Q.Match the solutions given in Column I and the colours given in Column II.
Column I (Aqueous solution of salt):

The key idea is that the colour of a transition-metal salt in aqueous solution depends on the number of unpaired $d$-electrons in the metal ion, which determines the specific wavelengths of light absorbed (d-d transitions).

Step 1 – Identify the metal ion and its $d^n$ configuration:

• $Fe^{2+}$ in $FeSO_4\cdot7H_2O$: $[Ar]3d^6$ (high spin, 4 unpaired electrons) → pale green.
• $Ni^{2+}$ in $NiCl_2\cdot4H_2O$: $[Ar]3d^8$ (2 unpaired) → green.
• $Mn^{2+}$ in $MnCl_2\cdot4H_2O$: $[Ar]3d^5$ (5 unpaired) → very pale pink (light pink).
• $Co^{2+}$ in $CoCl_2\cdot6H_2O$: $[Ar]3d^7$ (3 unpaired) → pink.
• $Cu^+$ in $Cu_2Cl_2$: $[Ar]3d^{10}$ (no unpaired electrons) → colourless. …

Why salts have colour — the d-orbital story

Most of us remember that transition metal compounds are colourful, but the reason is beautiful. When a transition metal ion is surrounded by water molecules (or other ligands), its five d-orbitals — which were all equal in energy in the free ion — split into two sets with different energies. Electrons can absorb visible light to jump from the lower set to the higher set. The colour we see is the complement of the light absorbed.

The exact colour depends on:

• The metal ion and its oxidation state
• The number of d-electrons
• The nature of the ligands (here, all are water or chloride)

Let's go salt by salt.

1. $FeSO_4 \cdot 7H_2O$ — Iron(II) sulphate heptahydrate

   Iron in this salt is in the +2 oxidation state. The electron configuration of $Fe^{2+}$ is $[Ar]\,3d^6$. In an octahedral field of water molecules, the d-orbitals split into $t_{2g}$ (lower) and $e_g$ (higher). With six d-electrons, the lower set is completely filled ($t_{2g}^6$) and the upper set is empty. The d-d transition requires a moderate amount of energy, corresponding to absorption in the red region. The complementary colour is a pale green.

   Tip

   $Fe^{2+}$ solutions are often described as pale green, but they can appear almost colourless if very dilute. The "pale" is important — it's not the deep green of $Ni^{2+}$.

   So (i) matches (d) Pale green.

2. $NiCl_2 \cdot 4H_2O$ — Nickel(II) chloride tetrahydrate

   Nickel is in the +2 state: $Ni^{2+}$ has $[Ar]\,3d^8$. In an octahedral environment, the eight d-electrons fill the $t_{2g}$ set completely and put two electrons in the $e_g$ set ($t_{2g}^6 e_g^2$). The d-d transitions absorb light in the blue and red regions, leaving green as the transmitted colour. This is a distinctive, bright green — think of the colour of nickel salts you might have seen in a lab.

   So (ii) matches (a) Green.

3. $MnCl_2 \cdot 4H_2O$ — Manganese(II) chloride tetrahydrate

   $Mn^{2+}$ has a $[Ar]\,3d^5$ configuration. This is a special case: with five d-electrons, each orbital gets one electron (Hund's rule), giving a half-filled, highly stable $d^5$ arrangement. The d-d transitions are spin-forbidden — they require an electron to flip its spin, which is much less likely. As a result, the absorption is very weak, and the colour is a very faint light pink (almost colourless in dilute solutions, but distinctly pink when concentrated).

   Watch out

   Don't confuse $Mn^{2+}$ (light pink) with $Mn^{7+}$ in permanganate (deep purple). The oxidation state changes everything — $Mn^{7+}$ has no d-electrons and gets its colour from charge transfer, not d-d transitions.

   So (iii) matches (b) Light pink.

4. $CoCl_2 \cdot 6H_2O$ — Cobalt(II) chloride hexahydrate …

Method: Colour of Hydrated Transition Metal Ions

This is a memory-based matching problem that relies on knowing the characteristic colours of common hydrated transition metal ions in aqueous solution.

Steps:

1. Identify the metal ion in each compound (ignore the water of crystallisation and anion for colour purposes — the colour comes from the hydrated cation).

2. Recall the standard colour of each hydrated transition metal ion in aqueous solution:

3. Match each compound from Column I to its colour in Column II:

Here are the most common mistakes students make when matching hydrated salt solutions to their colours, along with how to avoid each.

Mistake 1: Confusing the colours of $Fe^{2+}$ and $Fe^{3+}$ ions

• The Error: Students often think all iron compounds are reddish-brown (rust colour). They incorrectly match $FeSO_4 \cdot 7H_2O$ with a brown or dark colour, or confuse it with the yellow/brown of $Fe^{3+}$.
• The Correct Fact: $Fe^{2+}$ (ferrous) in aqueous solution is pale green. $Fe^{3+}$ (ferric) is yellow/brown.
• How to Avoid: Memorise the oxidation states. Remember: Ferrous ($+2$) = Pale green; Ferric ($+3$) = Yellow/Brown. For this question, (i) $FeSO_4 \cdot 7H_2O$ matches with (d) Pale green.

Mistake 2: Forgetting that $Cu^+$ is colourless (diamagnetic)

• The Error: Students assume all copper compounds are blue. They match $Cu_2Cl_2$ (copper(I) chloride) with blue.
• The Correct Fact: $Cu^+$ has a $d^{10}$ configuration (fully filled d-orbital). It has no unpaired electrons and cannot undergo d-d transitions. Therefore, its solutions are colourless (or white if solid). The blue colour comes from $Cu^{2+}$ ($d^9$).
• How to Avoid: Remember the diamagnetic rule: Ions with $d^{10}$ or $d^0$ configurations are colourless. $Cu^+$ ($d^{10}$), $Zn^{2+}$ ($d^{10}$), $Sc^{3+}$ ($d^0$) are all colourless. For this question, (v) $Cu_2Cl_2$ matches with (f) Colourless.

Mistake 3: Mixing up the colours of $Ni^{2+}$ and $Co^{2+}$

• The Error: Students swap the colours of Nickel and Cobalt salts. They match $NiCl_2$ with pink and $CoCl_2$ with green.
• The Correct Fact:
  • $Ni^{2+}$ (aqueous) is typically green (or apple green).
  • $Co^{2+}$ (aqueous) is typically pink (or rose red).
• How to Avoid: Use a simple mnemonic: "Nickel is Nice and Green" ; "Cobalt is Cool and Pink" . For this question:
  • (ii) $NiCl_2 \cdot 4H_2O$ → (a) Green
  • (iv) $CoCl_2 \cdot 6H_2O$ → (e) Pink

Mistake 4: Ignoring the hydration state or confusing it with the anhydrous form

• The Error: Students think $MnCl_2 \cdot 4H_2O$ is colourless (like $KMnO_4$) or dark. They also confuse the hydrated $CoCl_2 \cdot 6H_2O$ (pink) with anhydrous $CoCl_2$ (blue).
• The Correct Fact: …