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Chemical Bonding and Molecular Structure question

2009 · Shift 0 · Q22
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Chemical Bonding and Molecular Structure question

2009 · Shift 0 · Q22

JEE MainChemistryChemical Bonding and Molecular StructureMCQ+4 / −1
Using MO theory, predict which of the following species has the shortest bond length?
  1. A
    O2+O_2^+O2+​
  2. B
    O2−O_2^-O2−​
  3. C
    O22−O_2^{2-}O22−​
  4. D
    O22+O_2^{2+}O22+​
View written solutionFree

Correct answer: D

  1. Principle used

For the same pair of atoms, shorter bond length corresponds to higher bond order.

So we will find the bond order of each species using MO theory.

  1. MO configuration of oxygen species

For O2O_2O2​ and related species, the relevant MO ordering is:

σ(2s), σ∗(2s), σ(2pz), π(2px)=π(2py), π∗(2px)=π∗(2py), σ∗(2pz)\sigma(2s),\ \sigma^*(2s),\ \sigma(2p_z),\ \pi(2p_x)=\pi(2p_y),\ \pi^*(2p_x)=\pi^*(2p_y),\ \sigma^*(2p_z)σ(2s), σ∗(2s), σ(2pz​), π(2px​)=π(2py​), π∗(2px​)=π∗(2py​), σ∗(2pz​)

Neutral O2O_2O2​ has 12 valence electrons, with configuration:

σ(2s)2 σ∗(2s)2 σ(2pz)2 (π(2px))2 (π(2py))2 (π∗(2px))1 (π∗(2py))1\sigma(2s)^2\,\sigma^*(2s)^2\,\sigma(2p_z)^2\,(\pi(2p_x))^2\,(\pi(2p_y))^2\,(\pi^*(2p_x))^1\,(\pi^*(2p_y))^1σ(2s)2σ∗(2s)2σ(2pz​)2(π(2px​))2(π(2py​))2(π∗(2px​))1(π∗(2py​))1

Bond order is:

B.O.=Nb−Na2\text{B.O.} = \frac{N_b - N_a}{2}B.O.=2Nb​−Na​​

where NbN_bNb​ = bonding electrons, NaN_aNa​ = antibonding electrons.

For O2O_2O2​:

B.O.=2\text{B.O.} = 2B.O.=2
  1. Now evaluate each species

(A) O2+O_2^+O2+​

Removing one electron from the highest occupied MO, i.e. from π∗\pi^*π∗.

So antibonding electrons decrease by 1.

B.O.(O2+)=2+12=2.5\text{B.O.}(O_2^+) = 2 + \frac{1}{2} = 2.5B.O.(O2+​)=2+21​=2.5

(B) O2−O_2^-O2−​

Adding one electron to π∗\pi^*π∗ increases antibonding electrons by 1.

B.O.(O2−)=2−12=1.5\text{B.O.}(O_2^-) = 2 - \frac{1}{2} = 1.5B.O.(O2−​)=2−21​=1.5

(C) O22−O_2^{2-}O22−​

Adding two electrons to π∗\pi^*π∗ increases antibonding electrons by 2.

B.O.(O22−)=2−1=1\text{B.O.}(O_2^{2-}) = 2 - 1 = 1B.O.(O22−​)=2−1=1

(D) O22+O_2^{2+}O22+​

Removing two electrons from π∗\pi^*π∗ decreases antibonding electrons by 2.

B.O.(O22+)=2+1=3\text{B.O.}(O_2^{2+}) = 2 + 1 = 3B.O.(O22+​)=2+1=3
  1. Compare bond orders
O22+(3)>O2+(2.5)>O2(2)>O2−(1.5)>O22−(1)O_2^{2+} (3) > O_2^+ (2.5) > O_2 (2) > O_2^- (1.5) > O_2^{2-} (1)O22+​(3)>O2+​(2.5)>O2​(2)>O2−​(1.5)>O22−​(1)

Highest bond order means shortest bond length.

Therefore, the species with the shortest bond length is:

O22+\boxed{O_2^{2+}}O22+​​

So the correct option is D.

  1. Comparison with stored correct answer

Stored correct answer: D

My derived answer: D

They agree.

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