Q.CuSO4.5H2O is blue in colour whereas anhydrous CuSO4 is colourless. Why?

Showing the 12 most recent of 15 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.Which of the following ion is colourless in aqueous solution-

   (a)

   (i) Fe²⁺

   (b)

   (ii) Mn²⁺

   (c)

   (iii) Zn²⁺

   (d)

   (iv) Cu²⁺

   ›Reveal solutionSolution

   $\text{Zn}^{2+}$ has a fully filled $3d^{10}$ configuration, so it cannot undergo d–d transitions and is colourless. Correct option: (iii).

   Concept. A transition-metal ion is coloured when it can absorb visible light by promoting an electron between the split $t_{2g}$ and $e_g$ d-orbitals (a d–d transition). This is only possible if the d-subshell is partially filled ($d^1$ to $d^9$).

   Steps (configurations of the ions).

   • $\text{Fe}^{2+}$: $3d^{6}$ — partly filled $\Rightarrow$ coloured (pale green).
   • $\text{Mn}^{2+}$: $3d^{5}$ — partly filled $\Rightarrow$ coloured (pale pink). …

   🔒 Start your 14-day free trial to see the full solution →
2. CBSE 2025Set 56/4/11 markMCQ
   Q.Out of $Fe^{3+}$, $Sc^{3+}$, $Cr^{3+}$ and $Co^{3+}$ ions, the one which is colourless in aqueous solution is : (A) $Sc^{3+}$ (B) $Fe^{3+}$ (C) $Cr^{3+}$ (D) $Co^{3+}$ [Atomic number : Fe = 26, Sc = 21, Cr = 24, Co = 27]

   ›Reveal solutionSolution

   Colour in transition metal ions arises from d–d transitions, which require unpaired electrons in the d‑orbitals. $Sc^{3+}$ has a $3d^0$ configuration (no d‑electrons), so it cannot undergo d–d transitions and is colourless. The correct option is (A).

   The question asks which of the given trivalent ions is colourless in aqueous solution. Colour in transition metal ions is almost always due to the absorption of visible light by electrons moving between d‑orbitals — the famous d–d transition. For this to happen, the ion must have at least one electron in its d‑orbitals (a partially filled d‑subshell). If the d‑subshell is completely empty ($d^0$) or completely filled ($d^{10}$), no d–d transition is possible, and the ion appears colourless (or white) in solution.

   Let’s check the electronic configuration of each ion.

   1. $Sc^{3+}$ (Atomic number 21)

      Sc ground state: $[Ar]\, 3d^1 4s^2$.

      Removing three electrons (the two 4s electrons and the one 3d electron) gives $Sc^{3+}$: $[Ar]\, 3d^0$.

      No d‑electrons at all → no d–d transitions → colourless.

   2. $Fe^{3+}$ (Atomic number 26)

      Fe ground state: $[Ar]\, 3d^6 4s^2$.

      Removing three electrons gives $Fe^{3+}$: $[Ar]\, 3d^5$.

      Five unpaired electrons (half‑filled d‑subshell) → d–d transitions possible → coloured (typically yellow‑brown in aqueous solution).

   3. $Cr^{3+}$ (Atomic number 24)

      Cr ground state: $[Ar]\, 3d^5 4s^1$ (exception to the usual filling order).

      Removing three electrons gives $Cr^{3+}$: $[Ar]\, 3d^3$.

      Three d‑electrons → d–d transitions possible → coloured (violet or green depending on ligands). …

   🔒 Start your 14-day free trial to see the full solution →
3. CBSE 2025Set D1 markMCQ
   Q.Which of the following compounds can be coloured?

   (a) Ag2SO4

   (b) CuF2

   (c) Cu2Cl2

   (d) MgF2

   ›Reveal solutionSolution

   A compound is coloured only if the metal ion has a partially filled d subshell; CuF2 (Cu2+ = d9) qualifies.

   Colour in transition-metal compounds arises from d-d electronic transitions, which require a partially filled d subshell. Examining each cation:

   • Ag2SO4: Ag+ is [Kr] 4d10 -> filled d -> colourless. …

   🔒 Start your 14-day free trial to see the full solution →
4. CBSE 2025Set ANNUAL1 markMCQ
   Q.Which of the following transition elements does not give coloured salt?

   (a) Cr

   (b) Mn

   (c) Cu

   (d) Zn

   ›Reveal solutionSolution

   Colour in transition-metal ions arises from d-d electronic transitions, which require partially filled d orbitals; Zn2+ has a full d10 shell with no d-d transition possible, so its salts are colourless.

   Transition metal ions are typically coloured because of d-d transitions: an electron in a lower-energy d orbital (in the crystal field of surrounding ligands/anions) absorbs visible light and jumps to a higher-energy d orbital; the colour seen is complementary to the wavelength absorbed. This requires the d subshell to be partially filled (neither completely empty nor completely full), so a d-d transition is possible.

   …

   🔒 Start your 14-day free trial to see the full solution →
5. CBSE 2025Set ANNUAL1 markMCQ
   Q.Colourless metal ion in aqueous solution is -

   (a) Cu2+

   (b) Zn2+

   (c) Mn2+

   (d) V2+

   ›Reveal solutionSolution

   Colour in transition-metal ions arises from d-d transitions of unpaired electrons; Zn2+ (3d10, fully filled) has none, so it is colourless.

   Electronic configurations:

   • Cu2+: [Ar]3d9 - one unpaired electron -> coloured (blue)
   • Zn2+: [Ar]3d10 - all d orbitals completely filled, no unpaired electron and no vacant d orbital to promote an electron into -> colourless
   • Mn2+: [Ar]3d5 - unpaired electrons -> coloured (pale pink)
   • V2+: [Ar]3d3 - unpaired electrons -> coloured (violet) …

   🔒 Start your 14-day free trial to see the full solution →
6. CBSE 2024Set ANNUAL1 markMCQ
   Q.Direction: In this part of Question No.1, there are two statements labelled as Assertion (A) and Reason (R). From the following options select the correct answer.

   (i) Both A and R are correct and R is the correct explanation of A.

   (ii) Both A and R are correct but R is not the correct explanation of A.

   (iii) A is correct but R is incorrect.

   (iv) Both A and R are incorrect. Assertion (A): The elements of d and f blocks produce coloured ions. Reason (R): Unpaired electrons are present in these elements.

   (a) Both A and R are correct and R is the correct explanation of A.

   (b) Both A and R are correct but R is not the correct explanation of A.

   (c) A is correct but R is incorrect.

   (d) Both A and R are incorrect.

   ›Reveal solutionSolution

   Unpaired d/f electrons undergo d-d (or f-f) transitions, which is exactly why these elements form coloured ions.

   d- and f-block elements have partially filled d or f orbitals in their ions, i.e. unpaired electrons. When such an ion absorbs visible light, an electron is excited from a lower energy d (or f) orbital to a higher energy one within the same subshell (a d-d or f-f transition); the wavelength of visible light absorbed corresponds to the complementary colour observed. Elements/ions with no unpaired d or …

   🔒 Start your 14-day free trial to see the full solution →
7. CBSE 2024Set ANNUAL1 markMCQ
   Q.Which of the following statement about transition element is not correct?

   (a) They show variable oxidation states

   (b) They exhibit diamagnetic and paramagnetic properties

   (c) All ions are coloured

   (d) They exhibit catalytic property

   ›Reveal solutionSolution

   Colour requires a partially filled d-subshell for a d-d transition; ions with an empty (d0) or full (d10) d-subshell are colourless.

   Transition elements do show variable oxidation states (a) and both paramagnetic and diamagnetic behaviour depending on the number of unpaired d-electrons (b), and they are well known as catalysts (d) — all true statements. But 'all ions are coloured' is false: colour arises only from a d–d electronic transition, which requires a partially-filled d-subshell. Ions …

   🔒 Start your 14-day free trial to see the full solution →
8. CBSE 2023Set A1 mark
   Q.Fill in the blank: The colour of Ni^2+ is ______.

   ›Reveal solutionSolution

   Transition metal ions are typically coloured because of d-d electronic transitions; the hydrated Ni²⁺ ion is green.

   Ni²⁺ has the configuration [Ar]3d⁸ — it has partially filled d-orbitals, so it can absorb visible light to promote an electron from a lower-energy d-orbital to a higher-energy one (d-d transition). The colour observed is complementary to the colou …

   🔒 Start your 14-day free trial to see the full solution →
9. CBSE 2023Set ANNUAL1 markMCQ
   Q.Colours of transition metal ions are due to absorption of light of some wavelength. This results in

   (a) d-d transition

   (b) p-p transition

   (c) s-s transition

   (d) d-f transition

   ›Reveal solutionSolution

   Ligand-field splitting of the d-orbitals creates a small energy gap that visible light can bridge, so absorption/transmission of specific wavelengths gives transition-metal compounds their colour.

   In the presence of ligands, the five degenerate d-orbitals of a transition metal ion split into two (or more) sets of different energy (crystal/ligand field splitting, e.g. t2g and eg in an octahedral field). If the ion has partially filled d-orbitals, an electron can absorb a photon of visible light and jump from the lower-energy set to the higher-energy set - a d-d transition. The c …

   🔒 Start your 14-day free trial to see the full solution →
10. CBSE 2022Set E1 markMCQ
    Q.Which of the following ions is colourless ?

    (a) Cu+

    (b) Co2+

    (c) Ni2+

    (d) Fe3+

    ›Reveal solutionSolution

    Cu+ is 3d10 (no d-d transitions possible) → colourless; Co2+, Ni2+ and Fe3+ have partly filled d subshells → coloured.

    Colour in transition-metal ions arises from d-d electronic transitions, which need a partly filled d subshell.

    • Cu+: Cu is +1 → 3d10 (completely filled) → no d-d transition → colourless. …

    🔒 Start your 14-day free trial to see the full solution →
11. CBSE 2019Set HE1 mark
    Q.Answer in one word: The salts of transition metals are generally coloured. Why?

    ›Reveal solutionSolution

    Coloured salts of transition metals arise because their partially filled d-orbitals allow d–d electronic transitions on absorption of visible light; the observed colour is complementary to the wavelength absorbed.

    In a free transition-metal ion, all 5 d-orbitals are degenerate (equal energy). But in a compound/complex, the surrounding ligands (or ions/water molecules) split the d-orbitals into two sets of slightly different energies (crystal field splitting). If the ion has partially filled d-orbitals (i.e. neither completely empty nor completely full), an electron can be promoted from the lower-energy set of d-orbitals to the higher-energy set by absorbing a photon of visible light of just the right energy — this is called a d–d transition.

    The wavelengths NOT absorbed (transmitted/reflected) reach our eye and are perceived as the salt's colour — the colour we observe is complementary to the colour actually absorbed.

    …

    🔒 Start your 14-day free trial to see the full solution →
12. CBSE 2019Set ANNUAL1 mark
    Q.Predict which of the following will be coloured in aqueous solution: $Sc^{3+}$, $Fe^{3+}$, $Ti^{4+}$, $V^{3+}$ (Atomic nos. of Sc, Fe, Ti and V are 21, 26, 22 and 23 respectively)

    ›Reveal solutionSolution

    Colour in transition-metal ions arises from d–d electronic transitions, which require partly-filled d-orbitals; ions with a $d^0$ configuration have no such transition available and are colourless.

    Work out each ion's d-electron count:

    • $Sc^{3+}$ (Z = 21): Sc is $[Ar]3d^1 4s^2$; removing 3 electrons ($4s^2$, $3d^1$) gives $Sc^{3+} = [Ar]3d^0$ — no d electrons.
    • $Ti^{4+}$ (Z = 22): Ti is $[Ar]3d^2 4s^2$; removing all 4 gives $Ti^{4+} = [Ar]3d^0$ — no d electrons.
    • $Fe^{3+}$ (Z = 26): Fe is $[Ar]3d^6 4s^2$; removing 3 electrons ($4s^2$, one $3d$) gives $Fe^{3+} = [Ar]3d^5$ — 5 unpaired d electrons.
    • $V^{3+}$ (Z = 23): V is $[Ar]3d^3 4s^2$; removing 3 electrons ($4s^2$, one $3d$) gives $V^{3+} = [Ar]3d^2$ — 2 unpaired d electrons. …

    🔒 Start your 14-day free trial to see the full solution →