Among the statements (A)-(D), the incorrect ones are
(A) octahedral Co(III) complexes with strong fi

Q: Why does $Co^{3+}$ have a $d^6$ configuration?

Cobalt has an atomic number of 27, so its neutral ground state configuration is $[Ar] 4s^2 3d^7$. When it loses three electrons to form the $Co^{3+}$ ion, it first loses the two $4s$ electrons, and then one $3d$ electron, leaving it with a $3d^6$ configuration.

Q: Why is the magnetic moment zero for a strong field $d^6$ complex?

In a strong octahedral field, the energy gap ($\Delta_o$) between the $t_{2g}$ and $e_g$ orbitals is very large—greater than the energy required to pair electrons ($P$). Therefore, all six electrons prefer to pair up in the three lower-energy $t_{2g}$ orbitals ($t_{2g}^6$). Since there are no unpaired electrons, the magnetic moment is zero.

Q: How do we know that 'en' is a stronger ligand than $F^-$?

This is determined by the spectrochemical series, an experimentally derived list that ranks ligands based on the size of the crystal field splitting they produce. Halides like $F^-$ are weak field ligands, while nitrogen-donors like ethylenediamine ($en$) are strong field ligands.

Q: Why is $\Delta_t$ smaller than $\Delta_o$?

There are two main reasons: First, a tetrahedral complex only has 4 ligands compared to the 6 in an octahedral complex, so the overall electrostatic field is weaker. Second, in a tetrahedral geometry, the ligands do not point directly at any of the d-orbitals, leading to less direct repulsion and a smaller splitting.

Q: What exactly does $\Delta_o < P$ mean?

It means the Crystal Field Splitting Energy ($\Delta_o$) is less than the Pairing Energy ($P$). In this scenario, it takes less energy for an electron to jump up to the higher $e_g$ orbitals than it does to overcome the repulsion of pairing up with another electron in the lower $t_{2g}$ orbitals. This results in a 'high spin' complex.

Question

Among the statements (A)-(D), the incorrect ones are
(A) octahedral Co(III) complexes with strong, field ligands have very high magnetic moments
(B) When $\Delta_0 < P$, the d-electron configuration of Co(III) in an octahedral complex is $t_{2g}^4 e_g^2$,
(C) Wavelength of light absorbed by $[Co(en)_3]^{3+}$ is lower than that of $[CoF_6]^{3-}$
(D) If the $\Delta_0$ for an octahedral complex of Co(III) is $18,000 	ext{ cm}^{-1}$, the $\Delta_t$ for its tetrahedral complex with the same ligand will be $16,000 	ext{ cm}^{-1}$

Options:

(A) B and C only

(B) A and B only

(C) C and D only

(D) A and D only

Sigma Percentile · Accepted Answer

### Analyzing the Statements

In this problem, we are tasked with evaluating four distinct statements regarding the crystal field theory of coordination compounds. Let's break them down one by one to uncover the truth.

### Statement A: The Magnetic Moment of $Co(III)$ in a Strong Field

Statement A claims that octahedral $Co(III)$ complexes with strong field ligands have very high magnetic moments. Let's test this. Cobalt has an atomic number of 27. The $Co^{3+}$ ion has a $d^6$ electronic configuration.

When placed in an octahedral field created by strong field ligands, the crystal field splitting energy ($\Delta_o$) is greater than the pairing energy ($P$). This means it is energetically more favorable for the electrons to pair up in the lower energy $t_{2g}$ orbitals rather than jump to the higher energy $e_g$ orbitals.

Consequently, all six electrons pair up, resulting in the configuration $t_{2g}^6 e_g^0$. Because there are zero unpaired electrons, the magnetic moment ($\mu$) is exactly $0$. The complex is diamagnetic, not highly paramagnetic! Therefore, **Statement A is incorrect**.

### Statement B: The Weak Field Configuration

Statement B discusses the scenario where $\Delta_o < P$, which corresponds to a weak field ligand. In this case, the pairing energy is higher than the splitting energy.

Following Hund's rule, the electrons will first singly occupy all five d-orbitals ($t_{2g}$ and $e_g$) before any pairing occurs. For a $d^6$ ion, the first five electrons go into $t_{2g}^3 e_g^2$. The sixth electron must then pair up in one of the $t_{2g}$ orbitals.

This gives us the final configuration of $t_{2g}^4 e_g^2$. This perfectly matches the claim in the statement. Thus, **Statement B is correct**.

### Statement C: Spectrochemical Series and Wavelength

Statement C compares the wavelength of light absorbed by $[Co(en)_3]^{3+}$ and $[CoF_6]^{3-}$. To evaluate this, we must consult the spectrochemical series.

Ethylenediamine ($en$) is a strong field ligand, whereas the fluoride ion ($F^-$) is a weak field ligand. Therefore, the crystal field splitting energy for the $en$ complex is much larger: $\Delta_o([Co(en)_3]^{3+}) > \Delta_o([CoF_6]^{3-})$.

The energy of the absorbed light is directly proportional to $\Delta_o$, and inversely proportional to its wavelength ($E = \frac{hc}{\lambda}$). Because the $en$ complex has a larger energy gap, it will absorb light of a **lower wavelength**. This confirms that **Statement C is correct**.

### Statement D: The Relationship Between $\Delta_t$ and $\Delta_o$

Finally, Statement D provides the octahedral splitting energy $\Delta_o = 18,000 	ext{ cm}^{-1}$ and claims the tetrahedral splitting energy $\Delta_t$ for the same metal and ligand will be $16,000 	ext{ cm}^{-1}$.

We know the fundamental relationship between these two splitting energies is given by:

$$\Delta_t \approx \frac{4}{9} \Delta_o$$

Let's substitute the given value into our master equation:

$$\Delta_t = \frac{4}{9} 	imes 18,000 	ext{ cm}^{-1} = 8,000 	ext{ cm}^{-1}$$

The calculated value is $8,000 	ext{ cm}^{-1}$, which is vastly different from the claimed $16,000 	ext{ cm}^{-1}$. Therefore, **Statement D is incorrect**.

### Final Conclusion

The question asks us to identify the **incorrect** statements. Based on our rigorous analysis, Statements (A) and (D) are incorrect. This leads us to the final answer: **Option (d)**.

Select Answer:

Visualized Solution

$C o (I I I)$ \text{ in Strong Field}

\text{Magnetic Moment of } Co^{3+}

\text{Weak Field Configuration}

\text{Electron Distribution}

\text{Spectrochemical Series}

\text{Wavelength Comparison}

\text{Tetrahedral vs Octahedral Splitting}

\text{Calculating } \Delta_t

\text{Final Conclusion}

The Sigma Insight: Bonding and Crystal field

Analyzing the Statements

Statement A

Statement B

Statement C

Statement D

Final Conclusion

Similar Questions

In which of the following octahedral complexes of Co (At no. 27), will the magnitude of $Δ_{o}$ be the highest?

The incorrect statement is

Two complexes $[C r (H_{2} O)_{6}] C l_{3}$ (A) and $[C r (N H_{3})_{6}] C l_{3}$ (B) are violet and yellow coloured, respectively. The incorrect statement regarding them is

For a $d^{4}$ metal ion in an octahedral field, the correct electronic configuration is

The octahedral complex of a metal ion $M^{3 +}$ with four monodentate ligands $L_{1}, L_{2}, L_{3}$ and $L_{4}$ absorb wavelengths in the region of red, green, yellow and blue, respectively. The increasing order of ligand strength of the four ligands is

Homoleptic octahedral complexes of a metal ion $M^{3 +}$ with three monodentate ligands $L_{1}, L_{2}$ and $L_{3}$ absorb wavelengths in the region of green, blue and red respectively. The increasing order of the ligand strength is

The correct order of the wavelength maxima of the absorption band in the ultraviolet-visible region for the given complexes is

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

Consider that a $d^{6}$ metal ion ( $M^{2 +}$ ) forms a complex with aqua ligands and the spin only magnetic moment of the complex is $4.90 BM$ . The geometry and the crystal field stabilisation energy of the complex is