Q.Why is $Cr^{2+}$ reducing and $Mn^{3+}$ oxidising when both have $d^4$ configuration?

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). …

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) …

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. …

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. …

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: $t_{2g}^6 e_g^1$. …

The key idea is simple: a substance is paramagnetic if it contains unpaired electrons. Diamagnetic substances have all electrons paired. So to decide which of these molecules or ions is not paramagnetic, we need to determine the number of unpaired electrons in each. The most reliable tool for small diatomic molecules and ions is molecular orbital (MO) theory, which gives the electron configuration and tells us exactly how many electrons are unpaired.

For second-period homonuclear diatomics (like $\mathrm{O_2}$ and its ions), the MO energy order is:

For heteronuclear diatomics like $\mathrm{CO}$ and $\mathrm{NO}$, the same general filling pattern applies, with the $\pi_{2p}$ orbitals lying below $\sigma_{2p_z}$.

Let's work through each species step by step.

1. $\mathrm{O_2^-}$ (superoxide ion) Neutral $\mathrm{O_2}$ has 16 electrons. Its MO configuration is:

That gives two unpaired electrons in the $\pi^*$ orbitals — $\mathrm{O_2}$ is paramagnetic.

$\mathrm{O_2^-}$ has one extra electron (total 17). That electron goes into one of the $\pi^*$ orbitals, giving:

So there is one unpaired electron. Hence $\mathrm{O_2^-}$ is paramagnetic.

2. $\mathrm{O_2^+}$

   Neutral $\mathrm{O_2}$ has the configuration $\pi_{2p_x}^{*1}\pi_{2p_y}^{*1}$ in its antibonding $\pi^*$ orbitals (16 electrons total). Removing one electron to form $\mathrm{O_2^+}$ (15 electrons) removes one of these $\pi^*$ electrons, leaving $\pi_{2p_x}^{*1}\pi_{2p_y}^{*0}$ (or the reverse). This leaves one unpaired electron, so $\mathrm{O_2^+}$ is paramagnetic.

3. $\mathrm{CO}$ (carbon monoxide)

   $\mathrm{CO}$ has 14 electrons total (6 from C, 8 from O). Its MO configuration is:

   $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^2$ …

We need to match each complex with the correct $t_{2g}$ and $e_g$ electron configuration of the central metal ion. This requires knowing:

• The oxidation state of the metal (to get the d-electron count).
• The geometry (octahedral or tetrahedral) and the ligand field strength (strong-field vs. weak-field), which determines whether the complex is low-spin or high-spin.

Let’s work through each complex step by step.

1. Complex A: $[\text{Co(NH}_3)_6]^{3+}$

   • Cobalt is in the +3 oxidation state (each NH₃ is neutral). Co atomic number = 27, so Co³⁺ has $3d^6$ configuration.
   • NH₃ is a strong-field ligand (spectrochemical series: NH₃ > H₂O > F⁻). In an octahedral field, strong-field ligands cause a large splitting, so electrons pair up in the lower $t_{2g}$ orbitals.
   • For $d^6$ in a strong octahedral field: all six electrons fill the $t_{2g}$ set (three orbitals, each with two electrons) → $t_{2g}^6 e_g^0$.
   • This matches II (and also IV, but note II and IV are identical in the list; we’ll see which one fits elsewhere). So A → II.

2. Complex B: $[\text{CoF}_6]^{3-}$

   • Again Co is +3 (F⁻ is –1, total charge –3 gives Co³⁺). So again $d^6$.
   • F⁻ is a weak-field ligand. In a weak octahedral field, the splitting is small, so electrons occupy all five d-orbitals singly before pairing (Hund’s rule).
   • For $d^6$ weak-field: first five electrons go one per orbital (three $t_{2g}$, two $e_g$), then the sixth electron pairs in a $t_{2g}$ orbital → $t_{2g}^4 e_g^2$.
   • This matches III. So B → III.

3. Complex C: $[\text{Ni(CO)}_4]$

   • CO is a neutral ligand, and the complex is neutral. Nickel is in the 0 oxidation state. Ni atomic number = 28, so Ni⁰ has $3d^8 4s^2$, but in complexes the 4s electrons are lost first; effectively Ni⁰ is $d^{10}$ (since 4s² electrons are promoted or involved in bonding, but the common treatment: Ni(0) in carbonyls is $d^{10}$).
   • CO is a strong-field ligand, but more importantly, Ni(CO)₄ is tetrahedral (not octahedral). In tetrahedral geometry, the splitting is reversed and smaller: the $e$ set (lower energy) and $t_2$ set (higher energy).
   • For a $d^{10}$ ion, all orbitals are completely filled regardless of geometry. In tetrahedral notation: $e^4 t_2^6$. But the given options use octahedral notation $t_{2g}$ and $e_g$. For a tetrahedral complex, we often still use the same labels but with the understanding that the $t_2$ set is higher. However, the problem lists only octahedral-style configurations.
   • Since Ni(CO)₄ is diamagnetic and $d^{10}$, the only way to represent it in the given notation is $t_{2g}^6 e_g^0$ (all electrons paired in lower set? That doesn’t fit). Wait — careful: For a $d^{10}$ system, all d-orbitals are full. In octahedral field, that would be $t_{2g}^6 e_g^4$, but that’s not an option. The options only have $t_{2g}^x e_g^y$ with $x+y = 6$ or 5 or 4. So something is off.
   • Actually, Ni(CO)₄ is tetrahedral, and the metal is Ni(0) with $d^{10}$. In tetrahedral splitting, the $e$ set (lower) holds 4 electrons, the $t_2$ set (higher) holds 6. But the problem’s options are clearly in octahedral notation. The trick: For $d^{10}$, the configuration is completely filled, so it is often written as $t_{2g}^6 e_g^4$ in octahedral, but that’s not listed. However, look at the options: IV is $t_{2g}^6 e_g^0$. That would imply only 6 d-electrons, not 10.
   • This suggests that the problem expects us to consider the effective d-electron count after considering the strong-field CO ligands causing pairing? No, that’s not right.
   • Let’s re-evaluate: Ni(CO)₄ is a classic example of a $d^{10}$ tetrahedral complex. But the given configurations in List II all sum to 6 electrons (since $t_{2g}$ and $e_g$ together hold 6 in octahedral). So they are not meant for $d^{10}$. Perhaps the problem considers Ni in a different oxidation state? No, Ni(CO)₄ is Ni(0).
   • Wait — maybe the complex is actually $[\text{Ni(CO)}_4]$ but the metal is considered as Ni²⁺? That would be wrong.
   • Let’s check the options: C is matched with IV in options (A) and (D). IV is $t_{2g}^6 e_g^0$. That is a $d^6$ low-spin configuration. Could it be that Ni(CO)₄ is actually diamagnetic and the 10 d-electrons are all paired? In tetrahedral, the $e$ set holds 4, $t_2$ holds 6, so the configuration is $e^4 t_2^6$. If we relabel $e$ as $e_g$ and $t_2$ as $t_{2g}$ (which is not correct but sometimes done loosely), then it would be $e_g^4 t_{2g}^6$, which is not $t_{2g}^6 e_g^0$. …

Concept & Intuition

The spin-only magnetic moment is $\mu=\sqrt{n(n+2)}$ Bohr magnetons, where $n$ is the number of unpaired electrons. It depends only on $n$, so two ions match when their high-spin $d$-configurations give the same number of unpaired electrons. A weak ligand is a weak-field ligand, so electrons fill orbitals following Hund's rule (high-spin) before pairing.

Step-by-step

Count unpaired electrons for each ion (high-spin):

• $\mathrm{Sc^{3+}}$: $[\text{Ar}]3d^0 \Rightarrow n=0$
• $\mathrm{Ti^{2+}}$: $3d^2 \Rightarrow n=2$
• $\mathrm{V^{2+}}$: $3d^3 \Rightarrow n=3$
• $\mathrm{Co^{2+}}$: $3d^7$ (high-spin $t_{2g}^5 e_g^2$) $\Rightarrow n=3$
• $\mathrm{Ni^{2+}}$: $3d^8 \Rightarrow n=2$
• $\mathrm{Cu^{2+}}$: $3d^9 \Rightarrow n=1$

Now test the pairs: …

The question asks which set of elements, in their +2 oxidation state, have a completely filled $d^{10}$ electronic configuration. This is a classic test of your understanding of how transition and post-transition metals lose electrons, and which ones have a stable, full $d$-subshell after losing two electrons.

Let’s break it down.

1. What does $d^{10}$ in the +2 state mean?

   For an element to have a $d^{10}$ configuration when it is in the +2 oxidation state, the neutral atom must have exactly two more electrons than the $d^{10}$ core. Those two electrons are almost always the outermost $s$-electrons, because $s$-electrons are lost before $d$-electrons in transition metals. So the neutral atom’s electron configuration ends in $...\,d^{10}s^2$. When it loses both $s$-electrons, the ion becomes $...\,d^{10}$.

2. Which elements have a $d^{10}s^2$ configuration in their ground state?

   This is the hallmark of group 12 elements: zinc (Zn), cadmium (Cd), and mercury (Hg). Their configurations are:

   • Zn: $[\text{Ar}]\,3d^{10}4s^2$
   • Cd: $[\text{Kr}]\,4d^{10}5s^2$
   • Hg: $[\text{Xe}]\,4f^{14}5d^{10}6s^2$ In each case, the $d$-subshell is already full, and the two $s$-electrons are the valence electrons. Losing both gives a $d^{10}$ ion: $\text{Zn}^{2+}$, $\text{Cd}^{2+}$, $\text{Hg}^{2+}$.

3. Check the other elements mentioned in the options.

   • Cu (copper): Neutral Cu is $[\text{Ar}]\,3d^{10}4s^1$. In the +2 state, it loses the $4s$ electron and one $3d$ electron, giving $3d^9$ — not $d^{10}$.
   • Ni (nickel): Neutral Ni is $[\text{Ar}]\,3d^8 4s^2$. In the +2 state, it loses both $4s$ electrons, giving $3d^8$ — not $d^{10}$.
   • Au (gold): Neutral Au is $[\text{Xe}]\,4f^{14}5d^{10}6s^1$. In the +2 state, it loses the $6s$ electron and one $5d$ electron, giving $5d^9$ — not $d^{10}$. …

The magnetic moment of a transition‑metal complex tells us directly how many unpaired electrons the metal ion has. The formula that connects them is the spin‑only formula:

where $n$ is the number of unpaired electrons and $\mu$ is the magnetic moment in Bohr magnetons (BM). This formula works well for first‑row transition metals because orbital angular momentum is often quenched by the ligand field.

Given $\mu = 5.92\ \text{BM}$, we solve for $n$:

Squaring both sides: