Answer the following: Write the electronic configuration of the following: (1) (high spin…

(1) $d^4$ (high spin octahedral): $t_{2g}^3 e_g^1$ (2) $d^6$ (low spin octahedral): $t_{2g}^6 e_g^0$

Formula: $\text{CFSE} = [-0\cdot4 \times n(t_{2g}) + 0\cdot6 \times n(e_g)] \Delta_o$ (1) For $d^4$ high spin ($t_{2g}^3 e_g^1$): $\text{CFSE} = [-0\cdot4(3) + 0\cdot6(1)] \Delta_o = -1\cdot2\Delta_o + 0\cdot6\Delta_o = -0\cdot6\Delta_o$ (2) For $d^6$ low spin ($t_{2g}^6 e_g^0$): $\text{CFSE} = [-0\cdot4(6) + 0\cdot6(0)] \Delta_o = -2\cdot4\Delta_o$ Final answer: -0.6 \Delta_o, -2.4 \Delta_o

(1) Linkage isomerism (due to the ambidentate ligand $\text{NO}_2^-$, coordinating through $\text{O}$ as $\text{ONO}^-$ or through $\text{N}$ as $\text{NO}_2^-$). (2) Ionisation isomerism (due to exchange of $\text{Cl}^-$ and $\text{Br}^-$ ions between the coordination sphere and the ionization sphere).

$[\text{Fe}(\text{H}_2\text{O})_6]^{2+}$ is an outer orbital complex. Explanation: In both complexes, iron is in the +2 oxidation state with configuration $[\text{Ar}] 3d^6$. 1. In $[\text{Fe}(\text{H}_2\text{O})_6]^{2+}$, $\text{H}_2\text{O}$ is a weak field ligand. The crystal field splitting energy $\Delta_o$ is less than the pairing energy $P$ ($\Delta_o < P$). Therefore, pairing of $3d$ electrons does not occur, leaving no empty inner $3d$ orbitals. The complex utilizes outer $4s$, $4p$, and $4d$ orbitals to undergo $sp^3d^2$ hybridization, forming an outer orbital (high spin) complex. 2. In contrast, in $[\text{Fe}(\text{CN})_6]^{4-}$, $\text{CN}^-$ is a strong field ligand ($\Delta_o > P$) that forces pairing of $3d$ electrons to vacate two inner $3d$ orbitals, undergoing $d^2sp^3$ hybridization to form an inner orbital (low spin) complex.