Question

The lanthanide ion having four unpaired electrons is
(Given : Atomic numbers of Ce = 58, Nd = 60, Tb = 65 and Ho = 67)

• A. $\text{Ho}^{3+}$
• B. $\text{Nd}^{3+}$
• C. $\text{Ce}^{3+}$
• D. $\text{Tb}^{3+}$

Hint:

To easily find the number of unpaired electrons in a $4f^n$ subshell: - If $n \le 7$, the number of unpaired electrons is simply equal to $n$. - If $n \gt 7$, the number of unpaired electrons is equal to $14 - n$. For $\text{Ho}^{3+}$ ($4f^{10}$), the number of unpaired electrons is $14 - 10 = 4$.

Answer 1

The Correct Option is D

Concept:

 

Lanthanides belong to the \(f\)-block elements and are characterized by the progressive filling of the \(4f\) subshell. The most stable oxidation state exhibited by lanthanides is \(+3\). When a lanthanide atom forms a \(\mathrm{Ln}^{3+}\) ion, three electrons are removed from the outermost orbitals, generally the two \(6s\) electrons and one electron from the \(5d\) or \(4f\) subshell.

The number of unpaired electrons in a \(4f^n\) configuration can be determined using Hund's rule of maximum multiplicity:

• For \(n \le 7\), the number of unpaired electrons is equal to \(n\).
• For \(n > 7\), pairing begins and the number of unpaired electrons becomes \(14-n\).

Since the \(4f\) subshell contains seven orbitals and can accommodate a maximum of fourteen electrons, this rule provides a quick method to determine the number of unpaired electrons in lanthanide ions.

Step 1: Determination of the electronic configuration of Ce³⁺.

Cerium has atomic number 58.

Its ground-state electronic configuration is:

\(\mathrm{Ce}=[\mathrm{Xe}]\,4f^1 5d^1 6s^2\)

Formation of Ce³⁺ involves removal of three electrons (two from 6s and one from 5d):

\(\mathrm{Ce}^{3+}=[\mathrm{Xe}]\,4f^1\)

Thus, the number of unpaired electrons is:

\(1\)

Ce³⁺ has 1 unpaired electron.

Step 2: Determination of the electronic configuration of Nd³⁺.

Neodymium has atomic number 60.

Its electronic configuration is:

\(\mathrm{Nd}=[\mathrm{Xe}]\,4f^4 6s^2\)

On forming the trivalent ion:

\(\mathrm{Nd}^{3+}=[\mathrm{Xe}]\,4f^3\)

The three electrons occupy three different \(4f\) orbitals according to Hund's rule.

Number of unpaired electrons = 3

Nd³⁺ has 3 unpaired electrons.

Step 3: Determination of the electronic configuration of Tb³⁺.

Terbium has atomic number 65.

Its electronic configuration is:

\(\mathrm{Tb}=[\mathrm{Xe}]\,4f^9 6s^2\)

On losing three electrons:

\(\mathrm{Tb}^{3+}=[\mathrm{Xe}]\,4f^8\)

For a \(4f^8\) configuration:

Number of unpaired electrons = 14 − 8 = 6

Tb³⁺ has 6 unpaired electrons.

Step 4: Determination of the electronic configuration of Ho³⁺.

Holmium has atomic number 67.

Its electronic configuration is:

\(\mathrm{Ho}=[\mathrm{Xe}]\,4f^{11}6s^2\)

When three electrons are removed:

\(\mathrm{Ho}^{3+}=[\mathrm{Xe}]\,4f^{10}\)

Since the configuration contains ten electrons in the \(4f\) subshell:

Number of unpaired electrons = 14 − 10 = 4

Ho³⁺ has 4 unpaired electrons.

Step 5: Comparison of all the given ions.

Ion	Configuration	Unpaired Electrons
Ce3+	4f1	1
Nd3+	4f3	3
Tb3+	4f8	6
Ho3+	4f10	4

From the comparison table, it is evident that only Ho³⁺ possesses exactly four unpaired electrons.

Therefore, Ho³⁺ is the lanthanide ion having four unpaired electrons.

Hence, the correct answer is (1) Ho³⁺.