Arrange the following metal complex/compounds in the increasing order of spin only magnetic moment.

Q: Why does Gadolinium have an electron in the 5d subshell?

Gadolinium (atomic number 64) has the configuration $[\text{Xe}] 4f^7 5d^1 6s^2$ instead of $[\text{Xe}] 4f^8 6s^2$. This happens because a half-filled $4f$ subshell ($4f^7$) is exceptionally stable due to high exchange energy and symmetrical electron distribution. The next electron prefers to enter the $5d$ orbital rather than disrupt this stability.

Q: The book's solution says Gadolinium has 8 unpaired electrons. Is that possible?

No, that is a chemical impossibility for an f-subshell, which only has 7 orbitals and can hold a maximum of 7 unpaired electrons. The book incorrectly assumes the compound is $\text{Gd}^{2+}$ with a $4f^7 5d^1$ configuration. However, in $\text{Gd}(\text{NO}_3)_3$, Gadolinium is clearly in a +3 oxidation state, giving it a $4f^7$ configuration with exactly 7 unpaired electrons.

Q: Why do we use the spin-only formula for lanthanides? I thought orbital angular momentum is significant for them.

You are absolutely right! For most lanthanides, the orbital angular momentum is not quenched, and we must use the formula $\mu = g\sqrt{J(J+1)}$. However, the question explicitly instructs us to "Presume all the three, high spin system" and asks for the "spin only magnetic moment." In JEE, we must follow the specific constraints given in the problem.

Q: How did we find the oxidation state of Cerium in the first complex?

The complex is $(\text{NH}_4)_2[\text{Ce}(\text{NO}_3)_6]$. We know the ammonium ion $(\text{NH}_4^+)$ has a +1 charge, and the nitrate ion $(\text{NO}_3^-)$ has a -1 charge. Setting up the equation for neutrality: $2(+1) + x + 6(-1) = 0$. Solving this gives $x = +4$.

Q: Does the nature of the ligand (strong or weak field) affect the pairing in f-block complexes?

Generally, no. The $4f$ orbitals are deeply buried within the atom and are shielded by the $5s$ and $5p$ electrons. Therefore, they interact very weakly with ligands, and crystal field splitting is usually too small to cause electron pairing. They almost always form high-spin complexes.

Question

Arrange the following metal complex/compounds in the increasing order of spin only magnetic moment. Presume all the three, high spin system.
(Atomic numbers $	ext{Ce} = 58$, $	ext{Gd} = 64$ and $	ext{Eu} = 63$.)
A. $(	ext{NH}_4)_2[	ext{Ce}(	ext{NO}_3)_6]$
B. $	ext{Gd}(	ext{NO}_3)_3$ and
C. $	ext{Eu}(	ext{NO}_3)_3$

Options:

(A) (B) < (A) < (C)

(B) (C) < (A) < (B)

(C) (A) < (B) < (C)

(D) (A) < (C) < (B)

Sigma Percentile · Accepted Answer

### The Mystery of the f-Block

When we dive into the world of coordination chemistry, the $d$-block elements usually steal the spotlight. But the $f$-block elements—the lanthanides and actinides—hold their own beautiful symmetry secrets.

In this problem, we are tasked with finding the spin-only magnetic moment for three lanthanide complexes. The magnetic moment is a direct window into the quantum soul of an atom, revealing exactly how many unpaired electrons are dancing in its outermost orbitals.

The formula for the spin-only magnetic moment is elegantly simple:

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

where $n$ is the number of unpaired electrons. Let's decode each complex one by one.

### Decoding Cerium

Our first complex is $(	ext{NH}_4)_2[	ext{Ce}(	ext{NO}_3)_6]$. To find the number of unpaired electrons, we first need the oxidation state of Cerium.

We know that the ammonium ion $(	ext{NH}_4^+)$ carries a $+1$ charge, and the nitrate ion $(	ext{NO}_3^-)$ carries a $-1$ charge. Setting up our neutrality equation:

$2(+1) + x + 6(-1) = 0$

$x = +4$

Cerium (atomic number 58) has a neutral electronic configuration of $[	ext{Xe}] 4f^1 5d^1 6s^2$. When it loses four electrons to become $	ext{Ce}^{4+}$, it loses all its valence electrons, leaving it with a noble gas core: $[	ext{Xe}] 4f^0$.

With **zero unpaired electrons**, its magnetic moment is exactly **$0 	ext{ BM}$**.

### The Europium Enigma

Next up is $	ext{Eu}(	ext{NO}_3)_3$. With three nitrate ions, Europium is clearly in a $+3$ oxidation state.

Europium (atomic number 63) has a neutral configuration of $[	ext{Xe}] 4f^7 6s^2$. Notice how it perfectly half-fills its $f$-subshell before filling the $d$-subshell!

When it loses three electrons to form $	ext{Eu}^{3+}$, it loses the two $6s$ electrons and one $4f$ electron, leaving it with a $[	ext{Xe}] 4f^6$ configuration.

This gives us **$6$ unpaired electrons**. Plugging this into our formula:

$\mu = \sqrt{6(6+2)} = \sqrt{48} \approx 6.93 	ext{ BM}$

### The Gadolinium Correction

Finally, we have $	ext{Gd}(	ext{NO}_3)_3$. Just like Europium, Gadolinium is in a $+3$ oxidation state.

Gadolinium (atomic number 64) is a classic exception in electronic configurations. To maintain the incredible stability of a half-filled $f$-subshell, its neutral configuration is $[	ext{Xe}] 4f^7 5d^1 6s^2$.

When it forms $	ext{Gd}^{3+}$, it loses the two $6s$ electrons and the single $5d$ electron, leaving a pristine, half-filled $[	ext{Xe}] 4f^7$ core.

*(Note: Some textbooks mistakenly claim this forms a $4f^7 5d^1$ configuration with 8 unpaired electrons. This is chemically impossible for an $f$-subshell, which maxes out at 7 unpaired electrons!)*

With exactly **$7$ unpaired electrons**, its magnetic moment is:

$\mu = \sqrt{7(7+2)} = \sqrt{63} \approx 7.93 	ext{ BM}$

### The Final Verdict

Now, we simply arrange them in increasing order of their magnetic moments:

$\mu_A = 0 	ext{ BM}$

$\mu_C = 6.93 	ext{ BM}$

$\mu_B = 7.93 	ext{ BM}$

The correct increasing order is **$A < C < B$**, which corresponds perfectly to option (d).

Always trust the quantum mechanics, and don't let textbook typos throw you off your game!

Select Answer:

Visualized Solution

$Spin-Only Magnetic Moment$

$Oxidation State of Ce$

$Magnetic Moment of Ce^{4 +}$

$Oxidation State of Eu$

$Magnetic Moment of Eu^{3 +}$

$Oxidation State of Gd$

$Magnetic Moment of Gd^{3 +}$

$Final Order$

The Sigma Insight: Inner Transition Elements

The Mystery of the f-Block

Decoding Cerium

The Europium Enigma

The Gadolinium Correction

The Final Verdict

Similar Questions

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

The pair(s) of diamagnetic ions is(are)

Which one of the following lanthanides exhibits $+ 2$ oxidation state with diamagnetic nature ? (Given, $Z$ for $Nd = 60$ , $Yb = 70$ , $La = 57$ , $Ce = 58$ )

The electronic configuration of bivalent europium and trivalent cerium are (atomic number : $Xe = 54, Ce = 58, Eu = 63$ )

The $Eu^{2 +}$ ion is a strong reducing agent in spite of its ground state electronic configuration (outermost) : [Atomic number of Eu = 63]

The outer electron configuration of Gd (At. no. 64) is

The number of $4 f$ electrons in the ground state electronic configuration of $Gd^{2 +}$ is ...... . [Atomic number of Gd is 64.]

Given below are two statements. Statement I The $E^{\circ}$ value of $C e^{4 +} / C e^{3 +}$ is $+ 1.74 V$ . Statement II Ce is more stable in $C e^{4 +}$ state than $C e^{3 +}$ state. In the light of the above statements, choose the most appropriate answer from the options given below.

Which one of the following lanthanoids does not form $MO_{2}$ ? [ $M$ is lanthanoid metal]

Most common oxidation states of Ce (Cerium) are

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Spin-Only Magnetic Moment

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Oxidation State of Ce

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnetic Moment of Ce4+

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Oxidation State of Eu

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnetic Moment of Eu3+

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Oxidation State of Gd

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnetic Moment of Gd3+

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Order