Q.Aqueous copper sulphate solution (blue in colour) gives:

Concept: Complex Formation – The colour change arises from the replacement of water ligands around $\text{Cu}^{2+}$ by different halide ions, altering the crystal field splitting and hence the absorbed/transmitted wavelengths.

Reasoning:

1. In aqueous solution, $\text{Cu}^{2+}$ exists as the blue hexaaqua complex $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$.
2. With $\text{KF}$, fluoride ions ($\text{F}^-$) are small, hard anions but weak-field ligands ($\text{F}^-$ lies below water in the spectrochemical series). They displace water and precipitate sparingly soluble $\text{CuF}_2$; the changed ligand field shifts the d–d transition energy, giving a green colour. …

The key to understanding these observations lies in complex formation — the ability of transition metal ions like $\ce{Cu^{2+}}$ to bind with ligands (ions or molecules that donate electron pairs). In aqueous solution, $\ce{Cu^{2+}}$ is surrounded by six water molecules, giving the familiar pale blue colour of $\ce{[Cu(H2O)6]^{2+}}$. When other ligands are added, they can replace water, altering the electronic environment around the copper ion and thus changing the colour of the solution.

But not all ligands behave the same way. The halide ions $\ce{F-}$ and $\ce{Cl-}$ differ significantly in size, charge density, and ability to form stable complexes. This leads to two very different outcomes.

1. Reaction with aqueous potassium fluoride ($\ce{KF}$) Fluoride ion is small and has a high charge density. As a ligand it is nevertheless weak-field ($\ce{F-}$ sits below water in the spectrochemical series — $[\ce{CoF6}]^{3-}$ is the textbook's own high-spin example), and it has a strong tendency to form an insoluble salt with $\ce{Cu^{2+}}$. When $\ce{KF}$ is added to blue $\ce{CuSO4}$ solution, the fluoride ions displace water and immediately precipitate as copper(II) fluoride, $\ce{CuF2}$, which is green. The reaction is:

The green colour of $\ce{CuF2}$ arises because the fluoride ligand produces a different crystal field splitting than water, shifting the absorption of light to a different wavelength. No soluble complex forms because $\ce{CuF2}$ is sparingly soluble.

2. Reaction with aqueous potassium chloride ($\ce{KCl}$) Chloride ion is larger and has a lower charge density. It does not form an insoluble salt with $\ce{Cu^{2+}}$ under these conditions. Instead, chloride ions gradually replace water molecules to form a soluble complex ion. The most stable species in concentrated chloride solution is the tetrachlorocuprate(II) ion, $\ce{[CuCl4]^{2-}}$, which has a distorted tetrahedral geometry. The reaction is: $\ce{[Cu(H2O)6]^{2+} + 4Cl- -> [CuCl4]^{2-} + 6H2O}$ …

Concept: Complex Formation — Crystal Field Theory & Ligand Effects

Method: Ligand Field Analysis & Spectrochemical Series

This method explains colour changes by comparing how different ligands split the d‑orbitals of the central metal ion.

Step 1: Identify the central metal ion and its initial state

• Aqueous copper sulphate contains the hexaaquacopper(II) ion: $\ce{[Cu(H2O)6]^{2+}}$
• $\ce{Cu^{2+}}$ has a $d^9$ configuration.
• Water ($\ce{H2O}$) is a moderate field ligand — it causes a certain amount of d‑orbital splitting.
• The observed blue colour comes from $d$–$d$ transitions in this aqua complex.

Step 2: Analyse reaction with aqueous potassium fluoride ($\ce{KF}$)

• Fluoride ion ($\ce{F^-}$) is a weak field ligand (low in the spectrochemical series).
• $\ce{F^-}$ does not remain in solution as a stable soluble fluoro complex here: being small, with a high lattice energy for its salts, it precipitates sparingly soluble copper(II) fluoride — $\ce{[Cu(H2O)6]^{2+} + 2F^- -> CuF2 (s) + 6H2O}$
• Effect on splitting: Weak field → smaller $\Delta$ (crystal field splitting energy).
• For a $d^9$ ion, smaller $\Delta$ shifts the absorption to longer wavelengths (red‑shift).
• The complementary colour shifts from blue toward green.

Result: Green precipitate — due to weak field ligand causing a smaller d‑orbital splitting and altered colour.

Step 3: Analyse reaction with aqueous potassium chloride ($\ce{KCl}$)

• Chloride ion ($\ce{Cl^-}$) is also a weak field ligand — in fact even weaker than $\ce{F^-}$ in the spectrochemical series ($\ce{I^- < Br^- < SCN^- < Cl^- < S^{2-} < F^- < OH^- ...}$).
• The complex formed is likely: $\ce{[CuCl4]^{2-}}$ (tetrachlorocuprate(II) ion).
• Effect on splitting: $\ce{Cl^-}$ produces a different splitting pattern (often tetrahedral or distorted geometry).
• For $\ce{Cu^{2+}}$ in a tetrahedral field, the splitting is smaller than in octahedral, and the $d$–$d$ transitions absorb in the red‑orange region.
• The complementary colour is bright green.

Result: Bright green solution — due to formation of a soluble chloro‑complex with distinct geometry and absorption.

Step 4: Summarise the key principle …

Here are the common mistakes students make when explaining the complex formation in this question, along with how to avoid each.

Mistake 1: Forgetting the Role of Ligand Field Strength

The Mistake: Students often treat all halides (F⁻, Cl⁻) as identical ligands. They fail to realize that the color change depends on which ligand replaces water and how strongly that ligand splits the d-orbitals.

How to Avoid:

• Remember the spectrochemical series: I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < NO₂⁻ < CN⁻ < CO.
• Key fact: F⁻ is a weak field ligand (lower in the series than H₂O), while Cl⁻ is an even weaker field ligand.
• Apply the logic:
  • When F⁻ replaces H₂O, the ligand field strength decreases slightly → change in color (blue → green precipitate).
  • When Cl⁻ replaces H₂O, the ligand field strength decreases significantly → different color (bright green solution).

Mistake 2: Confusing Precipitate vs. Solution

The Mistake: Students write that both reactions give a solution, or they don’t explain why one gives a precipitate and the other a solution.

How to Avoid:

• Check solubility rules:
  • CuF₂ is sparingly soluble in water → forms a green precipitate.
  • [CuCl₄]²⁻ is a soluble complex ion → forms a bright green solution.
• Write the correct products:
  • With KF: $\text{CuSO}_4(aq) + 2\text{KF}(aq) \rightarrow \text{CuF}_2(s) + \text{K}_2\text{SO}_4(aq)$
  • With KCl: $\text{CuSO}_4(aq) + 4\text{KCl}(aq) \rightarrow \text{K}_2[\text{CuCl}_4](aq) + \text{K}_2\text{SO}_4(aq)$

Mistake 3: Ignoring Coordination Number Differences

The Mistake: Students assume both F⁻ and Cl⁻ form the same type of complex (e.g., both form tetrahedral [CuX₄]²⁻).

How to Avoid:

• Know the typical coordination numbers:
  • F⁻ (small, hard base) usually forms octahedral complexes (e.g., [CuF₆]⁴⁻) or simple salts like CuF₂.
  • Cl⁻ (larger, softer base) can form tetrahedral complexes like [CuCl₄]²⁻.
• In this case:
  • With F⁻: No complex ion forms; instead, CuF₂ precipitate forms (Cu²⁺ with 2 F⁻).
  • With Cl⁻: A tetrahedral complex [CuCl₄]²⁻ forms (Cu²⁺ with 4 Cl⁻).

Mistake 4: Not Explaining the Color Change Mechanism

The Mistake: Students simply state “color changes” without linking it to d-orbital splitting and energy of absorbed light.

How to Avoid:

• Use the crystal field theory:
  • In [Cu(H₂O)₆]²⁺ (blue), H₂O is a moderate field ligand → absorbs orange-red light → appears blue.
  • In CuF₂ (green precipitate), F⁻ is a weaker field than H₂O → smaller splitting → absorbs lower energy (redder) light → appears green.
  • In [CuCl₄]²⁻ (bright green solution), Cl⁻ is even weaker → even smaller splitting → absorbs even redder light → appears bright green.
• Summary table:

Mistake 5: Writing Incorrect Chemical Equations

The Mistake: Students write unbalanced equations or use wrong formulas (e.g., writing CuF instead of CuF₂).

How to Avoid:

• Always balance charges:
  • Cu²⁺ + 2F⁻ → CuF₂ (correct)
  • Cu²⁺ + 4Cl⁻ → [CuCl₄]²⁻ (correct)
• Include state symbols: (aq) for solutions, (s) for precipitate. …