The crystal field stabilisation energy (CFSE) and magnetic moment (spin-only) of an octahedral aqua

Q: Why is water ($H_2O$) considered a weak field ligand here?

According to the spectrochemical series, $H_2O$ lies in the middle but generally acts as a weak field ligand for $3d$ series metal ions in $+2$ oxidation states (like $Co^{2+}$). This means the crystal field splitting energy $\Delta_o$ is less than the pairing energy $P$, leading to a high-spin complex.

Q: What would happen if the ligand was a strong field ligand like $CN^-$?

If it were a strong field ligand, $\Delta_o$ would be greater than the pairing energy $P$. The 7 electrons of $Co^{2+}$ would fill as $t_{2g}^6 e_g^1$. This would result in only 1 unpaired electron (giving a magnetic moment of $\approx 1.73 \text{ BM}$) and a different CFSE of $[-0.4 \times 6 + 0.6 \times 1]\Delta_o = -1.8 \Delta_o$.

Q: How did we instantly know that $\sqrt{15}$ means 3 unpaired electrons?

It's helpful to memorize the magnetic moments for 1 to 5 unpaired electrons: $n=1 \implies 1.73$, $n=2 \implies 2.83$, $n=3 \implies 3.87$, $n=4 \implies 4.90$, $n=5 \implies 5.92$. Notice how the digit before the decimal point always matches the number of unpaired electrons!

Q: Why is the CFSE value negative?

The negative sign indicates stabilization. The energy of the $t_{2g}$ orbitals is lowered by $0.4 \Delta_o$ relative to the hypothetical spherical field (barycenter), while the $e_g$ orbitals are raised by $0.6 \Delta_o$. A net negative CFSE means the complex is more stable than the hypothetical un-split state.

Q: Is it safe to just pick option (d) because it's the only $+2$ ion?

In a high-pressure exam like JEE, if you are short on time, yes, it's a very smart and valid test-taking strategy. However, if you have time, it is always recommended to quickly verify the magnetic moment and CFSE to ensure there isn't a typo in the question or a deeper trick.

Question

The crystal field stabilisation energy (CFSE) and magnetic moment (spin-only) of an octahedral aqua complex of a metal ion ($M^{2+}$) are $-0.8 \Delta_o$ and $3.87$ BM, respectively.
Identify ($M^{2+}$).

Options:

(A) $V^{3+}$

(B) $Cr^{3+}$

(C) $Mn^{4+}$

(D) $Co^{2+}$

Sigma Percentile · Accepted Answer

### Decoding the Crystal Field: Finding the Mystery Metal Ion

Sometimes, competitive exams like JEE throw a massive curveball that is actually a disguised gift. This question is a perfect example of that. We are given the Crystal Field Stabilisation Energy (CFSE) and the spin-only magnetic moment of an octahedral aqua complex of a metal ion, denoted as $M^{2+}$.

Before we even pick up a pen to calculate anything, let's look at the options provided:
(a) $V^{3+}$
(b) $Cr^{3+}$
(c) $Mn^{4+}$
(d) $Co^{2+}$

Notice anything interesting? The question explicitly asks us to identify an **$M^{2+}$** ion. Out of all the four options, only option (d) features a metal ion in the $+2$ oxidation state! Technically, you could mark $Co^{2+}$ and move on to the next question in 5 seconds. However, as rigorous students of chemistry, we must verify this mathematically to ensure it's not a trick.

### Unveiling the Unpaired Electrons

Let's start with the magnetic moment. We are given that the spin-only magnetic moment $\mu$ is $3.87 	ext{ BM}$. The formula relating magnetic moment to the number of unpaired electrons ($n$) is:

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

Substituting the given value:

$$ 3.87 = \sqrt{n(n+2)} $$

We know that $\sqrt{15}$ is approximately $3.87$ (since $\sqrt{16} = 4$). Therefore:

$$ n(n+2) = 15 $$

Solving this simple quadratic equation mentally, we find that **$n = 3$**. Our mystery metal ion must have exactly 3 unpaired electrons.

### Testing the Suspect: $Co^{2+}$

Let's put $Co^{2+}$ to the test. Cobalt has an atomic number of 27, which gives it a ground state electronic configuration of $[Ar] 3d^7 4s^2$. When it loses two electrons to form the $Co^{2+}$ ion, its configuration becomes **$[Ar] 3d^7$**.

The complex is an "aqua" complex, meaning the ligands are water ($H_2O$) molecules. According to the spectrochemical series, water is a **weak field ligand**. This means the crystal field splitting energy ($\Delta_o$) is relatively small compared to the pairing energy ($P$). Consequently, the electrons will prefer to occupy the higher energy $e_g$ orbitals before pairing up in the lower energy $t_{2g}$ orbitals, resulting in a **high-spin complex**.

Let's fill the 7 electrons into the split $d$-orbitals:
1. The first three electrons go into the $t_{2g}$ orbitals: $t_{2g}^3$
2. The next two electrons go into the $e_g$ orbitals: $t_{2g}^3 e_g^2$
3. The remaining two electrons must now pair up in the $t_{2g}$ orbitals: **$t_{2g}^5 e_g^2$**

Looking at this configuration, we have 1 unpaired electron in the $t_{2g}$ set and 2 unpaired electrons in the $e_g$ set, giving us a total of **3 unpaired electrons**. This perfectly matches our calculation from the magnetic moment!

### The Final Verification: CFSE

To be absolutely certain, let's calculate the Crystal Field Stabilisation Energy (CFSE) for this $t_{2g}^5 e_g^2$ configuration. The formula for an octahedral complex is:

$$ 	ext{CFSE} = [-0.4 	imes n(t_{2g}) + 0.6 	imes n(e_g)] \Delta_o $$

Substituting our electron counts:

$$ 	ext{CFSE} = [-0.4 	imes 5 + 0.6 	imes 2] \Delta_o $$
$$ 	ext{CFSE} = [-2.0 + 1.2] \Delta_o $$
$$ 	ext{CFSE} = -0.8 \Delta_o $$

This result is a flawless match with the value given in the question. The evidence is conclusive: the metal ion is indeed **$Co^{2+}$**.

Animated Solution for Chemistry - Coordination Compounds: The crystal field stabilisation energy (CFSE) and magnetic moment (spin-only) of an octahedral aqua complex of a metal ion ( $M^{2 +}$ ) are $- 0.8 Δ_{o}$ and $3.87$ BM, respectively. Identify ( $M^{2 +}$ ).

Select Answer:

Visualized Solution

\text{The Hidden Clue}

\text{Magnetic Moment}

\text{Number of Unpaired Electrons}

\text{Testing } Co^{2+}

\text{Crystal Field Splitting}

\text{Calculating CFSE}

The Sigma Insight: Bonding and Crystal field

Decoding the Crystal Field

Unveiling the Unpaired Electrons

Testing the Suspect: $C o^{2 +}$

The Final Verification

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