JEE Main 2025 — D And F Block Elements Question with Solution

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.