Consider the figure given below, where M is a metal and L is a monodentate ligand. The $\sigma$-bonding ligand group orbital (LGO) having same symmetry with $d_{z^2}$ orbital of M in the octahedral coordination geometry is
To determine the \(\sigma\)-bonding ligand group orbital (LGO) that has the same symmetry as the \(d_{z^2}\) orbital of an octahedral metal complex, we need to consider the symmetry properties of the orbitals involved.
In an octahedral complex, the \(d_{z^2}\) orbital points directly along the z-axis and aligns with the ligands \(\sigma_1\) and \(\sigma_2\) which are positioned along the z-axis. The symmetry of the \(d_{z^2}\) orbital is such that it is symmetric with respect to the z-axis and has no nodal plane perpendicular to the z-axis.
The correct ligand group orbital will have the greatest contribution from the ligands along the z-axis (\(\sigma_1\) and \(\sigma_2\)) and smaller contributions from the other ligands. Therefore, the correct symmetry adapted linear combination (SALC) of the ligand orbitals should reflect these characteristics. Let's analyze the options:
Thus, the correct answer is:
\(\frac{1}{\sqrt{12}}(2\sigma_1 + 2\sigma_2 - \sigma_3 - \sigma_4 - \sigma_5 - \sigma_6)\)
An aqueous solution of $Co(ClO_4)_2 \cdot 6H_2O$ is light pink in colour. Addition of conc. HCl results in an intense blue coloured solution due to the formation of a new species. The new species among the following is
[Given: Atomic number of Co = 27]
The rates of substitution for the following reaction vary with L in the order
