Chemistry - Coordination Chemistry Question with Solution | TestHub

$\left(\mathrm{A}\right)$ ${\left[{\mathrm{FeCl}}_4\right]}^{-}$

Oxidation number of Fe atom $=+3$

$\mathrm{Fe}\to 3{\mathrm{d}}^{6}4{\mathrm{s}}^2$

${\mathrm{Fe}}^{3+}\to 3{\mathrm{d}}^5$

$\therefore$ ${\left[{\mathrm{FeCl}}_4\right]}^{-}$ is having tetrahedral geometry

$\left(\mathrm{B}\right)$ ${\left[\mathrm{Co}\left(\mathrm{en}\right){\left({\mathrm{NH}}_3\right)}_2\right]}^{+}$ have three geometrical isomers.

$\left(\mathrm{C}\right)$ ${\left[{\mathrm{FeCl}}_4\right]}^{-}$ is ${\mathrm{sp}}^3$ hybridized with 5 -unpaired electrons. While $\left[\mathrm{Co}\left(\mathrm{en}\right){\left({\mathrm{NH}}_3\right)}_2{\mathrm{Cl}}_2\right]+$ is ${\mathrm{d}}^2{\mathrm{sp}}^3$hybridised with zero unpaired electrons.

$\left(\mathrm{D}\right)$ ${\left[\mathrm{Co}\left(\mathrm{en}\right){\left({\mathrm{NH}}_3\right)}_2{\mathrm{Cl}}_2\right]}^{+}$ has ${\mathrm{d}}^2{\mathrm{sp}}^3$ hybridization.