Q.(i) Solution of $[Ni(H_2O)_6]^{2+}$ is green, but solution of $[Ni(CN)_4]^{2-}$ is colourless. Explain.

Showing the 12 most recent of 21 on this concept.

1. CBSE 2026Set ANNUAL1 mark
   Q.Write any one example of low spin complex.

   ›Reveal solutionSolution

   A low-spin complex forms when a strong-field ligand causes the d electrons to pair up in the lower-energy t2g set rather than spreading into eg, reducing the number of unpaired electrons.

   …

   🔒 Start your 14-day free trial to see the full solution →
2. CBSE 2026Set ANNUAL1 markMCQ
   Q.Assertion [A]: [Ni(CN)4]2- is a square-planar and diamagnetic. Reason [R]: It has no unpaired electrons due to presence of strong field.

   (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.

   ›Reveal solutionSolution

   $[Ni(CN)_4]^{2-}$ is indeed square planar and diamagnetic, and this is correctly explained by CN⁻ being a strong field ligand that forces electron pairing, leaving no unpaired electrons.

   In $[Ni(CN)_4]^{2-}$, nickel is in the $+2$ oxidation state: $Ni^{2+}$ has configuration $3d^8$ (8 electrons: $t_{2g}^6 e_g^2$ in a free-ion sense, or $3d^8 = \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\ \uparrow$).

   …

   🔒 Start your 14-day free trial to see the full solution →
3. CBSE 2026Set ANNUAL1 mark
   Q.Which one is an inner-orbital complex? $[Co(NH_3)_6]^{3+}$ or $[CoF_6]^{3-}$

   ›Reveal solutionSolution

   Because $NH_3$ is a strong-field ligand, $Co^{3+}$'s $d$-electrons pair up and the complex uses the inner $(n-1)d$ orbitals for hybridisation — making $[Co(NH_3)_6]^{3+}$ the inner-orbital complex, unlike $[CoF_6]^{3-}$.

   Analysis

   $Co^{3+}$ has the configuration $3d^6$ in both complexes; the difference lies in the field strength of the ligand.

   • In $[Co(NH_3)_6]^{3+}$: $NH_3$ is a strong-field ligand. It forces all 6 $d$-electrons to pair up within three $3d$ orbitals ($t_{2g}^6$), freeing the other two $3d$ orbitals for hybridisation. Cobalt then hybridises as $d^2sp^3$, using inner $(n-1)d$, i.e. $3d$, orbitals — this is an inner-orbital (low-spin) complex, diamagnetic. …

   🔒 Start your 14-day free trial to see the full solution →
4. CBSE 2026Set ANNUAL1 mark
   Q.The oxidation number of all the alkali metals in their compounds is ________.

   ›Reveal solutionSolution

   [!TLDR]

   +1

   Method

   Alkali metals (Group 1) have one valence electron and invariably show a +1 oxida …

   🔒 Start your 14-day free trial to see the full solution →
5. CBSE 2025Set 56/5/11 markMCQ
   Q.In which of the following groups are both ions coloured in aqueous solution ? I. $Cu^{+}$ II. $Ti^{4+}$ III. $Co^{2+}$ IV. $Fe^{2+}$ [Atomic number : Cu = 29, Ti = 22, Co = 27, Fe = 26] (A) I and II (B) II and III (C) III and IV (D) I and IV

   ›Reveal solutionSolution

   The colour of a transition metal ion in aqueous solution depends on the presence of unpaired d-electrons, which allow d-d transitions. Both $Co^{2+}$ and $Fe^{2+}$ have unpaired d-electrons and are coloured, while $Cu^{+}$ and $Ti^{4+}$ have fully filled or empty d-subshells and are colourless. The correct pair is III and IV, i.e., option (C).

   The question asks which two ions among the given four are coloured in aqueous solution. Colour in transition metal ions arises from the absorption of visible light due to electronic transitions between split d-orbitals — the famous d-d transition. But this only happens if the d-subshell is partially filled (i.e., has at least one unpaired electron and at least one vacant orbital). If the d-subshell is completely empty ($d^0$) or completely filled ($d^{10}$), no d-d transition is possible, and the ion is colourless (or white) in solution.

   Let’s examine each ion one by one.

   1. $Cu^{+}$ (Copper(I))

      Atomic number of Cu = 29. Neutral Cu has configuration $[Ar]\,3d^{10}4s^1$.

      $Cu^{+}$ loses the 4s electron, so its configuration becomes $[Ar]\,3d^{10}$.

      The d-subshell is completely filled. No d-d transitions possible.

      Result: Colourless in aqueous solution.

   2. $Ti^{4+}$ (Titanium(IV))

      Atomic number of Ti = 22. Neutral Ti has $[Ar]\,3d^{2}4s^2$.

      $Ti^{4+}$ loses all four valence electrons (two from 4s and two from 3d), so its configuration becomes $[Ar]\,3d^{0}$.

      The d-subshell is completely empty. No d-d transitions possible.

      Result: Colourless in aqueous solution.

   3. $Co^{2+}$ (Cobalt(II))

      Atomic number of Co = 27. Neutral Co has $[Ar]\,3d^{7}4s^2$.

      $Co^{2+}$ loses the two 4s electrons, giving $[Ar]\,3d^{7}$.

      The d-subshell is partially filled (7 electrons in 5 orbitals — there are unpaired electrons). In aqueous solution, $Co^{2+}$ forms the pink $[Co(H_2O)_6]^{2+}$ complex.

      Result: Coloured (pink) in aqueous solution. …

   🔒 Start your 14-day free trial to see the full solution →
6. CBSE 2025Set D1 markMCQ
   Q.The structure of complex ion [Ni(CN)4]2- is

   (a) Linear

   (b) Tetrahedral

   (c) Square planar

   (d) Octahedral

   ›Reveal solutionSolution

   Ni2+ (d8) with strong-field CN- gives dsp2 hybridisation -> square planar [Ni(CN)4]2-.

   Step 1 - oxidation state: In [Ni(CN)4]2-, four CN- (each -1) give -4; overall charge -2, so Ni is +2.

   Step 2 - configuration: Ni2+ is 3d8.

   Step 3 - ligand strength: CN- is a strong-field ligand. It pairs up the d electrons, freeing one 3d orbital. …

   🔒 Start your 14-day free trial to see the full solution →
7. CBSE 2025Set ANNUAL1 mark
   Q.CO is stronger ligand than Cl⁻¹. (True / False)

   ›Reveal solutionSolution

   True — CO lies far above Cl⁻ in the spectrochemical series, so it is a much stronger field ligand.

   The spectrochemical series arranges ligands in order of increasing crystal-field splitting ($\Delta_o$) they cause:

   $I^- < Br^- < S^{2-} < SCN^- < Cl^- < ... < NH_3 < en < CN^- < CO$

   …

   🔒 Start your 14-day free trial to see the full solution →
8. CBSE 2025Set ANNUAL1 mark
   Q.Draw a figure to show the splitting of d-orbitals in an octahedral crystal field.

   ›Reveal solutionSolution

   

   Ligands approaching along the axes in an octahedral complex raise the energy of orbitals pointing along the axes more than those pointing between the axes, splitting the 5 degenerate d-orbitals into two sets separated by $\Delta_o$.

   Description of the splitting (energy-level diagram in words)

   In a free (gaseous) metal ion, all five $d$-orbitals ($d_{xy}, d_{yz}, d_{zx}, d_{x^2-y^2}, d_{z^2}$) are degenerate (equal energy). When 6 ligands approach the metal ion symmetrically along the $\pm x, \pm y, \pm z$ axes to form an octahedral complex, the orbitals lying along the axes experience more electrostatic repulsion from the approaching ligand electron pairs than the orbitals lying between the axes. This splits the 5 orbitals into two sets:

   • $e_g$ set (higher energy): $d_{x^2-y^2}$ and $d_{z^2}$ — these point directly at the ligands along the axes, so they are raised in energy above the mean (barycentre) by $+0.6\Delta_o$ (i.e. $+\tfrac{3}{5}\Delta_o$).
   • $t_{2g}$ set (lower energy): $d_{xy}, d_{yz}, d_{zx}$ — these point between the axes (away from the ligand directions), so they are lowered below the barycentre by $-0.4\Delta_o$ (i.e. $-\tfrac{2}{5}\Delta_o$).

   Schematically (energy increasing upward):

               ____  ____        <- e_g (d(x2-y2), d(z2))   +0.6(Delta_o) …
   

   🔒 Start your 14-day free trial to see the full solution →
9. CBSE 2024Set 56/3/11 markMCQ
   Q.Which of the following is diamagnetic in nature ? (A) $Co^{3+}$, octahedral complex with strong field ligand (B) $Co^{3+}$, octahedral complex with weak field ligand (C) $Co^{3+}$, in a square planar complex (D) $Co^{3+}$, in a tetrahedral complex [ Atomic number : Co = 27 ]

   ›Reveal solutionSolution

   The key is to determine the number of unpaired electrons in $Co^{3+}$ ($3d^6$) under each geometry and ligand field. Only the octahedral strong-field (low-spin) case gives zero unpaired electrons, making it diamagnetic. The correct option is (A).

   Let’s start with the core idea. A substance is diamagnetic when all its electrons are paired — no unpaired electrons means no net magnetic moment. For transition metal complexes, this depends entirely on how the $d$-orbitals split in energy under the influence of the surrounding ligands (Crystal Field Splitting) and how electrons fill those orbitals.

   Cobalt has atomic number 27. Its ground state electron configuration is $[Ar] 3d^7 4s^2$. When it forms $Co^{3+}$, it loses three electrons — typically the two $4s$ electrons and one $3d$ electron. So $Co^{3+}$ has a $3d^6$ configuration.

   Now, six $d$-electrons can arrange themselves in different ways depending on the geometry of the complex and the strength of the ligand field. The geometry determines the splitting pattern of the $d$-orbitals, and the ligand field strength decides whether electrons pair up in lower orbitals or spread out (Hund’s rule) into higher ones.

   Let’s examine each option one by one.

   1. Option (A): Octahedral complex with strong field ligand

      In an octahedral field, the five $d$-orbitals split into two sets: the lower-energy $t_{2g}$ (three orbitals) and the higher-energy $e_g$ (two orbitals). The energy gap $\Delta_o$ is large when the ligand is strong (like CN⁻, CO).

      For $3d^6$, a strong field forces electrons to pair up in the $t_{2g}$ set before any electron goes to $e_g$. So the filling is: $t_{2g}^6$ — all six electrons paired in three orbitals. That gives zero unpaired electrons.

      Tip

      Strong field = low spin = maximum pairing. For $d^6$, low-spin octahedral is always diamagnetic.

   2. Option (B): Octahedral complex with weak field ligand

      Here $\Delta_o$ is small. Electrons follow Hund’s rule: they occupy all five orbitals singly before pairing. For $d^6$, the first five electrons go one each into $t_{2g}$ and $e_g$ (actually $t_{2g}^3 e_g^2$), and the sixth electron must pair in a $t_{2g}$ orbital. So the configuration is $t_{2g}^4 e_g^2$ — that’s four electrons in $t_{2g}$ (one pair, two unpaired) and two unpaired in $e_g$. Total unpaired electrons = 4. Hence paramagnetic.

   3. Option (C): Square planar complex …

   🔒 Start your 14-day free trial to see the full solution →
10. CBSE 2024Set ANNUAL1 mark
    Q.What is crystal field splitting energy?

    ›Reveal solutionSolution

    When ligands approach a metal ion, electrostatic repulsion splits the previously degenerate d-orbitals into two energy sets; the gap between them is the crystal field splitting energy, $\Delta$.

    In an isolated (gas-phase) transition-metal ion, all five d-orbitals are degenerate (equal energy). When ligands approach to form a complex, their electron pairs create an electric field that repels electrons in the d-orbitals unequally, depending on each orbital's spatial orientation relative to the ligand positions.

    In an octahedral field, the d-orbitals split into two sets:

    • $t_{2g}$ ($d_{xy}, d_{yz}, d_{xz}$) — lower energy, point between the ligand axes
    • $e_g$ ($d_{x^2-y^2}, d_{z^2}$) — higher energy, point directly at the ligands …

    🔒 Start your 14-day free trial to see the full solution →
11. CBSE 2024Set ANNUAL1 markMCQ
    Q.A coordination compound is colourless due to –

    (a) the absence of ligand

    (b) loss of water molecules

    (c) d-d transition of the electron

    (d) energy of crystal field splitting energy

    ›Reveal solutionSolution

    A coordination compound is colourless when it cannot undergo d-d electronic transitions — either because it has no d electrons or a completely filled d-subshell.

    Colour in most coordination compounds arises from d–d transitions, where an electron is excited from a lower-energy $d$-orbital ($t_{2g}$) to a higher-energy one ($e_g$) after crystal field splitting, absorbing a specific wavelength of visible light (and transmitting/reflecting the complementary colour).

    …

    🔒 Start your 14-day free trial to see the full solution →
12. CBSE 2023Set 56/1/11 markMCQ
    Q.Assertion (A) : Low spin tetrahedral complexes are rarely observed. Reason (R) : Crystal field splitting energy is less than pairing energy for tetrahedral complexes. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.

    ›Reveal solutionSolution

    The assertion is true — low-spin tetrahedral complexes are rare — and the reason is also true: for tetrahedral complexes, the crystal field splitting energy $\Delta_t$ is much smaller than the pairing energy $P$, making low-spin configurations energetically unfavourable. The reason correctly explains the assertion, so option (A) is correct.

    Why this question hinges on crystal field splitting

    In coordination chemistry, the spin state of a complex (high-spin vs low-spin) depends on a tug-of-war between two energies: the crystal field splitting energy ($\Delta$) and the pairing energy ($P$). If $\Delta > P$, electrons prefer to pair up in the lower-energy orbitals (low-spin). If $\Delta < P$, electrons spread out to avoid pairing (high-spin).

    For tetrahedral complexes, the splitting pattern is the inverse of octahedral — the $d_{xy}, d_{yz}, d_{zx}$ orbitals (called $t_2$) are higher in energy, and the $d_{x^2-y^2}, d_{z^2}$ orbitals (called $e$) are lower. But the key number is the magnitude of $\Delta_t$ (tetrahedral splitting).

    $\Delta_t \approx \frac{4}{9} \Delta_o$

    For the same metal ion and ligands, tetrahedral splitting is only about 44% of octahedral splitting.

    Since $\Delta_o$ itself is often comparable to or smaller than $P$ for many metal ions (especially first-row transition metals), $\Delta_t$ ends up being much smaller than $P$ in almost all cases. That means the energy cost of pairing electrons is never recovered by the splitting — so electrons always occupy orbitals singly before pairing, giving high-spin configurations.

    Watch out

    A common mistake is to think that low-spin tetrahedral complexes are impossible. They are not — they are just rare. With very strong-field ligands (like CN⁻) and heavy metals (where $\Delta$ is larger), a few examples exist. But for typical exam contexts (first-row transition metals, common ligands), the statement holds.

    Step-by-step reasoning

    1. Understand the assertion: "Low spin tetrahedral complexes are rarely observed." This is a factual statement about coordination chemistry. For a tetrahedral complex to be low-spin, the splitting $\Delta_t$ must exceed the pairing energy $P$. But because $\Delta_t$ is inherently small (about $4/9$ of $\Delta_o$), this condition is seldom met. …

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