Read the passage carefully and answer the Questions.

According to Valence Bond theory, the metal atom or ion under the influence of ligands can use its (n-1)d, ns, np or ns, np, nd orbitals for hybridisation to yield a set of equivalent orbitals of definite geometry such as octahedral, tetrahedral, square planar and so on. These hybridised orbitals are allowed to overlap with ligand orbitals that can donate electron pairs for bonding. While the VB theory, to a larger extent, explains the formation, structures and magnetic behaviour of coordination compounds, it suffers from the many shortcomings. The crystal field theory (CFT) is an electrostatic model which considers the metal-ligand bond to be ionic arising purely from electrostatic interactions between the metal ion and the ligand. The five d orbitals in an isolated gaseous metal atom/ion have same energy, i.e., they are degenerate. This degeneracy is maintained if a spherically symmetrical field of negative charges surrounds the metal atom/ion. However, when this negative field is due to ligands in a complex, it becomes asymmetrical and the degeneracy of the d orbitals is lifted. It results in splitting of the d orbitals. The pattern of splitting depends upon the nature of the crystal field.

Which of the following represents the correct pair?

inner orbital complex - low spin complex

The correct answer is Option (2) → inner orbital complex - low spin complex **

Statement: A complex showing $d^2sp^3$ hybridisation is an inner orbital complex and is low spin.

Reason: In Valence Bond Theory, $d^2sp^3$ hybridisation uses the inner $(n-1)d$ orbitals of the metal ion. For these inner d orbitals to become vacant, the electrons in them must pair up first. This electron pairing reduces the number of unpaired electrons, producing a low spin complex.

Step-by-step understanding:

• In an isolated metal ion, the five d orbitals are degenerate.
• When ligands approach, they create a crystal field that splits these d orbitals.
• Strong field ligands (like $CN^-$, $NH_3$) cause large splitting, so electrons prefer pairing rather than occupying higher orbitals.

Two possibilities in octahedral complexes

1. Inner orbital complex ($d^2sp^3$)

• Uses inner $(n-1)d$ orbitals
• Requires electron pairing first
• Fewer unpaired electrons
• Hence low spin
• Example: $[Fe(CN)_6]^{4-}$

2. Outer orbital complex ($sp^3d^2$)

• Uses outer $nd$ orbitals
• No need for prior pairing
• More unpaired electrons
• Hence high spin
• Example: $[FeF_6]^{3-}$

Conclusion:

Because $d^2sp^3$ hybridisation involves inner $d$ orbitals and electron pairing, it always corresponds to an inner orbital, low spin complex.