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MCQEasyJEE 2023Electronic Configuration

JEE Chemistry 2023 Question with Solution

Which one of the following sets of ions represents a collection of isoelectronic species? (Given : Atomic Number : F :99, Cl : 1717, Na = 1111, Mg = 1212, Al = 1313, K = 1919, Ca = 2020, Sc = 2121)

  • A

    (Li+\text{Li}^{+}, Na+\text{Na}^{+}, Mg2+\text{Mg}^{2+}, Ca2+\text{Ca}^{2+})

  • B

    (Ba2+\text{Ba}^{2+}, Sr2+\text{Sr}^{2+}, K+\text{K}^{+}, Ca2+\text{Ca}^{2+})

  • C

    (N3−\text{N}^{3-}, O2−\text{O}^{2-}, F−\text{F}^{-}, S2−\text{S}^{2-})

  • D

    (K+\text{K}^{+}, Cl−\text{Cl}^{-}, Ca2+\text{Ca}^{2+}, Sc3+\text{Sc}^{3+})

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: We must identify the set of ions that are isoelectronic.

Find: The option in which all species have the same number of electrons.

Isoelectronic species have the same number of electrons.

Check each set using atomic number and ionic charge:

Li+=3−1=2,Na+=11−1=10,Mg2+=12−2=10,Ca2+=20−2=18\text{Li}^{+} = 3 - 1 = 2, \quad \text{Na}^{+} = 11 - 1 = 10, \quad \text{Mg}^{2+} = 12 - 2 = 10, \quad \text{Ca}^{2+} = 20 - 2 = 18

So option A is not isoelectronic.

Ba2+=56−2=54,Sr2+=38−2=36,K+=19−1=18,Ca2+=20−2=18\text{Ba}^{2+} = 56 - 2 = 54, \quad \text{Sr}^{2+} = 38 - 2 = 36, \quad \text{K}^{+} = 19 - 1 = 18, \quad \text{Ca}^{2+} = 20 - 2 = 18

So option B is not isoelectronic.

N3−=7+3=10,O2−=8+2=10,F−=9+1=10,S2−=16+2=18\text{N}^{3-} = 7 + 3 = 10, \quad \text{O}^{2-} = 8 + 2 = 10, \quad \text{F}^{-} = 9 + 1 = 10, \quad \text{S}^{2-} = 16 + 2 = 18

So option C is not isoelectronic.

K+=19−1=18,Cl−=17+1=18,Ca2+=20−2=18,Sc3+=21−3=18\text{K}^{+} = 19 - 1 = 18, \quad \text{Cl}^{-} = 17 + 1 = 18, \quad \text{Ca}^{2+} = 20 - 2 = 18, \quad \text{Sc}^{3+} = 21 - 3 = 18

All species in option D have 1818 electrons.

Therefore, the isoelectronic set is option D.

The solution states "The Correct Option is B", but the worked electron counts clearly show that option D is correct. Hence the defensible answer is D.

Electron Counting Shortcut

Given: Each option contains ions.

Find: The set with the same electron count throughout.

Shortcut idea: For cations, subtract the charge from atomic number. For anions, add the magnitude of charge.

Directly inspect the last set:

K+=19−1=18,Cl−=17+1=18,Ca2+=20−2=18,Sc3+=21−3=18\text{K}^{+} = 19 - 1 = 18, \quad \text{Cl}^{-} = 17 + 1 = 18, \quad \text{Ca}^{2+} = 20 - 2 = 18, \quad \text{Sc}^{3+} = 21 - 3 = 18

All four ions give 1818 electrons, so this set is immediately identified as isoelectronic.

Therefore, the correct option is D.

Common mistakes

  • Counting electrons of ions using atomic number without adjusting for charge. A positive charge means electrons are lost and a negative charge means electrons are gained. Always subtract for cations and add for anions.

  • Trusting the listed option label in the solution without checking the actual working. Here the heading says option B, but the electron counts clearly match option D. Verify with calculation.

  • Assuming that if three species in a set are isoelectronic, the whole set must be correct. Every species in the set must have the same number of electrons, so each one must be checked.

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