The crystal field stabilization energy energy for low spin \(d^6\) octahedral complex is

\(-2.4 \Delta _O\)

The correct answer is option 2. \(-2.4 \Delta _O\).

In an octahedral field, the five degenerate \(d\) orbitals split into two sets:

\(e_g\) set (higher energy, includes \(d_{z^2}\) and \(d_{x^2-y^2}\))

\(t_{2g}\) set (lower energy, includes \(d_{xy}\), \(d_{xz}\), and \(d_{yz}\))

The energy difference between these sets is denoted as \(\Delta_O\) (octahedral crystal field splitting energy).

Low-Spin configuration for \(d^6\) Complex:

In a low-spin octahedral complex, the \(d^6\) electron configuration will fill the \(t_{2g}\) orbitals first and then the \(e_g\) orbitals, but with fewer unpaired electrons due to strong field ligands causing pairing of electrons. Thus, the electronic configuration will be \(t_{2g}^6e_g^0\). The splitting can be represented as :

Thus, the crystal field stabilization energy (C.F.S.E.) is given by

\(C.F.S.E.\, \ = \, \ (-0.4 \times \text{ number of electrons in }t_{2g})\Delta _O + (0.6 \times \text{ number of electrons in }e_g) \Delta _O\)

\(⇒ C.F.S.E.\, \ = \, \ (-0.4 \times 6) \Delta _O+ (0.6 \times 0)\Delta _O\)

\(⇒ C.F.S.E.\, \ = \, \ -2.4 \Delta _O\)

The correct answer for the CFSE of a low-spin \(d^6\) octahedral complex is: 2. \(-2.4 \Delta _O\).