Question

In which of the following complexes the CFSE, \(\Delta_0\) will be equal to zero?

• A. \([Fe(NH_3)_6]Br_2\)
• B. \([Fe(en)_3]Cl_3\)
• C. \(K_4[Fe(CN)_6]\)
• D. \(K_3[Fe(SCN)_6]\)

Hint:

To determine CFSE, consider the electronic configuration of the metal ion and use the crystal field theory. If the d-electron configuration is such that the electrons are evenly distributed between the \(e_g\) and \(t_2\)g orbitals, the CFSE will be zero.

Answer 1

The Correct Option is D

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 t_2g and e_g orbitals. In the case of a \(d^5\) configuration, the orbitals are often half-filled, leading to a uniform distribution of electrons in both t_2g and e_g\ orbitals. For high spin configurations, like Fe\(^{3+}\) in \([Fe(SCN)_6]^{3-}\), there are three electrons in t_2g and two in e_{g\), 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.