Select the correct combination of coordination number and type of hyb…

The correct answer is Option (2) → 5 and \(sp^3d\)

In valence bond theory the metal ion uses a set of hybrid orbitals, one for each donor pair of the ligands. The number of hybrid orbitals is therefore the coordination number, and the kind of hybridisation decides the shape:

• 4, \( sp^{3} \): tetrahedral

• 4, \( dsp^{2} \): square planar

• 5, \( sp^{3}d \): trigonal bipyramidal

• 6, \( sp^{3}d^{2} \) or \( d^{2}sp^{3} \): octahedral

A trigonal bipyramid has five positions (three in a plane at 120° and two along the axis), so it needs five hybrid orbitals: one s, three p and one d orbital, that is \( sp^{3}d \). An example is \( \ce{[Fe(CO)5]} \). So the coordination number is 5 and the hybridisation \( sp^{3}d \).

Note: In the original NTA paper, Option (3) was printed as “5 and \( dsp^{3} \)”. This was also correct, because \( dsp^{3} \) is another name for the trigonal bipyramidal hybridisation \( sp^{3}d \). So that students are not confused and the question has a single answer, Option (3) has been changed to “5 and \( sp^{3}d^{2} \)”.

NCERT (2026-27 edition):
Valence Bond Theory

Table 5.2: Number of Orbitals and Types of Hybridisations (row)