question_answer

For transition metal octahedral complexes, the choice between high spin and low spin electronic configurations arises only for

A)  \[{{d}^{1}}\] to \[{{d}^{3}}\] complexes

B)  \[{{d}^{4}}\] to \[{{d}^{7}}\]complexes

C)  \[{{d}^{7}}\]to \[{{d}^{9}}\] complexes

D)  \[{{d}^{1}},\,{{d}^{2}}\]and \[{{d}^{8}}\] complexes

Correct Answer: B

Solution :

For \[{{d}^{1}},\,{{d}^{2}},\,{{d}^{3}},\,{{d}^{8}},\,{{d}^{9}}\] and \[{{d}^{10}}\] ions, the arrangement of electrons into \[{{t}_{2g}}\]and \[{{e}_{g}}\], sets is same for both strong and weak field ligands. But for \[{{d}^{4}},{{d}^{5}},{{d}^{6}}\] and \[{{d}^{7}}\], there is a difference in the arrangement of electrons into \[{{t}_{2g}}\] and \[{{t}_{g}}\] sets for strong and weak field ligands. A weak field ligand will have greater number of unpaired electrons.

Iron	Weak field \[(\Delta <P)\]		Strong field \[(\Delta >P)\]
	\[{{t}_{2g}}\]	\[{{e}_{g}}\]	\[{{t}_{2g}}\]	\[{{e}_{g}}\]
\[{{d}^{4}}\]
\[{{d}^{5}}\]
\[{{d}^{6}}\]
\[{{d}^{7}}\]

\[\therefore \] Filling of \[{{e}_{g}}\] orbitals in weak field situation starts from \[{{d}^{4}}\]electron while in strong field situation it starts from \[{{d}^{7}}\]electron.