Consider the following low-spin complexes K₃[Co(NO₂)₆]$K_{3}[Co(NO_{2})_{6}]$, K₄[Fe(CN)₆]$K_{4}[Fe(CN)_{6}]$, K₃[Fe(CN)₆]$K_{3}[Fe(CN)_{6}]$, Cu₂[Fe(CN)₆]$Cu_{2}[Fe(CN)_{6}]$ and Zn₂[Fe(CN)₆]$Zn_{2}[Fe(CN)_{6}]$. The sum of the spin-only magnetic moment values of complexes having yellow colour is ________ B.M. (answer is nearest integer)

Core Logic

In K₃[Co(CO₃)₃]$K_3[Co(CO_3)_3]$, the oxidation state of Co is +3$+3$ (3d⁶$3d^6$ configuration). Since carbonate is a weak field ligand, electrons do not pair up. Hybrdisation is sp³d²$sp^3d^2$ (octahedral), with 4$4$ unpaired electrons.

Thus, Statement I is true.

For Statement II: [Ni(CN)₄]²⁻$[Ni(CN)_4]^{2-}$: Ni²⁺$Ni^{2+}$ (3d⁸$3d^8$). CN^-$CN^-$ is a strong field ligand. Pairing occurs. Hybridisation is dsp²$dsp^2$ (square planar), 0$0$ unpaired electrons, μ = 0 B.M.$\mu = 0 \text{ B.M.}$ [MnBr₄]²⁻$[MnBr_4]^{2-}$: Mn²⁺$Mn^{2+}$ (3d⁵$3d^5$). Br^-$Br^-$ is a weak field ligand. sp³$sp^3$ hybridised (tetrahedral), 5$5$ unpaired electrons, μ = 5.9 B.M.$\mu = 5.9 \text{ B.M.}$ [CoF₆]³⁻$[CoF_6]^{3-}$: Co³⁺$Co^{3+}$ (3d⁶$3d^6$). F^-$F^-$ is a weak field ligand. sp³d²$sp^3d^2$ hybridised (octahedral), 4$4$ unpaired electrons, μ = 4.9 B.M.$\mu = 4.9 \text{ B.M.}$ Thus, Statement II is also true.

Step 1: Final Conclusion

Both Statement I and Statement II are true.

Pattern Recognition

Check the ligand strength: CO₃²⁻$CO_3^{2-}$ and Halogens are WFL (outer orbital complexes, high spin), while CN^-$CN^-$ is SFL (inner orbital, low spin). Calculate unpaired electrons n$n$, then use √(n(n+2))$\sqrt{n(n+2)}$.

Chapter Mix

Class 12 Chemistry: Coordination Compounds

Core Logic

The 4 L$4 \text{ L}$ solution contains: 0.4 mol$0.4 \text{ mol}$ of [Co(NH₃)₅SO₄]Br$[Co(NH_3)_5SO_4]Br$ $\implies$ Gives 0.4 mol Br^-$0.4 \text{ mol } Br^-$ in solution. 0.4 mol$0.4 \text{ mol}$ of [Co(NH₃)₅Br]SO₄$[Co(NH_3)_5Br]SO_4$ $\implies$ Gives 0.4 mol SO₄²⁻$0.4 \text{ mol } SO_4^{2-}$ in solution.

The solution is divided into two 2 L$2 \text{ L}$ portions. Each 2 L$2 \text{ L}$ portion will contain exactly half the moles of each species: Moles of Br^-$Br^-$ in 2 L = 0.2 mol$2 \text{ L} = 0.2 \text{ mol}$ Moles of SO₄²⁻$SO_4^{2-}$ in 2 L = 0.2 mol$2 \text{ L} = 0.2 \text{ mol}$

Step 1: Reaction with Silver Nitrate

When the first 2 L$2 \text{ L}$ of the mixture reacts with excess AgNO₃$AgNO_3$:

Since there are 0.2 mol$0.2 \text{ mol}$ of free Br^-$Br^-$ ions, 0.2 mol$0.2 \text{ mol}$ of AgBr$AgBr$ (Y) will be formed.

Step 2: Reaction with Barium Chloride

When the remaining 2 L$2 \text{ L}$ of the mixture reacts with excess BaCl₂$BaCl_2$:

Since there are 0.2 mol$0.2 \text{ mol}$ of free SO₄²⁻$SO_4^{2-}$ ions, 0.2 mol$0.2 \text{ mol}$ of BaSO₄$BaSO_4$ (Z) will be formed.

Step 3: Final Conclusion

Checking the options: (1) 0.2 mol$0.2 \text{ mol}$ of 'Z' is formed $\implies$ True. (2) 'Y' is BaSO₄$BaSO_4$ and 'Z' is AgBr$AgBr$ $\implies$ False, they are reversed. (3) 0.4 mol$0.4 \text{ mol}$ of 'Z' is formed $\implies$ False, it's 0.2 mol$0.2 \text{ mol}$. (4) 0.1 mol$0.1 \text{ mol}$ of 'Y' is formed $\implies$ False, it's 0.2 mol$0.2 \text{ mol}$.

Pattern Recognition

Only the counter-ions (outside the coordination sphere) ionize in water and participate in precipitation reactions. Always account for volume fractions when a solution is split.

Chapter Mix

Class 12 Chemistry: Coordination Compounds Class 11 Chemistry: Some Basic Concepts of Chemistry

Core Logic

Analyzing Statement I: [CoF₆]³⁻$[CoF_6]^{3-}$: Co³⁺$Co^{3+}$ is 3d⁶$3d^6$. F is WFL $\implies$ high spin, 4$4$ unpaired electrons (paramagnetic). [TiF₆]³⁻$[TiF_6]^{3-}$: Ti³⁺$Ti^{3+}$ is 3d¹$3d^1$. 1$1$ unpaired electron (paramagnetic). V₂O₅$V_2O_5$: V⁵⁺$V^{5+}$ is 3d⁰$3d^0$. 0$0$ unpaired electrons (diamagnetic). [Fe(CN)₆]³⁻$[Fe(CN)_6]^{3-}$: Fe³⁺$Fe^{3+}$ is 3d⁵$3d^5$. CN is SFL $\implies$ low spin, 1$1$ unpaired electron (paramagnetic). Total paramagnetic species = 3. Statement I is true.

Analyzing Statement II: K₄[Fe(CN)₆]$K_4[Fe(CN)_6]$: Fe²⁺$Fe^{2+}$ is 3d⁶$3d^6$. SFL t2g⁶ eg⁰$\implies t_{2g}^6 e_g^0$, 0$0$ unpaired electrons. K₃[Fe(CN)₆]$K_3[Fe(CN)_6]$: Fe³⁺$Fe^{3+}$ is 3d⁵$3d^5$. SFL t2g⁵ eg⁰$\implies t_{2g}^5 e_g^0$, 1$1$ unpaired electron. [Fe(H₂O)₆]SO₄ · H₂O$[Fe(H_2O)_6]SO_4 \cdot H_2O$: Fe²⁺$Fe^{2+}$ is 3d⁶$3d^6$. H₂O$H_2O$ is WFL t2g⁴ eg²$\implies t_{2g}^4 e_g^2$, 4$4$ unpaired electrons. [Fe(H₂O)₆]Cl₃$[Fe(H_2O)_6]Cl_3$: Fe³⁺$Fe^{3+}$ is 3d⁵$3d^5$. H₂O$H_2O$ is WFL t2g³ eg²$\implies t_{2g}^3 e_g^2$, 5$5$ unpaired electrons. The order 0 < 1 < 4 < 5$0 < 1 < 4 < 5$ is correct. Statement II is true.

Step 1: Final Conclusion

Both Statement I and Statement II are true.

Pattern Recognition

Identify the oxidation state, count d$d$-electrons, apply spectrochemical series to determine pairing, then count unpaired electrons to establish diamagnetic (n=0$n=0$) vs paramagnetic (n>0$n>0$).

Chapter Mix

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

Related Formula

Core Logic

Wavelength of light absorbed increases as the Crystal Field Splitting Energy (C.F.S.E, Δₒ$\Delta_o$) of the complex decreases. According to the spectrochemical series, the ligand field strength follows the order: CN^- > NH₃ > H₂O > F^-$CN^- > NH_3 > H_2O > F^-$

Based on this: Ligand field strength $\uparrow$ ; C.F.S.E $\uparrow$ ; Absorbed wavelength (λ$\lambda$) $\downarrow$

[Co(CN)₆]³⁻$[Co(CN)_{6}]^{3-}$ (III) has the strongest ligand, so maximum CFSE and minimum λ$\lambda$. [CoF₆]³⁻$[CoF_{6}]^{3-}$ (V) has the weakest ligand, so least CFSE and maximum λ$\lambda$.

Comparing (NH₃)₆$(NH_3)_6$ (I) vs (NH₃)₅(H₂O)$(NH_3)_5(H_2O)$ (IV) vs (H₂O)₆$(H_2O)_6$ (II): Since NH₃ > H₂O$NH_3 > H_2O$, the field strength decreases as we replace NH₃$NH_3$ with H₂O$H_2O$. Thus, field strength order: III > I > IV > II > V Correct absorbed wavelength (λ$\lambda$) order will be the reverse: III < I < IV < II < V

Pattern Recognition

Strong field ligands cause high splitting (large Δ$\Delta$), meaning they absorb higher energy photons, which correspond to shorter wavelengths. Remember: λ ∝ 1/field strength$\lambda \propto 1/\text{field strength}$.

Chapter Mix

Class 12 Chemistry: Coordination Compounds

Core Logic

Since the complex behaves as a 1:2 electrolyte, it must dissociate into 3 ions: 1 complex cation and 2 counter anions. This means two chloride ions (Cl^-$Cl^-$) must reside outside the coordination sphere. The structural formula is: [Cr(H₂O)₅Cl]Cl₂ ₂O$[\mathrm{Cr}(\mathrm{H}_{2}\mathrm{O})_{5}\mathrm{Cl}]\mathrm{Cl}_{2}\cdot\mathrm{H}_{2}\mathrm{O}$. Reaction with silver nitrate: [Cr(H₂O)₅Cl]Cl₂ ₂O + 2AgNO₃ arrow 2AgCl + complex nitrate$[\mathrm{Cr}(\mathrm{H}_{2}\mathrm{O})_{5}\mathrm{Cl}]\mathrm{Cl}_{2}\cdot\mathrm{H}_{2}\mathrm{O} + 2\mathrm{AgNO}_{3} \rightarrow 2\mathrm{AgCl} \downarrow + \text{complex nitrate}$

Step 1: Calculate Moles of Complex

Molar mass of CrCl₃· 6H₂O = 52 + 3(35.5) + 6(18) = 52 + 106.5 + 108 = 266.5 g/mol$CrCl_3\cdot 6H_2O = 52 + 3(35.5) + 6(18) = 52 + 106.5 + 108 = 266.5 \text{ g/mol}$. Moles of the complex = (2.75)/(266.5) 0.01031 moles$\frac{2.75}{266.5} \simeq 0.01031 \text{ moles}$.

Step 2: Calculate Mass of Precipitate

Since each mole of complex gives 2 moles of AgCl$AgCl$: Moles of AgCl$AgCl$ formed = 2 × 0.01031 = 0.02063 moles$2 \times 0.01031 = 0.02063 \text{ moles}$. Molar mass of AgCl = 108 + 35.5 = 143.5 g/mol$AgCl = 108 + 35.5 = 143.5 \text{ g/mol}$. Mass of AgCl = 0.02063 × 143.5 2.96 gm$AgCl = 0.02063 \times 143.5 \simeq 2.96 \text{ gm}$. Rounding to the nearest integer yields 3.

Pattern Recognition

A 1:2 electrolyte contains 3 total ions. For CrCl₃ · 6H₂O$CrCl_3 \cdot 6H_2O$, this strictly dictates 2 Cl^-$2 Cl^-$ are outside the coordination sphere. Hence n$n$ factor for chloride precipitation is exactly 2.

Chapter Mix

Class 12 Chemistry: Coordination Compounds Class 11 Chemistry: Some Basic Concepts of Chemistry