn+dil. HNO3\mathrm{Zn+dil.~HNO_3}Zn+dil. HNO3​ (R) N2O\mathrm{N_2O}N2​O
(D) Zn+conc. HNO3\mathrm{Zn+conc.~HNO_3}Zn+conc. HNO3​ (S) Cu(NO3)2\mathrm{Cu(NO_3)_2}Cu(NO3​)2​
(T) Zn(NO3)2\mathrm{Zn(NO_3)_2}Zn(NO3​)2​

  1. A
    (A) →\to→(P), (S); (B) →\to→(R), (S); (C) →\to→(R), (T); (D) →\to→(R), (T)
  2. B
    (A) →\to→(T), (S); (B) →\to→(P), (S); (C) →\to→(R), (T); (D) →\to→(S), (T)
  3. C
    (A) →\to→(P), (S); (B) →\to→(Q), (S); (C) →\to→(R), (T); (D) →\to→(Q), (T)
  4. D
    (A) →\to→(R), (S); (B) →\to→(Q), (S); (C) →\to→(R), (T); (D) →\to→(S), (T)
View written solutionFree

Correct answer: C

  1. Identify the products of metals with nitric acid

Nitric acid is an oxidizing acid, so metals generally form their nitrates and nitrogen oxides instead of evolving H2H_2H2​.

We match each case:

  1. Reaction (A): Cu+dil. HNO3\mathrm{Cu + dil.\ HNO_3}Cu+dil. HNO3​

Copper with dilute nitric acid gives NO\mathrm{NO}NO and copper nitrate:

3Cu+8HNO3(dil)→3Cu(NO3)2+2NO+4H2O3\mathrm{Cu} + 8\mathrm{HNO_3(dil)} \rightarrow 3\mathrm{Cu(NO_3)_2} + 2\mathrm{NO} + 4\mathrm{H_2O}3Cu+8HNO3​(dil)→3Cu(NO3​)2​+2NO+4H2​O

So,

  • product gas: (P)  NO(P)\; \mathrm{NO}(P)NO
  • salt: (S)  Cu(NO3)2(S)\; \mathrm{Cu(NO_3)_2}(S)Cu(NO3​)2​

Thus,

(A)→(P),(S)(A) \to (P), (S)(A)→(P),(S)

  1. Reaction (B): Cu+conc. HNO3\mathrm{Cu + conc.\ HNO_3}Cu+conc. HNO3​

Copper with concentrated nitric acid gives NO2\mathrm{NO_2}NO2​ and copper nitrate:

Cu+4HNO3(conc)→Cu(NO3)2+2NO2+2H2O\mathrm{Cu} + 4\mathrm{HNO_3(conc)} \rightarrow \mathrm{Cu(NO_3)_2} + 2\mathrm{NO_2} + 2\mathrm{H_2O}Cu+4HNO3​(conc)→Cu(NO3​)2​+2NO2​+2H2​O

So,

  • product gas: (Q)  NO2(Q)\; \mathrm{NO_2}(Q)NO2​
  • salt: (S)  Cu(NO3)2(S)\; \mathrm{Cu(NO_3)_2}(S)Cu(NO3​)2​

Thus,

(B)→(Q),(S)(B) \to (Q), (S)(B)→(Q),(S)

  1. Reaction (C): Zn+dil. HNO3\mathrm{Zn + dil.\ HNO_3}Zn+dil. HNO3​

For active metals like zinc, with dilute nitric acid, the reduction product can be N2O\mathrm{N_2O}N2​O (among other possibilities depending on conditions). The standard match here is:

Zn+dil. HNO3→Zn(NO3)2+N2O+⋯\mathrm{Zn} + \mathrm{dil.\ HNO_3} \rightarrow \mathrm{Zn(NO_3)_2} + \mathrm{N_2O} + \cdotsZn+dil. HNO3​→Zn(NO3​)2​+N2​O+⋯

So,

  • product gas: (R)  N2O(R)\; \mathrm{N_2O}(R)N2​O
  • salt: (T)  Zn(NO3)2(T)\; \mathrm{Zn(NO_3)_2}(T)Zn(NO3​)2​

Thus,

(C)→(R),(T)(C) \to (R), (T)(C)→(R),(T)

  1. Reaction (D): Zn+conc. HNO3\mathrm{Zn + conc.\ HNO_3}Zn+conc. HNO3​

Zinc with concentrated nitric acid gives NO2\mathrm{NO_2}NO2​ and zinc nitrate:

Zn+4HNO3(conc)→Zn(NO3)2+2NO2+2H2O\mathrm{Zn} + 4\mathrm{HNO_3(conc)} \rightarrow \mathrm{Zn(NO_3)_2} + 2\mathrm{NO_2} + 2\mathrm{H_2O}Zn+4HNO3​(conc)→Zn(NO3​)2​+2NO2​+2H2​O

So,

  • product gas: (Q)  NO2(Q)\; \mathrm{NO_2}(Q)NO2​
  • salt: (T)  Zn(NO3)2(T)\; \mathrm{Zn(NO_3)_2}(T)Zn(NO3​)2​

Thus,

(D)→(Q),(T)(D) \to (Q), (T)(D)→(Q),(T)

  1. Final matching

Collecting all:

  • (A)→(P),(S)(A) \to (P), (S)(A)→(P),(S)
  • (B)→(Q),(S)(B) \to (Q), (S)(B)→(Q),(S)
  • (C)→(R),(T)(C) \to (R), (T)(C)→(R),(T)
  • (D)→(Q),(T)(D) \to (Q), (T)(D)→(Q),(T)

This corresponds to Option C.

  1. Comparison with stored answer

Stored correct answer: C

My derived answer: C

So they agree.

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