The spin only magnetic moment of a divalent ion in aqueous solution (atomic number = 29) is ........

Q: Why do we remove electrons from the 4s orbital first when ionizing Copper, even though 3d is filled last?

This is a classic trap! While the 4s orbital is filled before the 3d orbital (following the Aufbau principle), once the orbitals are populated, the 4s orbital actually lies at a higher energy level and is further from the nucleus than the 3d orbital. Therefore, when an atom ionizes, the outermost electrons (in the 4th shell) are always the first to be removed.

Q: Why is Copper's ground state configuration $[Ar] 4s^1 3d^{10}$ instead of $[Ar] 4s^2 3d^9$?

Nature loves stability, and stability in electron configurations comes from symmetry and exchange energy. A fully filled d-subshell ($3d^{10}$) is highly symmetrical and allows for maximum exchange energy between electrons of the same spin. This extra stability more than compensates for the energy required to promote an electron from the 4s to the 3d orbital.

Q: What exactly is a Bohr Magneton (BM)?

The Bohr Magneton is the fundamental unit of magnetic moment for an electron. It represents the magnetic moment caused by the orbital or spin angular momentum of a single electron. Think of it as the 'quantum currency' of magnetism at the atomic scale.

Q: The question mentions 'in aqueous solution'. Does water change the magnetic moment?

For the first transition series (3d metals), the orbital contribution to the magnetic moment is largely 'quenched' or suppressed by the electric fields of surrounding molecules (like water ligands in an aqueous solution). This is exactly why we can safely rely on the 'spin-only' formula $\mu = \sqrt{n(n+2)}$ without worrying about the orbital angular momentum!

Q: Why did we round off $1.732$ to $2$? Isn't it closer to $1.5$?

The question specifically asks for the answer as the 'Nearest integer'. An integer is a whole number without a fractional part. The integers closest to $1.732$ are $1$ and $2$. Since $1.732$ is greater than $1.5$, it rounds up to the nearest whole number, which is $2$.

Question

The spin only magnetic moment of a divalent ion in aqueous solution (atomic number = 29) is ……… BM.

Sigma Percentile · Accepted Answer

### The Enigma of Transition Metals

When we dive into the world of transition metals, we are greeted by a fascinating array of colors, catalytic properties, and most intriguingly, magnetism. The root of this magnetism lies deep within the atomic structure, specifically in the behavior of electrons residing in the d-orbitals. In this problem, we are tasked with finding the spin-only magnetic moment of a divalent ion with an atomic number of 29. Let's embark on this quantum journey.

### Copper: The Rule Breaker

The first step is to identify our protagonist. An atomic number of $Z = 29$ unmistakably points to Copper (Cu). If you recall your Aufbau principle, you might be tempted to write its electronic configuration as $[Ar] 4s^2 3d^9$. However, Copper is a classic rebel.

Nature favors stability, and stability in the quantum realm is often achieved through symmetry and exchange energy. A fully filled d-subshell ($3d^{10}$) is highly symmetrical and allows for maximum exchange energy between electrons of the same spin. To achieve this zen state, Copper borrows an electron from the 4s orbital, resulting in the anomalous ground state configuration:

$$Cu = [Ar] 4s^1 3d^{10}$$

### The Art of Ionization: Who Leaves First?

The question specifically asks for a **divalent ion**, which means our Copper atom has lost two electrons to become $Cu^{2+}$. This brings us to a critical trap where many students stumble: *Which electrons are removed first?*

Even though the 3d orbitals are filled after the 4s orbital, the 4s orbital is physically further from the nucleus (it belongs to the 4th principal quantum shell). When ionization occurs, the outermost electrons are always the first to be stripped away. Therefore, we remove the single 4s electron first, and then one electron from the 3d subshell.

$$Cu \xrightarrow{-2e^-} Cu^{2+}$$
$$Cu^{2+} = [Ar] 4s^0 3d^9$$

### Hund's Rule and the Lone Electron

Now that we have our $3d^9$ configuration, we need to visualize how these nine electrons occupy the five available d-orbitals. Enter Hund's Rule of Maximum Multiplicity.

We first place one electron in each of the five orbitals with parallel spins. That accounts for five electrons. We then begin pairing them up with the remaining four electrons. After filling four orbitals with pairs, we are left with exactly **one unpaired electron** in the final orbital.

Thus, the number of unpaired electrons, denoted by $n$, is exactly $1$.

### The Spin-Only Magnetic Moment

In transition metal complexes, especially those of the 3d series in aqueous solutions, the orbital contribution to the magnetic moment is largely "quenched" by the surrounding electric fields of the water ligands. This allows us to rely entirely on the **spin-only magnetic moment** formula:

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

Here, $\mu$ is the magnetic moment and BM stands for Bohr Magneton, the fundamental unit of atomic magnetism.

### The Final Calculation

Substituting our value of $n = 1$ into the master equation:

$$\mu = \sqrt{1(1+2)}$$
$$\mu = \sqrt{3} 	ext{ BM}$$

We know that the square root of 3 is approximately $1.732$. The question, however, demands the answer rounded off to the **nearest integer**.

Since $1.732$ is closer to $2$ than it is to $1$, we round it up.

**Final Answer: 2**

This problem is a beautiful symphony of exceptions in electronic configuration, the rules of ionization, and the quantum mechanics of magnetism. Always remember to watch out for those anomalous configurations!

Animated Solution for Chemistry - Atomic Structure: The spin only magnetic moment of a divalent ion in aqueous solution (atomic number = 29) is ……… BM.

Enter Numerical Value:

Visualized Solution

$Z = 29 Configuration$

$Divalent Ion (C u^{2 +})$

$Orbital Filling (3 d^{9})$

$Unpaired Electrons (n)$

$Magnetic Moment Formula$

$\text{Substitution & Calculation}$

$Final Answer$

$The Way Forward$

The Sigma Insight: Quantum Mechanical Model

The Enigma of Transition Metals

Copper

The Art of Ionization

Hund's Rule and the Lone Electron