Paper: CSIR-NET December 2014 · Subject: Inorganic Chemistry · Chapter: Chemical Bonding · Topic: Chemical Bonding – General · Marks: 2 · Difficulty: Easy
The $\delta$-bond is formed via the overlap of
(a)$\mathrm{d}_{\mathrm{x}^{2}-\mathrm{y}^{2}}$ and $\mathrm{d}_{\mathrm{x}^{2}-\mathrm{y}^{2}}$ orbitals
(b)$\mathrm{d}_{\mathrm{xz}}$ and $\mathrm{d}_{\mathrm{xz}}$ orbitals
(c)$\mathrm{d}_{\mathrm{xy}}$ and $\mathrm{d}_{\mathrm{xy}}$ orbitals
(d)$\mathrm{d}_{\mathrm{yz}}$ and $\mathrm{d}_{\mathrm{yz}}$ orbitals
Answer
Answer: C ✓ checked by 4AB · confidence medium
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
With the M–M axis along z, the δ bond of a quadruple bond arises from face-to-face overlap of the d_xy orbitals (d_x²−y² is used for the M–L σ bonds); d_xz and d_yz give the two π bonds.