Question

What is the hybridization of \([ {CrF}_6 ]^{3-}\)?

• A. sp\(^3d\)
• B. d\(^2\)sp\(^3\)
• C. d\(^2\)sp
• D. d\(^2\)sp

Hint:

In coordination complexes with 6 ligands (like \([ {CrF}_6 ]^{3-}\)), the hybridization is typically \( d^2sp^3 \), which involves the mixing of two \( d \)-orbitals, one \( s \)-orbital, and three \( p \)-orbitals.

Answer 1

The Correct Option is C

The complex ion \([ {CrF}_6 ]^{3-}\) involves chromium in the +3 oxidation state, i.e., \( {Cr}^{3+} \). The electron configuration of \( {Cr}^{3+} \) is: \[ {Cr}^{3+}: [Ar] 3d^3 \] In this case, chromium has three electrons in its \( 3d \)-orbitals available for bonding. To form a coordination complex with six fluoride ions (F\(^-\)), chromium must undergo hybridization to accommodate six bonding pairs. For six ligands (such as fluoride ions) to bond to chromium, six hybrid orbitals are required. This involves the promotion of electrons from the \( 3d \)-orbitals to the higher energy \( 4s \) and \( 4p \)-orbitals, resulting in the \( d^2sp^3 \) hybridization. In this hybridization, two \( 3d \)-orbitals, one \( 4s \)-orbital, and three \( 4p \)-orbitals combine to form six equivalent hybrid orbitals, which then overlap with the fluoride ions. 
Thus, the hybridization of \([ {CrF}_6 ]^{3-}\) is \( \boxed{{d}^2{sp}^3} \).