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Chemical Bonding and Molecular Structure question

2023 · 6 Apr · Shift 1 · Q7
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Chemical Bonding and Molecular Structure question

2023 · 6 Apr · Shift 1 · Q7

JEE MainChemistryChemical Bonding and Molecular StructureMCQ+4 / −1

Match List I with List II

LIST I
Oxide		 LIST II
Type of bond
A. N2O4\mathrm{N_2O_4}N2​O4​ I. 1 N = O bond
B. NO2\mathrm{NO_2}NO2​ II. 1 N −-− O −-− N bond
C. N2O5\mathrm{N_2O_5}N2​O5​ III. 1 N −-− N bond
D. N2O\mathrm{N_2O}N2​O IV. 1 N=N / N ≡\equiv≡ N bond

Choose the correct answer from the options given below:

  1. A
    A-II, B-I, C-III, D-IV
  2. B
    A-III, B-I, C-II, D-IV
  3. C
    A-III, B-I, C-IV, D-II
  4. D
    A-II, B-IV, C-III, D-I
View written solutionFree

Correct answer: B

  1. Identify the bonding present in each oxide

We match each compound with the characteristic bond type listed.

  1. Compound A: N2O4\mathrm{N_2O_4}N2​O4​

N2O4\mathrm{N_2O_4}N2​O4​ is the dimer of NO2\mathrm{NO_2}NO2​ and has the structure:

O2N−NO2\mathrm{O_2N-NO_2}O2​N−NO2​

So it contains one N−N\mathrm{N-N}N−N bond.

Thus,

A→IIIA \to IIIA→III

  1. Compound B: NO2\mathrm{NO_2}NO2​

In NO2\mathrm{NO_2}NO2​, nitrogen is bonded to two oxygens, and due to resonance one can identify the presence of one N=O\mathrm{N=O}N=O bond in a canonical structure.

Thus,

B→IB \to IB→I

  1. Compound C: N2O5\mathrm{N_2O_5}N2​O5​

N2O5\mathrm{N_2O_5}N2​O5​ has the structure:

O2N−O−NO2\mathrm{O_2N-O-NO_2}O2​N−O−NO2​

So it contains one bridging N−O−N\mathrm{N-O-N}N−O−N bond.

Thus,

C→IIC \to IIC→II

  1. Compound D: N2O\mathrm{N_2O}N2​O

Nitrous oxide has resonance structures such as:

N≡N−O↔N=N=O\mathrm{N\equiv N-O} \leftrightarrow \mathrm{N=N=O}N≡N−O↔N=N=O

Hence it is characterized by one N=N\mathrm{N=N}N=N / N≡N\mathrm{N\equiv N}N≡N bond.

Thus,

D→IVD \to IVD→IV

  1. Final matching

A−III,  B−I,  C−II,  D−IVA-III,\; B-I,\; C-II,\; D-IVA−III,B−I,C−II,D−IV

This corresponds to Option B.

  1. Comparison with stored answer

Stored correct answer = B

Our derived answer = B

So, they agree.

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