Identify the homoleptic complexes with odd number of d electrons in the central metal. (A) [FeO₄]²⁻$[FeO_{4}]^{2-}$ (B) [Fe(CN)₆]³⁻$[Fe(CN)_{6}]^{3-}$ (C) [Fe(CN)₅NO]²⁻$[Fe(CN)_{5}NO]^{2-}$ (D) [CoCl₄]²⁻$[CoCl_{4}]^{2-}$ (E) [Co(H₂O)₃F₃]$[Co(H_{2}O)_{3}F_{3}]$ Choose the correct answer from the options given below:

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