Q.Why do compounds having similar geometry have different magnetic moment?

Showing the 12 most recent of 34 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.Consider the following $\mathrm{Ti, V, Cr, Mn, Fe}$ The spin only magnetic moment (in BM) of the metal with lowest melting point in its +3 oxidation state is (A) $\sqrt{15}$ (B) $\sqrt{24}$ (C) $\sqrt{35}$ (D) $\sqrt{3}$

   ›Reveal solutionSolution

   This tests recognising Mn's anomalous melting point and computing a spin-only magnetic moment; the answer is $\sqrt{24}$ BM.

   Concept and Intuition

   Across the 3d series, melting points generally rise then fall with the number of unpaired d-electrons available for metallic bonding, but Mn is a famous outlier: its complex crystal structure (with several inequivalent Mn sites) leads to unusually weak metallic bonding, so Mn has by far the LOWEST melting point among Ti–Fe. Once the correct metal is identified, the spin-only formula $\mu=\sqrt{n(n+2)}$ BM gives the magnetic moment from the number of unpaired electrons in the specified oxidation state.

   Step-by-Step Solution

   1. Compare melting points of Ti, V, Cr, Mn, Fe: Ti≈1668°C, V≈1910°C, Cr≈1907°C, Mn≈1246°C (anomalously low), Fe≈1538°C. Mn has the lowest.
   2. Mn (Z=25): ground state $[Ar]3d^5 4s^2$.
   3. $Mn^{3+}$: remove 2 electrons from 4s and 1 from 3d ⇒ configuration $3d^4$.
   4. By Hund's rule, $3d^4$ places 4 electrons in 5 d-orbitals all unpaired (free ion, no ligand field specified): $n=4$. …

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2. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.In which of the following, elements are correctly arranged in the increasing order of unpaired electrons? (A) Fe < Co < Ni < Mn (B) Ni < Co < Mn < Fe (C) Mn < Fe < Co < Ni (D) Ni < Co < Fe < Mn

   ›Reveal solutionSolution

   Counting unpaired 3d electrons in Mn, Fe, Co, Ni (Hund's rule) gives Mn=5, Fe=4, Co=3, Ni=2, so the increasing order is Ni < Co < Fe < Mn.

   Concept and Intuition

   For first-row transition metals, the $4s$ orbital fills before $3d$ but electrons in the (n−1)d subshell obey Hund's rule of maximum multiplicity — electrons singly occupy all five d orbitals before any pairing begins. As we move across the row adding one more d-electron at a time past the half-filled $d^5$ configuration, pairing begins and the number of unpaired electrons decreases even though the electron count increases, until $d^{10}$ (all paired).

   Step-by-Step Solution

   1. Write ground-state configurations (outer shells): Mn = $[Ar]3d^5 4s^2$; Fe = $[Ar]3d^6 4s^2$; Co = $[Ar]3d^7 4s^2$; Ni = $[Ar]3d^8 4s^2$.
   2. Apply Hund's rule to count unpaired d-electrons:
      • Mn, $3d^5$: all five orbitals singly occupied → 5 unpaired.
      • Fe, $3d^6$: one orbital now doubly occupied, four still singly occupied → 4 unpaired.
      • Co, $3d^7$: two orbitals doubly occupied, three singly occupied → 3 unpaired.
      • Ni, $3d^8$: three orbitals doubly occupied, two singly occupied → 2 unpaired. …

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3. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.A transition metal ion $X^{3+}$ has a magnetic moment of $\sqrt{15}\ BM$. The atomic number of the metal X is (A) $24$ (B) $25$ (C) $26$ (D) $27$

   ›Reveal solutionSolution

   $\sqrt{15}\ BM$ means 3 unpaired electrons; the only ion among the choices that is unambiguously $d^3$ is $Cr^{3+}$, atomic number 24.

   Concept and Intuition

   The spin-only magnetic moment formula $\mu = \sqrt{n(n+2)}\ BM$ connects the number of unpaired electrons $n$ to the measured moment. Working backwards from a given moment tells you $n$ directly, and then you match $n$ to the $d$-electron count of the ion.

   Step-by-Step Solution

   1. $\mu = \sqrt{n(n+2)} = \sqrt{15} \Rightarrow n(n+2) = 15 \Rightarrow n^2+2n-15=0 \Rightarrow (n+5)(n-3)=0 \Rightarrow n=3$.
   2. So $X^{3+}$ has 3 unpaired electrons — i.e. it is a $d^3$ ion (three electrons singly occupying the three $t_{2g}$ orbitals with no possibility of pairing, regardless of ligand field).
   3. Check each candidate atomic number as the neutral atom, then remove 3 electrons for the $3+$ ion:
      • $Z=24$, Cr: $[Ar]3d^5 4s^1 \Rightarrow Cr^{3+} = [Ar]3d^3$. Exactly 3 unpaired electrons. ✓
      • $Z=25$, Mn: $[Ar]3d^5 4s^2 \Rightarrow Mn^{3+} = [Ar]3d^4$, which has 4 unpaired electrons (high spin). ✗ …

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4. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.How many of the following complex ions contain 4 unpaired electrons? $[Cr(H_2O)_6]^{2+},\ [Mn(H_2O)_6]^{2+},\ [Fe(H_2O)_6]^{2+},\ [Co(H_2O)_6]^{3+},\ [Cu(H_2O)_6]^{2+},\ [CoF_6]^{3-},\ [Cr(CN)_6]^{4-},\ [MnCl_4]^{2-}$ The correct answer is (A) $3$ (B) $4$ (C) $2$ (D) $5$

   ›Reveal solutionSolution

   Working out the $d$-electron count and spin state for each of the eight complex ions, exactly four of them ($[Cr(H_2O)_6]^{2+}$, $[Fe(H_2O)_6]^{2+}$, $[Co(H_2O)_6]^{3+}$, $[CoF_6]^{3-}$) have 4 unpaired electrons.

   Concept and Intuition

   The number of unpaired $d$-electrons in a complex depends on (a) the metal's oxidation state and $d^n$ configuration, and (b) whether the ligand field is strong enough to force pairing (low spin) or not (high spin). $H_2O$, $F^-$ and $Cl^-$ are weak-to-intermediate field ligands, so octahedral complexes with them are high spin; $CN^-$ is a strong field ligand, forcing low spin; tetrahedral complexes have such a small crystal field splitting that they are essentially always high spin regardless of the ligand.

   Step-by-Step Solution

   Go through each ion (metal ion configuration, spin state, unpaired count):

   1. $[Cr(H_2O)_6]^{2+}$: $Cr^{2+} = d^4$, weak field (HS) $\Rightarrow t_{2g}^3 e_g^1 \Rightarrow$ 4 unpaired.
   2. $[Mn(H_2O)_6]^{2+}$: $Mn^{2+} = d^5$, HS $\Rightarrow t_{2g}^3 e_g^2 \Rightarrow$ 5 unpaired.
   3. $[Fe(H_2O)_6]^{2+}$: $Fe^{2+} = d^6$, HS $\Rightarrow t_{2g}^4 e_g^2 \Rightarrow$ 4 unpaired (one paired orbital, four singly occupied).
   4. $[Co(H_2O)_6]^{3+}$: $Co^{3+} = d^6$; $H_2O$ is too weak a field to pair $Co^{3+}$ electrons (only very strong ligands like $NH_3$/$CN^-$ do that), so it is HS like Fe above $\Rightarrow$ 4 unpaired.
   5. $[Cu(H_2O)_6]^{2+}$: $Cu^{2+}=d^9 \Rightarrow$ 1 unpaired. …

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5. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.Which of the following pairs of ions are not paramagnetic in nature? (Atomic Number: La=57, Ce=58, Eu=63, Gd=64, Tb=65, Yb=70, Lu=71) I. $La^{3+}, Ce^{4+}$ II. $Eu^{2+}, Ce^{3+}$ III. $Lu^{3+}, Yb^{2+}$ IV. $Tb^{4+}, Gd^{3+}$ The correct answer is (A) I & III (B) II & III (C) III & IV (D) I & IV

   ›Reveal solutionSolution

   Determine the 4f-electron count of each lanthanide ion; a pair is diamagnetic (not paramagnetic) only if both ions have a fully empty (4f⁰) or fully filled (4f¹⁴) f-subshell.

   Concept and Intuition

   Lanthanide ions are paramagnetic whenever they have unpaired 4f electrons. The two 'magic' configurations with zero unpaired electrons are 4f⁰ (empty) and 4f¹⁴ (completely filled) — both diamagnetic. Any partially filled 4f subshell (4f¹ through 4f¹³, excluding the special stable half/fully filled cases which still can have unpaired electrons unless exactly f0/f14) gives unpaired electrons and hence paramagnetism.

   Step-by-Step Solution

   1. $La$ (Z=57): $[Xe]5d^16s^2$; $La^{3+}$ removes all 3 outer electrons → $[Xe]4f^0$ — diamagnetic.
   2. $Ce$ (Z=58): $[Xe]4f^15d^16s^2$; $Ce^{4+}$ removes all 4 → $[Xe]4f^0$ — diamagnetic. So pair I ($La^{3+}, Ce^{4+}$) is NOT paramagnetic.
   3. $Ce^{3+}$ retains one f-electron: $4f^1$ — paramagnetic. $Eu$ (Z=63): $[Xe]4f^76s^2$; $Eu^{2+}$ → $4f^7$ — paramagnetic. So pair II IS paramagnetic.
   4. $Yb$ (Z=70): $[Xe]4f^{14}6s^2$; $Yb^{2+}$ → $4f^{14}$ — diamagnetic. $Lu$ (Z=71): $[Xe]4f^{14}5d^16s^2$; $Lu^{3+}$ → $4f^{14}$ — diamagnetic. So pair III is NOT paramagnetic. …

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6. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.In which of the following given sets, complexes are correctly arranged in the increasing order of their spin only magnetic moment values? I. $[Fe(CN)_6]^{4-} < [Fe(CN)_6]^{3-} < [Fe(H_2O)_6]^{3+}$ II. $[Co(NH_3)_6]^{3+} < [Ni(H_2O)_6]^{2+} < [Cr(H_2O)_6]^{3+}$ III. $[V(H_2O)_6]^{3+} < [Cr(CN)_6]^{3-} < [Fe(H_2O)_6]^{2+}$ The correct answer is (A) I, II only (B) I, II, III (C) II, III only (D) I, III only

   ›Reveal solutionSolution

   Working out the $d$-electron count, spin state (from field strength of the ligand) and hence unpaired-electron count for each complex confirms all three given orderings of spin-only magnetic moment are correct.

   Concept and Intuition

   Spin-only magnetic moment is $\mu = \sqrt{n(n+2)}$ BM, where $n$ = number of unpaired electrons. The number of unpaired electrons depends on the metal's oxidation state (which fixes the $d^n$ configuration) and whether the ligand is weak-field (high spin, e.g. $H_2O$ usually) or strong-field (low spin, e.g. $CN^-$, $NH_3$ for many metals) — strong field ligands pair electrons into $t_{2g}$ before populating $e_g$.

   Step-by-Step Solution

   Set I:

   • $[Fe(CN)_6]^{4-}$: $Fe^{2+}$, $d^6$, $CN^-$ strong field $\Rightarrow$ low spin, $t_{2g}^6 e_g^0$, 0 unpaired, $\mu=0$.
   • $[Fe(CN)_6]^{3-}$: $Fe^{3+}$, $d^5$, $CN^-$ strong field $\Rightarrow$ low spin, $t_{2g}^5$, 1 unpaired, $\mu=\sqrt3=1.73$.
   • $[Fe(H_2O)_6]^{3+}$: $Fe^{3+}$, $d^5$, $H_2O$ weak field $\Rightarrow$ high spin, $t_{2g}^3e_g^2$, 5 unpaired, $\mu=\sqrt{35}=5.92$.
   • Order $0<1.73<5.92$ matches I. True.

   Set II:

   • $[Co(NH_3)_6]^{3+}$: $Co^{3+}$, $d^6$, $NH_3$ strong field for Co$^{3+}$ $\Rightarrow$ low spin, 0 unpaired, $\mu=0$.
   • $[Ni(H_2O)_6]^{2+}$: $Ni^{2+}$, $d^8$, always $t_{2g}^6e_g^2$ regardless of field, 2 unpaired, $\mu=\sqrt8=2.83$.
   • $[Cr(H_2O)_6]^{3+}$: $Cr^{3+}$, $d^3$, $t_{2g}^3$, 3 unpaired, $\mu=\sqrt{15}=3.87$.
   • Order $0<2.83<3.87$ matches II. True.

   Set III:

   • $[V(H_2O)_6]^{3+}$: $V^{3+}$, $d^2$, $t_{2g}^2$, 2 unpaired, $\mu=\sqrt8=2.83$. …

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7. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
   Q.Which of the following sets are not correctly matched? I) $\mathrm{O_2^{2+}}, \mathrm{O_2^{2-}}$ - diamagnetic II) $\mathrm{O_2^{+}}, \mathrm{O_2}$ - paramagnetic III) $\mathrm{O_2^{-}}, \mathrm{O_2^{2-}}$ - diamagnetic IV) $\mathrm{O_2^{+}}, \mathrm{O_2^{2-}}$ - paramagnetic (A) II & III (B) I & II (C) III & IV (D) I & III

   ›Reveal solutionSolution

   Working out the MO electron configuration and unpaired-electron count for each oxygen species shows sets III and IV are mismatched (superoxide O$_2^-$ is actually paramagnetic, and peroxide O$_2^{2-}$ is actually diamagnetic).

   Concept and Intuition

   For O$_2$ and its ions, molecular orbital theory places 12 valence electrons (for neutral O$_2$) into the sequence $\sigma2s,\ \sigma^*2s,\ \sigma2p_z,\ \pi2p_x=\pi2p_y,\ \pi^*2p_x=\pi^*2p_y$. Removing or adding electrons changes how many go into the degenerate $\pi^*$ pair, which determines whether unpaired electrons (paramagnetism) exist.

   Step-by-Step Solution

   1. O$_2$ (12 valence e$^-$): fills up to $\pi^*2p_x^1\pi^*2p_y^1$ — 2 unpaired electrons ⟹ paramagnetic.
   2. O$_2^{2+}$ (10 e$^-$, remove 2 from $\pi^*$): $\pi^*$ is empty ⟹ all electrons paired ⟹ diamagnetic.
   3. O$_2^+$ (11 e$^-$, remove 1 from $\pi^*$): one $\pi^*$ orbital has 1 electron ⟹ 1 unpaired electron ⟹ paramagnetic.
   4. O$_2^-$ (13 e$^-$, add 1 to $\pi^*$): $\pi^*2p_x^2\pi^*2p_y^1$ ⟹ 1 unpaired electron ⟹ paramagnetic (not diamagnetic).
   5. O$_2^{2-}$ (14 e$^-$, add 2 to $\pi^*$): $\pi^*2p_x^2\pi^*2p_y^2$ ⟹ all paired ⟹ diamagnetic.
   6. Now check each set:

   • I) O$_2^{2+}$ (diamagnetic ✓), O$_2^{2-}$ (diamagnetic ✓) → correctly matched.
   • II) O$_2^+$ (paramagnetic ✓), O$_2$ (paramagnetic ✓) → correctly matched. …

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8. AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
   Q.Match the following List-I (Complex) List-II (Number of unpaired electrons) A) $[\mathrm{MnCl_6}]^{3-}$ I) 5 B) $[\mathrm{FeF_6}]^{3-}$ II) 2 C) $[\mathrm{Mn(CN)_6}]^{3-}$ III) 0 D) $[\mathrm{Co(C_2O_4)_3}]^{3-}$ IV) 4 The correct answer is (A) A-II, B-IV, C-III, D-I (B) A-IV, B-II, C-I, D-III (C) A-III, B-I, C-IV, D-II (D) A-IV, B-I, C-II, D-III

   ›Reveal solutionSolution

   Working out the oxidation state and $d$-electron count of the central metal in each complex, then applying crystal field theory (weak-field ligands give high spin, strong-field ligands give low spin) gives the unpaired-electron counts: A=4, B=5, C=2, D=0 — matching option (D).

   Concept and Intuition

   For octahedral transition-metal complexes, the number of unpaired electrons depends on (i) the metal's oxidation state and resulting $d^n$ configuration, and (ii) whether the ligand is weak-field (high spin, electrons spread out over $t_{2g}$ and $e_g$ following Hund's rule) or strong-field (low spin, electrons pair up in $t_{2g}$ before occupying $e_g$). Halide ligands ($\mathrm{Cl^-}$, $\mathrm{F^-}$) are weak field; $\mathrm{CN^-}$ is strong field; oxalate is a chelating ligand that with $\mathrm{Co^{3+}}$ specifically gives a well-known diamagnetic (low-spin) complex.

   Step-by-Step Solution

   1. A) $[\mathrm{MnCl_6}]^{3-}$: Mn is $+3$ here ($d^4$ since Mn is group 7, $d^4$ for $\mathrm{Mn^{3+}}$). $\mathrm{Cl^-}$ is weak field $\Rightarrow$ high spin: $t_{2g}^3 e_g^1$ $\Rightarrow$ 4 unpaired electrons $\Rightarrow$ matches IV.
   2. B) $[\mathrm{FeF_6}]^{3-}$: $\mathrm{Fe^{3+}}$ is $d^5$. $\mathrm{F^-}$ is weak field $\Rightarrow$ high spin: $t_{2g}^3 e_g^2$, all 5 electrons unpaired $\Rightarrow$ matches I.
   3. C) $[\mathrm{Mn(CN)_6}]^{3-}$: $\mathrm{Mn^{3+}}$ is $d^4$ again, but $\mathrm{CN^-}$ is strong field $\Rightarrow$ low spin: $t_{2g}^4 e_g^0$ $\Rightarrow$ 2 unpaired electrons $\Rightarrow$ matches II. …

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9. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
   Q.Consider the following complex ions (only = only) I) $[\mathrm{Fe(CN)_6}]^{3-}$ II) $[\mathrm{Co(CN)_6}]^{3-}$ III) $[\mathrm{Mn(CN)_6}]^{4-}$ IV) $[\mathrm{Fe(CN)_6}]^{4-}$ Identify the complex ion/s with the least spin only magnetic moment (in BM). (A) II & IV only (B) I only (C) III only (D) I & III only

   ›Reveal solutionSolution

   With the strong-field ligand $\mathrm{CN^-}$, $d^6$ metal ions ($\mathrm{Co^{3+}}$, $\mathrm{Fe^{2+}}$) become perfectly diamagnetic (low-spin $t_{2g}^6$), giving the least possible spin-only moment of $0$ BM.

   Concept and Intuition

   Whether a complex is high spin or low spin depends on the ligand field strength versus the pairing energy. $\mathrm{CN^-}$ sits at the strong end of the spectrochemical series, so it always forces low spin in octahedral 3d complexes. Once you know the metal's $d^n$ configuration, low-spin filling (fill $t_{2g}$ completely before touching $e_g$) tells you the unpaired electron count directly.

   Step-by-Step Solution

   1. $[\mathrm{Fe(CN)_6}]^{3-}$: $\mathrm{Fe^{3+}}$ is $3d^5$. Low spin: $t_{2g}^5 e_g^0$ → 3 orbitals hold 5 electrons (2+2+1) → 1 unpaired → $\mu=\sqrt{1\times3}=1.73$ BM.
   2. $[\mathrm{Co(CN)_6}]^{3-}$: $\mathrm{Co^{3+}}$ is $3d^6$. Low spin: $t_{2g}^6 e_g^0$ (all three $t_{2g}$ orbitals doubly filled) → 0 unpaired → $\mu = 0$ BM.
   3. $[\mathrm{Mn(CN)_6}]^{4-}$: $\mathrm{Mn^{2+}}$ is $3d^5$ (same electron count as $\mathrm{Fe^{3+}}$). Low spin $t_{2g}^5$ → 1 unpaired → $\mu = 1.73$ BM. …

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10. AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
    Q.In 3d series, a metal 'X' has highest second ionisation enthalpy. The spin only magnetic moment (in BM) of $X^+$ ion is (A) $1.73$ (B) $0.0$ (C) $2.84$ (D) $5.92$

    ›Reveal solutionSolution

    The 3d-series metal with the highest second ionisation enthalpy is copper, because Cu⁺ has an extra-stable filled $3d^{10}$ configuration; this ion is diamagnetic, so its spin-only moment is 0.0 BM.

    Concept and Intuition

    Ionisation enthalpies show characteristic anomalies in the 3d transition series tied to especially stable electron configurations — half-filled ($d^5$) and fully-filled ($d^{10}$) subshells resist further electron removal. While Cr (which forms the stable $3d^5$ configuration in Cr⁺) is often the first anomaly students recall, it is actually copper whose second ionisation enthalpy is exceptionally and uniquely high across the whole row: Cu already achieves the special $3d^{10}4s^0$ (Cu⁺) configuration on losing just its first electron, so knocking out a second electron means breaking into this very stable, fully-filled d-subshell — requiring markedly more energy than for its neighbours.

    Step-by-Step Solution

    1. Ground state of Cu: $[Ar]3d^{10}4s^1$.
    2. First ionisation removes the 4s electron: $Cu^+=[Ar]3d^{10}$ — a fully-filled, extra-stable d-subshell.
    3. The second ionisation enthalpy of Cu (removing an electron from this stable $3d^{10}$ to give Cu²⁺, $3d^9$) is measurably higher than that of its 3d-series neighbours, making Cu the metal 'X' with the highest second ionisation enthalpy. …

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11. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
    Q.Identify the complex ion with spin only magnetic moment of 4.90 BM. (A) $[\mathrm{Co(NH_3)_6}]^{3+}$ (B) $[\mathrm{Cr(NH_3)_6}]^{3+}$ (C) $[\mathrm{Mn(CN)_6}]^{3-}$ (D) $[\mathrm{MnCl_6}]^{3-}$

    ›Reveal solutionSolution

    This tests computing the number of unpaired electrons in transition-metal complexes based on ligand field strength (CFT) and matching to a given magnetic moment. The answer is $[\mathrm{MnCl_6}]^{3-}$.

    Concept and Intuition

    Spin-only magnetic moment is given by $\mu=\sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons. To find $n$ for a complex, determine the metal's oxidation state and d-electron count, then decide whether the ligand is strong-field (causes pairing → low spin) or weak-field (electrons stay unpaired → high spin) using the spectrochemical series.

    Step-by-Step Solution

    1. Given $\mu=4.90$ BM. Solve $\sqrt{n(n+2)}=4.90 \Rightarrow n(n+2)=24.01 \Rightarrow n=4$ (since $4\times6=24$).
    2. (A) $[\mathrm{Co(NH_3)_6}]^{3+}$: Co is +3, d$^6$. NH$_3$ is a strong field ligand → low spin: $t_{2g}^6 e_g^0$, 0 unpaired electrons. Not a match.
    3. (B) $[\mathrm{Cr(NH_3)_6}]^{3+}$: Cr is +3, d$^3$. $t_{2g}^3$, always 3 unpaired electrons (no pairing possible in only 3 orbitals with 3 electrons). $\mu=\sqrt{15}=3.87$ BM. Not a match. …

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12. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
    Q.Identify the ion (hydrated in solution) which is not correctly matched with its spin only magnetic moment (in BM) given in brackets (A) $\mathrm{Cr}^{3+}$ (4.90) (B) $\mathrm{Cu}^{2+}$ (1.73) (C) $\mathrm{Co}^{3+}$ (4.90) (D) $\mathrm{Fe}^{2+}$ (4.90)

    ›Reveal solutionSolution

    Spin-only magnetic moment $\mu=\sqrt{n(n+2)}$ BM depends on the number of unpaired d-electrons. $\mathrm{Cr^{3+}}$ ($d^3$) always has 3 unpaired electrons giving $\mu\approx3.87$ BM, so pairing it with 4.90 BM (which needs 4 unpaired electrons) is the incorrect match.

    Concept and Intuition

    The spin-only formula $\mu=\sqrt{n(n+2)}$ BM depends only on the number of unpaired electrons $n$ in the ion's d-subshell. For a $d^3$ configuration, all three electrons must occupy separate $t_{2g}$ orbitals by Hund's rule — there is no possible arrangement (high-spin or low-spin) that changes this, so $d^3$ always gives exactly 3 unpaired electrons. Ions like $d^6$ or $d^9$ can vary in unpaired count depending on ligand field strength, so those need checking against the specific hydrated (weak-field, high-spin) case.

    Step-by-Step Solution

    1. $\mathrm{Cr^{3+}}$: configuration $[Ar]3d^3$. In any octahedral environment (weak or strong field), $d^3$ always has 3 unpaired electrons ($t_{2g}^3$). $\mu=\sqrt{3(3+2)}=\sqrt{15}\approx3.87$ BM — not 4.90 BM as option (A) states. This is the mismatch.
    2. $\mathrm{Cu^{2+}}$: $[Ar]3d^9$, always 1 unpaired electron regardless of field. $\mu=\sqrt{1(1+2)}=\sqrt3\approx1.73$ BM — matches option (B), correctly matched.
    3. $\mathrm{Co^{3+}}$ (hydrated): $[Ar]3d^6$. The hydrated (aqua) $\mathrm{Co^{3+}}$ ion is a well-known high-spin exception ($t_{2g}^4 e_g^2$), giving 4 unpaired electrons: $\mu=\sqrt{4(4+2)}=\sqrt{24}\approx4.90$ BM — matches option (C), correctly matched. …

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