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Chemical Bonding and Molecular Structure question

2022 · 27 Jul · Shift 2 · Q2
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Chemical Bonding and Molecular Structure question

2022 · 27 Jul · Shift 2 · Q2

JEE MainChemistryChemical Bonding and Molecular StructureMCQ+4 / −1

Match List - I with List - II.

List - I List - II
(A) ψMO=ψA−ψB\psi_{\mathrm{MO}}=\psi_{\mathrm{A}}-\psi_{\mathrm{B}}ψMO​=ψA​−ψB​ (I) Dipole moment
(B) μ=Q×r\mu=Q \times rμ=Q×r (II) Bonding molecular orbital
(C) Nb−Na2\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}2Nb​−Na​​ (III) Anti-bonding molecular orbital
(D) ψMO=ψA+ψB\psi_{\mathrm{MO}}=\psi_{\mathrm{A}}+\psi_{\mathrm{B}}ψMO​=ψA​+ψB​ (IV) Bond order

Choose the correct answer from the options given below :

  1. A
    (A)−(II),(B)−(I),(C)−(IV),(D)−(III)(\mathrm{A})-(\mathrm{II}),(\mathrm{B})-(\mathrm{I}),(\mathrm{C})-(\mathrm{IV}),(\mathrm{D})-(\mathrm{III})(A)−(II),(B)−(I),(C)−(IV),(D)−(III)
  2. B
    (A)−(III),(B)−(IV),(C)−(I),(D)−(II)(\mathrm{A})-(\mathrm{III}),(\mathrm{B})-(\mathrm{IV}),(\mathrm{C})-(\mathrm{I}),(\mathrm{D})-(\mathrm{II})(A)−(III),(B)−(IV),(C)−(I),(D)−(II)
  3. C
    (A)−(III),(B)−(I),(C)−(IV),(D)−(II)(\mathrm{A})-(\mathrm{III}),(\mathrm{B})-(\mathrm{I}),(\mathrm{C})-(\mathrm{IV}),(\mathrm{D})-(\mathrm{II})(A)−(III),(B)−(I),(C)−(IV),(D)−(II)
  4. D
    (A)−(III),(B)−(IV),(C)−(II),(D)−(I)(\mathrm{A})-(\mathrm{III}),(\mathrm{B})-(\mathrm{IV}),(\mathrm{C})-(\mathrm{II}),(\mathrm{D})-(\mathrm{I})(A)−(III),(B)−(IV),(C)−(II),(D)−(I)
View written solutionFree

Correct answer: C

  1. We match each item in List-I with its correct meaning in List-II.

  2. Analyze each expression:

    (A) ψMO=ψA−ψB\psi_{\mathrm{MO}} = \psi_A - \psi_BψMO​=ψA​−ψB​

    • Subtraction of atomic wave functions gives destructive overlap.
    • This forms an anti-bonding molecular orbital.
    • So, (A) →\to→ (III).

  3. (B) μ=Q×r\mu = Q \times rμ=Q×r

    • This is the formula for dipole moment.
    • So, (B) →\to→ (I).

  4. (C) Nb−Na2\dfrac{N_b - N_a}{2}2Nb​−Na​​

    • This is the formula for bond order.
    • Here, NbN_bNb​ = number of electrons in bonding orbitals, and NaN_aNa​ = number of electrons in antibonding orbitals.
    • So, (C) →\to→ (IV).

  5. (D) ψMO=ψA+ψB\psi_{\mathrm{MO}} = \psi_A + \psi_BψMO​=ψA​+ψB​

    • Addition of atomic wave functions gives constructive overlap.
    • This forms a bonding molecular orbital.
    • So, (D) →\to→ (II).

  6. Final matching:

(A)−(III),(B)−(I),(C)−(IV),(D)−(II)(A)-(III),\quad (B)-(I),\quad (C)-(IV),\quad (D)-(II)(A)−(III),(B)−(I),(C)−(IV),(D)−(II)
  1. Compare with options:
  • This corresponds to Option C.

Therefore, the correct answer is C.

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