Two complexes [Cr(H2O)6]Cl3 (A) and [Cr(NH3)6]Cl3 (B) are violet and yellow coloured, respectively.

Q: Why do transition metal complexes show color?

They show color due to d-d transitions. When light falls on the complex, an electron in a lower energy d-orbital absorbs a photon of specific energy and jumps to a higher energy d-orbital. The transmitted light, which lacks this absorbed color, reaches our eyes as the complementary color.

Q: How do I remember the complementary colors?

A simple way is to use the VIBGYOR color wheel. Draw a circle, divide it into 6 sectors, and label them V, B, G, Y, O, R. Opposite sectors are complementary (e.g., Violet and Yellow, Blue and Orange, Green and Red).

Q: Why does ammonia cause a larger splitting than water?

Ammonia is a stronger field ligand than water according to the spectrochemical series. It interacts more strongly with the metal's d-orbitals, causing a larger energy gap ($\Delta_o$) between the $t_{2g}$ and $e_g$ sets.

Q: Does the oxidation state of Chromium change between the two complexes?

No, in both $[Cr(H_2O)_6]Cl_3$ and $[Cr(NH_3)_6]Cl_3$, the ligands ($H_2O$ and $NH_3$) are neutral. The three chloride ions outside the coordination sphere give a -3 charge, meaning the complex ion has a +3 charge. Thus, Chromium is in the +3 oxidation state in both.

Q: Why are there exactly 3 unpaired electrons in $Cr^{3+}$?

Chromium's atomic number is 24, so its neutral configuration is $[Ar] 3d^5 4s^1$. Losing 3 electrons (one from 4s, two from 3d) leaves a $3d^3$ configuration. In an octahedral field, these 3 electrons singly occupy the three lower-energy $t_{2g}$ orbitals according to Hund's rule, resulting in 3 unpaired electrons.

Question

Two complexes $[Cr(H_2O)_6]Cl_3$ (A) and $[Cr(NH_3)_6]Cl_3$ (B) are violet and yellow coloured, respectively. The incorrect statement regarding them is

Options:

(A) $\Delta_o$ value for (A) is less than that of (B)

(B) both absorb energies corresponding to their complementary colours

(C) $\Delta_o$ values of (A) and (B) are calculated from the energies of violet and yellow light, respectively

(D) both are paramagnetic with three unpaired electrons

Sigma Percentile · Accepted Answer

The phenomenon of color in coordination compounds is one of the most beautiful and visually striking aspects of transition metal chemistry. When we look at a solution of a transition metal complex and see a vibrant color, we are actually witnessing quantum mechanics in action! Let's dive into this problem and unravel the mystery behind the colors of these two chromium complexes.

### Analyzing the Setup

We are given two octahedral complexes of chromium:
1. **Complex A:** $[Cr(H_2O)_6]Cl_3$, which appears **violet**.
2. **Complex B:** $[Cr(NH_3)_6]Cl_3$, which appears **yellow**.

Our goal is to identify the incorrect statement among the given options. To do this, we need to understand the relationship between the color we observe, the light the complex absorbs, and the nature of the ligands attached to the metal center.

### The Concept of Complementary Colors

Why do these complexes have color in the first place? It's due to **d-d transitions**. When white light (which contains all colors of the visible spectrum) passes through the solution, the complex absorbs a specific wavelength of light. The energy of this absorbed photon is used to excite an electron from the lower energy $t_{2g}$ orbitals to the higher energy $e_g$ orbitals.

The crucial point here is that **the color we see is NOT the color that was absorbed**. Instead, we see the light that is transmitted or reflected, which is the **complementary color** to the absorbed light.

We can easily determine complementary colors using a simple VIBGYOR color wheel. In this wheel, colors opposite to each other are complementary. 
- Violet is opposite to Yellow.
- Blue is opposite to Orange.
- Green is opposite to Red.

Applying this to our complexes:
- **Complex A** appears violet, which means it must be absorbing **yellow** light from the spectrum.
- **Complex B** appears yellow, which means it must be absorbing **violet** light.

### Crystal Field Splitting and Energy

The energy of the absorbed light corresponds exactly to the energy gap between the d-orbitals, known as the crystal field splitting energy ($\Delta_o$). The relationship is given by the Planck-Einstein equation:
$$E = \Delta_o = \frac{hc}{\lambda}$$

From the electromagnetic spectrum, we know that violet light has a shorter wavelength ($\lambda$) and therefore **higher energy** than yellow light. 
Since Complex B absorbs the higher-energy violet light, its splitting energy must be greater than that of Complex A (which absorbs lower-energy yellow light).
$$\Delta_o(B) > \Delta_o(A)$$

Does this align with our chemical knowledge? Let's check the **spectrochemical series**. Ammonia ($NH_3$) is a stronger field ligand than water ($H_2O$). A stronger ligand causes a larger splitting of the d-orbitals. Therefore, it makes perfect sense that the ammonia complex (B) has a larger $\Delta_o$ than the water complex (A). This confirms that statements (a) and (b) are absolutely correct.

### Evaluating the Statements

Now, let's look closely at statement (c): *"$\Delta_o$ values of (A) and (B) are calculated from the energies of violet and yellow light, respectively."*

This is a classic trap! The splitting energy $\Delta_o$ is calculated from the energy of the **absorbed** photon. 
- For Complex A, the absorbed light is yellow, so its $\Delta_o$ is calculated from the energy of yellow light.
- For Complex B, the absorbed light is violet, so its $\Delta_o$ is calculated from the energy of violet light.

Statement (c) incorrectly states that $\Delta_o$ is calculated from the *observed* colors (violet for A and yellow for B). Therefore, **statement (c) is the incorrect statement**.

### Final Check: Paramagnetism

Just to be thorough, let's verify statement (d). Both complexes contain the chromium ion in a $+3$ oxidation state. The neutral chromium atom has an electron configuration of $[Ar] 3d^5 4s^1$. Removing three electrons gives the $Cr^{3+}$ ion a $3d^3$ configuration.

In an octahedral crystal field, these three electrons will occupy the lower energy $t_{2g}$ orbitals singly, following Hund's rule. This results in a $t_{2g}^3 e_g^0$ configuration, leaving exactly **three unpaired electrons**. Because they have unpaired electrons, both complexes are indeed paramagnetic. Statement (d) is correct.

By carefully distinguishing between absorbed and observed colors, we successfully navigated the trap and found the right answer!

Animated Solution for Chemistry - Coordination Compounds: 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

Select Answer:

Visualized Solution

$O b ser v e d C o l or s$

$C o m pl e m e n t a r y C o l or s$

$A b sor b e d C o l or s$

$E n er g y an d W a v e l e n g t h$

$L i g an d F i e l d S t r e n g t h$

$E v a l u a t in g S t a t e m e n t (c)$

$C h ec k in g P a r ama g n e t i s m$

The Sigma Insight: Bonding and Crystal field

Analyzing the Setup

The Concept of Complementary Colors

Crystal Field Splitting and Energy

Evaluating the Statements

Final Check

Similar Questions

Which of the following facts about the complex $[C r (N H_{3})_{6}] C l_{3}$ is wrong?

The incorrect statement is

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

The correct order of the spin only magnetic moments of the following complexes is (I) $[Cr (H_{2} O)_{6}] Br_{2}$ (II) $Na_{4} [Fe (CN)_{6}]$ (III) $Na_{3} [Fe (C_{2} O_{4})_{3}] (Δ_{0} > P)$ (IV) $(Et_{4} N)_{2} [CoCl_{4}]$

Which one of the following cyano complexes would exhibit the lowest value of paramagnetic behaviour ? (At. no. of Cr = 24, Mn = 25, Fe = 26, Co = 27)

The total number of unpaired electrons present in $[C o (N H_{3})_{6}] C l_{2}$ and $[C o (N H_{3})_{6}] C l_{3}$ is ....... .

The correct order of the spin only magnetic moment of metal ions in the following low spin complexes, $[V (C N)_{6}]^{4 -}$ , $[F e (C N)_{6}]^{4 -}$ , $[R u (N H_{3})_{6}]^{3 +}$ , and $[C r (N H_{3})_{6}]^{2 +}$ , is