Crystal field splitting in octahedral vs tetrahedral complexes — color of transition metal compounds

Question

A coordination compound $[\text{CoCl}_4]^{2-}$ appears blue, while $[\text{Co(NH}_3)_6]^{3+}$ appears yellow-orange. Given that the crystal field splitting energy in an octahedral complex is $\Delta_o$ , the corresponding value for a tetrahedral complex with the same ligands is approximately $\frac{4}{9}\Delta_o$ .

(a) Explain why $[\text{CoCl}_4]^{2-}$ absorbs in a different region than $[\text{Co(NH}_3)_6]^{3+}$ .

(b) If $[\text{Co(NH}_3)_6]^{3+}$ has $\Delta_o = 23{,}000\ \text{cm}^{-1}$ , estimate the $\Delta_t$ for a hypothetical $[\text{Co(NH}_3)_4]^{n+}$ tetrahedral complex.

Solution — Step by Step

The color we see is the complement of the color absorbed. When $d$ - $d$ transitions occur, the complex absorbs light of energy $\Delta$ (the crystal field splitting). The absorbed wavelength corresponds to a specific color, and we see what’s left.

$[\text{Co(NH}_3)_6]^{3+}$ absorbs violet light (~430 nm) → we see yellow-orange. $[\text{CoCl}_4]^{2-}$ absorbs red light (~700 nm) → we see blue.

The tetrahedral splitting is always smaller than octahedral for two reasons: fewer ligands (4 vs 6) and the geometry means no ligand points directly at a $d$ orbital along an axis.

\Delta_t = \frac{4}{9} \Delta_o

For $[\text{Co(NH}_3)_6]^{3+}$ with $\Delta_o = 23{,}000\ \text{cm}^{-1}$ :

\Delta_t = \frac{4}{9} \times 23{,}000 = \frac{92{,}000}{9} \approx 10{,}222\ \text{cm}^{-1}

For $[\text{CoCl}_4]^{2-}$ vs $[\text{Co(NH}_3)_6]^{3+}$ , there are two compounding factors: geometry (tetrahedral vs octahedral) AND the spectrochemical series position of the ligand.

Cl⁻ is a weak field ligand — it causes smaller $\Delta$ even in octahedral geometry. NH₃ is a strong field ligand. So $[\text{CoCl}_4]^{2-}$ has a doubly reduced $\Delta$ : tetrahedral geometry + weak ligand.

Energy and wavelength are inversely related:

E = h\nu = \frac{hc}{\lambda} \implies \lambda = \frac{hc}{\Delta}

Larger $\Delta$ → shorter wavelength absorbed (higher energy, toward violet/UV). Smaller $\Delta$ → longer wavelength absorbed (lower energy, toward red/IR).

$[\text{CoCl}_4]^{2-}$ has small $\Delta_t$ → absorbs red (~700 nm) → appears blue. $[\text{Co(NH}_3)_6]^{3+}$ has large $\Delta_o$ → absorbs violet (~430 nm) → appears yellow-orange.

(a) The difference arises from two combined effects: (i) $[\text{CoCl}_4]^{2-}$ is tetrahedral so $\Delta_t \approx \frac{4}{9}\Delta_o$ , and (ii) Cl⁻ is a weak-field ligand while NH₃ is strong-field. Together, $[\text{CoCl}_4]^{2-}$ has a much smaller splitting energy, absorbing at longer wavelengths (red region), so it appears blue.

(b) $\Delta_t \approx 10{,}222\ \text{cm}^{-1}$ for the hypothetical tetrahedral NH₃ complex.

Why This Works

The $\frac{4}{9}$ ratio comes directly from the geometry. In an octahedral complex, 6 ligands approach along $\pm x$ , $\pm y$ , $\pm z$ axes — directly toward the $d_{x^2-y^2}$ and $d_{z^2}$ orbitals ( $e_g$ set). This maximizes repulsion and maximizes $\Delta_o$ .

In a tetrahedral complex, the 4 ligands approach along alternate corners of a cube — they never point directly at any $d$ orbital. The closest they get is bisecting the axes. This indirect overlap means repulsion is weaker, and $\Delta_t$ is inherently smaller. This is why tetrahedral complexes are almost always high-spin — $\Delta_t$ rarely exceeds the spin-pairing energy.

The spectrochemical series layered on top of geometry explains the full color spectrum of transition metal compounds. A strong-field ligand in tetrahedral geometry ( $\Delta_t$ from strong field) can sometimes rival a weak-field ligand in octahedral geometry ( $\Delta_o$ from weak field). Knowing both factors is the key to predicting color and spin state.

Alternative Method

You can work backwards from the observed color using the color wheel.

Absorbed Color	Absorbed $\lambda$ (nm)	Observed Color
Violet	400–430	Yellow-green
Blue	430–480	Orange
Green	490–560	Red-purple
Yellow	560–600	Violet
Orange	600–625	Blue
Red	625–750	Blue-green

$[\text{CoCl}_4]^{2-}$ appears blue → absorbs orange-red region (~625–700 nm). Converting:

E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34})(3 \times 10^{10}\ \text{cm/s})}{700 \times 10^{-7}\ \text{cm}} \approx 14{,}300\ \text{cm}^{-1}

This matches what we expect for a tetrahedral weak-field cobalt(II) complex — rough agreement with back-calculating $\Delta_t$ from typical $\Delta_o$ values for Co²⁺ with Cl⁻.

In JEE, you’ll often be asked to predict color from the spectrochemical series position. Remember the order: $\text{I}^- < \text{Br}^- < \text{Cl}^- < \text{F}^- < \text{OH}^- < \text{H}_2\text{O} < \text{NH}_3 < \text{en} < \text{CO} \approx \text{CN}^-$ . Moving right = larger $\Delta$ = shorter $\lambda$ absorbed = shift toward blue/violet in absorption.

Common Mistake

Students often confuse the absorbed color with the observed color — these are complements, not the same. If a student sees ” $[\text{Co(NH}_3)_6]^{3+}$ absorbs violet light” and writes “the complex looks violet,” that’s wrong. It looks yellow-orange because that’s what remains after violet is absorbed. Always use the complementary color relationship. In JEE MCQs, one of the wrong options will always be the absorbed color itself — don’t fall for it.

A second trap: applying $\Delta_t = \frac{4}{9}\Delta_o$ when the metal or oxidation state is also different. The $\frac{4}{9}$ ratio holds when comparing the same metal-ligand pair in two geometries. If both the geometry AND the ligand change (like comparing $[\text{CoCl}_4]^{2-}$ directly to $[\text{Co(NH}_3)_6]^{3+}$ ), you cannot use just the $\frac{4}{9}$ factor — you need to account for the spectrochemical series difference separately.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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