Crystal Field Stabilization Energy (CFSE) is a measure of the stability provided by a ligand to a transition metal ion in a coordination complex. It depends on the distribution of electrons in the d-orbitals of the metal ion, which is influenced by the geometry and the type of ligands.
To determine in which of the complexes the CFSE, \(\Delta_0\), will be equal to zero, we need to examine the oxidation state and the d-electron configuration of iron in each complex:
\([Fe(NH_3)_6]Br_2\):
Oxidation State: +2
d-electron Configuration: \(d^6\)
\([Fe(en)_3]Cl_3\):
Oxidation State: +3
d-electron Configuration: \(d^5\)
\(K_4[Fe(CN)_6]\):
Oxidation State: +2
d-electron Configuration: \(d^6\)
\(K_3[Fe(SCN)_6]\):
Oxidation State: +3
d-electron Configuration: \(d^5\)
The CFSE depends on the arrangement of d-electrons in the t2g and eg orbitals. In the case of a \(d^5\) configuration, the orbitals are often half-filled, leading to a uniform distribution of electrons in both t2g and eg\ orbitals. For high spin configurations, like Fe\(^{3+}\) in \([Fe(SCN)_6]^{3-}\), there are three electrons in t2g and two in eg\), which results in a CFSE of zero:
d⁵=(0×t2g+0×eg
Therefore, the complex \(K_3[Fe(SCN)_6]\) has a CFSE, \(\Delta_0\), of zero.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
| List I (Substances) | List II (Element Present) |
| (A) Ziegler catalyst | (I) Rhodium |
| (B) Blood Pigment | (II) Cobalt |
| (C) Wilkinson catalyst | (III) Iron |
| (D) Vitamin B12 | (IV) Titanium |
| List-I (Complex ion) | List-II (Spin only magnetic moment in B.M.) |
|---|---|
| (A) [Cr(NH$_3$)$_6$]$^{3+}$ | (I) 4.90 |
| (B) [NiCl$_4$]$^{2-}$ | (II) 3.87 |
| (C) [CoF$_6$]$^{3-}$ | (III) 0.0 |
| (D) [Ni(CN)$_4$]$^{2-}$ | (IV) 2.83 |
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,