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Structure of Atom question

2017 · Shift 0 · Q20
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Structure of Atom question

2017 · Shift 0 · Q20

JEE MainChemistryStructure of AtomMCQ+4 / −1
The group having isoelectronic species is:
  1. A
    O2−O^{2-}O2− , F−F^-F− , Na+Na^+Na+, Mg2+Mg^{2+}Mg2+
  2. B
    O2−O^{2-}O2− , F−F^-F− , NaNaNa, Mg2+Mg^{2+}Mg2+
  3. C
    O−O^-O− , F−F^-F− , Na+Na^+Na+, Mg2+Mg^{2+}Mg2+
  4. D
    O−O^-O− , F−F^-F− , NaNaNa, Mg+Mg^+Mg+
View written solutionFree

Correct answer: A

  1. Idea of isoelectronic species

    Species are isoelectronic if they have the same number of electrons.

  2. Count electrons in each species

    Atomic numbers:

    • O=8O = 8O=8
    • F=9F = 9F=9
    • Na=11Na = 11Na=11
    • Mg=12Mg = 12Mg=12

    Rules:

    • For anions, add electrons equal to the negative charge.
    • For cations, subtract electrons equal to the positive charge.

  3. Check Option A

    • O2−O^{2-}O2−: 8+2=108 + 2 = 108+2=10 electrons
    • F−F^{-}F−: 9+1=109 + 1 = 109+1=10 electrons
    • Na+Na^{+}Na+: 11−1=1011 - 1 = 1011−1=10 electrons
    • Mg2+Mg^{2+}Mg2+: 12−2=1012 - 2 = 1012−2=10 electrons

    All have 101010 electrons, so these are isoelectronic.

  4. Check Option B

    • O2−=10O^{2-} = 10O2−=10
    • F−=10F^- = 10F−=10
    • Na=11Na = 11Na=11
    • Mg2+=10Mg^{2+} = 10Mg2+=10

    NaNaNa has 111111 electrons, so not all are isoelectronic.

  5. Check Option C

    • O−:8+1=9O^-: 8 + 1 = 9O−:8+1=9
    • F−:9+1=10F^-: 9 + 1 = 10F−:9+1=10
    • Na+:11−1=10Na^+: 11 - 1 = 10Na+:11−1=10
    • Mg2+:12−2=10Mg^{2+}: 12 - 2 = 10Mg2+:12−2=10

    O−O^-O− has 999 electrons, so not all are isoelectronic.

  6. Check Option D

    • O−=9O^- = 9O−=9
    • F−=10F^- = 10F−=10
    • Na=11Na = 11Na=11
    • Mg+:12−1=11Mg^+: 12 - 1 = 11Mg+:12−1=11

    Electron counts are different, so not isoelectronic.

  7. Conclusion

    The correct group is: A\boxed{A}A​

  8. Comparison with stored answer

    Stored correct answer = AAA.

    My derived answer also = AAA, so they agree.

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