Question

Which one of the following sets of ions represents the collection of isoelectronic species? (a) K⁺, Cl⁻, Mg²⁺, Sc³⁺ (b) Na⁺, Ca²⁺, Sc³⁺, F⁻(c) K⁺, Ca²⁺, Sc³⁺, Cl⁻ (d) Na⁺, Mg²⁺, Al³⁺, Cl⁻(Atomic nos. : F = 9, Cl = 17, Na = 11, Mg = 12, Al = 13, K = 19, Ca = 20, Sc = 21)

Answer 1

The correct answer is option C they all have electronic configuration of Ar

Answer 2

Answer:

The answer is (c) $K^+, Ca^2^+, Sc^3^+, Cl^-$

Explanation:

Isoelectronic species are atoms or ions having the same number of electrons in

(i) In the first set a)$K^+, Cl^-, Mg^2^+, Sc^3^+$

• The atomic number of K=19, So $K^+$  the ion has 18 electrons
• The atomic number of Cl=17, So $Cl^-$ the ion has 18 electrons
• The atomic number of Mg=12, So $Mg^2^+$ the ion has 10 electrons
• The atomic number of Sc=21, So $Sc^3^+$ the ion has 18 electrons

(ii) In the second set  (b)$Na^+, Ca^2^+, Sc^3^+, F^-$

• The atomic number of Na= 11, So $Na^+$ the ion has 10 electrons
• The atomic number of  Ca=20, so $Ca^2^+$ the ion has 18 electrons
• The atomic number of Sc=21, So $Sc^3^+$ the ion has 18 electrons
• The atomic number of F=9, so $F^-$ the ion has 10 electrons

(iii) In the third set (c) $K^+, Ca^2^+, Sc^3^+, Cl^-$

• The atomic number of K=19, So $K^+$  the ion has 18 electrons
• The atomic number of  Ca=20, so $Ca^2^+$ the ion has 18 electrons
• The atomic number of Sc=21, So $Sc^3^+$ the ion has 18 electrons
• The atomic number of Cl=17, So $Cl^-$ the ion has 18 electrons

(iv) In the fourth set  (d) $Na^+, Mg^2^+, Al^3^+, Cl^-$

• The atomic number of Na= 11, So $Na^+$ the ion has 10 electrons
• The atomic number of Mg=12, So $Mg^2^+$ the ion has 10 electrons
• The atomic number of Al=13, So $Al^3^+$ the ion has 10 electrons
• The atomic number of Cl=17, So $Cl^-$ the ion has 18 electrons

Hence the ions in the third set have the same number of electrons.

Therefore the (c) $K^+, Ca^2^+, Sc^3^+ ,Cl^-$ are isoelectronic species.