JEE Challenger
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ChemMedMarks to AllAdv 2017 • Paper 1

Count of Diamagnetic Molecular Species Among Given Molecules

Find the total count of diamagnetic species present among the following: H2\text{H}_2, He2+\text{He}_2^+, Li2\text{Li}_2, Be2\text{Be}_2, B2\text{B}_2, C2\text{C}_2, N2\text{N}_2, O2−\text{O}_2^-, and F2\text{F}_2.

(Atomic numbers: H=1\text{H} = 1, He=2\text{He} = 2, Li=3\text{Li} = 3, Be=4\text{Be} = 4, B=5\text{B} = 5, C=6\text{C} = 6, N=7\text{N} = 7, O=8\text{O} = 8, F=9\text{F} = 9)

Official Notice: Marks Awarded to All

This question was dropped / full marks were awarded to all candidates in the official answer key by the exam conducting body due to an ambiguity or error in the question or options.

Topics & Concepts

Chemical Bonding and Molecular StructureMolecular Orbital Theory

Step-by-Step Solution

To determine the total count of diamagnetic species among the given molecules and ions, we analyze their electronic configurations using Molecular Orbital (MO) Theory. A species is diamagnetic if all of its electrons are paired (zero unpaired electrons), and paramagnetic if it contains at least one unpaired electron.

Detailed Analysis of Each Species

  1. H2\text{H}_2 (Total electrons = 22):

    • MO Configuration: (σ1s)2(\sigma 1s)^2
    • Unpaired electrons: 00
    • Nature: Diamagnetic

  2. He2+\text{He}_2^+ (Total electrons = 33):

    • MO Configuration: (σ1s)2(σ∗1s)1(\sigma 1s)^2 (\sigma^* 1s)^1
    • Unpaired electrons: 11
    • Nature: Paramagnetic

  3. Li2\text{Li}_2 (Total electrons = 66):

    • MO Configuration: (σ1s)2(σ∗1s)2(σ2s)2(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2
    • Unpaired electrons: 00
    • Nature: Diamagnetic

  4. Be2\text{Be}_2 (Total electrons = 88):

    • MO Configuration: (σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2
    • Bond Order: Nb−Na2=4−42=0\frac{N_b - N_a}{2} = \frac{4 - 4}{2} = 0
    • Unpaired electrons: 00
    • Nature: Purely based on electronic configuration, all electrons are paired (Diamagnetic). However, because its bond order is 00, Be2\text{Be}_2 is considered an unstable species under standard conditions.

  5. B2\text{B}_2 (Total electrons = 1010):

    • MO Configuration: (σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(π2px)1(π2py)1(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^1 (\pi 2p_y)^1
    • Unpaired electrons: 22
    • Nature: Paramagnetic

  6. C2\text{C}_2 (Total electrons = 1212):

    • MO Configuration: (σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(π2px)2(π2py)2(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2
    • Unpaired electrons: 00
    • Nature: Diamagnetic

  7. N2\text{N}_2 (Total electrons = 1414):

    • MO Configuration: (σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)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
    • Unpaired electrons: 00
    • Nature: Diamagnetic

  8. O2−\text{O}_2^- (Total electrons = 1717):

    • MO Configuration: (σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(σ2pz)2(π2px)2(π2py)2(π∗2px)2(π∗2py)1(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^2 (\pi^* 2p_y)^1
    • Unpaired electrons: 11
    • Nature: Paramagnetic

  9. F2\text{F}_2 (Total electrons = 1818):

    • MO Configuration: (σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(σ2pz)2(π2px)2(π2py)2(π∗2px)2(π∗2py)2(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^2 (\pi^* 2p_y)^2
    • Unpaired electrons: 00
    • Nature: Diamagnetic

Conclusion

  • Definite diamagnetic molecules: H2,Li2,C2,N2,F2\text{H}_2, \text{Li}_2, \text{C}_2, \text{N}_2, \text{F}_2 (Total = 55).
  • Including Be2\text{Be}_2 (due to fully paired configuration despite zero bond order): Total = 66.

Due to the ambiguity regarding whether Be2\text{Be}_2 should be counted as a valid molecular species, Marks to All (+3) was officially awarded for this question.

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