Identify the diamagnetic octahedral complex ions from below; A. [mathrmMn(mathrmCN)_6]^3-$[\mathrm{Mn}(\mathrm{CN})_6]^{3-}$ B. [mathrmCo(mathrmNH_3)_6]^3+$[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$ C. [mathrmFe(mathrmCN)_6]^4-$[\mathrm{Fe}(\mathrm{CN})_6]^{4-}$ D. [mathrmCo(mathrmH_2mathrmO)_3mathrmF_3]$[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_3\mathrm{F}_3]$ Choose the correct answer from the options given below :

### Core Logic Let us systematically determine the hybridization configuration of each species: 1. **[textCo(NH_3)_6]^3+$[\text{Co(NH}_3)_6]^{3+}$**: textCo^3+$\text{Co}^{3+}$ has a 3d^6$3d^6$ configuration. Ammonia (textNH_3$\text{NH}_3$) acts as a Strong Field Ligand (SFL), forcing the pairing of 3d$3d$ electrons. This leaves two internal 3d$3d$ orbitals vacant, leading to an **inner orbital** d^2sp^3$d^2sp^3$ hybridization. 2. **textSF_6$\text{SF}_6$**: Central sulfur has 6 valence electrons, forming 6 single bonds. Steric number = 6$= 6$, resulting in a regular outer octahedral **sp^3d^2$sp^3d^2$** hybridization. 3. **[textCrF_6]^3-$[\text{CrF}_6]^{3-}$**: textCr^3+$\text{Cr}^{3+}$ has a 3d^3$3d^3$ configuration. The t_2g$t_{2g}$ subshell holds 3 unpaired electrons, leaving the two e_g$e_g$ orbitals empty regardless of ligand field strength. This results in a d^2sp^3$d^2sp^3$ hybridization. 4. **[textCoF_6]^3-$[\text{CoF}_6]^{3-}$**: textCo^3+$\text{Co}^{3+}$ has a 3d^6$3d^6$ configuration. Fluoride (F^-$F^-$) is a Weak Field Ligand (WFL) and cannot induce spin pairing. Thus, the complex utilizes outer shell 4d$4d$ orbitals, yielding an **outer orbital** **sp^3d^2$sp^3d^2$** configuration. 5. **[textMn(CN)_6]^3-$[\text{Mn(CN)}_6]^{3-}$**: textMn^3+$\text{Mn}^{3+}$ has a 3d^4$3d^4$ configuration. Cyanide (textCN^-$\text{CN}^-$) is a Strong Field Ligand (SFL), inducing pairing to leave two 3d$3d$ slots vacant, giving a d^2sp^3$d^2sp^3$ hybridization. 6. **[textMnCl_6]^3-$[\text{MnCl}_6]^{3-}$**: textMn^3+$\text{Mn}^{3+}$ has a 3d^4$3d^4$ configuration. Chloride (textCl^-$\text{Cl}^-$) is a Weak Field Ligand (WFL) and cannot cause pairing. It utilizes the outer 4d$4d$ shell, resulting in an **outer orbital** **sp^3d^2$sp^3d^2$** hybridization. Counting the outer-orbital sp^3d^2$sp^3d^2$ species: textSF_6$\text{SF}_6$, [textCoF_6]^3-$[\text{CoF}_6]^{3-}$, and [textMnCl_6]^3-$[\text{MnCl}_6]^{3-}$. Total count = 3$= 3$. ### Pattern Recognition Outer orbital complexes (sp^3d^2$sp^3d^2$) require weak field ligands (like F^-, Cl^-$F^-, Cl^-$) paired with metal configurations where internal d$d$-orbitals cannot be cleared by pairing (d^4, d^5, d^6$d^4, d^5, d^6$). SF_6$SF_6$ is a primary group molecule that always uses outer-shell d-orbitals. This brings our total to 3. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds Class 11 Chemistry: Chemical Bonding and Molecular Structure

### Core Logic Let us apply Valence Bond Theory (VBT) and crystal field rules to evaluate each coordination complex: * **A. [textNi(CO)_4]$[\text{Ni(CO)}_4]$**: Nickel is in the 0$0$ oxidation state (3d^8 4s^2$3d^8 4s^2$). Carbon monoxide (textCO$\text{CO}$) is a strong field ligand, forcing the 4s$4s$ electrons into the 3d$3d$ shell to produce a fully paired 3d^10$3d^{10}$ configuration. The vacant 4s$4s$ and three 4p$4p$ orbitals hybridize into an **sp^3$sp^3$ tetrahedral** geometry. All spins are paired, so mu = 0 text BM$\mu = 0 \text{ BM}$. Thus, textA rightarrow textIII$\text{A} \rightarrow \text{III}$.

* **B. [textNi(CN)_4]^2-$[\text{Ni(CN)}_4]^{2-}$**: Nickel is in the +2$+2$ state (3d^8$3d^8$). Cyanide (textCN^-$\text{CN}^-$) is a strong field ligand, forcing the pairing of the two unpaired 3d$3d$ electrons. This leaves one internal 3d$3d$ orbital vacant, leading to **dsp^2$dsp^2$ square planar** hybridization with zero unpaired electrons (mu = 0 text BM$\mu = 0 \text{ BM}$). Thus, textB rightarrow textII$\text{B} \rightarrow \text{II}$.

* **C. [textNiCl_4]^2-$[\text{NiCl}_4]^{2-}$**: Nickel is in the +2$+2$ state (3d^8$3d^8$). Chloride (textCl^-$\text{Cl}^-$) is a weak field ligand, leaving the two 3d$3d$ electrons unpaired (n = 2$n = 2$). The system adopts **sp^3$sp^3$ tetrahedral** hybridization with a spin-only moment of mu = sqrt2(2+2) = sqrt8 approx 2.8 text BM$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.8 \text{ BM}$. Thus, textC rightarrow textI$\text{C} \rightarrow \text{I}$.

* **D. [textMnBr_4]^2-$[\text{MnBr}_4]^{2-}$**: Manganese is in the +2$+2$ state (3d^5$3d^5$). Bromide (textBr^-$\text{Br}^-$) is a weak field ligand, preserving five unpaired parallel spins (n = 5$n = 5$). The geometry is **sp^3$sp^3$ tetrahedral** with a maximum spin-only moment of mu = sqrt5(5+2) = sqrt35 approx 5.9 text BM$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.9 \text{ BM}$. Thus, textD rightarrow textIV$\text{D} \rightarrow \text{IV}$.

### Step 1: Alignment Summary Consolidating our results: textA-III, B-II, C-I, D-IV$$\text{A-III, B-II, C-I, D-IV}$$ This matches Option (3). ### Pattern Recognition Nickel complexes provide classic benchmarks: Nickel zero tetracarbonyl is always tetrahedral diamagnetic. Nickel +2$+2$ tetracyanide is square planar diamagnetic due to strong ligand field pairing. Spotting these properties cuts down the problem solving time significantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds

### Related Formula mu = sqrtn(n+2)text B.M.$$mu = sqrt{n(n+2)}\text{ B.M.}$$ ### Core Logic In K_3[Fe(OH)_6]$K_{3}[Fe(OH)_{6}]$, iron is in the +3$+3$ oxidation state (Fe^3+ = 3d^5$Fe^{3+} = 3d^{5}$). Since OH^-$OH^{-}$ is a weak field ligand, no pairing of electrons takes place. The number of unpaired electrons (n$n$) is 5$5$. mu = sqrt5(5+2) = sqrt35 approx 5.92text B.M.$$mu = sqrt{5(5+2)} = sqrt{35} approx 5.92\text{ B.M.}$$ In K_4[Fe(OH)_6]$K_{4}[Fe(OH)_{6}]$, iron is in the +2$+2$ oxidation state (Fe^2+ = 3d^6$Fe^{2+} = 3d^{6}$). Since OH^-$OH^{-}$ is a weak field ligand, no pairing occurs. The number of unpaired electrons (n$n$) is 4$4$. mu = sqrt4(4+2) = sqrt24 approx 4.90text B.M.$$mu = sqrt{4(4+2)} = sqrt{24} approx 4.90\text{ B.M.}$$ ### Pattern Recognition Identify the ligand field strength first. OH^-$OH^{-}$ is a weak field ligand in the spectrochemical series, so it does not cause pairing in either Fe^2+$Fe^{2+}$ or Fe^3+$Fe^{3+}$ configurations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds

### Core Logic A complex is homoleptic if the metal is bound to only one kind of donor ligand group. * (A) [FeO_4]^2-$[FeO_4]^{2-}$ is homoleptic, but Fe^+6$Fe^{+6}$ corresponds to a 3d^2$3d^2$ (even) electronic configuration. * (B) [Fe(CN)_6]^3-$[Fe(CN)_6]^{3-}$ is homoleptic. Fe^+3$Fe^{+3}$ corresponds to a 3d^5$3d^5$ (odd) configuration. * (C) [Fe(CN)_5NO]^2-$[Fe(CN)_5NO]^{2-}$ is heteroleptic (contains two types of ligands). * (D) [CoCl_4]^2-$[CoCl_4]^{2-}$ is homoleptic. Co^+2$Co^{+2}$ corresponds to a 3d^7$3d^7$ (odd) configuration. * (E) [Co(H_2O)_3F_3]$[Co(H_2O)_3F_3]$ is heteroleptic. ### Pattern Recognition Filter by 'homoleptic' first to instantly eliminate multi-ligand mixed structures like options (C) and (E). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds

### Core Logic From the given list, the complexes exhibiting a distinct yellow color are K_3[Co(NO_2)_6]$K_{3}[Co(NO_{2})_{6}]$ and K_4[Fe(CN)_6]$K_{4}[Fe(CN)_{6}]$. Let's calculate the spin-only magnetic moments for these low-spin configurations: 1) For K_3[Co(NO_2)_6]$K_{3}[Co(NO_{2})_{6}]$, cobalt is in +3$+3$ oxidation state (Co^3+ = 3d^6$Co^{3+} = 3d^{6}$). In the presence of the strong ligand field (NO_2^-$NO_2^-$), all six electrons pair up completely in the t_2g$t_{2g}$ orbitals: t_2g^6 e_g^0 implies n = 0 text unpaired electrons implies mu = 0text BM$$t_{2g}^6 e_g^0 implies n = 0 \text{ unpaired electrons} implies mu = 0\text{ BM}$$

2) For K_4[Fe(CN)_6]$K_{4}[Fe(CN)_{6}]$, iron is in +2$+2$ oxidation state (Fe^2+ = 3d^6$Fe^{2+} = 3d^{6}$). In the strong field of cyanide ligands (CN^-$CN^-$), pairing is complete: t_2g^6 e_g^0 implies n = 0 text unpaired electrons implies mu = 0text BM$$t_{2g}^6 e_g^0 implies n = 0 \text{ unpaired electrons} implies mu = 0\text{ BM}$$ Therefore, the sum of their spin-only magnetic moments is 0 + 0 = 0$0 + 0 = 0$. ### Pattern Recognition Low-spin d^6$d^6$ octahedral complexes always yield a fully closed-shell t_2g^6$t_{2g}^6$ arrangement with zero unpaired electrons, leading deterministically to a magnetic moment of 0 BM. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds