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Structure of Atom question

2023 · 1 Feb · Shift 2 · Q10
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  5. /2023 · 1 Feb · Shift 2 · Q10

Structure of Atom question

2023 · 1 Feb · Shift 2 · Q10

JEE MainChemistryStructure of AtomMCQ+4 / −1
Which one of the following sets of ions represents a collection of isoelectronic species? (Given : Atomic Number : F:9,Cl:17,Na=11,Mg=12,Al=13,K=19,Ca=20,Sc=21\mathrm{F:9,Cl:17,Na=11,Mg=12,Al=13,K=19,Ca=20,Sc=21}F:9,Cl:17,Na=11,Mg=12,Al=13,K=19,Ca=20,Sc=21)
  1. A
    N3−,O2−,F−,S2−\mathrm{N}^{3-}, \mathrm{O}^{2-}, \mathrm{F}^{-}, \mathrm{S}^{2-}N3−,O2−,F−,S2−
  2. B
    Ba2+,Sr2+,K+,Ca2+\mathrm{Ba}^{2+}, \mathrm{Sr}^{2+}, \mathrm{K}^{+}, \mathrm{Ca}^{2+}Ba2+,Sr2+,K+,Ca2+
  3. C
    Li+,Na+,Mg2+,Ca2+\mathrm{Li}^{+}, \mathrm{Na}^{+}, \mathrm{Mg}^{2+}, \mathrm{Ca}^{2+}Li+,Na+,Mg2+,Ca2+
  4. D
    K+,Cl−,Ca2+,Sc3+\mathrm{K}^{+}, \mathrm{Cl}^{-}, \mathrm{Ca}^{2+}, \mathrm{Sc}^{3+}K+,Cl−,Ca2+,Sc3+
View written solutionFree

Correct answer: D

  1. Concept: Isoelectronic species

    Isoelectronic species are atoms/ions having the same number of electrons.

    So, we calculate the number of electrons in each ion:

    • For a cation: electrons =Z−= Z -=Z− positive charge
    • For an anion: electrons =Z+= Z +=Z+ magnitude of negative charge

  2. Check each option

Option A: N3−,O2−,F−,S2−\mathrm{N}^{3-}, \mathrm{O}^{2-}, \mathrm{F}^{-}, \mathrm{S}^{2-}N3−,O2−,F−,S2−

  • N3−\mathrm{N}^{3-}N3−: 7+3=107 + 3 = 107+3=10 electrons
  • O2−\mathrm{O}^{2-}O2−: 8+2=108 + 2 = 108+2=10 electrons
  • F−\mathrm{F}^{-}F−: 9+1=109 + 1 = 109+1=10 electrons
  • S2−\mathrm{S}^{2-}S2−: 16+2=1816 + 2 = 1816+2=18 electrons

These are not all equal, so not isoelectronic.

Option B: Ba2+,Sr2+,K+,Ca2+\mathrm{Ba}^{2+}, \mathrm{Sr}^{2+}, \mathrm{K}^{+}, \mathrm{Ca}^{2+}Ba2+,Sr2+,K+,Ca2+

  • Ba2+\mathrm{Ba}^{2+}Ba2+: 56−2=5456 - 2 = 5456−2=54 electrons
  • Sr2+\mathrm{Sr}^{2+}Sr2+: 38−2=3638 - 2 = 3638−2=36 electrons
  • K+\mathrm{K}^{+}K+: 19−1=1819 - 1 = 1819−1=18 electrons
  • Ca2+\mathrm{Ca}^{2+}Ca2+: 20−2=1820 - 2 = 1820−2=18 electrons

These are not all equal, so not isoelectronic.

Option C: Li+,Na+,Mg2+,Ca2+\mathrm{Li}^{+}, \mathrm{Na}^{+}, \mathrm{Mg}^{2+}, \mathrm{Ca}^{2+}Li+,Na+,Mg2+,Ca2+

  • Li+\mathrm{Li}^{+}Li+: 3−1=23 - 1 = 23−1=2 electrons
  • Na+\mathrm{Na}^{+}Na+: 11−1=1011 - 1 = 1011−1=10 electrons
  • Mg2+\mathrm{Mg}^{2+}Mg2+: 12−2=1012 - 2 = 1012−2=10 electrons
  • Ca2+\mathrm{Ca}^{2+}Ca2+: 20−2=1820 - 2 = 1820−2=18 electrons

These are not all equal, so not isoelectronic.

Option D: K+,Cl−,Ca2+,Sc3+\mathrm{K}^{+}, \mathrm{Cl}^{-}, \mathrm{Ca}^{2+}, \mathrm{Sc}^{3+}K+,Cl−,Ca2+,Sc3+

  • K+\mathrm{K}^{+}K+: 19−1=1819 - 1 = 1819−1=18 electrons
  • Cl−\mathrm{Cl}^{-}Cl−: 17+1=1817 + 1 = 1817+1=18 electrons
  • Ca2+\mathrm{Ca}^{2+}Ca2+: 20−2=1820 - 2 = 1820−2=18 electrons
  • Sc3+\mathrm{Sc}^{3+}Sc3+: 21−3=1821 - 3 = 1821−3=18 electrons

All have 18 electrons, so these are isoelectronic species.

  1. Final Answer

The correct option is:

D\boxed{\text{D}}D​

  1. Comparison with stored answer

Stored correct answer: D\mathrm{D}D

My derived answer: D\mathrm{D}D

They match.

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