Question

According to valence bond theory the metal atom or ion can make use of which of the following orbitals to yield hybrid orbitals, that can form bonds with ligands.

• A. \((n-1)d,\ (n-1)s,\ np\)
• B. \((n-1)d,\ ns,\ np\)
• C. \((n-1)d,\ ns,\ (n-1)p\)
• D. \(nd,\ (n-1)s,\ (n-1)p\)

Hint:

In Valence Bond Theory, hybridization in coordination compounds generally involves: \[ (n-1)d,\ ns,\ np \] orbitals of the central metal ion. These combine to form hybrid orbitals that accommodate ligand lone pairs.

Answer 1

The Correct Option is B

Valence bond theory pictures a metal ion offering empty hybrid orbitals for ligand lone pairs to slot into, so the real question is which shells actually have empty orbitals of comparable energy available for hybridization.

For a transition metal ion, the (n-1)d orbitals are only partly filled or empty, and the ns and np orbitals of the outer shell are also empty and close in energy to the (n-1)d set. These three sets mix together to give hybrid orbitals such as dsp2, sp3d2, or d2sp3, depending on the geometry.

The (n-1)s and (n-1)p orbitals, by contrast, are already completely filled with core electrons from earlier shells, they are not available to accept anything from a ligand, so any option involving them can be ruled out immediately.

That leaves (n-1)d, ns, np as the orbital set actually used for hybridization in valence bond theory.

So the correct choice is option (2).