QJEEM25Ch264b: JEE Main 2025 Chemistry - Class 12 JEEM25Ch264b Solution: Lanthanoid ions with configuration are : (A) (B) (C) (D) (E)…

Let's analyze the electron configurations of the given lanthanide ions.

Europium ( $\mathrm{Eu}$ )

The neutral atom of Eu (atomic number 63) typically has the configuration:

$[\mathrm{Xe}]\,4f^7\,6s^2.$

For $\mathrm{Eu}^{2+}$ , two electrons are removed, usually from the $6s$ orbital, resulting in:

$[\mathrm{Xe}]\,4f^7.$

Thus, $\mathrm{Eu}^{2+}$ has a $4f^7$ configuration.

For $\mathrm{Eu}^{3+}$ , three electrons are removed (the two $6s$ electrons and one $4f$ electron), giving:

$[\mathrm{Xe}]\,4f^6.$

Therefore, $\mathrm{Eu}^{3+}$ does not have a $4f^7$ configuration.

Gadolinium ( $\mathrm{Gd}$ )

The neutral atom of Gd (atomic number 64) is usually represented as:

$[\mathrm{Xe}]\,4f^7\,5d^1\,6s^2.$

For $\mathrm{Gd}^{3+}$ , three electrons are removed (typically the $5d^1$ and the two $6s$ electrons), resulting in:

$[\mathrm{Xe}]\,4f^7.$

So, $\mathrm{Gd}^{3+}$ has a $4f^7$ configuration.

Terbium ( $\mathrm{Tb}$ )

The neutral atom of Tb (atomic number 65) typically has the configuration:

$[\mathrm{Xe}]\,4f^9\,6s^2.$

For $\mathrm{Tb}^{3+}$ , removal of three electrons gives:

$[\mathrm{Xe}]\,4f^8.$

Hence, $\mathrm{Tb}^{3+}$ does not have a $4f^7$ configuration.

Samarium ( $\mathrm{Sm}$ )

The neutral atom of Sm (atomic number 62) generally has:

$[\mathrm{Xe}]\,4f^6\,6s^2.$

For $\mathrm{Sm}^{2+}$ , after removal of two electrons (likely from the $6s$ orbital), the configuration is:

$[\mathrm{Xe}]\,4f^6.$

Therefore, $\mathrm{Sm}^{2+}$ does not have a $4f^7$ configuration.

Thus, the ions with a $4f^7$ configuration are:

(A) $\mathrm{Eu}^{2+}$

(B) $\mathrm{Gd}^{3+}$

The correct answer is:

Option D – (A) and (B) only.