Q.Why are $Mn^{2+}$ compounds more stable than $Fe^{2+}$ towards oxidation to their +3 state?

Concept & Intuition

In coordination chemistry, "outer orbital" complexes use the metal’s outer d-orbitals (nd) for hybridisation, typically sp³d², and are formed with weak-field ligands. These complexes tend to be high-spin, meaning they retain unpaired electrons and are paramagnetic. In contrast, "inner orbital" complexes use inner d-orbitals ((n−1)d) for d²sp³ hybridisation, usually with strong-field ligands, and are often low-spin (fewer or no unpaired electrons). So we need to check each complex for: (1) the ligand field strength, (2) the resulting electron configuration, and (3) whether the hybridisation uses outer or inner d-orbitals.

Step-by-step reasoning

1. Identify the metal oxidation states and d-electron counts

   • (A) $\left[\mathrm{Co}(\mathrm{C}_2\mathrm{O}_4)_3\right]^{3-}$: Oxalate is a bidentate ligand with charge −2. Let Co be $x$: $x + 3(-2) = -3 \Rightarrow x = +3$. Co³⁺ has electron configuration $[\mathrm{Ar}]\,3d^6$.
   • (B) $\left[\mathrm{MnCl}_6\right]^{3-}$: Cl⁻ has charge −1. Let Mn be $x$: $x + 6(-1) = -3 \Rightarrow x = +3$. Mn³⁺ has $[\mathrm{Ar}]\,3d^4$.
   • (C) $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}$: CN⁻ is −1, same calculation gives Mn³⁺, $3d^4$.
   • (D) $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$: CN⁻ is −1, so Fe is +3. Fe³⁺ has $[\mathrm{Ar}]\,3d^5$.

2. Determine ligand field strength and spin state

   • Oxalate ($\mathrm{C}_2\mathrm{O}_4^{2-}$) is a moderate-field ligand — it can cause pairing but not as strongly as CN⁻. For Co³⁺ ($3d^6$), oxalate typically gives a low-spin configuration (all electrons paired) because Co³⁺ has a high pairing energy and oxalate is strong enough to cause pairing. So $\left[\mathrm{Co}(\mathrm{C}_2\mathrm{O}_4)_3\right]^{3-}$ is diamagnetic (no unpaired electrons).
   • Cl⁻ is a weak-field ligand — it cannot cause pairing. For Mn³⁺ ($3d^4$), weak field gives high-spin: $t_{2g}^3 e_g^1$ (4 unpaired electrons).
   • CN⁻ is a strong-field ligand — it forces pairing. For Mn³⁺ ($3d^4$), strong field gives low-spin: $t_{2g}^4$ (2 unpaired electrons).
   • CN⁻ with Fe³⁺ ($3d^5$) gives low-spin: $t_{2g}^5$ (1 unpaired electron).

3. Decide inner vs. outer orbital hybridisation

   • Inner orbital (d²sp³) uses two inner (n−1)d orbitals, so it requires at least two empty d-orbitals in the inner set. This happens when electrons are paired to vacate d-orbitals.
   • Outer orbital (sp³d²) uses outer nd orbitals, so it occurs when the inner d-orbitals are too filled to allow d²sp³ — typically in high-spin configurations.
   • For (A): Low-spin Co³⁺ ($t_{2g}^6$) has all three $t_{2g}$ orbitals filled, but the $e_g$ orbitals are empty. It uses two $e_g$ (inner d) orbitals → inner orbital (d²sp³).
   • For (B): High-spin Mn³⁺ ($t_{2g}^3 e_g^1$) has only one empty inner d-orbital (one $e_g$ is empty, but we need two). So it cannot form d²sp³; it uses outer 4d orbitals → outer orbital (sp³d²). It has 4 unpaired electrons → paramagnetic.
   • For (C): Low-spin Mn³⁺ ($t_{2g}^4$) has two empty $e_g$ orbitals → inner orbital (d²sp³). It has 2 unpaired electrons → paramagnetic, but it is inner orbital.
   • For (D): Low-spin Fe³⁺ ($t_{2g}^5$) has one empty $e_g$ orbital (need two for d²sp³) — actually, careful: $t_{2g}^5$ means one $t_{2g}$ orbital is half-filled, but the $e_g$ orbitals are empty. So two empty inner d-orbitals exist → inner orbital (d²sp³). It has 1 unpaired electron → paramagnetic, but again inner orbital.

4. Select the only outer orbital paramagnetic complex

   Only (B) $\left[\mathrm{MnCl}_6\right]^{3-}$ uses outer d-orbitals (sp³d²) and has unpaired electrons.

Concept & Intuition

Paramagnetism occurs when a molecule has one or more unpaired electrons. The key tool is the molecular orbital (MO) diagram for homonuclear diatomic molecules (and closely related heteronuclear ones like NO). For second-period elements, the order of MOs is:

(For $O_2$ and beyond, the $\sigma_{2p_z}$ is lower than the $\pi_{2p}^*$ orbitals, but the key is filling electrons in order and applying Hund’s rule.)

We will determine the electron configuration for each species and check for unpaired electrons.

Step-by-step analysis

1. $N_2$ (14 electrons)

   Configuration: $(\sigma_{1s})^2 (\sigma_{1s}^*)^2 (\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\pi_{2p})^4 (\sigma_{2p_z})^2$

   All electrons paired → Diamagnetic.

2. $O_2$ (16 electrons)

   Configuration: … $(\sigma_{2p_z})^2 (\pi_{2p}^*)^2$

   The two $\pi^*$ electrons occupy separate orbitals (Hund’s rule) → 2 unpaired electrons → Paramagnetic.

3. $F_2$ (18 electrons)

   Configuration: … $(\pi_{2p}^*)^4$ → all paired → Diamagnetic.

4. $NO$ (15 electrons)

   Similar to $O_2^+$ (15 e⁻). MO diagram: … $(\pi_{2p}^*)^1$ → 1 unpaired electron → Paramagnetic.

5. $CO$ (14 electrons)

   Isoelectronic with $N_2$ → all paired → Diamagnetic.

6. $O_2^+$ (15 electrons)

   Remove one electron from $O_2$’s $\pi^*$ orbital → $(\pi_{2p}^*)^1$ → 1 unpaired electron → Paramagnetic.

7. $O_2^-$ (17 electrons)

   Add one electron to $O_2$’s $\pi^*$ → $(\pi_{2p}^*)^3$ → one orbital doubly occupied, one singly → 1 unpaired electron → Paramagnetic.

8. $O_2^{2-}$ (18 electrons)

   $(\pi_{2p}^*)^4$ → all paired → Diamagnetic.

9. $N_2^+$ (13 electrons)

   Remove one electron from $N_2$’s $\sigma_{2p_z}$ → one unpaired electron in $\sigma_{2p_z}$ → Paramagnetic.

10. $NO^+$ (14 electrons)

    Isoelectronic with $N_2$ → all paired → Diamagnetic.

11. $NO^-$ (16 electrons)

    Isoelectronic with $O_2$ → $(\pi_{2p}^*)^2$ → 2 unpaired electrons → Paramagnetic.

12. $H_2$ (2 electrons)

    $\sigma_{1s}^2$ → all paired → Diamagnetic.

Count of paramagnetic species:

$O_2$, $NO$, $O_2^+$, $O_2^-$, $N_2^+$, $NO^-$ → 6 species.

The key to this question lies in two decisions: the oxidation state of cobalt, and whether the ligand (fluoride) is strong-field or weak-field. Fluoride is a weak-field ligand — it does not force pairing of electrons. That choice determines the entire electronic arrangement and hence the hybridization and magnetic properties.

Let’s walk through it step by step.

1. Find the oxidation state of cobalt. The complex is $[\mathrm{CoF}_6]^{3-}$. Fluoride has a charge of $-1$ each, so six fluorides contribute $-6$. The overall charge is $-3$. Let the oxidation state of Co be $x$:

So cobalt is in the $+3$ oxidation state.

2. Write the electronic configuration of $\mathrm{Co}^{3+}$.

   Cobalt (atomic number 27) has ground-state configuration $[\mathrm{Ar}]\,3d^7 4s^2$. Removing three electrons (the two 4s electrons and one 3d electron) gives $\mathrm{Co}^{3+}$: $[\mathrm{Ar}]\,3d^6$.

3. Consider the ligand field.

   Fluoride ($\mathrm{F}^-$) is a weak-field ligand. In the spectrochemical series, $\mathrm{F}^-$ lies far to the left (small $\Delta$). This means the crystal field splitting energy $\Delta$ is small, so electrons prefer to occupy all five $d$ orbitals singly before pairing — the high-spin configuration.

4. Arrange the six $d$ electrons in the octahedral field.

   In an octahedral field, the $d$ orbitals split into $t_{2g}$ (lower energy, three orbitals) and $e_g$ (higher energy, two orbitals). With weak field, Hund’s rule applies:

   • First three electrons go into $t_{2g}$ with parallel spins.
   • Next two electrons go into $e_g$ with parallel spins.
   • The sixth electron goes into $t_{2g}$, pairing with one electron there. The result: $t_{2g}^4 e_g^2$ — four electrons in $t_{2g}$ (one pair, two unpaired) and two unpaired in $e_g$. Total unpaired electrons = 4.
   Watch out

   A common mistake is to assume $\mathrm{Co}^{3+}$ always forms low-spin $d^6$ (like in $[\mathrm{Co(NH_3)_6}]^{3+}$). That only happens with strong-field ligands like $\mathrm{NH_3}$ or $\mathrm{CN}^-$. Fluoride is weak-field, so the high-spin arrangement is correct.

5. Determine hybridization.

   With six ligands, the complex is octahedral. The central ion uses six orbitals for bonding. Since the $d$ electrons are in the $3d$ orbitals and the complex is high-spin, the inner $d$ orbitals are occupied and not available for hybridization. Therefore, the hybridization uses outer orbitals: one $4s$, three $4p$, and two $4d$ orbitals — giving $sp^3d^2$ hybridization.

   Tip

   If the complex were low-spin (strong-field ligand), two $3d$ orbitals would be vacant (after pairing), allowing $d^2sp^3$ hybridization (inner orbital complex). Here, with weak field, it’s outer orbital — $sp^3d^2$.

6. Determine magnetic nature.

   With four unpaired electrons, the complex is paramagnetic (attracted to a magnetic field). Diamagnetic would require all electrons paired, which is not the case.

The key idea here is that the same metal ion can have different numbers of unpaired electrons depending on the ligand field — weak field (high spin) vs strong field (low spin). The magnetic moment tells us exactly how many unpaired electrons are present, and we match that to the possible $d$-electron configurations.

The spin-only formula is $\mu = \sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons. Let’s decode what the given values mean.

1. For $\mu = 3.87$ BM:

   $\sqrt{n(n+2)} = 3.87$

   Squaring: $n(n+2) = 15$

   Solving $n^2 + 2n - 15 = 0$ gives $n = 3$ (the positive root). So complex X has 3 unpaired electrons.

2. For $\mu = 1.73$ BM:

   $\sqrt{n(n+2)} = 1.73$

   Squaring: $n(n+2) = 3$

   Solving $n^2 + 2n - 3 = 0$ gives $n = 1$. So complex Y has 1 unpaired electron.

So the same metal ion M$^{n+}$ can exist in one complex with 3 unpaired electrons and in another with just 1 unpaired electron. That means it must have a $d$-electron count that allows both a high-spin (weak field) and a low-spin (strong field) configuration.

Now check each option:

• (A) Mn$^{2+}$: $d^5$ configuration. High spin: 5 unpaired electrons ($\mu = \sqrt{35} \approx 5.92$ BM). Low spin: 1 unpaired electron ($\mu = \sqrt{3} \approx 1.73$ BM). But 3.87 BM (3 unpaired) is not possible for $d^5$ — it’s either 5 or 1. So not this.

• (B) Co$^{2+}$: $d^7$ configuration. High spin: 3 unpaired electrons ($\mu = \sqrt{15} \approx 3.87$ BM). Low spin: 1 unpaired electron ($\mu = \sqrt{3} \approx 1.73$ BM). This matches both values perfectly! But let’s check the others to be sure.

• (C) Fe$^{2+}$: $d^6$ configuration. High spin: 4 unpaired electrons ($\mu = \sqrt{24} \approx 4.90$ BM). Low spin: 0 unpaired electrons ($\mu = 0$ BM). Neither 3.87 nor 1.73 matches. So not this.

• (D) Cr$^{3+}$: $d^3$ configuration. Always 3 unpaired electrons ($\mu = \sqrt{15} \approx 3.87$ BM). Cannot give 1.73 BM. So not this.

So only Co$^{2+}$ ($d^7$) can give both 3.87 BM (high-spin, 3 unpaired) and 1.73 BM (low-spin, 1 unpaired).

The problem requires us to evaluate the correctness of four statements related to inorganic chemistry, covering topics like gemstone composition, metal carbonyl structures, and coordination complex properties (hybridization, geometry, and orbital type). The core concepts involve understanding oxidation states, $d$-electron configurations, ligand field strength, and how these factors dictate the electronic structure and geometry of coordination compounds.

Here's a detailed evaluation of each option:

1. Evaluating Option (A): Ruby composition

• Statement: Ruby is $\mathrm{Al_2O_3}$ containing $5\%$ $\mathrm{Cr^{3+}}$ ions.
• Concept: Ruby is a variety of the mineral corundum, which is aluminium oxide ($\mathrm{Al_2O_3}$). Its characteristic red color is indeed due to the presence of trace amounts of chromium(III) ions ($\mathrm{Cr^{3+}}$) substituting for $\mathrm{Al^{3+}}$ ions in the crystal lattice.
• Analysis: While the chemical formula $\mathrm{Al_2O_3}$ and the presence of $\mathrm{Cr^{3+}}$ ions are correct, the concentration of $\mathrm{Cr^{3+}}$ is typically very low in natural rubies, usually ranging from $0.5\%$ to $2\%$. A concentration of $5\%$ $\mathrm{Cr^{3+}}$ would be unusually high for a natural ruby and would likely result in a much darker, almost opaque, or brownish-red color, rather than the vibrant red typically associated with ruby.
• Conclusion: The statement is incorrect due to the specified concentration of $\mathrm{Cr^{3+}}$ ions.

2. Evaluating Option (B): $\mathrm{Mn_2(CO)_{10}}$ structure

• Statement: $\mathrm{Mn_2(CO)_{10}}$ contains two bridged carbonyl groups.
• Concept: This statement relates to the structure of dimeric metal carbonyls, which can often be predicted using the 18-electron rule. The 18-electron rule states that stable organometallic compounds often have 18 valence electrons around the central metal atom.
• Analysis:
  • Manganese (Mn) is a Group 7 element, so it has 7 valence electrons.
  • A carbonyl (CO) ligand is a 2-electron donor.
  • In $\mathrm{Mn_2(CO)_{10}}$, each Mn atom is bonded to five CO ligands.
  • Electrons contributed by ligands to one Mn atom: $5 \times 2 = 10$ electrons.
  • Total electrons around one Mn atom from its own valence electrons and ligands: $7 + 10 = 17$ electrons.
  • To satisfy the 18-electron rule, each Mn atom needs one more electron. This is achieved by forming a single metal-metal bond between the two Mn atoms.
  • The structure of $\mathrm{Mn_2(CO)_{10}}$ is $\mathrm{(CO)_5Mn-Mn(CO)_5}$, which consists of two $\mathrm{Mn(CO)_5}$ units linked by a direct Mn-Mn bond. There are no bridging carbonyl groups in this molecule.
• Conclusion: The statement is incorrect.

3. Evaluating Option (C): Inner and outer orbital complexes

• Statement: $[\mathrm{Co(NH_3)_6}]^{3+}$ is an inner orbital complex whereas $[\mathrm{Ni(NH_3)_6}]^{3+}$ is an outer orbital complex.

• Concept: This involves determining the oxidation state of the metal, its $d$-electron configuration, the nature of the ligand (strong or weak field), and then predicting the hybridization and whether it's an inner ($d^2sp^3$) or outer ($sp^3d^2$) orbital complex.

  • Inner orbital complex: Uses inner $d$ orbitals (e.g., $3d$ for a $3d$ series metal) for hybridization, typically $d^2sp^3$. This requires two empty $d$ orbitals.
  • Outer orbital complex: Uses outer $d$ orbitals (e.g., $4d$ for a $3d$ series metal) for hybridization, typically $sp^3d^2$.
  • Ligand field strength: $\mathrm{NH_3}$ is generally considered a strong field ligand, causing electron pairing in $d$ orbitals for certain configurations.

• Analysis for $[\mathrm{Co(NH_3)_6}]^{3+}$:

  • Let the oxidation state of Co be $x$. Since $\mathrm{NH_3}$ is a neutral ligand, $x + 6(0) = +3 \implies x = +3$.
  • Electronic configuration of Co: $[\mathrm{Ar}] 3d^7 4s^2$.
  • Electronic configuration of $\mathrm{Co^{3+}}$: $[\mathrm{Ar}] 3d^6$.
  • In an octahedral field, the $d$ orbitals split into $t_{2g}$ (lower energy) and $e_g$ (higher energy) sets.
  • $\mathrm{NH_3}$ is a strong field ligand, causing maximum pairing of electrons in the $t_{2g}$ orbitals.
  • For $d^6$, all six electrons pair up in the $t_{2g}$ orbitals: $t_{2g}^6 e_g^0$.
  • This leaves two empty $3d$ orbitals (the $e_g$ orbitals) available for hybridization.
  • Hybridization: $d^2sp^3$. This is an inner orbital complex.

• Analysis for $[\mathrm{Ni(NH_3)_6}]^{3+}$:

  • Let the oxidation state of Ni be $y$. Since $\mathrm{NH_3}$ is a neutral ligand, $y + 6(0) = +3 \implies y = +3$.
  • Electronic configuration of Ni: $[\mathrm{Ar}] 3d^8 4s^2$.
  • Electronic configuration of $\mathrm{Ni^{3+}}$: $[\mathrm{Ar}] 3d^7$.
  • $\mathrm{NH_3}$ is a strong field ligand.
  • For $d^7$ in an octahedral field, even with strong field ligands, the electrons will fill the $t_{2g}$ orbitals first, then the $e_g$ orbitals.
  • The $t_{2g}$ orbitals can accommodate 6 electrons. The 7th electron must go into an $e_g$ orbital.
  • Configuration: $t_{2g}^6 e_g^1$.
  • Since there is an electron in an $e_g$ orbital, there are no two empty $3d$ orbitals available for $d^2sp^3$ hybridization.
  • Therefore, the complex must use the outer $4d$ orbitals for hybridization.
  • Hybridization: $sp^3d^2$. This is an outer orbital complex.

• Conclusion: Both parts of the statement are correct.

4. Evaluating Option (D): Geometry of Ni complexes

• Statement: $[\mathrm{Ni(CN)_4}]^{2-}$, $[\mathrm{NiCl_4}]^{2-}$ both have tetrahedral geometry.

• Concept: This involves determining the oxidation state of the metal, its $d$-electron configuration, the nature of the ligand (strong or weak field), and predicting the hybridization and geometry for coordination number 4 complexes.

  • For coordination number 4, common geometries are tetrahedral ($sp^3$) and square planar ($dsp^2$).
  • Strong field ligands often favor square planar geometry for $d^8$ ions.
  • Weak field ligands often favor tetrahedral geometry.

• Analysis for $[\mathrm{Ni(CN)_4}]^{2-}$:

  • Let the oxidation state of Ni be $x$. $x + 4(-1) = -2 \implies x = +2$.
  • Electronic configuration of Ni: $[\mathrm{Ar}] 3d^8 4s^2$.
  • Electronic configuration of $\mathrm{Ni^{2+}}$: $[\mathrm{Ar}] 3d^8$.
  • $\mathrm{CN^-}$ is a very strong field ligand.
  • For a $d^8$ ion with a strong field ligand, square planar geometry is typically observed. The strong field causes pairing of electrons in the $d$ orbitals, leaving one $3d$ orbital empty for hybridization.
  • Hybridization: $dsp^2$. Geometry: Square planar.

• Analysis for $[\mathrm{NiCl_4}]^{2-}$:

  • Let the oxidation state of Ni be $y$. $y + 4(-1) = -2 \implies y = +2$.
  • Electronic configuration of $\mathrm{Ni^{2+}}$: $[\mathrm{Ar}] 3d^8$.
  • $\mathrm{Cl^-}$ is a weak field ligand.
  • For a $d^8$ ion with a weak field ligand, electron pairing does not occur to create an empty $3d$ orbital for $dsp^2$ hybridization.
  • Instead, the complex uses $4s$ and $4p$ orbitals.
  • Hybridization: $sp^3$. Geometry: Tetrahedral.

• Conclusion: $[\mathrm{Ni(CN)_4}]^{2-}$ is square planar, while $[\mathrm{NiCl_4}]^{2-}$ is tetrahedral. Therefore, the statement that both have tetrahedral geometry is incorrect.