Q.Why are different colours observed in octahedral and tetrahedral complexes for the same metal and same ligands?

Showing the 12 most recent of 17 on this concept.

1. KCET 2026Set D31 markMCQ
   Q.Match List-I with List-II

   List-I (Complex)	List-II (Geometry)
   a. $[Co(NH_3)_6]^{3+}$	i. Trigonal bipyramidal
   b. $[NiCl_4]^{2-}$	ii. Octahedral
   c. $[Ni(CN)_4]^{2-}$	iii. Tetrahedral
   d. $[Fe(CO)_5]$	iv. Square planar

   Choose the correct answer from the options given below. (A) a - ii, b – iii, c – iv, d - i (B) a - ii, b - i, c - iii, d – iv (C) a – iii, b – ii, c – iv, d - i (D) a – i, b – iii, c – iv, d – ii

   ›Reveal solutionSolution

   Each complex's geometry is fixed by its coordination number together with the metal's oxidation state/d-electron count and the field strength of its ligands.

   Step 1 — $[Co(NH_3)_6]^{3+}$

   Cobalt here is $Co^{3+}$ ($d^6$), six-coordinate with $NH_3$, a moderately strong-field ligand. Six-coordinate complexes of this type adopt octahedral geometry, matching item ii.

   Step 2 — $[NiCl_4]^{2-}$

   Nickel here is $Ni^{2+}$ ($d^8$), four-coordinate with $Cl^-$, a weak-field ligand. A weak field is unable to pair up the $d^8$ electrons into a low-spin arrangement, so the complex uses $sp^3$ hybridization and adopts tetrahedral geometry, matching item iii.

   Step 3 — $[Ni(CN)_4]^{2-}$ …

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2. KCET 2025Set D-41 markMCQ
   Q.In the following pairs, the one in which both transition metal ions are colourless is (A) $\ce{Sc^{3+}, Zn^{2+}}$ (B) $\ce{V^{2+}, Ti^{3+}}$ (C) $\ce{Zn^{2+}, Mn^{2+}}$ (D) $\ce{Ti^{4+}, Cu^{2+}}$

   ›Reveal solutionSolution

   Colour in transition-metal ions comes from d–d transitions, which require a partially filled d-subshell; so find the pair where both ions are $d^0$ or $d^{10}$.

   Step 1 — The concept: why transition-metal ions are coloured

   In a complex (or in aqueous solution, where water acts as the ligand), the five degenerate d-orbitals are split by the ligand field into a lower set and an upper set, separated by the crystal-field splitting energy $\Delta$.

   $\Delta E = \Delta = h\nu = \frac{hc}{\lambda}$

   For most first-row transition-metal complexes $\Delta$ happens to correspond to a photon in the visible range. An electron in the lower set absorbs that photon and jumps to the upper set — a d–d transition — and the complementary colour of the absorbed light is what we see.

   The essential requirement: the d-subshell must be partially filled, i.e. $d^1$ to $d^9$.

   • If the ion is $\mathbf{d^0}$ — there is no electron to promote.
   • If the ion is $\mathbf{d^{10}}$ — the upper set is completely full, so there is no vacancy to promote into.

   Either way, no d–d transition ⇒ colourless.

   Step 2 — Work out the d-configuration of every ion offered

   Recall that for a transition metal we remove the $ns$ electrons first, then the $(n-1)d$ electrons.

   Ion	Atomic no.	Neutral atom	d-config of ion	Coloured?
   $\ce{Sc^{3+}}$	21	$[\text{Ar}]3d^1 4s^2$	$3d^{\mathbf{0}}$	Colourless
   $\ce{Zn^{2+}}$	30	$[\text{Ar}]3d^{10} 4s^2$	$3d^{\mathbf{10}}$	Colourless
   $\ce{V^{2+}}$	23	$[\text{Ar}]3d^3 4s^2$	$3d^{3}$	Coloured (violet)
   $\ce{Ti^{3+}}$	22	$[\text{Ar}]3d^2 4s^2$	$3d^{1}$	Coloured (purple)
   $\ce{Mn^{2+}}$	25	$[\text{Ar}]3d^5 4s^2$	$3d^{5}$	Coloured (pale pink)
   $\ce{Ti^{4+}}$	22	$[\text{Ar}]3d^2 4s^2$	$3d^{0}$	Colourless

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3. COMEDK 2025Set 2025-A1 markMCQ
   Q.A transition metal M forms 4 homoleptic octahedral coordination compounds, $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and D of the type $\left[\mathrm{MX}_6\right]^{z-}$ with monodentate ligands a, b, c and d respectively. These compounds absorb red. blue. yellow and blue-green light respectively. Which one of the options shows the correct order of decreasing ligand strength? (A) $\mathrm{B}>\mathrm{D}>\mathrm{C}>\mathrm{A}$ (B) $\mathrm{D}>\mathrm{C}>\mathrm{B}>\mathrm{A}$ (C) $\mathrm{A}>\mathrm{C}>\mathrm{D}>\mathrm{B}$ (D) $\mathrm{A}>\mathrm{B}>\mathrm{C}>\mathrm{D}$

   ›Reveal solutionSolution

   The colour absorbed by a complex is complementary to the colour we see; the energy of absorbed light (and thus the crystal field splitting Δₒ) increases from red to blue, so the ligand that causes the largest Δₒ is the strongest. The correct order of decreasing ligand strength is B > D > C > A, which corresponds to option (A).

   ---

   The key idea is the spectrochemical series: ligands are ranked by how strongly they split the d‑orbitals in an octahedral field. The stronger the ligand, the larger the crystal field splitting energy Δₒ. The colour we see is the complement of the colour absorbed — so the absorbed colour tells us the energy of the transition, and hence the relative Δₒ.

   Here, the complexes absorb:

   • A: red light
   • B: blue light
   • C: yellow light
   • D: blue‑green light

   We need to rank the ligands a, b, c, d from strongest to weakest.

   ---

   1. Recall the relationship between absorbed colour and energy. In the visible spectrum, red light has the longest wavelength (lowest energy), and blue/violet light has the shortest wavelength (highest energy). The order of increasing energy for the absorbed colours is:

   $\text{red} < \text{yellow} < \text{blue‑green} < \text{blue}$

   (Blue‑green is intermediate between green and blue, so it is higher in energy than yellow but lower than pure blue.)

   2. Connect absorbed energy to Δₒ.

      For an octahedral d‑complex, the energy of the d‑d transition (typically from $t_{2g}$ to $e_g$) is approximately equal to Δₒ. So a complex that absorbs higher‑energy light has a larger Δₒ, meaning its ligand is stronger in the spectrochemical series.

   3. Rank the complexes by Δₒ from largest to smallest.

      • B absorbs blue → highest energy → largest Δₒ → strongest ligand (b).
      • D absorbs blue‑green → next highest energy → next largest Δₒ.
      • C absorbs yellow → lower energy than blue‑green → smaller Δₒ. …

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4. COMEDK 2025Set 2025-E1 markMCQ
   Q.Which of the following compounds has electrons symmetrically distributed in both $\mathrm{t}_{2 \mathrm{~g}}$ and $\mathrm{e}_{\mathrm{g}}$ orbitals? (A) $\left[\mathrm{CoF}_6\right]^{3-}$ (B) $\left[M n(C N)_6\right]^{4-}$ (C) $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ (D) $\left[\mathrm{FeCl}_6\right]^{3-}$

   ›Reveal solutionSolution

   "Symmetric distribution in both $t_{2g}$ and $e_g$" requires each set to be half-filled (or full). Among the options only high-spin $d^5$, i.e. $[\mathrm{FeCl_6}]^{3-}$ with $t_{2g}^3 e_g^2$, satisfies this. The correct option is (D).

   Concept

   In an octahedral field the five $d$ orbitals split into the lower $t_{2g}$ ($d_{xy},d_{xz},d_{yz}$) and the higher $e_g$ ($d_{z^2},d_{x^2-y^2}$). Electrons are symmetrically distributed in a set when every orbital of that set holds the same number of electrons — one each (half-filled) or two each (full). For both sets to be symmetric with electrons present, the classic case is high-spin $d^5$: $t_{2g}^3 e_g^2$, one electron in each of the five orbitals.

   Solution

   1. Oxidation states and $d^n$.
      • (A) $[\mathrm{CoF_6}]^{3-}$: $\mathrm{Co^{3+}}=d^6$.
      • (B) $[\mathrm{Mn(CN)_6}]^{4-}$: $\mathrm{Mn^{2+}}=d^5$.
      • (C) $[\mathrm{Cr(NH_3)_6}]^{3+}$: $\mathrm{Cr^{3+}}=d^3$.
      • (D) $[\mathrm{FeCl_6}]^{3-}$: $\mathrm{Fe^{3+}}=d^5$.
   2. Field strength and filling.
      • $\mathrm{F^-}$ weak $\Rightarrow$ $\mathrm{Co^{3+}}$ high-spin $t_{2g}^4 e_g^2$ — $t_{2g}$ uneven. …

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5. KCET 2024Set B-21 markMCQ
   Q.Which of the following statements are true about $\left[\mathrm{CoF_6}\right]^{3-}$ ion? I. The complex has octahedral geometry. II. Coordination number of Co is 3 and oxidation state is +6. III. The complex is sp$^3$d$^2$ hybridised. IV. It is a high spin complex. (A) I, II and IV (B) I, III and IV (C) II and IV (D) II, III and IV

   ›Reveal solutionSolution

   $\mathrm{F^-}$ is a weak-field ligand, so $[\mathrm{CoF_6}]^{3-}$ is an outer-orbital, high-spin, octahedral $sp^3d^2$ complex — statements I, III and IV are true; only statement II (about CN and oxidation state) is false.

   1. Oxidation state and coordination number

   Let the oxidation state of Co be $x$. Each fluoride ligand is $\mathrm{F^-}$, and the overall charge is $-3$:

   $x + 6(-1) = -3 \quad\Longrightarrow\quad x = +3$

   There are six ligands directly bonded to the metal, so the coordination number is 6.

   $\text{Co: oxidation state } = \mathbf{+3}, \qquad \text{coordination number } = \mathbf{6}$

   Statement II says "coordination number of Co is 3 and oxidation state is +6" — this has the two numbers exactly swapped. II is FALSE. ✗

   This single deduction is enough to eliminate options (A) [I, II, IV], (C) [II and IV] and (D) [II, III, IV] — all of them include statement II. Only (B) survives. Let us confirm each of I, III, IV independently.

   2. Statement I — geometry

   A coordination number of 6 in Werner-type complexes means an octahedral arrangement (the six ligands at the vertices of a regular octahedron, minimising repulsion). I is TRUE. ✓

   3. The electronic configuration of $\mathrm{Co^{3+}}$

   $\mathrm{Co}\ (Z=27): [\mathrm{Ar}]\,3d^7 4s^2 \quad\Longrightarrow\quad \mathrm{Co^{3+}}: [\mathrm{Ar}]\,3d^6$

   4. Statement IV — high spin or low spin?

   The deciding factor is the crystal-field splitting energy $\Delta_o$ versus the pairing energy $P$:

   • If $\Delta_o > P$ (strong-field ligand, e.g. $\mathrm{CN^-}$, $\mathrm{NH_3}$, CO) ⇒ electrons pair up in $t_{2g}$ ⇒ low spin.
   • If $\Delta_o < P$ (weak-field ligand) ⇒ electrons spread out and stay unpaired ⇒ high spin.

   In the spectrochemical series,

   $\mathrm{I^-} < \mathrm{Br^-} < \mathrm{Cl^-} < \mathbf{F^-} < \mathrm{H_2O} < \mathrm{NH_3} < \mathrm{en} < \mathrm{CN^-} < \mathrm{CO}$

   $\mathrm{F^-}$ sits firmly on the weak-field end. Hence $\Delta_o < P$ and the $d^6$ electrons occupy the orbitals with maximum multiplicity:

   $t_{2g}^{4}\ e_g^{2} \qquad \Rightarrow \qquad \textbf{4 unpaired electrons — HIGH SPIN (paramagnetic)}$

   $\mu = \sqrt{n(n+2)} = \sqrt{4\times6} = \sqrt{24} \approx 4.9\ \mathrm{BM}$ …

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6. COMEDK 2024Set 2024-A1 markMCQ
   Q.On the basis of crystal field theory, electronic configuration of a low spin $\mathrm{d}^4$ complex is: (A) $t_{2 g}^1 e_g^3$ (B) $t_{2 g}^4 e_g$ (C) $t_{2 g}^3 e_g^1$ (D) $t_{2 g}^2 e_g^2$

   ›Reveal solutionSolution

   A low-spin $d^4$ octahedral complex places all four electrons in $t_{2g}$: $t_{2g}^4 e_g^0$.

   In crystal field theory for an octahedral complex, the $d$ orbitals split into lower $t_{2g}$ (three orbitals) and higher $e_g$ (two orbitals). In a low-spin (strong-field) case the pairing energy is less than $\Delta_o$, so electrons pair up in $t_{2g}$ before occupying $e_g$. For $d^4$ …

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7. COMEDK 2024Set 2024-E1 markMCQ
   Q.Based on Crystal Field theory, match the Complex ions listed in Column I with the electronic configuration in the d orbitals of the central metal ion listed in Column II. .tg {border-collapse:collapse;border-spacing:0;} .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-c3ow{border-color:inherit;text-align:center;vertical-align:top} .tg .tg-7btt{border-color:inherit;font-weight:bold;text-align:center;vertical-align:top} .tg .tg-0pky{border-color:inherit;text-align:left;vertical-align:top} No. Complexion No. d orbital configuration of central metal ion. (A) $\left[M n(C N)_6\right]^{4-}$ (P) $e_g^2 t_{2 g}^3$ (B) $\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ (Q) $t_{2 g}^4 e_g^2$ (C) $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ (R) $t_{2 g}^5$ (D) $\left[\mathrm{MnCl}_4\right]^{2-}$ (S) $t_{2 g}^5 e_g^2$ (A) $\mathrm{A}=\mathrm{S} \quad \mathrm{B}=\mathrm{R} \quad \mathrm{C}=\mathrm{P} \quad \mathrm{D}=\mathrm{Q}$ (B) $\mathrm{A}=\mathrm{R} \quad \mathrm{B}=\mathrm{S} \quad \mathrm{C}=\mathrm{Q} \quad \mathrm{D}=\mathrm{P}$ (C) $\mathrm{A}=\mathrm{Q} \quad \mathrm{B}=\mathrm{S} \quad \mathrm{C}=\mathrm{P} \quad \mathrm{D}=\mathrm{R}$ (D) $\mathrm{A}=\mathrm{R} \quad \mathrm{B}=\mathrm{S} \quad \mathrm{C}=\mathrm{P} \quad \mathrm{D}=\mathrm{Q}$

   ›Reveal solutionSolution

   A=R, B=S, C=Q, D=P.

   Work out each central-ion d-configuration:

   • A $[Mn(CN)_6]^{4-}$: Mn$^{2+}$ is d$^5$; $CN^-$ is strong-field $\Rightarrow$ low spin octahedral $\Rightarrow t_{2g}^5e_g^0=t_{2g}^5$ = R.
   • B $[Co(H_2O)_6]^{2+}$: Co$^{2+}$ is d$^7$; $H_2O$ weak-field $\Rightarrow$ high spin $\Rightarrow t_{2g}^5e_g^2$ = S.
   • C $[Fe(H_2O)_6]^{2+}$: Fe$^{2+}$ is d$^6$; high spin $\Rightarrow t_{2g}^4e_g^2$ = Q. …

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8. KCET 2023Set D-21 markMCQ
   Q.Match the column A (type of crystalline solid) with the column B (example for each type): A P. Molecular Solid Q. Ionic Solid R. Metallic Solid S. Network Solid B i. SiC ii. Mg iii. H$_2$O iv. MgO (A) P-iii, Q-i, R-ii, S-iv (B) P-iv, Q-iii, R-ii, S-i (C) P-ii, Q-iv, R-iii, S-i (D) P-iii, Q-iv, R-ii, S-i

   ›Reveal solutionSolution

   The question asks you to match each type of crystalline solid (molecular, ionic, metallic, network) with its correct example. The key is to identify the bonding and structure of each substance: $\ce{H2O}$ is a molecular solid, $\ce{MgO}$ is ionic, $\ce{Mg}$ is metallic, and $\ce{SiC}$ is a network covalent solid. The correct match is P-iii, Q-iv, R-ii, S-i, which corresponds to option (D).

   The concept here is classification of crystalline solids based on the nature of the bonding forces between their constituent particles. Each type has a distinct set of properties that you can use to identify examples.

   1. Molecular solids are held together by weak intermolecular forces (van der Waals, hydrogen bonding). They consist of discrete molecules. Water ($\ce{H2O}$) is a classic example — it forms ice crystals where individual $\ce{H2O}$ molecules are linked by hydrogen bonds. So P matches with iii.

   2. Ionic solids are composed of positive and negative ions held together by strong electrostatic (ionic) bonds. Magnesium oxide ($\ce{MgO}$) is an ionic compound: $\ce{Mg^{2+}}$ and $\ce{O^{2-}}$ ions arranged in a lattice. So Q matches with iv.

   3. Metallic solids consist of metal atoms held together by metallic bonding — a "sea" of delocalized electrons around positive ions. Magnesium ($\ce{Mg}$) is a metal, so R matches with ii.

   4. Network solids (also called covalent network solids) are giant molecules where atoms are bonded together by a continuous network of covalent bonds. Silicon carbide ($\ce{SiC}$) has a structure similar to diamond, with each Si atom covalently bonded to four C atoms. So S matches with i. …

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9. KCET 2023Set D-21 markMCQ
   Q.Which of the following system in an octahedral complex has maximum unpaired electrons? (A) $d^9$ (high spin) (B) $d^6$ (low spin) (C) $d^4$ (low spin) (D) $d^7$ (high spin)

   ›Reveal solutionSolution

   Fill the $t_{2g}$/$e_g$ levels for each configuration using the correct spin state and simply count the unpaired electrons.

   Step 1 — The concept: high spin vs low spin

   In an octahedral complex the five d orbitals split into a lower $t_{2g}$ set (3 orbitals) and an upper $e_g$ set (2 orbitals), separated by the crystal-field splitting energy $\Delta_o$.

   • High spin ($\Delta_o < P$, the pairing energy): electrons obey Hund's rule as far as possible — they occupy $e_g$ singly before pairing in $t_{2g}$. Maximum unpaired electrons.
   • Low spin ($\Delta_o > P$): electrons pair up in $t_{2g}$ before entering $e_g$. Minimum unpaired electrons.

   Step 2 — Fill each configuration

   (A) $d^9$ (high spin): $t_{2g}^{6}\,e_g^{3}$. Six paired in $t_{2g}$; in $e_g$ one orbital is doubled and one is singly filled.

   $\text{unpaired} = 1$

   (Note that $d^9$ has only one unpaired electron in either spin state — hence the Jahn–Teller distortion of $\mathrm{Cu^{2+}}$.)

   (B) $d^6$ (low spin): all six electrons pair into the three $t_{2g}$ orbitals: $t_{2g}^{6}\,e_g^{0}$.

   $\text{unpaired} = 0 \quad (\text{diamagnetic, e.g. } [\mathrm{Co(NH_3)_6}]^{3+})$ …

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10. KCET 2023Set D-21 markMCQ
    Q.If a didentate ligand ethane-1,2-diamine is progressively added in the molar ratio en : Ni :: 1 : 1, 2 : 1, 3 : 1 to [Ni(H$_2$O)$_6$]$^{2+}$ aq solution, following co-ordination entities are formed. I. [Ni(H$_2$O)$_4$en]$^{2+}_{(aq)}$ – pale blue II. [Ni(H$_2$O)$_2$(en)$_2$]$^{2+}_{(aq)}$ – blue/purple III. [Ni(en)$_3$]$^{2+}_{(aq)}$ – violet The wavelength in nm of light absorbed in case of I and III are respectively (A) 475 nm and 310 nm (B) 300 nm and 475 nm (C) 310 nm and 500 nm (D) 600 nm and 535 nm

    ›Reveal solutionSolution

    The colour of a coordination complex is the complement of the colour it absorbs. As the ligand field strength increases (en replaces H₂O), the crystal field splitting Δ₀ increases, so the absorbed light shifts to shorter wavelengths (blue-shift). For [Ni(H₂O)₆]²⁺ (green), the absorbed wavelength is around 600–650 nm; replacing water with the stronger-field en shifts absorption to shorter wavelengths. The pale blue complex I absorbs in the orange-red (~600 nm), and the violet complex III absorbs in the yellow-green (~535 nm). The correct option is (D).

    The question is about the relationship between the colour we see and the wavelength of light absorbed. A complex appears coloured because it absorbs a specific portion of visible light; the colour we see is the complement of the absorbed colour. The key variable here is the crystal field splitting energy $\Delta_o$, which determines which wavelength gets absorbed.

    Ethane-1,2-diamine (en) is a stronger field ligand than water. As you replace H₂O with en, $\Delta_o$ increases. A larger $\Delta_o$ means the energy gap between the $t_{2g}$ and $e_g$ orbitals is bigger, so the absorbed photon must have higher energy — that is, a shorter wavelength. So the sequence from I to III should show a progressive shift of the absorption band toward shorter wavelengths.

    Now, what do the observed colours tell us?

    1. Complex I — pale blue.

       Pale blue is the complementary colour of orange/red. A pale blue complex absorbs light in the orange-red region, roughly 600–650 nm. Since en is a stronger ligand than water, the absorption for I should be at a slightly shorter wavelength than for the original [Ni(H₂O)₆]²⁺ (which is green and absorbs around 650–700 nm). So ~600 nm is reasonable.

    2. Complex III — violet.

       Violet is the complementary colour of yellow-green. A violet complex absorbs light in the yellow-green region, roughly 530–560 nm. With three en ligands, the field is strongest, so the absorption is at the shortest wavelength among the three — around 535 nm fits perfectly.

    3. Complex II — blue/purple (intermediate).

       This falls between I and III, absorbing at an intermediate wavelength (~570–580 nm), consistent with two en ligands. …

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11. KCET 2022Set B-31 markMCQ
    Q.Crystal Field Splitting Energy (CFSE) for [CoCl_4]^{2-} is 18000 cm^{-1}. The Crystal Field Splitting Energy (CFSE) for [CoCl_4]^{2-} will be (A) 8000 cm^{-1} (B) 10,000 cm^{-1} (C) 18,000 cm^{-1} (D) 16,000 cm^{-1}

    ›Reveal solutionSolution

    Apply the standard crystal-field relation $\Delta_t=\dfrac{4}{9}\Delta_o$ for the same metal ion and the same ligands.

    Step 1 — Read the question correctly.

    The two complexes are $[\mathrm{CoCl_6}]^{4-}$ (octahedral, six chloride ligands, $\Delta_o = 18000\ \mathrm{cm^{-1}}$) and $[\mathrm{CoCl_4}]^{2-}$ (tetrahedral, four chloride ligands). Both contain $\mathrm{Co^{2+}}$ with the same ligand, $\mathrm{Cl^-}$ — only the geometry changes. (The stem as reproduced repeats the formula, but the accompanying data statement makes the intent explicit: the $18000\ \mathrm{cm^{-1}}$ value belongs to the octahedral hexachloro complex.)

    Step 2 — Why tetrahedral splitting is smaller.

    In crystal field theory the $d$-orbitals split because the ligand lone pairs repel the $d$-electrons. Two things weaken that repulsion in a tetrahedral field:

    1. Fewer ligands — only 4 instead of 6, so about $\tfrac{4}{6}=\tfrac23$ of the repulsive interaction;
    2. Poorer orbital alignment — in an octahedron the ligands point straight at the $e_g$ orbitals ($d_{z^2}$, $d_{x^2-y^2}$); in a tetrahedron no ligand points directly at any $d$-orbital. The $t_2$ set is merely the less badly oriented one, giving a further factor of about $\tfrac23$.

    Multiplying the two effects:

    $\Delta_t \approx \frac{2}{3}\times\frac{2}{3}\,\Delta_o=\frac{4}{9}\,\Delta_o\approx 0.45\,\Delta_o .$ …

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12. KCET 2021Set B-21 markMCQ
    Q.Which of the following does not represent property stated against it? (A) CO$^{+2}$ < Fe$^{+2}$ < Mn$^{+2}$ – Ionic size (B) Ti < V < Mn – Number of oxidation states (C) Cr$^{+2}$ < Mn$^{+2}$ < Fe$^{+2}$ – Paramagnetic behaviour (D) Sc > Cr > Fe – Density

    ›Reveal solutionSolution

    Mn²⁺ (3d⁵, half-filled) has the maximum unpaired electrons in this part of the series — Cr²⁺ and Fe²⁺ both have 4, so a strictly increasing order across all three is wrong.

    Electron configurations (3d series, +2 ions):

    • $Cr^{2+}$: [Ar] 3d⁴ → 4 unpaired electrons
    • $Mn^{2+}$: [Ar] 3d⁵ → 5 unpaired electrons (half-filled — the classic maximum-stability, maximum-paramagnetism case)
    • $Fe^{2+}$: [Ar] 3d⁶ → 4 unpaired electrons (one pair forms once past the half-filled point) …

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