Question

Identify the isoelectronic pair of ions from the following

• A. \(\text{Pr}^{3+},\ \text{Nd}^{3+}\)
• B. \(\text{Tb}^{3+},\ \text{Dy}^{2+}\)
• C. \(\text{Eu}^{2+},\ \text{Gd}^{3+}\)
• D. \(\text{Pr}^{3+},\ \text{Ce}^{4+}\)

Hint:

Isoelectronic species have the same number of electrons, calculated by subtracting the ion's charge from the atomic number.

Answer 1

The Correct Option is C

Isoelectronic ions have the same number of electrons. Let's calculate: - (1) \(\text{Pr}^{3+}\): Pr (59 electrons) - 3 = 56; \(\text{Nd}^{3+}\): Nd (60) - 3 = 57. Not isoelectronic.
- (2) \(\text{Tb}^{3+}\): Tb (65) - 3 = 62; \(\text{Dy}^{2+}\): Dy (66) - 2 = 64. Not isoelectronic.
- (3) \(\text{Eu}^{2+}\): Eu (63) - 2 = 61; \(\text{Gd}^{3+}\): Gd (64) - 3 = 61. Isoelectronic (61 electrons).
- (4) \(\text{Pr}^{3+}\): Pr (59) - 3 = 56; \(\text{Ce}^{4+}\): Ce (58) - 4 = 54. Not isoelectronic.
Thus, the isoelectronic pair is \(\text{Eu}^{2+},\ \text{Gd}^{3+}\).
Thus, the correct answer is option 3.