The complex ion that will lose its crystal field stabilisation energy upon oxidation of its metal to

Q: Why does phenanthroline act as a strong field ligand?

Phenanthroline is a bidentate ligand with nitrogen donor atoms. It has empty pi-antibonding orbitals that can accept electron density back from the metal's filled d-orbitals (pi-backbonding). This synergic effect significantly increases the crystal field splitting energy ($\Delta_o$), making it a strong field ligand.

Q: Why do we ignore pairing energy in our calculations here?

The question explicitly instructs us to 'Ignore pairing energy'. In comparative problems like this, the primary driving force for the change in stability is the $\Delta_o$ term. By ignoring the pairing energy $P$, we can focus purely on the geometric stabilisation provided by the crystal field.

Q: What exactly does 'losing CFSE' mean physically?

CFSE is a negative value representing the energy released (stabilisation) when ligands approach the metal ion. 'Losing CFSE' means the energy value becomes less negative (closer to zero). Physically, it means the complex is less thermodynamically stable in the $+3$ state compared to the $+2$ state purely from a crystal field perspective.

Q: Why did we jump straight to checking the Iron complex?

In coordination chemistry, the $d^6$ to $d^5$ transition in an octahedral strong field is a classic textbook example of CFSE loss. A $d^6$ strong field ion ($t_{2g}^6$) has the maximum possible CFSE ($-2.4\Delta_o$). Any removal of an electron from this perfect closed-shell-like $t_{2g}$ level will inevitably result in a loss of stabilisation.

Q: Would a weak field ligand cause the same effect for Iron?

No, the math changes. In a weak field, $Fe^{2+}$ ($d^6$) is $t_{2g}^4 e_g^2$ with a CFSE of $-0.4\Delta_o$. Upon oxidation to $Fe^{3+}$ ($d^5$), it becomes $t_{2g}^3 e_g^2$ with a CFSE of $0\Delta_o$. While it technically still 'loses' CFSE (goes from $-0.4$ to $0$), the magnitude and the physical reason (loss of the single paired electron) are different.

Question

The complex ion that will lose its crystal field stabilisation energy upon oxidation of its metal to $+3$ state is

Options:

(A) $[Co(phen)_3]^{2+}$

(B) $[Ni(phen)_3]^{2+}$

(C) $[Zn(phen)_3]^{2+}$

(D) $[Fe(phen)_3]^{2+}$

Sigma Percentile · Accepted Answer

### The Quest for Lost Energy

Imagine you are an electron residing in the $d$-orbitals of a transition metal. Your ultimate goal is to find the lowest energy state possible, a state of profound stability. This stability is quantified by the **Crystal Field Stabilisation Energy (CFSE)**. In this problem, we are on a hunt to find a complex that actually *loses* this precious stabilisation energy when it undergoes oxidation from a $+2$ to a $+3$ state.

### Analyzing the Setup

The first crucial step is to identify the environment our metal ion is in. The ligand provided is **Phenanthroline (phen)**. If you recall the spectrochemical series, phenanthroline is a robust, strong field ligand.

What does a strong field ligand do? It creates a massive energy gap ($\Delta_o$) between the lower $t_{2g}$ and upper $e_g$ orbitals. This gap is so large that it exceeds the pairing energy ($P$). Consequently, the electrons prefer to snuggle up and pair together in the lower $t_{2g}$ orbitals rather than making the exhausting jump to the $e_g$ level.

### The Master Equation

To calculate the stability, we use the CFSE formula for octahedral complexes:

$$	ext{CFSE} = (-0.4 n_{t_{2g}} + 0.6 n_{e_g})\Delta_o$$

Here, $n_{t_{2g}}$ is the number of electrons in the lower energy level, and $n_{e_g}$ is the number of electrons in the higher energy level. Every electron in the $t_{2g}$ level drops the energy by $0.4\Delta_o$, contributing to stability.

### The Iron Complex: A Classic Case

Let's put the iron complex, $[Fe(phen)_3]^{2+}$, under the microscope. In this state, iron is $Fe^{2+}$, which has a $3d^6$ electronic configuration.

Because phenanthroline is a strong field ligand, all six electrons will pair up in the lower $t_{2g}$ orbitals, leaving the $e_g$ orbitals completely empty. The splitting looks like this: $t_{2g}^6 e_g^0$.

Let's calculate its initial CFSE:

$$	ext{CFSE}_{initial} = (-0.4 	imes 6 + 0.6 	imes 0)\Delta_o = -2.4\Delta_o$$

This is a highly negative value, indicating a very stable, happy complex.

### The Oxidation Event

Now, we oxidize the complex. Oxidation means the loss of an electron. Our $Fe^{2+}$ transforms into $Fe^{3+}$, and the electronic configuration shifts from $3d^6$ to $3d^5$.

One electron is removed from the $t_{2g}$ level. The new splitting arrangement is $t_{2g}^5 e_g^0$. Let's calculate the new CFSE:

$$	ext{CFSE}_{final} = (-0.4 	imes 5 + 0.6 	imes 0)\Delta_o = -2.0\Delta_o$$

### Final Calculation and Conclusion

Let's compare the two states. The energy went from **$-2.4\Delta_o$** to **$-2.0\Delta_o$**.

Notice how the value became *less negative*? In the world of thermodynamics, a less negative energy means a decrease in stability. The complex has effectively **lost** $0.4\Delta_o$ of its crystal field stabilisation energy during the oxidation process.

Thus, the iron complex $[Fe(phen)_3]^{2+}$ perfectly satisfies the condition of the question, making it our correct answer!

Animated Solution for Chemistry - Coordination Compounds: The complex ion that will lose its crystal field stabilisation energy upon oxidation of its metal to $+ 3$ state is

Select Answer:

Visualized Solution

$Understanding the Objective$

$Nature of the Ligand$

$Analyzing [F e (p h e n)_{3}]^{2 +}$

$CFSE of F e^{2 +}$

$Oxidation to F e^{3 +}$

$CFSE of F e^{3 +}$

$Comparing CFSE$

$The Way Forward$

The Sigma Insight: Bonding and Crystal field

The Quest for Lost Energy

Analyzing the Setup

The Master Equation

The Iron Complex

The Oxidation Event

Final Calculation and Conclusion

Similar Questions

Arrange the following cobalt complexes in the order of increasing crystal field stabilisation energy (CFSE) value. $[C o F_{6}]^{3 -} (A), [C o (H_{2} O)_{6}]^{3 +} (B), [C o (N H_{3})_{6}]^{3 +} (C) and [C o (e n)_{3}]^{3 +} (D)$ Choose the correct option.

Which one of the following metal complexes is most stable?

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 ( $M^{2 +}$ ).

Which one of the following species responds to an external magnetic field

The crystal field stabilisation energy (CFSE) of $[C o F_{3} (H_{2} O)_{3}]$ ( $Δ_{0} < P$ ) is

The values of the crystal field stabilisation energies for a high spin $d^{6}$ metal ion in octahedral and tetrahedral fields respectively, are

Which one of the following complexes is an outer orbital complex? (At. no. of Mn = 25, Fe = 26, Co = 27, Ni = 28)

Match the electronic configurations in List-I with appropriate metal complex ions in List-II and choose the correct option. [Atomic Number: Fe = 26, Mn = 25, Co = 27]

List-I

List-II

In which of the following order the given complex ions are arranged correctly with respect to their decreasing spin only magnetic moment? (i) $[F e F_{6}]^{3 -}$ (ii) $[C o (N H_{3})_{6}]^{3 +}$ (iii) $[N i C l_{4}]^{2 -}$ (iv) $[C u (N H_{3})_{4}]^{2 +}$

The difference in the number of unpaired electrons of a metal ion in its high-spin and low-spin octahedral complexes is two. The metal ion is

Animated Solution for Chemistry - Coordination Compounds: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The complex ion that will lose its crystal field stabilisation energy upon oxidation of its metal to +3 state is