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Structure of Atom question

2004 · Shift 0 · Q58
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Structure of Atom question

2004 · Shift 0 · Q58

JEE MainChemistryStructure of AtomMCQ+4 / −1
Which one of the following sets of ions represents the collection of isoelectronic species? (Atomic nos. : F = 9, Cl = 17, Na = 11, Mg = 12, Al = 13, K = 19, Ca = 20, Sc = 21)
  1. A
    K+K^+K+, Cl−Cl^-Cl−, Mg2+Mg^{2+}Mg2+, Sc3+Sc^{3+}Sc3+
  2. B
    Na+Na^+Na+, Ca2+Ca^{2+}Ca2+, Sc3+Sc^{3+}Sc3+, F−F^-F−
  3. C
    K+K^+K+, Ca2+Ca^{2+}Ca2+, Sc3+Sc^{3+}Sc3+, Cl−Cl^-Cl−
  4. D
    Na+Na^+Na+, Mg2+Mg^{2+}Mg2+, Al3+Al^{3+}Al3+, Cl−Cl^-Cl−
View written solutionFree

Correct answer: C

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

  2. Let us calculate the number of electrons in each ion using: Number of electrons=Z−positive charge\text{Number of electrons} = Z - \text{positive charge}Number of electrons=Z−positive charge for cations, and Number of electrons=Z+negative charge\text{Number of electrons} = Z + \text{negative charge}Number of electrons=Z+negative charge for anions.


  1. Check each option:

Option A: K+K^+K+, Cl−Cl^-Cl−, Mg2+Mg^{2+}Mg2+, Sc3+Sc^{3+}Sc3+

  • K+K^+K+: 19−1=1819 - 1 = 1819−1=18
  • Cl−Cl^-Cl−: 17+1=1817 + 1 = 1817+1=18
  • Mg2+Mg^{2+}Mg2+: 12−2=1012 - 2 = 1012−2=10
  • Sc3+Sc^{3+}Sc3+: 21−3=1821 - 3 = 1821−3=18

Not all have the same number of electrons. So, A is incorrect.


Option B: Na+Na^+Na+, Ca2+Ca^{2+}Ca2+, Sc3+Sc^{3+}Sc3+, F−F^-F−

  • Na+Na^+Na+: 11−1=1011 - 1 = 1011−1=10
  • Ca2+Ca^{2+}Ca2+: 20−2=1820 - 2 = 1820−2=18
  • Sc3+Sc^{3+}Sc3+: 21−3=1821 - 3 = 1821−3=18
  • F−F^-F−: 9+1=109 + 1 = 109+1=10

Two ions have 10 electrons and two have 18 electrons. So, B is incorrect.


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

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

All have 18 electrons. So, C is correct.


Option D: Na+Na^+Na+, Mg2+Mg^{2+}Mg2+, Al3+Al^{3+}Al3+, Cl−Cl^-Cl−

  • Na+Na^+Na+: 11−1=1011 - 1 = 1011−1=10
  • Mg2+Mg^{2+}Mg2+: 12−2=1012 - 2 = 1012−2=10
  • Al3+Al^{3+}Al3+: 13−3=1013 - 3 = 1013−3=10
  • Cl−Cl^-Cl−: 17+1=1817 + 1 = 1817+1=18

Three ions have 10 electrons, but Cl−Cl^-Cl− has 18 electrons. So, D is incorrect.


  1. Therefore, the collection of isoelectronic species is: K+, Ca2+, Sc3+, Cl−\boxed{K^+,\ Ca^{2+},\ Sc^{3+},\ Cl^-}K+, Ca2+, Sc3+, Cl−​

So, the correct option is: C\boxed{\text{C}}C​

  1. Comparison with stored correct answer: Stored correct answer = C

My derived answer = C

Hence, they agree.

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