The calculated spin-only magnetic moments of K₃[Fe(OH)₆]$K_{3}[Fe(OH)_{6}]$ and K₄[Fe(OH)₆]$K_{4}[Fe(OH)_{6}]$ respectively are: (1) 4.90 and 4.90 B.M. (2) 5.92 and 4.90 B.M. (3) 3.87 and 4.90 B.M. (4) 4.90 and 5.92 B.M.

Related Formula

Core Logic

Let's find the oxidation state of Manganese in the complex [MnCl₆]³⁻$[\mathrm{MnCl}_6]^{3-}$:

Thus, manganese is in the +3$+3$ oxidation state: Mn³⁺ = [Ar]3d⁴$\mathrm{Mn^{3+}} = [\mathrm{Ar}]3\mathrm{d}^4$.

Step 1: Determine Ligand Splitting and Orbitals

Cl^-$\mathrm{Cl^-}$ is a weak-field ligand (WFL). Consequently, crystal field splitting is small (Δₒ < P$\Delta_o < P$), and no pairing of the 3d$3\mathrm{d}$ electrons occurs.

The distribution of the 4$4$ electrons in the 3d$3\mathrm{d}$ orbitals remains high-spin:

Because the inner 3d$3\mathrm{d}$ orbitals are not empty (since they contain 4 singly occupied orbitals), the complex must utilize the outer 4d$4\mathrm{d}$ orbitals for hybridization.

Step 2: Assign Hybridization

The vacant outer orbitals used for bonding are one 4s$4\mathrm{s}$, three 4p$4\mathrm{p}$, and two 4d$4\mathrm{d}$ orbitals, which hybridize to form six sp³d²$\mathrm{sp^3d^2}$ hybrid orbitals.

Since there are four unpaired electrons, the complex is paramagnetic with four unpaired electrons.

Pattern Recognition

Whenever WFL (like halides Cl^-$\mathrm{Cl^-}$, F^-$\mathrm{F^-}$) are present with octahedral transition metal complexes with d⁴$\mathrm{d^4}$ to d⁷$\mathrm{d^7}$ configuration, they always yield high-spin, outer orbital complexes with sp³d²$\mathrm{sp^3d^2}$ hybridization.

Chapter Mix

Class 12 Chemistry: Coordination Compounds

Related Formula

Core Logic

This question requires three distinct sequential conceptual steps:

• Determine the count n$n$ of diamagnetic complexes.
• Identify which of the metal ions (Ni, Zn, Mn, Cu) has the lowest enthalpy of atomisation.
• Compute the spin-only magnetic moment of that metal in its +n$+n$ state.

Step 1: Count Diamagnetic Complexes to find n

• K₂[NiCl₄]$\mathrm{K_2[NiCl_4]}$: Ni²⁺ = 3d⁸$\mathrm{Ni^{2+}} = 3\mathrm{d^8}$. Weak field chloride ligand leads to 2 unpaired electrons $\implies$ Paramagnetic.
• [Zn(H₂O)₆]Cl₂$\mathrm{[Zn(H_2O)_6]Cl_2}$: Zn²⁺ = 3 d¹⁰$\mathrm{Zn^{2+}} = 3\mathrm{d^{10}}$. Completely filled subshell $\implies$ Diamagnetic.
• K₃[Mn(CN)₆]$\mathrm{K_3[Mn(CN)_6]}$: Mn³⁺ = 3d⁴$\mathrm{Mn^{3+}} = 3\mathrm{d^4}$. Strong field cyanide ligand gives low-spin state with 2 unpaired electrons $\implies$ Paramagnetic.
• [Cu(PPh₃)₃I]$\mathrm{[Cu(PPh_3)_3I]}$: Cu^+$\mathrm{Cu^+}$ = 3 d¹⁰$3\mathrm{d^{10}}$. Completely filled subshell $\implies$ Diamagnetic.

Thus, there are exactly 2 diamagnetic complexes: n = 2$n = 2$.

Step 2: Identify Metal with Lowest Enthalpy of Atomisation

Among the transition metals of the 3d series (Ni, Zn, Mn, Cu), Zinc (Zn) has the lowest enthalpy of atomisation (126~ kJ~mol⁻¹$126~\mathrm{kJ~mol^{-1}}$). This is because Zinc has a fully occupied d-subshell (3 d¹⁰4s²$3\mathrm{d^{10}}4\mathrm{s^2}$) and lacks any unpaired d-electrons to participate in metallic bonding.

Step 3: Calculate Spin-only Magnetic Moment of M^n+

With M = Zinc$M = \text{Zinc}$ and n = 2$n = 2$, the ion is Zn²⁺$\mathrm{Zn^{2+}}$.

Electronic configuration of Zn²⁺$\mathrm{Zn^{2+}}$ is [Ar]3 d¹⁰$[\mathrm{Ar}]3\mathrm{d^{10}}$, which contains zero unpaired electrons (nunpaired = 0$n_{\text{unpaired}} = 0$).

Therefore, the spin-only magnetic moment is:

Pattern Recognition

Zinc is always a unique outlier in the d-block. Because it has a completely filled d¹⁰$d^{10}$ shell in both its atomic and +2$+2$ oxidation states, it exhibits no d-orbital metallic bonding (leading to lowest melting point, boiling point, and atomisation enthalpy in the 3d series) and is always diamagnetic.

Chapter Mix

Class 12 Chemistry: Coordination Compounds Class 12 Chemistry: d- and f-Block Elements

Related Formula

Crystal Field Splitting configuration rule for strong field ligands:

Core Logic

Let's isolate properties step-by-step:

• Among the given first-row elements (Mn, Cr, Co, Fe$\mathrm{Mn, Cr, Co, Fe}$), Cobalt (Co$\mathrm{Co}$) possesses the highest standard electrode potential value for the 3+/2+$3+/2+$ couple:

• The oxidation state configuration within complex [Co(CN)₆]⁴⁻$[\mathrm{Co}(\mathrm{CN})_6]^{4-}$ is Co²⁺$\mathrm{Co^{2+}}$, which has a d⁷$\mathrm{d^7}$ valence configuration.
• Cyanide (CN^-$\mathrm{CN^-}$) acts as a strong field ligand, forcing maximum electron pairing within the lower energy levels.

Step 1: Subshell Filling Matrix

Distribute 7 electrons across the split crystal field levels:

• First 6 electrons fill the lower t2g$\mathrm{t_{2g}}$ levels completely, forming paired tracks.
• The 7th electron has no choice but to step up to the higher level.

This structural splitting is visualized below:



Therefore, the number of electrons present in the eg$\mathrm{e_g}$ orbital block is exactly 1.

Pattern Recognition

Always double check the specific oxidation state value: [Co(CN)₆]⁴⁻$[\mathrm{Co}(\mathrm{CN})_6]^{4-}$ gives Co²⁺$\mathrm{Co^{2+}}$ (d⁷$\mathrm{d^7}$), whereas [Co(CN)₆]³⁻$[\mathrm{Co}(\mathrm{CN})_6]^{3-}$ would be Co³⁺$\mathrm{Co^{3+}}$ (d⁶$\mathrm{d^6}$), which has zero electrons in its eg$\mathrm{e_g}$ level.

Chapter Mix

Class 12 Chemistry: Coordination Compounds Class 12 Chemistry: d- and f-Block Elements

Related Formula

Crystal field splitting values relation for matching configuration choices:

Core Logic

Let's verify both rules based on Crystal Field Theory principles:

• Statement I: In octahedral configurations, if pairing penalty energy P$P$ exceeds field split magnitude Δ₀$\Delta_0$, electrons prefer moving to upper sub-shells, creating high-spin states. Conversely, if Δ₀ > P$\Delta_0 > P$, forced pairing occurs, creating low-spin complexes. (Statement I is accurate).
• Statement II: Because tetrahedral configurations separate by an extremely narrow gap magnitude Δₜ$\Delta_{\mathrm{t}}$ (about half of octahedral field splits), the value almost never exceeds standard pairing energy P$P$. Electrons consistently choose higher sub-levels rather than pairing up, meaning low-spin arrangements are extremely rare. (Statement II is accurate).

Step 1: Verdict

Therefore, both Statement I and Statement II are correct.

Pattern Recognition

Tetrahedral configurations are systematically assumed to be high-spin unless special structural properties dictate otherwise, due to the Δₜ < P$\Delta_{\mathrm{t}} < P$ constraint.

Chapter Mix

Class 12 Chemistry: Coordination Compounds

Related Formula

According to Crystal Field Theory (CFT):

• A complex is diamagnetic if all electrons are paired up (unpaired electrons, n=0$n=0$).
• Strong field ligands (like CN^-$\mathrm{CN}^-$, NH₃$\mathrm{NH}_3$ with Co³⁺$Co^{3+}$) cause pairing of electrons if Δₒ > P$\Delta_o > P$.

{{SOLUTION_IMG}}

Core Logic

Analyze each complex:

• A. [Mn(CN)₆]³⁻$[\mathrm{Mn}(\mathrm{CN})_6]^{3-}$:
• Mn³⁺$Mn^{3+}$ has d⁴$d^4$ configuration.
• Strong field ligand CN^-$\mathrm{CN}^-$ causes pairing in t2g$t_{2g}$ orbitals: t2g⁴ eg⁰$t_{2g}^4 e_g^0$.
• There are 2 unpaired electrons arrow$\rightarrow$ Paramagnetic.
• B. [Co(NH₃)₆]³⁺$[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$:
• Co³⁺$Co^{3+}$ has d⁶$d^6$ configuration.
• NH₃$\mathrm{NH}_3$ acts as strong field ligand with Co³⁺$Co^{3+}$, causing complete pairing: t2g⁶ eg⁰$t_{2g}^6 e_g^0$.
• No unpaired electrons arrow$\rightarrow$ Diamagnetic.

Step 1: Analyze complexes C and D

• C. [Fe(mathrmCN)₆]⁴⁻$[\mathrm{Fe}(mathrm{CN})_6]^{4-}$:
• Fe²⁺$Fe^{2+}$ has d⁶$d^6$ configuration.
• Strong field ligand CN^-$\mathrm{CN}^-$ causes complete pairing: t2g⁶ eg⁰$t_{2g}^6 e_g^0$.
• No unpaired electrons arrow$\rightarrow$ Diamagnetic.
• D. [Co(H₂O)₃F₃]$[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_3\mathrm{F}_3]$:
• Co³⁺$Co^{3+}$ has d⁶$d^6$ configuration.
• Weak field ligands (F^-$\mathrm{F}^-$, H₂O$\mathrm{H}_2\mathrm{O}$) do not cause pairing: t2g⁴ eg²$t_{2g}^4 e_g^2$.
• There are 4 unpaired electrons arrow$\rightarrow$ Paramagnetic.

Step 2: Conclusion

Only complexes B and C are diamagnetic, matching Option (4).

Pattern Recognition

Octahedral d⁶$d^6$ ions (such as Co³⁺$Co^{3+}$ or Fe²⁺$Fe^{2+}$) coupled with strong-field ligands are exceptionally stable and always form low-spin, fully paired, diamagnetic complexes (t2g⁶ eg⁰$t_{2g}^6 e_g^0$).

Chapter Mix

Class 12 Chemistry: Coordination Compounds