Which of the following linear combination of atomic orbitals will lead to formation of molecular orbitals in…

🧠 Symmetry along the bond axis decides overlap With the bond axis along $z$, only orbitals with the same symmetry about $z$ can give a non-zero overlap. $s$ and $p_z$ are both symmetric around the $z$-axis ($\sigma$-type). Things like $p_x$ or $d_{xy}$ are not — they have positive and negative lobes that cancel.

🗺️ Check each combination

• (A) $2p_z$ + $2p_x$ — one is $\sigma$ along $z$, the other is $\pi$. Symmetry mismatch.
• (B) $2s$ + $2p_x$ — $s$ is symmetric all around, but $p_x$ has $+/-$ lobes perpendicular to $z$ which average to zero. No net overlap.
• (C) $3d_{xy}$ + $3d_{x^2-y^2}$ — both are perpendicular to $z$, but with different angular shapes. Mismatch.
• (D) $2s$ + $2p_z$ — both symmetric around $z$. Good overlap.
• (E) $2p_z$ + $3d_{x^2-y^2}$ — $p_z$ is $\sigma$, $d_{x^2-y^2}$ has lobes in $xy$ plane. Mismatch.

Only D gives a true molecular orbital.

⚠️ Common mistake Many students think any pair of orbitals can combine. They cannot — the angular shapes must match around the bond axis, otherwise overlap is zero.

$\boxed{\text{Answer: (c)}}$