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:
Show Hint
The \(d_{z^2}\) orbital has large lobes on \(\pm z\) and a small negative torus in the \(xy\)-plane; weight the axial \(\sigma\)'s twice and the four equatorial \(\sigma\)'s with an opposite sign, then normalize.
Step 1: Reduce the sigma representation.
For six sigma-only ligands on an octahedron, the reducible representation of the six sigma bonds reduces as $\Gamma_\sigma = a_{1g} + e_g + t_{1u}$. The $e_g$ piece transforms exactly like the pair $(d_{z^2}, d_{x^2-y^2})$, so the LGO matching $d_{z^2}$ is one specific combination inside this $e_g$ block.
Step 2: Use the sign pattern of $d_{z^2}$ directly.
Write $d_{z^2}$ as proportional to $2z^2-x^2-y^2$. Each ligand contributes its sigma orbital weighted by this function along its bond direction: along $+z$ and $-z$ the value is $2(1)^2=2$, and along $\pm x,\pm y$ the value is $-(1)^2=-1$. Reading these off gives the unnormalized LGO $2\sigma_1+2\sigma_2-\sigma_3-\sigma_4-\sigma_5-\sigma_6$, with $\sigma_1,\sigma_2$ axial and $\sigma_3$ to $\sigma_6$ equatorial.
Step 3: Normalize by summing squared coefficients.
$2^2+2^2+(-1)^2+(-1)^2+(-1)^2+(-1)^2=12$, so divide by $\sqrt{12}$:
\[ \psi_{z^2} = \frac{1}{\sqrt{12}}(2\sigma_1+2\sigma_2-\sigma_3-\sigma_4-\sigma_5-\sigma_6) \]
Step 4: Check orthogonality with the other $e_g$ partner and with $a_{1g}$.
The $d_{x^2-y^2}$ partner is $\frac{1}{2}(\sigma_3-\sigma_4+\sigma_5-\sigma_6)$, with no axial terms at all since $x^2-y^2$ vanishes on the $z$-axis, and it is orthogonal to $\psi_{z^2}$ by inspection. The totally symmetric $a_{1g}$ combination is $\frac{1}{\sqrt{6}}(\sigma_1+\sigma_2+\sigma_3+\sigma_4+\sigma_5+\sigma_6)$, all-positive and equal, a different orbital (option D), not the $d_{z^2}$ match. This confirms only the $(2,2,-1,-1,-1,-1)$ pattern with $\frac{1}{\sqrt{12}}$ works.
Final Answer:
Option (A) is the correct $\sigma$-LGO of $e_g$ symmetry matching $d_{z^2}$.
\[\boxed{\frac{1}{\sqrt{12}}(2\sigma_1+2\sigma_2-\sigma_3-\sigma_4-\sigma_5-\sigma_6)}\]
