Question

While $T{i}^{3+},V^{3+},F{e}^{3+}$ and $C{o}^{2+}$ can afford a large number of tetrahedral complexes, $C{r}^{3+}$ can never have. What is the reason behind this?

• A. $C{r}^{3+}$ forces high crystal field splitting with a varieties of ligands.
• B. Crystal field stabilisation energy in octrahedral vis-a-vis tetrahedral $C{r}^{3+}$ system plays the deciding role.
• C. The ionic radius of $C{r}^{3+}$ is the largest among other $M^{3+}$ ions mentioned.
• D. Electronegativity of $C{r}^{3+}$ is the largest among these trivalent 3d-metals and so chromium prefers to be associated with a many ligands as its ionic radius permits.

Answer 1

The correct option is B Crystal field stabilisation energy in octrahedral vis-a-vis tetrahedral $C{r}^{3+}$ system plays the deciding role.

While $T{i}^{3+},V^{3+},F{e}^{3+}$ and $C{o}^{2+}$ afford a large number of tetrahedral complexes, $C{r}^{3+}$ never have. Crystal field stabilisation energy in octrahedral vis-a-vis tetrahedral, $C{r}^{3+}$ system plays the deciding role. $C{r}^{3+}$ has 3 electrons in 3d sub-shell. In octahedral field, the valence shell electronic configuration becomes $t_{2g}^3{e}_g^0$.

$CFSE=3\times 4Dq=12Dq$.

In tetrahedral field, the outer electronic configuration becomes $e_g^2{t}_{2g}^1$.

$CFSE=2\times 6Dq-4Dq=8Dq$.

Since, the CFSE in octahedral complexes is more, the octahedral complexes are formed for $C{r}^{3+}$.