The correct electronic configuration and spin-only magnetic moment (BM) of Gd3+ (Z = 64), respective

Q: Why is the configuration of Gd $[\text{Xe}] 4f^7 5d^1 6s^2$ and not $[\text{Xe}] 4f^8 6s^2$?

Because a half-filled $4f$ subshell ($4f^7$) provides extra exchange energy and symmetrical electron distribution, making it significantly more stable than a $4f^8$ configuration.

Q: Why do we remove electrons from $6s$ before $4f$?

Electrons are always removed from the outermost shell (highest principal quantum number $n$) first because they are the farthest from the nucleus and experience the least effective nuclear charge.

Q: How do I quickly estimate $\sqrt{63}$ without a calculator?

Find the nearest perfect squares. $8^2 = 64$ and $7^2 = 49$. Since 63 is very close to 64, its square root will be just under 8, making 7.9 a highly accurate and logical estimate.

Q: What does 'spin-only' magnetic moment mean?

It assumes that the magnetic moment arises entirely from the spin of the electrons, ignoring any contribution from their orbital angular momentum, which is often quenched or negligible in these specific complexes.

Q: Can $n$ ever be greater than 7 for an f-block element?

No, the $f$ subshell has exactly 7 orbitals, so it can hold a maximum of 7 unpaired electrons before pairing must begin according to the Pauli Exclusion Principle.

Question

The correct electronic configuration and spin-only magnetic moment (BM) of $	ext{Gd}^{3+}$ ($Z = 64$), respectively, are

Options:

(A) $[	ext{Xe}]4f^7$ and $7.9$

(B) $[	ext{Xe}]5f^7$ and $7.9$

(C) $[	ext{Xe}]5f^7$ and $8.9$

(D) $[	ext{Xe}]4f^7$ and $8.9$

Sigma Percentile · Accepted Answer

### The Gadolinium Anomaly

To solve this problem, we first need to understand the ground state of the Gadolinium (Gd) atom. Gadolinium, with an atomic number of $Z = 64$, is a member of the lanthanide series in the f-block of the periodic table.

Normally, we might expect its electronic configuration to follow a strict Aufbau filling order, leading to $[	ext{Xe}] 4f^8 6s^2$. However, Gadolinium is a classic exception. Because a half-filled $4f$ subshell ($4f^7$) provides immense exchange energy and symmetrical stability, one electron from the $4f$ orbital shifts to the $5d$ orbital.

Thus, the true ground state configuration of Gadolinium is:

$$ 	ext{Gd} = [	ext{Xe}] 4f^7 5d^1 6s^2 $$

### Stripping Electrons: Forming the Ion

The question asks for the configuration of the $	ext{Gd}^{3+}$ ion. To form a cation, we must remove electrons. A critical rule in chemistry is that **electrons are always removed from the outermost principal quantum shell first** (the shell with the highest $n$).

We need to remove three electrons in total. We start by taking two electrons from the outermost $6s$ orbital. We still need to remove one more, so we take it from the next outermost orbital, which is the $5d$ orbital.

After removing these three electrons, we are left with a pristine, highly stable half-filled $4f$ subshell:

$$ 	ext{Gd}^{3+} = [	ext{Xe}] 4f^7 $$

### Hund's Rule and Unpaired Electrons

Now, let's visualize this $4f^7$ configuration. The $f$-subshell consists of seven degenerate (equal energy) orbitals. According to **Hund's Rule of Maximum Multiplicity**, we must place one electron into each of these seven orbitals before any pairing occurs.

Since we have exactly seven electrons to place in seven orbitals, every single orbital gets exactly one electron. This means we have a total of **seven unpaired electrons** ($n = 7$). This is the absolute maximum number of unpaired electrons possible in an $f$-subshell.

### Calculating the Magnetic Moment

The second part of the question asks for the spin-only magnetic moment, denoted by $\mu$. The formula for this is beautifully simple:

$$ \mu = \sqrt{n(n+2)} 	ext{ BM} $$

Here, $n$ represents the number of unpaired electrons, and BM stands for Bohr Magnetons, the standard unit for magnetic moment. Let's substitute our value of $n = 7$ into the equation:

$$ \mu = \sqrt{7(7+2)} $$

$$ \mu = \sqrt{7 	imes 9} $$

$$ \mu = \sqrt{63} 	ext{ BM} $$

Now, we don't need a calculator to find the final answer. We know that $8^2 = 64$. Since $63$ is just slightly less than $64$, the square root of $63$ must be just slightly less than $8$. Looking at our options, $7.9$ is the perfect logical estimate.

Therefore, the correct electronic configuration is $[	ext{Xe}] 4f^7$ and the magnetic moment is $7.9 	ext{ BM}$, making option (a) the correct choice.

Animated Solution for Chemistry - d and f-Block Elements: The correct electronic configuration and spin-only magnetic moment (BM) of $Gd^{3 +}$ ( $Z = 64$ ), respectively, are

Select Answer:

Visualized Solution

\text{Gadolinium (Gd) Atom}

\text{Formation of } \text{Gd}^{3+}

\text{Unpaired Electrons in } 4f

\text{Number of Unpaired Electrons } (n)

\text{Spin-Only Magnetic Moment}

\text{Calculation}

\text{Final Value}

The Sigma Insight: Inner Transition Elements

The Gadolinium Anomaly

Stripping Electrons

Hund's Rule and Unpaired Electrons

Calculating the Magnetic Moment

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Animated Solution for Chemistry - d and f-Block Elements: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The correct electronic configuration and spin-only magnetic moment (BM) of Gd3+ (Z=64), respectively, are