ChemistryISC Class 12 · Coordination Compounds

Crystal Field Theory: ISC Class 12 Coordination Compounds

Q: What is crystal field splitting?

The separation of the five degenerate d orbitals of a metal ion into sets of different energy when ligands approach in a particular geometry, such as t₂g and e_g in an octahedral complex.

Q: What decides whether a complex is high spin or low spin?

The comparison of Δo with the pairing energy P: if Δo > P (strong field) the complex is low spin; if Δo < P (weak field) it is high spin.

Q: Why are tetrahedral complexes usually high spin?

Their splitting is only about four-ninths of the octahedral value, too small to force electrons to pair.

Q: How does crystal field theory explain colour?

An electron absorbs visible light to jump from a lower to a higher d level (a d-d transition); the complex shows the complementary colour.

Q: Why is anhydrous copper sulphate colourless?

With no water ligands, the d orbitals are not split, so no d-d transition occurs in the visible region.

Updated 15 October 2026 · Fact-checked

Crystal field theory treats ligands as point charges that split the five d orbitals. In an octahedral field they split into a lower t₂g set and a higher e_g set separated by Δo. Strong-field ligands make Δo larger than the pairing energy, giving low-spin complexes; weak-field ligands give high-spin ones. d-d transitions cause colour.

Understand Crystal Field Theory

The model. Crystal field theory (CFT) is an electrostatic model. The metal-ligand bond is treated as purely ionic: anionic ligands are point charges and neutral ligands (H₂O, NH₃) are point dipoles with their negative ends towards the metal. In a free ion all five d orbitals have the same energy (they are degenerate). In a spherical field of negative charge they would rise together, but ligands at definite positions repel some d orbitals more than others, so the degeneracy is lifted: crystal field splitting.

Octahedral splitting. Six ligands approach along the x, y and z axes. The $d_{x^{2} - y^{2}}$ and $d_{z^{2}}$ orbitals point straight at the ligands, are repelled more, and rise: the e_g set (2 orbitals), raised by $\frac{3}{5} Δ_{o}$ . The $d_{x y}$ , $d_{y z}$ and $d_{x z}$ orbitals point between the axes and fall: the t₂g set (3 orbitals), lowered by $\frac{2}{5} Δ_{o}$ . The energy gap is $Δ_{o}$ .

Tetrahedral splitting. Four ligands approach between the axes, so the pattern is inverted: the e set (2 orbitals) is lower and the t₂ set (3 orbitals) is higher. The splitting is smaller: $Δ_{t} = \frac{4}{9} Δ_{o}$ for the same metal, ligands and distances. Because $Δ_{t}$ is small, it is almost never bigger than the pairing energy, so low-spin tetrahedral complexes are rarely observed.

What decides Δo. The field strength of the ligand and the charge on the metal ion (a higher charge gives a larger splitting). The spectrochemical series, found by experiment from absorption spectra, lists ligands in order of increasing field strength:
I⁻ < Br⁻ < SCN⁻ < Cl⁻ < S²⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < edta⁴⁻ < NH₃ < en < CN⁻ < CO.

Filling t₂g and e_g: high spin and low spin. $d^{1}$ to $d^{3}$ ions put electrons singly into t₂g (Hund's rule). For $d^{4}$ to $d^{7}$ there is a choice. The extra electron can pair in t₂g (cost: the pairing energy P) or go up to e_g (cost: $Δ_{o}$ ). If $Δ_{o} < P$ (weak-field ligands), electrons spread out into e_g before pairing: high-spin (spin-free) complexes with many unpaired electrons. If $Δ_{o} > P$ (strong-field ligands), electrons pair in t₂g first: low-spin (spin-paired) complexes. Example for $d^{4}$ : weak field $t_{2 g}^{3} e_{g}^{1}$ (4 unpaired); strong field $t_{2 g}^{4} e_{g}^{0}$ (2 unpaired). For $d^{6}$ : weak $t_{2 g}^{4} e_{g}^{2}$ (4 unpaired); strong $t_{2 g}^{6} e_{g}^{0}$ (0, diamagnetic). $d^{8}$ , $d^{9}$ and $d^{10}$ have only one arrangement in an octahedral field. Calculations show that $d^{4}$ to $d^{7}$ complexes are more stable with strong-field ligands. NCERT's examples: $[M n (C N)_{6}]^{3 -}$ has two unpaired electrons but $[M n C l_{6}]^{3 -}$ four; $[F e (C N)_{6}]^{3 -}$ one but $[F e F_{6}]^{3 -}$ five; $[C o (C_{2} O_{4})_{3}]^{3 -}$ is diamagnetic but $[C o F_{6}]^{3 -}$ has four.

Colour. A complex removes part of white light and shows the complementary colour. In $[T i (H_{2} O)_{6}]^{3 +}$ (Ti³⁺, $3 d^{1}$ ), the single electron in t₂g absorbs blue-green light (about 498 nm) and is promoted to e_g ( $t_{2 g}^{1} e_{g}^{0} \to t_{2 g}^{0} e_{g}^{1}$ ), so the complex looks violet. This is a d-d transition. The energy of the light absorbed equals the splitting: $Δ = \frac{h c N _{A}}{λ}$ per mole. A stronger-field ligand gives a larger Δ and absorption at a shorter wavelength, so changing the ligand changes the colour. Without ligands there is no splitting: anhydrous CuSO₄ is white while CuSO₄·5H₂O is blue.

Limitations of CFT. Ligands are treated as point charges, so it cannot explain why neutral CO (with no charge) is a strong-field ligand while anionic I⁻ is weak; it ignores the covalent character of metal-ligand bonds.

Crystal Field Theory: formulas and key facts

Octahedral splitting: $e_{g} : + \frac{3}{5} Δ_{o}, t_{2 g} : - \frac{2}{5} Δ_{o}$; e_g = $d_{x^{2} - y^{2}}$ , $d_{z^{2}}$ ; t₂g = $d_{x y}$ , $d_{y z}$ , $d_{x z}$ .
Tetrahedral splitting: $Δ_{t} = \frac{4}{9} Δ_{o}$; Inverted pattern: e lower, t₂ higher; low spin rare.
High or low spin: $Δ_{o} < P \Rightarrow high spin; Δ_{o} > P \Rightarrow low spin$; P = pairing energy. Matters for $d^{4}$ – $d^{7}$ .
Splitting from absorbed light: $Δ = \frac{h c N _{A}}{λ}$; $h = 6.626 \times 1 0^{- 34}$ J s, $c = 3.0 \times 1 0^{8}$ m s⁻¹, λ in m; gives J mol⁻¹.
Spectrochemical series (increasing field): $I^{-} < B r^{-} < S C N^{-} < C l^{-} < S^{2 -} < F^{-} < O H^{-} < C_{2} O_{4}^{2 -} < H_{2} O < N C S^{-} < e d t a^{4 -} < N H_{3} < e n < C N^{-} < C O$; Experimental order from absorption spectra.

How to solve Crystal Field Theory questions

Almost every CFT question asks for an electron arrangement, a spin state, a number of unpaired electrons, a colour reason or a splitting energy. Use this sequence.

1Find the oxidation state of the metal and its d electron count.
2Decide the geometry (usually octahedral for CN 6, tetrahedral for most CN 4 with weak ligands).
3Place the ligand in the spectrochemical series: CN⁻, CO, en and NH₃ are strong field (for Co³⁺ and most 3d ions); halides and H₂O are weak field.
4Fill t₂g and e_g: weak field, fill all five singly before pairing; strong field, fill t₂g completely before e_g.
5Count unpaired electrons and, if asked, find $μ = n (n + 2)$ BM.
6For colour, name the d-d transition ( $t_{2 g} \to e_{g}$ ) and the complementary colour; for Δ from λ, use $Δ = h c N_{A} / λ$ and convert nm to m.

Crystal Field Theory in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Changing geometry and ligands changes colour

Suppose a student makes a home-made humidity indicator by soaking filter paper in cobalt(II) chloride solution and drying it. In humid air the paper is pink, because cobalt is present as octahedral $[C o (H_{2} O)_{6}]^{2 +}$ . When the paper is dried in a warm oven it turns blue, because tetrahedral chloro complexes such as $[C o C l_{4}]^{2 -}$ form.

Use crystal field theory to explain why the two forms have different colours, and why the blue tetrahedral form is high spin.

Try it first, then show the solutionSolution

Both forms contain Co²⁺ ( $3 d^{7}$ ); the colour comes from d-d transitions whose energy equals the crystal field splitting.
In the octahedral aqua complex the splitting is $Δ_{o}$ ; in the tetrahedral chloro complex it is only about $\frac{4}{9}$ as large, and Cl⁻ is also weaker than H₂O in the spectrochemical series.
A smaller splitting means lower-energy (longer-wavelength) light is absorbed, so the complementary colour seen changes from pink to blue.
$Δ_{t}$ is much smaller than the pairing energy, so electrons occupy the upper set before pairing: the tetrahedral complex is high spin.

Answer: Losing water changes octahedral $[C o (H_{2} O)_{6}]^{2 +}$ into tetrahedral chloro complexes with a much smaller splitting, so a different part of the spectrum is absorbed and the paper turns from pink to blue; small $Δ_{t}$ keeps the tetrahedral form high spin.

Memory hook: Smaller split, longer wavelength absorbed, different colour seen.

Tricks to remember Crystal Field Theory

Short memory aids. Say each one aloud once, then use it on a question the same day.

e_g points at the ligands
'e_g = edgy': $d_{x^{2} - y^{2}}$ and $d_{z^{2}}$ lie on the axes, poke the ligands and go up. t₂g sit between the axes and go down.
Use it when: Drawing the octahedral splitting diagram.
I Bet Some Clever Students Find Our Chemistry...
I⁻ Br⁻ SCN⁻ Cl⁻ S²⁻ F⁻ OH⁻ C₂O₄²⁻ H₂O NCS⁻ edta NH₃ en CN⁻ CO: 'I Bet Some Clever Students Find Our Chemistry Hard, Not Every Night; Even CN Copes'.
Use it when: Recalling the spectrochemical series in order.
Strong pairs, weak spreads
Strong-field ligands make electrons pair up (low spin); weak-field ligands let them spread out (high spin).
Use it when: Deciding the configuration for $d^{4}$ – $d^{7}$ .

Crystal Field Theory: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Strongest field ligand · 1 mark · MCQBoard level

Which of the following ligands produces the largest crystal field splitting with a given metal ion?
(a) Cl⁻
(b) H₂O
(c) NH₃
(d) CN⁻

Try it first, then show the solutionSolution

In the spectrochemical series: Cl⁻ < H₂O < NH₃ < CN⁻.
CN⁻ is the strongest of the four.

Answer: (d) CN⁻.

Example 2: A d³ ion needs no choice · 2 marks · Short answerBoard level

Write the t₂g/e_g configuration of Cr³⁺ (Z = 24) in an octahedral complex and the number of unpaired electrons. Does it matter whether the ligands are strong or weak field? Explain.

Try it first, then show the solutionSolution

Cr³⁺ is $3 d^{3}$ : $t_{2 g}^{3} e_{g}^{0}$ , 3 unpaired electrons (1 mark).
No: three electrons fit singly into the three t₂g orbitals, so no pairing or e_g occupation decision arises whatever the field strength (1 mark).

Answer: $t_{2 g}^{3} e_{g}^{0}$ , 3 unpaired; the same for strong and weak fields.

Example 3: Δo from an absorption wavelength · 3 marks · Short answerHarder board level

The complex $[C o C l (N H_{3})_{5}]^{2 +}$ absorbs light most strongly at 535 nm. Calculate the crystal field splitting energy in kJ mol⁻¹ and state the colour of the complex if the light absorbed is yellow. ( $h = 6.626 \times 1 0^{- 34}$ J s, $c = 3.0 \times 1 0^{8}$ m s⁻¹, $N_{A} = 6.022 \times 1 0^{23}$ mol⁻¹)

Try it first, then show the solutionSolution

$Δ = \frac{h c N _{A}}{λ}$ (½ mark).
$Δ = \frac{6.626 \times 1 0 ^{- 34} \times 3.0 \times 1 0 ^{8} \times 6.022 \times 1 0 ^{23}}{535 \times 1 0 ^{- 9}}$ (1 mark).
$Δ = 2.24 \times 1 0^{5}$ J mol⁻¹ $= 224$ kJ mol⁻¹ (1 mark).
Yellow light is absorbed, so the complementary colour, violet, is seen (½ mark).

Answer: About 224 kJ mol⁻¹; the complex is violet.

Example 4: Two manganese(III) complexes · 3 marks · Short answerHarder board level

$[M n (C N)_{6}]^{3 -}$ has two unpaired electrons while $[M n C l_{6}]^{3 -}$ has four. Using crystal field theory, write the t₂g/e_g configuration of each, identify which is low spin, and calculate the spin-only magnetic moment of each. (Mn Z = 25)

Try it first, then show the solutionSolution

Mn³⁺ is $3 d^{4}$ (½ mark).
CN⁻ is strong field ( $Δ_{o} > P$ ): $t_{2 g}^{4} e_{g}^{0}$ , 2 unpaired, low spin; $μ = 8 = 2.83$ BM (1 mark).
Cl⁻ is weak field ( $Δ_{o} < P$ ): $t_{2 g}^{3} e_{g}^{1}$ , 4 unpaired, high spin; $μ = 24 = 4.90$ BM (1 mark).
The cyanide complex is the low-spin one (½ mark).

Answer: $[M n (C N)_{6}]^{3 -}$ : $t_{2 g}^{4} e_{g}^{0}$ , low spin, 2.83 BM. $[M n C l_{6}]^{3 -}$ : $t_{2 g}^{3} e_{g}^{1}$ , high spin, 4.90 BM.

Example 5: Two cobalt(III) complexes and a tetrahedral comparison · 5 marks · Long answerHOTS / competency

$[C o F_{6}]^{3 -}$ is paramagnetic with four unpaired electrons, whereas $[C o (C_{2} O_{4})_{3}]^{3 -}$ is diamagnetic. (Co Z = 27)
(a) Write the t₂g/e_g configuration of each and relate it to Δo and P. [2]
(b) Which ligand is the stronger field ligand here? Is this consistent with the spectrochemical series? [1]
(c) If $Δ_{o}$ for a certain octahedral complex is 200 kJ mol⁻¹, estimate $Δ_{t}$ for the corresponding tetrahedral complex, and explain why tetrahedral complexes are almost always high spin. [2]

Try it first, then show the solutionSolution

(a) Co³⁺ is $3 d^{6}$ . $[C o F_{6}]^{3 -}$ : $t_{2 g}^{4} e_{g}^{2}$ , 4 unpaired, so $Δ_{o} < P$ (weak field, high spin) (1 mark). $[C o (C_{2} O_{4})_{3}]^{3 -}$ : $t_{2 g}^{6} e_{g}^{0}$ , no unpaired electrons, so $Δ_{o} > P$ (low spin) (1 mark).
(b) Oxalate is the stronger field ligand; this agrees with the series, where F⁻ < C₂O₄²⁻ (1 mark).
(c) $Δ_{t} = \frac{4}{9} \times 200 = 88.9$ kJ mol⁻¹ (1 mark). Tetrahedral splitting is less than half the octahedral value and is nearly always smaller than the pairing energy, so electrons occupy the upper level rather than pair: high spin (1 mark).

Answer: (a) $t_{2 g}^{4} e_{g}^{2}$ (Δo < P) and $t_{2 g}^{6} e_{g}^{0}$ (Δo > P). (b) Oxalate; consistent. (c) About 89 kJ mol⁻¹; small Δt rarely exceeds P.

Where students lose marks in Crystal Field Theory

The slips examiners see most often, and what to write instead.

Mistake: Labelling the wrong complex as low spin
ISC examiners reported that in a 2025 question candidates called the $t_{2 g}^{4} e_{g}^{2}$ complex high spin but failed to identify the $t_{2 g}^{6} e_{g}^{0}$ complex as the low-spin one.
Do this: Low spin = electrons paired in t₂g before e_g is used. $t_{2 g}^{6} e_{g}^{0}$ is low spin; $t_{2 g}^{4} e_{g}^{2}$ is high spin.
Usually costs: 1 mark.
Mistake: Miscounting unpaired electrons from a t₂g/e_g configuration
Examiners recorded 2 or 0 unpaired for $t_{2 g}^{4} e_{g}^{2}$ and 2 or 4 for $t_{2 g}^{6} e_{g}^{0}$ (2025).
Wrong working
$t_{2 g}^{4} e_{g}^{2}$ : 2 unpaired.
Right working
$t_{2 g}^{4}$ : one pair + 2 single; $e_{g}^{2}$ : 2 single. Total 4 unpaired.
Do this: Draw three t₂g boxes and two e_g boxes and fill them; then count.
Usually costs: 1 mark.
Mistake: Confusing strong and weak field ligands
Examiners noted that candidates could not say whether a high-spin complex has strong- or weak-field ligands (2025).
Do this: High spin (more unpaired electrons) means Δo < P: weak-field ligands. Low spin means strong-field ligands.
Usually costs: 1 mark.
Mistake: Forgetting to convert nm to m in Δ calculations
Students divide by 535 instead of $535 \times 1 0^{- 9}$ .
Do this: 1 nm = $1 0^{- 9}$ m; check that Δ comes out in the range of about 100 to 400 kJ mol⁻¹ for visible light.
Usually costs: The answer mark.

Exam tips

Draw the splitting diagram with the barycentre: e_g up by $\frac{3}{5} Δ_{o}$ , t₂g down by $\frac{2}{5} Δ_{o}$ .
Always state the d count first; most errors start with the wrong oxidation state.
Say 'strong field ligand, Δo > P, low spin' as one chain of reasoning.
For colour, use the words 'd-d transition' and 'complementary colour'.
Use NCERT's matched pairs as examples: $[F e (C N)_{6}]^{3 -}$ / $[F e F_{6}]^{3 -}$ and $[C o (C_{2} O_{4})_{3}]^{3 -}$ / $[C o F_{6}]^{3 -}$ .

Practice questions on Crystal Field Theory

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
In an octahedral crystal field, the d orbitals that are raised in energy are:
(a) $d_{x y}$ and $d_{y z}$
(b) $d_{x^{2} - y^{2}}$ and $d_{z^{2}}$
(c) $d_{x y}$ , $d_{y z}$ and $d_{x z}$
(d) all five equally
Hint
Which orbitals point along the axes?
Answer
(b), the e_g set.
Q2 · 2 marks · Short answerBoard level
What is the spectrochemical series? Arrange F⁻, en, H₂O and I⁻ in order of increasing crystal field strength.
Hint
Halides are at the weak end.
Answer
A series of ligands arranged in order of increasing field strength (increasing Δo), found experimentally from absorption spectra. I⁻ < F⁻ < H₂O < en.
Q3 · 2 marks · Short answerHarder board level
Write the t₂g/e_g configurations and the number of unpaired electrons for a $d^{7}$ ion in (a) a weak octahedral field and (b) a strong octahedral field.
Hint
Fill singly first in a weak field; fill t₂g first in a strong field.
Answer
(a) $t_{2 g}^{5} e_{g}^{2}$ , 3 unpaired (high spin). (b) $t_{2 g}^{6} e_{g}^{1}$ , 1 unpaired (low spin).
Q4 · 3 marks · Short answerHarder board level
$[C o (N H_{3})_{6}]^{3 +}$ absorbs at 475 nm. (a) Calculate Δo in kJ mol⁻¹. (b) $[C o C l (N H_{3})_{5}]^{2 +}$ absorbs at a longer wavelength. Explain this using the spectrochemical series. ( $h = 6.626 \times 1 0^{- 34}$ J s, $c = 3.0 \times 1 0^{8}$ m s⁻¹, $N_{A} = 6.022 \times 1 0^{23}$ mol⁻¹)
Hint
Longer wavelength means lower energy.
Answer
(a) $Δ_{o} = \frac{6.626 \times 1 0 ^{- 34} \times 3.0 \times 1 0 ^{8} \times 6.022 \times 1 0 ^{23}}{475 \times 1 0 ^{- 9}} = 2.52 \times 1 0^{5}$ J mol⁻¹ $= 252$ kJ mol⁻¹. (b) Replacing one NH₃ by Cl⁻, a weaker-field ligand, lowers the splitting, so light of lower energy (longer wavelength) is absorbed.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Low-spin tetrahedral complexes are rarely observed.
Reason (R): The tetrahedral splitting, $Δ_{t} = \frac{4}{9} Δ_{o}$ , is usually smaller than the pairing energy.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Compare the energy gap with the cost of pairing.
Answer
(a).
Q6 · 3 marks · Short answerHarder board level
Using crystal field theory, explain why $[T i (H_{2} O)_{6}]^{3 +}$ is coloured, and predict what would happen to the colour if titanium were oxidised to Ti⁴⁺ with the same ligands. (Ti Z = 22)
Hint
Count the d electrons in each.
Answer
Ti³⁺ is $3 d^{1}$ : $t_{2 g}^{1} e_{g}^{0}$ . The electron absorbs blue-green visible light and is promoted to e_g (d-d transition), so the complex shows the complementary violet colour. Ti⁴⁺ is $3 d^{0}$ : no d electron to promote, no d-d transition, so the complex would be colourless.
Q7 · 3 marks · Short answerHOTS / competency
Predict the number of unpaired electrons and the spin-only magnetic moments of $[C o (H_{2} O)_{6}]^{2 +}$ (weak field) and $[C o (C N)_{6}]^{4 -}$ (strong field). Which would show the larger crystal field splitting? (Co Z = 27)
Hint
Co²⁺ is $3 d^{7}$ .
Answer
$[C o (H_{2} O)_{6}]^{2 +}$ : $t_{2 g}^{5} e_{g}^{2}$ , 3 unpaired, $μ = 3.87$ BM. $[C o (C N)_{6}]^{4 -}$ : $t_{2 g}^{6} e_{g}^{1}$ , 1 unpaired, $μ = 1.73$ BM. The cyanide complex has the larger splitting (CN⁻ is far stronger than H₂O).
Q8 · 3 marks · Case-basedHOTS / competency
Read and answer.
Suppose a student adds ethane-1,2-diamine (en) a little at a time to a green solution of $[N i (H_{2} O)_{6}]^{2 +}$ . The colour gradually shifts towards blue-violet as water ligands are replaced to form $[N i (e n)_{3}]^{2 +}$ . (Ni Z = 28)
(i) Write the t₂g/e_g configuration of Ni²⁺ in these octahedral complexes and the number of unpaired electrons. [1]
(ii) Why does the colour change? [1]
(iii) Does the number of unpaired electrons change when en replaces water? Explain. [1]
Hint
Ni²⁺ is $3 d^{8}$ .
Answer
(i) $t_{2 g}^{6} e_{g}^{2}$ , 2 unpaired. (ii) en is a stronger-field ligand than H₂O, so Δo increases; higher-energy (shorter-wavelength) light is absorbed and the complementary colour seen changes. (iii) No. A $d^{8}$ ion has only one possible arrangement in an octahedral field ( $t_{2 g}^{6} e_{g}^{2}$ ), so it always has 2 unpaired electrons.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Which d orbitals form the t₂g set?Show answer
$d_{x y}$ , $d_{y z}$ and $d_{x z}$ .
2.Relation between Δt and Δo?Show answer
$Δ_{t} = \frac{4}{9} Δ_{o}$ .
3.Configuration of a low-spin $d^{6}$ octahedral complex?Show answer
$t_{2 g}^{6} e_{g}^{0}$ (diamagnetic).
4.Strongest ligand in the spectrochemical series?Show answer
CO (then CN⁻).
5.Why is $[T i (H_{2} O)_{6}]^{3 +}$ violet?Show answer
Its $t_{2 g}^{1}$ electron absorbs blue-green light to move to e_g; violet is the complementary colour.
6.One limitation of CFT?Show answer
It treats ligands as point charges, so it cannot explain why neutral CO is a strong-field ligand.

Crystal Field Theory: frequently asked questions

What is crystal field splitting?

The separation of the five degenerate d orbitals of a metal ion into sets of different energy when ligands approach in a particular geometry, such as t₂g and e_g in an octahedral complex.

What decides whether a complex is high spin or low spin?

The comparison of Δo with the pairing energy P: if Δo > P (strong field) the complex is low spin; if Δo < P (weak field) it is high spin.

Why are tetrahedral complexes usually high spin?

Their splitting is only about four-ninths of the octahedral value, too small to force electrons to pair.

How does crystal field theory explain colour?

An electron absorbs visible light to jump from a lower to a higher d level (a d-d transition); the complex shows the complementary colour.

Why is anhydrous copper sulphate colourless?

With no water ligands, the d orbitals are not split, so no d-d transition occurs in the visible region.