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Chemical Bonding and Molecular Structure question

2021 · 26 Feb · Shift 2 · Q1
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Chemical Bonding and Molecular Structure question

2021 · 26 Feb · Shift 2 · Q1

JEE MainChemistryChemical Bonding and Molecular StructureMCQ+4 / −1
Match List - I with List - II.

List - I (Molecule) List - II (Bond order)
(a) Ne2N{e_2}Ne2​ (i) 1
(b) N2{N_2}N2​ (ii) 2
(c) F2{F_2}F2​ (iii) 0
(d) O2{O_2}O2​ (iv) 3

Choose the correct answer from the options given below :
  1. A
    (a) →\to→ (i), (b) →\to→ (ii), (c) →\to→ (iii), (d) →\to→ (iv)
  2. B
    (a) →\to→ (iv), (b) →\to→ (iii), (c) →\to→ (ii), (d) →\to→ (i)
  3. C
    (a) →\to→ (iii), (b) →\to→ (iv), (c) →\to→ (i), (d) →\to→ (ii)
  4. D
    (a) →\to→ (ii), (b) →\to→ (i), (c) →\to→ (iv), (d) →\to→ (iii)
View written solutionFree

Correct answer: C

  1. Use MO theory bond order formula
Bond order=Nb−Na2\text{Bond order} = \frac{N_b - N_a}{2}Bond order=2Nb​−Na​​

where NbN_bNb​ = number of bonding electrons and NaN_aNa​ = number of antibonding electrons.

  1. Find bond order of each molecule

(a) Ne2Ne_2Ne2​

Each Ne atom has 10 electrons, so Ne2Ne_2Ne2​ has 20 electrons.

For Ne2Ne_2Ne2​, all bonding and antibonding molecular orbitals up to 2p2p2p are completely filled, so the number of bonding and antibonding electrons becomes equal.

Bond order=Nb−Na2=0\text{Bond order} = \frac{N_b - N_a}{2} = 0Bond order=2Nb​−Na​​=0

So,

(a)→(iii)(a) \to (iii)(a)→(iii)

(b) N2N_2N2​

Each N atom has 7 electrons, so N2N_2N2​ has 14 electrons.

MO configuration for N2N_2N2​:

(σ1s)2(σ1s∗)2(σ2s)2(σ2s∗)2(π2px)2(π2py)2(σ2pz)2(\sigma_{1s})^2(\sigma_{1s}^*)^2(\sigma_{2s})^2(\sigma_{2s}^*)^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\sigma_{2p_z})^2(σ1s​)2(σ1s∗​)2(σ2s​)2(σ2s∗​)2(π2px​​)2(π2py​​)2(σ2pz​​)2

Bonding electrons = 10, antibonding electrons = 4

Bond order=10−42=3\text{Bond order} = \frac{10-4}{2} = 3Bond order=210−4​=3

So,

(b)→(iv)(b) \to (iv)(b)→(iv)

(c) F2F_2F2​

Each F atom has 9 electrons, so F2F_2F2​ has 18 electrons.

For F2F_2F2​, bond order is known to be:

Bond order=1\text{Bond order} = 1Bond order=1

So,

(c)→(i)(c) \to (i)(c)→(i)

(d) O2O_2O2​

Each O atom has 8 electrons, so O2O_2O2​ has 16 electrons.

For O2O_2O2​, bond order is:

Bond order=2\text{Bond order} = 2Bond order=2

So,

(d)→(ii)(d) \to (ii)(d)→(ii)
  1. Final matching
(a)→(iii),(b)→(iv),(c)→(i),(d)→(ii)(a) \to (iii), \quad (b) \to (iv), \quad (c) \to (i), \quad (d) \to (ii)(a)→(iii),(b)→(iv),(c)→(i),(d)→(ii)

This corresponds to Option C.

  1. Comparison with stored correct answer

Stored correct answer = C

My derived answer = C

So they agree.

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