Question

Identify the isoelectronic pair of ions from the following

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

Hint:

To check whether ions are isoelectronic, subtract the ionic charge from the atomic number and compare the total number of electrons.

Answer 1

The Correct Option is C

Step 1: Recall the meaning of isoelectronic species.
Isoelectronic species are those species which contain the same number of electrons.

Step 2: Find the number of electrons in each ion.
Atomic number of europium: \[ Eu=63 \] For \[ Eu^{2+} \] number of electrons is \[ 63-2=61 \] Atomic number of gadolinium: \[ Gd=64 \] For \[ Gd^{3+} \] number of electrons is \[ 64-3=61 \] Thus, \[ Eu^{2+} \] and \[ Gd^{3+} \] contain equal number of electrons.

Step 3: Verify the remaining options.
For \[ Pr^{3+} \] \[ 59-3=56 \] For \[ Nd^{3+} \] \[ 60-3=57 \] Hence, they are not isoelectronic.
For \[ Tb^{3+} \] \[ 65-3=62 \] For \[ Dy^{2+} \] \[ 66-2=64 \] Not isoelectronic.
For \[ Ce^{4+} \] \[ 58-4=54 \] which is not equal to electrons in \[ Pr^{3+}=56 \] Thus, these are also not isoelectronic.

Step 4: Final conclusion.
Therefore, the isoelectronic pair is \[ \boxed{Eu^{2+},\,Gd^{3+}} \]