question_answer

When \[(n-1)d\] orbitals are used in bond formation, the complex formed is called inner orbital complex. Sometimes, in place of \[(n-1)d\] orbitals, outturned orbitals are used for hybridization. The complex thus formed is called out orbital complexes. Hybridisation involved in \[{{\left[ Fe{{(CN)}_{5}}({{H}_{2}}O) \right]}^{3+}}\] is

A)  \[s{{p}^{3}}{{d}^{2}}\]                  

B)  \[{{d}^{2}}s{{p}^{3}}\]

C)  \[s{{p}^{3}}{{d}^{3}}\]                  

D)  None of these

Correct Answer: B

Solution :

CN is a strong ligand hence it causes the formation of inner orbital complex due to pairing. Oxidation sate of Fe in \[{{[Fe{{(CN)}_{5}}\,({{H}_{2}}O)]}^{3\,+}}\] is \[\operatorname{x}-5+0=+3\] \[\operatorname{x}=+ 2\] According to VBT hybridisation of \[{{[Fe{{(CN)}_{5}}({{H}_{2}}O)]}^{3\,+}}\] is determined as Vacant orbital electronic configuration of Fe in \[{{\left[ Fe{{(CN)}_{5}}({{H}_{2}}O) \right]}^{3\,+}}\] \[\operatorname{Hybridisation}={{d}^{2}}s{{p}^{3}}\]