Q.Which of the following complexes formed by $Cu^{2+}$ ions is most stable?

Showing the 12 most recent of 14 on this concept.

1. COMEDK 2026Set 2026-M1 markMCQ
   Q.The ion that has a spin only magnetic moment of 5.9 BM is: (A) $\mathrm{Fe}^{3+}$ (B) $\mathrm{Mn}^{3+}$ (C) $\mathrm{Ni}^{2+}$ (D) $\mathrm{Co}^{3+}$

   ›Reveal solutionSolution

   The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons. A value of 5.9 BM corresponds to $n = 5$ unpaired electrons. Among the options, only $\mathrm{Fe}^{3+}$ has 5 unpaired electrons, so the correct option is (A).

   The key concept here is the spin-only magnetic moment formula:

   $\mu = \sqrt{n(n+2)} \ \text{BM}$

   where $n$ is the number of unpaired electrons. This formula works well for first-row transition metal ions because orbital angular momentum is often "quenched" by the crystal field. The experimental value 5.9 BM is very close to the theoretical value for $n=5$, which is $\sqrt{5 \times 7} = \sqrt{35} \approx 5.92$ BM. So we need to find which ion has exactly 5 unpaired electrons in its ground state.

   Let’s check each option step by step:

   1. Determine the electronic configuration of each ion

      • $\mathrm{Fe}^{3+}$: Atomic number of Fe = 26. Neutral Fe: $[\mathrm{Ar}]\, 3d^6 4s^2$. Removing three electrons (first from 4s, then from 3d) gives $[\mathrm{Ar}]\, 3d^5$.
      • $\mathrm{Mn}^{3+}$: Atomic number of Mn = 25. Neutral Mn: $[\mathrm{Ar}]\, 3d^5 4s^2$. Removing three electrons gives $[\mathrm{Ar}]\, 3d^4$.
      • $\mathrm{Ni}^{2+}$: Atomic number of Ni = 28. Neutral Ni: $[\mathrm{Ar}]\, 3d^8 4s^2$. Removing two electrons gives $[\mathrm{Ar}]\, 3d^8$.
      • $\mathrm{Co}^{3+}$: Atomic number of Co = 27. Neutral Co: $[\mathrm{Ar}]\, 3d^7 4s^2$. Removing three electrons gives $[\mathrm{Ar}]\, 3d^6$.

   2. Count unpaired electrons using Hund’s rule

      • For $\mathrm{Fe}^{3+}$ ($3d^5$): All five d-orbitals are half-filled, each with one electron → 5 unpaired electrons.
      • For $\mathrm{Mn}^{3+}$ ($3d^4$): Four electrons occupy four orbitals singly (Hund’s rule) → 4 unpaired electrons.
      • For $\mathrm{Ni}^{2+}$ ($3d^8$): The d-orbitals fill as: ↑↓ ↑↓ ↑ ↑ ↑ (two paired, two singly, one singly) → 2 unpaired electrons.
      • For $\mathrm{Co}^{3+}$ ($3d^6$): In the high-spin state (common for weak-field ligands), it would be ↑ ↑ ↑ ↑↓ ↑ → 4 unpaired electrons. In low-spin (strong field), it would be ↑↓ ↑↓ ↑↓ → 0 unpaired. But the magnetic moment 5.9 BM is far too large for either case.

   3. Match with the magnetic moment

      • $n=5$ gives $\mu = \sqrt{35} \approx 5.92$ BM, which matches 5.9 BM.
      • $n=4$ gives $\mu = \sqrt{24} \approx 4.90$ BM.
      • $n=2$ gives $\mu = \sqrt{8} \approx 2.83$ BM.
      • $n=0$ gives $\mu = 0$ BM. Only $\mathrm{Fe}^{3+}$ has $n=5$.

   Watch out

   A common mistake is to forget that for $\mathrm{Co}^{3+}$, the $3d^6$ configuration can be low-spin (0 unpaired) in many complexes, but even high-spin gives only 4 unpaired electrons — never 5. So it cannot match 5.9 BM.

   Tip

   Memorize the "magic numbers": $\sqrt{3} \approx 1.73$ (1 e⁻), $\sqrt{8} \approx 2.83$ (2 e⁻), $\sqrt{15} \approx 3.87$ (3 e⁻), $\sqrt{24} \approx 4.90$ (4 e⁻), $\sqrt{35} \approx 5.92$ (5 e⁻). A value near 5.9 BM is a dead giveaway for 5 unpaired electrons.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

2. KCET 2026Set D31 markMCQ
   Q.The calculated spin only magnetic moment of $Cr^{2+}$ ion is (A) 3.87 BM (B) 4.90 BM (C) 5.92 BM (D) 2.84 BM

   ›Reveal solutionSolution

   The spin-only magnetic moment depends on the number of unpaired electrons in the ion's d-configuration.

   Step 1 — Electron configuration of Cr²⁺

   Neutral Cr (Z = 24) is $[Ar]3d^5 4s^1$. Forming $Cr^{2+}$ removes the single 4s electron first, then one 3d electron, giving $Cr^{2+} = [Ar]3d^4$.

   Step 2 — Counting unpaired electrons

   By Hund's rule, the 4 electrons in the 3d subshell occupy 4 different orbitals singly before any pairing occurs, so $n = 4$ unpaired electrons.

   Step 3 — Applying the spin-only formula

   $\mu = \sqrt{n(n+2)}$ BM $= \sqrt{4 \times 6} = \sqrt{24} \approx 4.90$ BM.

   ✓Final answer

   The correct option is (B) — 4.90 BM.

3. COMEDK 2025Set 2025-A1 markMCQ
   Q.Which is the correct order of increasing number of unpaired electrons in the following ions? $\mathrm{A}=\mathrm{Cr}^{2+}(\mathrm{Z}=24) \mathrm{B}=\mathrm{Cu}^{2+}(\mathrm{Z}=29) \quad \mathrm{C}=\mathrm{Ni}^{2+}(\mathrm{Z}=28) \mathrm{D}=\mathrm{Fe}^{3+}(\mathrm{Z}=26)$ (A) $\mathrm{B}<\mathrm{C}<\mathrm{A}<\mathrm{D}$ (B) $\mathrm{D}<\mathrm{C}<\mathrm{A}<\mathrm{B}$ (C) $\mathrm{B}<\mathrm{C}<\mathrm{D}<\mathrm{A}$ (D) $\mathrm{C}<\mathrm{B}<\mathrm{D}<\mathrm{A}$

   ›Reveal solutionSolution

   The number of unpaired electrons depends on the electronic configuration of each ion, considering the d-orbital filling and Hund's rule. The correct order is B (Cu²⁺, 1 unpaired) < C (Ni²⁺, 2 unpaired) < A (Cr²⁺, 4 unpaired) < D (Fe³⁺, 5 unpaired), which corresponds to option (A).

   Concept & Intuition

   The key is to write the ground-state electron configurations for each ion, focusing on the d-electrons. Transition metal ions lose s-electrons first, then d-electrons if needed. Unpaired electrons arise from half-filled or partially filled d-orbitals following Hund's rule (each orbital gets one electron before pairing). The more unpaired electrons, the higher the magnetic moment. Here, we count them directly.

   Step-by-step reasoning

   1. Cr²⁺ (Z = 24)

      • Neutral Cr: [Ar] 4s¹ 3d⁵ (exception: half-filled d gives stability).
      • Remove two electrons: first from 4s, then from 3d → Cr²⁺: [Ar] 3d⁴.
      • Hund's rule: four d-orbitals each get one electron, one orbital empty → 4 unpaired electrons.

   2. Cu²⁺ (Z = 29)

      • Neutral Cu: [Ar] 4s¹ 3d¹⁰ (exception: fully filled d).
      • Remove two electrons: one from 4s, one from 3d → Cu²⁺: [Ar] 3d⁹.
      • One d-orbital has a single electron (the rest are paired) → 1 unpaired electron.

   3. Ni²⁺ (Z = 28)

      • Neutral Ni: [Ar] 4s² 3d⁸.
      • Remove two 4s electrons → Ni²⁺: [Ar] 3d⁸.
      • Filling: five orbitals get 2, 2, 2, 1, 1 (two unpaired) → 2 unpaired electrons.

   4. Fe³⁺ (Z = 26)

      • Neutral Fe: [Ar] 4s² 3d⁶.
      • Remove three electrons: two from 4s, one from 3d → Fe³⁺: [Ar] 3d⁵.
      • Half-filled d: each orbital has one electron → 5 unpaired electrons.

   Tip

   Remember: For Cr and Cu, the neutral atoms have anomalous configurations (4s¹ not 4s²) due to half/full d-subshell stability. This carries over to their ions.

   Ordering by increasing unpaired electrons:

   Cu²⁺ (1) < Ni²⁺ (2) < Cr²⁺ (4) < Fe³⁺ (5).

   This matches option (A).

   Watch out

   A common mistake is to forget that Cr²⁺ has 4 unpaired electrons (not 2 or 6). Also, Fe³⁺ has 5, not 3 — losing three electrons from Fe leaves a half-filled d⁵.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

4. COMEDK 2025Set 2025-A1 markMCQ
   Q.The correct order of spin only magnetic moments among the following is: [Given: Atomic numbers: $\mathrm{Mn}=25, \mathrm{Fe}=26, \mathrm{Co}=27$ ] (A) $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}>\left[\mathrm{MnCl}_4\right]^{2-}>\left[\mathrm{CoCl}_4\right]^{2-}$ (B) $\left[\mathrm{MnCl}_4\right]^{2-}>\left[\mathrm{CoCl}_4\right]^{2-}>\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$ (C) $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}>\left[\mathrm{CoCl}_4\right]^{2-}>\left[\mathrm{MnCl}_4\right]^{2-}$ (D) $\left[\mathrm{MnCl}_4\right]^{2-}>\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}>\left[\mathrm{CoCl}_4\right]^{2-}$

   ›Reveal solutionSolution

   The spin-only magnetic moment depends on the number of unpaired electrons, which is determined by the metal’s oxidation state and the ligand field (strong-field CN⁻ vs. weak-field Cl⁻). The correct order is $\left[\mathrm{MnCl}_4\right]^{2-} > \left[\mathrm{CoCl}_4\right]^{2-} > \left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$, corresponding to option (B).

   Concept & Intuition

   The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)}$ Bohr magnetons, where $n$ is the number of unpaired electrons. So the problem reduces to finding $n$ for each complex. The key twist: ligands like CN⁻ are strong-field (low-spin), causing maximum pairing, while Cl⁻ is weak-field (high-spin), favoring unpaired electrons. Also, geometry matters — here all complexes are either octahedral or tetrahedral, which affects the d-orbital splitting and thus the electron configuration.

   Step-by-step reasoning

   1. Determine oxidation states and d-electron counts

      • $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$: CN⁻ is −1 each, so Fe must be +2 (since 6×(−1) + x = −4 → x = +2). Fe (Z=26): [Ar]3d⁶. Fe²⁺: 3d⁶.
      • $\left[\mathrm{MnCl}_4\right]^{2-}$: Cl⁻ is −1 each, so Mn is +2 (4×(−1) + x = −2 → x = +2). Mn (Z=25): [Ar]3d⁵. Mn²⁺: 3d⁵.
      • $\left[\mathrm{CoCl}_4\right]^{2-}$: Cl⁻ is −1 each, so Co is +2 (4×(−1) + x = −2 → x = +2). Co (Z=27): [Ar]3d⁷. Co²⁺: 3d⁷.

   2. Analyze $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$ — octahedral, strong-field

      CN⁻ is a strong-field ligand, causing large splitting. For Fe²⁺ (d⁶) in an octahedral field, strong-field means low-spin: all electrons pair in the $t_{2g}$ set first. Configuration: $t_{2g}^6 e_g^0$. All 6 electrons are paired → 0 unpaired electrons.

      $\mu = \sqrt{0(0+2)} = 0$ BM.

   3. Analyze $\left[\mathrm{MnCl}_4\right]^{2-}$ — tetrahedral, weak-field

      Cl⁻ is a weak-field ligand. Tetrahedral splitting is smaller than octahedral, so always high-spin. Mn²⁺ is d⁵. In tetrahedral geometry, the $e$ set is lower in energy than $t_2$. High-spin d⁵: each of the five d-orbitals gets one electron (Hund’s rule) → 5 unpaired electrons.

      $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92$ BM.

   4. Analyze $\left[\mathrm{CoCl}_4\right]^{2-}$ — tetrahedral, weak-field

      Co²⁺ is d⁷. In a tetrahedral weak field, the configuration is high-spin: $e^4 t_2^3$. The $e$ set holds 4 electrons (two pairs), and the $t_2$ set holds 3 unpaired electrons (one each) → 3 unpaired electrons.

      $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87$ BM.

   5. Compare the magnetic moments

      • $\left[\mathrm{MnCl}_4\right]^{2-}$: ~5.92 BM (largest)
      • $\left[\mathrm{CoCl}_4\right]^{2-}$: ~3.87 BM (middle)
      • $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$: 0 BM (smallest) So the order is: $\left[\mathrm{MnCl}_4\right]^{2-} > \left[\mathrm{CoCl}_4\right]^{2-} > \left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$.

   Watch out

   A common mistake is to forget that CN⁻ is strong-field and forces low-spin on Fe²⁺, giving zero unpaired electrons. If you assumed high-spin for Fe(CN)₆⁴⁻, you’d get 4 unpaired electrons and the wrong order.

   Tip

   For tetrahedral complexes, always assume high-spin because the splitting is too small to overcome pairing energy — even with strong-field ligands, tetrahedral low-spin is extremely rare.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

5. COMEDK 2025Set 2025-E1 markMCQ
   Q.Arrange the complex ions in the increasing order of their magnetic moments A. $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ B. $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$ C. $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ D. $\left[\mathrm{FeF}_6\right]^{3-}$ (A) $\mathrm{D}<\mathrm{A}<\mathrm{B}<\mathrm{C}$ (B) $\mathrm{B}<\mathrm{C}<\mathrm{A}<\mathrm{D}$ (C) $\mathrm{D}<\mathrm{C}<\mathrm{B}<\mathrm{A}$ (D) $\mathrm{A}<\mathrm{D}<\mathrm{C}<\mathrm{B}$

   ›Reveal solutionSolution

   The magnetic moment depends on the number of unpaired electrons, which is determined by the metal’s oxidation state, ligand field strength (weak vs. strong field), and the resulting high-spin or low-spin configuration. The correct increasing order is B < C < A < D, corresponding to option (B).

   Concept & Intuition

   Magnetic moment (μ) for a transition metal complex is given by the spin-only formula:

   $\mu = \sqrt{n(n+2)} \ \text{BM}$

   where $n$ is the number of unpaired electrons. So the order of magnetic moments is simply the order of number of unpaired electrons.

   To find $n$, we need:

   • The oxidation state of iron in each complex.
   • Whether the ligand is weak-field (high-spin) or strong-field (low-spin).
   • The d-electron count and how electrons fill the $t_{2g}$ and $e_g$ orbitals.

   Let’s work through each complex step by step.

   ---

   1. Determine oxidation states and d-electron counts

   • A: $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ — Water is neutral, so Fe is +2. Fe²⁺ has electron configuration [Ar] 3d⁶.
   • B: $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$ — CN⁻ is -1 each, total −6 from ligands, so Fe must be +2 to give overall −4. Again, Fe²⁺, 3d⁶.
   • C: $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ — CN⁻ gives −6, so Fe is +3. Fe³⁺ is 3d⁵.
   • D: $\left[\mathrm{FeF}_6\right]^{3-}$ — F⁻ is -1 each, total −6, so Fe is +3. Also 3d⁵.

   So we have two Fe²⁺ (d⁶) and two Fe³⁺ (d⁵) complexes.

   ---

   2. Identify ligand field strength

   • H₂O is a weak-field ligand (causes small splitting).
   • F⁻ is also a weak-field ligand.
   • CN⁻ is a strong-field ligand (causes large splitting).

   For weak-field ligands, electrons fill according to Hund’s rule (high-spin). For strong-field ligands, pairing occurs before occupying higher orbitals (low-spin).

   ---

   3. Determine unpaired electrons for each

   • A: Fe²⁺ (d⁶), weak field (H₂O) → high-spin.

     In an octahedral field: $t_{2g}^4 e_g^2$ → 4 unpaired electrons.

     $n = 4$.

   • B: Fe²⁺ (d⁶), strong field (CN⁻) → low-spin.

     All 6 electrons pair in $t_{2g}$: $t_{2g}^6 e_g^0$ → 0 unpaired electrons.

     $n = 0$.

   • C: Fe³⁺ (d⁵), strong field (CN⁻) → low-spin.

     Five electrons fill $t_{2g}$ with one pair: $t_{2g}^5 e_g^0$ → 1 unpaired electron.

     $n = 1$.

   • D: Fe³⁺ (d⁵), weak field (F⁻) → high-spin.

     Hund’s rule: $t_{2g}^3 e_g^2$ → 5 unpaired electrons.

     $n = 5$.

   ---

   4. Order by increasing magnetic moment

   Since μ increases with $n$, the order from smallest to largest $n$ is:

   B (0) < C (1) < A (4) < D (5).

   Thus the increasing order of magnetic moments is:

   $\boxed{\text{B} < \text{C} < \text{A} < \text{D}}$

   ---

   Watch out

   A common mistake is to forget that CN⁻ is a strong-field ligand for Fe²⁺ and Fe³⁺, leading to low-spin configurations. Another pitfall is assuming all Fe³⁺ complexes have 5 unpaired electrons — but with strong-field ligands, pairing occurs.

   Tip

   Remember the spectrochemical series: I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < NO₂⁻ < CN⁻ ≈ CO. Ligands to the right cause larger splitting and favor low-spin.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

6. COMEDK 2025Set 2025-M1 markMCQ
   Q.Match the coordination compounds in Column I having the given type of hybridisation of $\mathrm{M}^{\mathrm{n}+}$ ion and magnetic moment as given 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-baqh{text-align:center;vertical-align:top} .tg .tg-0lax{text-align:left;vertical-align:top} Coordination compounds Hybridisation & Magnetic nature A. $\mathrm{Ni}(\mathrm{CO})_4$ P. $\mathrm{sp}^3, \quad \mu=5.92 \mathrm{BM}$ B. $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ Q. $\mathrm{sp}^3, \quad \mu=2.84 \mathrm{BM}$ C. $\left[\mathrm{Ni}(\mathrm{Cl})_4\right]^{2-}$ R. $\mathrm{sp}^3, \quad \mu=0$ D. $\left[\mathrm{MnBr}_4\right]^{2-}$ S. $\mathrm{dsp}^2, \quad \mu=0$ (A) $\mathrm{A}=\mathrm{R} \quad \mathrm{B}=\mathrm{S} \quad \mathrm{C}=\mathrm{Q} \quad \mathrm{D}=\mathrm{P}$ (B) $A=S \quad B=R \quad C=P \quad D=Q$ (C) $\mathrm{A}=\mathrm{S} \quad \mathrm{B}=\mathrm{P} \quad \mathrm{C}=\mathrm{R} \quad \mathrm{D}=\mathrm{Q}$ (D) $\mathrm{A}=\mathrm{Q} \quad \mathrm{B}=\mathrm{P} \quad \mathrm{C}=\mathrm{S} \quad \mathrm{D}=\mathrm{R}$

   ›Reveal solutionSolution

   The key is to determine the oxidation state, geometry, and number of unpaired electrons for each complex, then match the hybridisation and magnetic moment. The correct matches are A→R, B→S, C→Q, D→P, so option (A) is correct.

   We need to match each coordination compound with its hybridisation and magnetic moment. The magnetic moment (in Bohr magnetons, BM) is given by $\mu = \sqrt{n(n+2)}$ where $n$ is the number of unpaired electrons. Let’s analyse each complex step by step.

   1. Complex A: $\mathrm{Ni(CO)_4}$

      • Nickel is in zero oxidation state (CO is neutral).
      • CO is a strong field ligand, causing pairing of electrons.
      • Ni(0) has atomic number 28: $[Ar] 3d^8 4s^2$. In the presence of strong field CO, the 4s electrons are used for bonding, and the 3d electrons pair up.
      • The complex is tetrahedral (4 ligands), so hybridisation is $sp^3$.
      • All electrons are paired → $\mu = 0$ BM.
      • So A matches R ($sp^3, \mu=0$).

   2. Complex B: $\left[\mathrm{Ni(CN)_4}\right]^{2-}$

      • CN⁻ is a strong field ligand.
      • Oxidation state: Ni + 4(−1) = −2 → Ni is +2.
      • Ni²⁺: $[Ar] 3d^8$. Strong field CN⁻ causes pairing of 3d electrons, leaving one d-orbital empty.
      • Geometry is square planar (4 ligands), hybridisation is $dsp^2$.
      • All electrons paired → $\mu = 0$ BM.
      • So B matches S ($dsp^2, \mu=0$).

   3. Complex C: $\left[\mathrm{Ni(Cl)_4}\right]^{2-}$

      • Cl⁻ is a weak field ligand.
      • Ni is again +2 (same as B).
      • Weak field → no pairing; Ni²⁺ has 8 d-electrons, which in a tetrahedral field give 2 unpaired electrons (since $t_2$ orbitals are higher energy, electrons occupy them singly first).
      • Geometry is tetrahedral → hybridisation $sp^3$.
      • $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.84$ BM.
      • So C matches Q ($sp^3, \mu=2.84$ BM).

   4. Complex D: $\left[\mathrm{MnBr_4}\right]^{2-}$

      • Br⁻ is a weak field ligand.
      • Oxidation state: Mn + 4(−1) = −2 → Mn is +2.
      • Mn²⁺: $[Ar] 3d^5$. Weak field → high spin, all five d-orbitals singly occupied.
      • Tetrahedral geometry → hybridisation $sp^3$.
      • Number of unpaired electrons = 5 → $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92$ BM.
      • So D matches P ($sp^3, \mu=5.92$ BM).

   Thus the correct mapping is:

   A→R, B→S, C→Q, D→P.

   Watch out

   A common mistake is to forget that $\mathrm{Ni(CO)_4}$ has Ni(0) and that CO is a strong field ligand, leading to a diamagnetic tetrahedral complex — not paramagnetic.

   Tip

   For tetrahedral complexes, strong field vs weak field doesn’t change the number of unpaired electrons as much as in octahedral, because the crystal field splitting is smaller. But here, the key is the oxidation state and ligand strength.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

7. COMEDK 2025Set 2025-M1 markMCQ
   Q.What is the spin only magnetic moment of the metal ion in $\mathbf{P}$ and the oxidation number of Sulphur in the oxidised product Z ? (A) $\mu=0 \quad$ Oxidation number $=+4$ (B) $\mu=5.00 \mathrm{BM} \quad$ Oxidation number $=0$ (C) $\mu=5.92 \mathrm{BM} \quad$ Oxidation number $=+2$ (D) $\mu=1.732 \mathrm{BM} \quad$ Oxidation number $=+6$

   ›Reveal solutionSolution

   Pyrolusite ($\mathrm{MnO_2}$) fused with KOH and $\mathrm{KNO_3}$ gives the manganate(VI) ion in $\mathbf{P}=\mathrm{K_2MnO_4}$. The metal ion $\mathrm{Mn^{6+}}$ is $d^1$, so $\mu=\sqrt{1(1+2)}=1.732$ BM. Permanganate then oxidises thiosulphate in neutral medium to sulphate, in which sulphur is $+6$. The correct option is (D).

   Concept

   In an oxidising alkaline fusion, $\mathrm{Mn(IV)}$ in pyrolusite is raised to $\mathrm{Mn(VI)}$, giving the green manganate ion $\mathrm{MnO_4^{2-}}$. The spin-only magnetic moment depends only on the number of unpaired electrons through $\mu=\sqrt{n(n+2)}$ BM. Separately, the oxidation number of an atom is read from the formula of the species it ends up in.

   Solution

   1. Identify P. $\mathrm{MnO_2}+2\mathrm{KOH}+\mathrm{KNO_3}\rightarrow \mathrm{K_2MnO_4}+\mathrm{KNO_2}+\mathrm{H_2O}$. So $\mathbf{P}=\mathrm{K_2MnO_4}$ and the metal ion is $\mathrm{Mn^{6+}}$.
   2. Magnetic moment of $\mathrm{Mn^{6+}}$. Configuration $[\mathrm{Ar}]3d^1$ gives $n=1$, so $\mu=\sqrt{1(1+2)}=\sqrt{3}=1.732$ BM.
   3. Fate in acid. $3\mathrm{MnO_4^{2-}}+4\mathrm{H^+}\rightarrow 2\mathrm{MnO_4^-}+\mathrm{MnO_2}+2\mathrm{H_2O}$ (disproportionation), giving permanganate and $\mathrm{MnO_2}$.
   4. Oxidation of thiosulphate. In neutral medium permanganate oxidises $\mathrm{S_2O_3^{2-}}$ to sulphate $\mathrm{SO_4^{2-}}$ ($\mathbf{Z}$); sulphur in $\mathrm{SO_4^{2-}}$ has oxidation number $+6$.
   5. Match. $\mu=1.732$ BM with sulphur $=+6$ is option (D).

   Watch out

   Only $\mathrm{Mn^{6+}}$ ($d^1$) gives $\mu=1.732$ BM; do not use $\mathrm{Mn^{2+}}$ ($d^5$, $5.92$ BM) here, because the metal ion asked about is the one in $\mathbf{P}$.

   ✓Final answer

   The correct option is (D): $\mu=1.732$ BM and oxidation number of sulphur $=+6$.

8. KCET 2024Set B-21 markMCQ
   Q.Match the following: I. $\mathrm{Zn}^{2+}$ II. $\mathrm{Cu}^{2+}$ III. $\mathrm{Ni}^{2+}$ i. $d^8$ configuration ii. colourless iii. $\mu = 1.73\,\mathrm{BM}$ I II III (A) i ii iii (B) ii iii i (C) ii i iii (D) i iii ii

   ›Reveal solutionSolution

   Write the $d^n$ configuration of each ion, then read off colour (needs a partly-filled $d$ shell) and magnetic moment from the spin-only formula.

   Step 1 — Get the $d$-configurations.

   For a transition-metal cation, remove the $4s$ electrons first, then the $3d$ electrons.

   Ion	Atomic no.	Neutral atom	$M^{2+}$ configuration
   $\mathrm{Zn}^{2+}$	30	$[\mathrm{Ar}]3d^{10}4s^2$	$[\mathrm{Ar}]\,3d^{10}$
   $\mathrm{Cu}^{2+}$	29	$[\mathrm{Ar}]3d^{10}4s^1$	$[\mathrm{Ar}]\,3d^{9}$
   $\mathrm{Ni}^{2+}$	28	$[\mathrm{Ar}]3d^{8}4s^2$	$[\mathrm{Ar}]\,3d^{8}$

   Step 2 — Match III: $\mathrm{Ni}^{2+}$ → (i) $d^8$.

   Read straight off the table above. This is the most direct match, and it already eliminates options (A) and (C), which both send $d^8$ to $\mathrm{Zn}^{2+}$.

   Step 3 — Match I: $\mathrm{Zn}^{2+}$ → (ii) colourless.

   Colour in transition-metal ions arises from $d$–$d$ electronic transitions: an electron absorbs visible light and is promoted from the lower ($t_{2g}$) to the higher ($e_g$) set. This demands a partially filled $d$ subshell. $\mathrm{Zn}^{2+}$ is $d^{10}$ — completely full, no vacancy to promote into — so no $d$–$d$ transition is possible and the ion is colourless.

   Step 4 — Match II: $\mathrm{Cu}^{2+}$ → (iii) $\mu = 1.73$ BM.

   Use the spin-only formula, where $n$ = number of unpaired electrons:

   $\mu = \sqrt{n(n+2)}\ \text{BM}$

   $\mathrm{Cu}^{2+}$ is $d^9$: filling five $d$ orbitals with nine electrons leaves exactly one unpaired electron, so $n = 1$:

   $\mu = \sqrt{1(1+2)} = \sqrt{3} = 1.73\ \text{BM} \ \checkmark$

   (Cross-check: $\mathrm{Ni}^{2+}$, $d^8$, has $n = 2 \Rightarrow \mu = \sqrt{8} = 2.83$ BM, and $\mathrm{Zn}^{2+}$, $d^{10}$, has $n = 0 \Rightarrow \mu = 0$ — so $1.73$ BM belongs uniquely to $\mathrm{Cu}^{2+}$.)

   Step 5 — Assemble.

   $\text{I} \rightarrow \text{ii}, \qquad \text{II} \rightarrow \text{iii}, \qquad \text{III} \rightarrow \text{i}$

   ✓Final answer

   The correct option is (B) — ii iii i.

   ANSWER: B

9. COMEDK 2024Set 2024-A1 markMCQ
   Q.The metallic ions that have almost same spin only magnetic moment are :

   (i) $\mathrm{Co}^{2+}$

   (ii) $\mathrm{Mn}^{2+}$

   (iii) $\mathrm{Cr}^{2+}$

   (iv) $\mathrm{Cr}^{3+}$ (A) $$ \text {

   (iii) and

   (iv) } $(B)$ \text {

   (i) and

   (ii) } $(C)$ \text {

   (i) and

   (iv) } $(D)$ \text {

   (i) and

   (iii) } $$

   ›Reveal solutionSolution

   Counting unpaired electrons: Co$^{2+}$ (d$^7$) = 3, Mn$^{2+}$ (d$^5$) = 5, Cr$^{2+}$ (d$^4$) = 4, Cr$^{3+}$ (d$^3$) = 3. Co$^{2+}$ and Cr$^{3+}$ share $n=3$, hence the same $\mu=3.87$ BM → (i) and (iv).

   Spin-only moment $\mu = \sqrt{n(n+2)}$ BM, where $n$ = number of unpaired electrons:

   • (i) Co$^{2+}$: $3d^7$ → 3 unpaired → $\mu=\sqrt{15}=3.87$ BM
   • (ii) Mn$^{2+}$: $3d^5$ → 5 unpaired → $\mu=\sqrt{35}=5.92$ BM
   • (iii) Cr$^{2+}$: $3d^4$ → 4 unpaired → $\mu=\sqrt{24}=4.90$ BM
   • (iv) Cr$^{3+}$: $3d^3$ → 3 unpaired → $\mu=\sqrt{15}=3.87$ BM

   Co$^{2+}$ and Cr$^{3+}$ have equal $n=3$ and therefore almost the same magnetic moment.

   ✓Final answer

   The correct option is (C) — (i) and (iv)

10. COMEDK 2024Set 2024-E1 markMCQ
    Q.Based on Valence Bond Theory, match the complexes listed in Column I with the number of unpaired electrons on the central metal ion, given 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. Complex ions No. Number of unpaired electrons (A) $\left[\mathrm{Fe} \mathrm{F}_6\right]^{3-}$ (P) 0 (B) $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$ (P) 1 (C) $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ (R) 5 (D) $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ (S) 4 (A) $\mathrm{A}=\mathrm{S} \quad \mathrm{B}=\mathrm{Q} \quad \mathrm{C}=\mathrm{R} \quad \mathrm{D}=\mathrm{P}$ (B) $\mathrm{A}=\mathrm{S} \quad \mathrm{B}=\mathrm{Q} \quad \mathrm{C}=\mathrm{P} \quad \mathrm{D}=\mathrm{R}$ (C) $\mathrm{A}=\mathrm{R} \quad \mathrm{B}=\mathrm{P} \quad \mathrm{C}=\mathrm{S} \quad \mathrm{D}=\mathrm{Q}$ (D) $\mathrm{A}=\mathrm{R} \quad \mathrm{B}=\mathrm{S} \quad \mathrm{C}=\mathrm{Q} \quad \mathrm{D}=\mathrm{P}$

    ›Reveal solutionSolution

    [!TLDR]

    Using d-electron counts and strong/weak-field ligands, the unpaired electrons are 5, 0, 4, 1 for A-D, matching option (C).

    Concept

    In Valence Bond Theory a strong-field ligand (like CN-) forces electron pairing, giving a low-spin inner-orbital complex, while a weak-field ligand (F-, H2O) leaves the electrons unpaired (high spin). Always count the d-electrons on the metal ion first (CBSE/NCERT Class 12, Coordination Compounds).

    Solution

    • (A) $[\mathrm{FeF_6}]^{3-}$: Fe$^{3+}$ is $3d^5$; F$^-$ is weak-field $\Rightarrow$ high spin $\Rightarrow$ 5 unpaired $\Rightarrow$ R.
    • (B) $[\mathrm{Fe(CN)_6}]^{4-}$: Fe$^{2+}$ is $3d^6$; CN$^-$ strong-field $\Rightarrow$ low spin ($t_{2g}^6$) $\Rightarrow$ 0 unpaired $\Rightarrow$ P.
    • (C) $[\mathrm{Fe(H_2O)_6}]^{2+}$: Fe$^{2+}$ is $3d^6$; H$_2$O weak-field $\Rightarrow$ high spin ($t_{2g}^4 e_g^2$) $\Rightarrow$ 4 unpaired $\Rightarrow$ S.
    • (D) $[\mathrm{Fe(CN)_6}]^{3-}$: Fe$^{3+}$ is $3d^5$; CN$^-$ strong-field $\Rightarrow$ low spin ($t_{2g}^5$) $\Rightarrow$ 1 unpaired $\Rightarrow$ Q.

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

    [!ANSWER]

    (C) A=R, B=P, C=S, D=Q

11. COMEDK 2024Set 2024-M1 markMCQ
    Q.A d - block metal $\mathrm{X}(\mathrm{Z}=26)$ forms a compound $[\mathrm{X}(\mathrm{CN})_2(\mathrm{CO})_4]^{+}$. Calculate its spin magnetic moment value. (A) 2.83 BM (B) 1.73 BM (C) 3.87 BM (D) 5.92 BM

    ›Reveal solutionSolution

    The metal is iron (Z=26); in $[\mathrm{Fe(CN)_2(CO)_4}]^+$ it is Fe$^{3+}$ (d$^5$), and since CN$^-$ and CO are both strong-field ligands, the complex is low-spin with 1 unpaired electron, giving $\mu = 1.73$ BM — option (B).

    Step-by-step reasoning

    1. Identify the metal. $Z=26$ is iron (Fe).

    2. Find the oxidation state.

       CN$^-$ contributes $-1$ each (×2), CO is neutral (×4), overall complex charge $=+1$:

    $x + 2(-1) + 4(0) = +1 \quad\Rightarrow\quad x = +3$

    So Fe is in the $+3$ state.

    3. Find the d-electron count.

       Neutral Fe is $[Ar]3d^64s^2$; Fe$^{3+}$ is $[Ar]3d^5$.

    4. Apply the ligand field.

       Both CN$^-$ and CO are strong-field ligands, favouring a low-spin configuration. For low-spin d$^5$ in an octahedral field: $t_{2g}^5e_g^0$ → 1 unpaired electron.

    5. Compute the magnetic moment.

    $\mu = \sqrt{n(n+2)} = \sqrt{1\times3} = \sqrt3 \approx 1.73\ \text{BM}$

    Watch out

    Forgetting the complex's $+1$ overall charge would incorrectly give Fe$^{2+}$ (d$^6$) instead of Fe$^{3+}$ (d$^5$) — always balance the charge first.

    ✓Final answer

    The correct option is (B): 1.73 BM.

12. COMEDK 2024Set 2024-M1 markMCQ
    Q.The permanganate ion in acid medium acts as an oxidant and gets converted to its lower oxidation state. What would be the spin only magnetic moment of such reduced manganese ion? (A) 2.84 BM (B) 4.90 BM (C) 3.87 BM (D) 5.92 BM

    ›Reveal solutionSolution

    In acidic medium, permanganate ($\text{MnO}_4^-$) is reduced to $\text{Mn}^{2+}$, which has five unpaired $3d$ electrons; the spin-only magnetic moment is $\sqrt{5(5+2)} = \sqrt{35} \approx 5.92$ BM, so the correct option is (D).

    The key concept here is the relationship between oxidation state, electronic configuration, and magnetic moment. The permanganate ion ($\text{MnO}_4^-$) contains manganese in the +7 oxidation state. In acidic medium, it is a powerful oxidant and gets reduced to a lower oxidation state — specifically, to the manganous ion ($\text{Mn}^{2+}$). The question asks for the spin-only magnetic moment of this reduced ion. The spin-only formula is $\mu = \sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons. So we need to find $n$ for $\text{Mn}^{2+}$.

    1. Determine the electronic configuration of the reduced manganese ion.

       Manganese (atomic number 25) has the ground-state configuration $[\text{Ar}]\, 3d^5 4s^2$.

       In the $\text{Mn}^{2+}$ ion, two electrons are removed — first from the $4s$ orbital (as is standard for transition metals), giving $[\text{Ar}]\, 3d^5$.

    2. Find the number of unpaired electrons in $\text{Mn}^{2+}$.

       The $3d$ subshell has five orbitals. For a $d^5$ configuration, Hund’s rule tells us that electrons occupy all five orbitals singly with parallel spins before any pairing occurs. Thus, all five $3d$ electrons are unpaired.

       So $n = 5$.

    3. Apply the spin-only magnetic moment formula.

       The formula is:

    $\mu = \sqrt{n(n+2)} \ \text{BM}$

    Substituting $n = 5$:

    $\mu = \sqrt{5 \times 7} = \sqrt{35} \ \text{BM}$

    Numerically, $\sqrt{35} \approx 5.916$, which rounds to $5.92$ BM.

    Tip

    A common shortcut: for $d^5$ (high-spin), the magnetic moment is always $\sqrt{35} \approx 5.92$ BM. This is a classic value to remember for $\text{Mn}^{2+}$ and $\text{Fe}^{3+}$.

    Watch out

    A frequent mistake is to forget that in acidic medium, permanganate reduces all the way to $\text{Mn}^{2+}$, not to $\text{Mn}^{4+}$ or $\text{Mn}^{6+}$. In neutral/alkaline medium, the product is $\text{MnO}_2$ ($\text{Mn}^{4+}$, $n=3$, $\mu \approx 3.87$ BM), which is a distractor here.

    4. Match with the options. The calculated value $5.92$ BM corresponds exactly to option (D).

    ✓Final answer

    The correct option is (D).

    ANSWER: D