Question

Choose the correct statement(s) among the following:

• A. ${\left[{\mathrm{FeCl}}_4\right]}^{-}$has tetrahedral geometry.
• B. ${\left[\mathrm{Co}\left(\mathrm{en}\right){\left({\mathrm{NH}}_3\right)}_2{\mathrm{Cl}}_2\right]}^{+}$has $2$ geometrical isomers.
• C. ${\left[{\mathrm{FeCl}}_4\right]}^{-}$has higher spin-only magnetic moment than${\left[\mathrm{Co}\left(\mathrm{en}\right){\left({\mathrm{NH}}_3\right)}_2{\mathrm{Cl}}_2\right]}^{+}$.
• D. The cobalt ion in ${\left[\mathrm{Co}\left(\mathrm{en}\right){\left({\mathrm{NH}}_3\right)}_2{\mathrm{Cl}}_2\right]}^{+}$ has ${\mathrm{sp}}^3{\mathrm{d}}^2$ hybridization.

Answer 1

The correct option is C

${\left[{\mathrm{FeCl}}_4\right]}^{-}$has higher spin-only magnetic moment than${\left[\mathrm{Co}\left(\mathrm{en}\right){\left({\mathrm{NH}}_3\right)}_2{\mathrm{Cl}}_2\right]}^{+}$.

The explanation for the correct option:

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

1. Since ${\mathrm{Cl}}^{-}$ is a weak field ligand.
2. So it can't pair the free electrons.
3. Hence ${\left[{\mathrm{FeCl}}_4\right]}^{-}$ is ${\mathrm{sp}}^3$ hybridized and tetrahedral.
4. 

(C) As we can see from the above diagram ${\left[{\mathrm{FeCl}}_4\right]}^{-}$ has $\mathrm{n}=5$ unpaired electrons, hence

spin only magnetic moment $$\begin{gathered} =\sqrt{\mathrm{n}(\mathrm{n}+2} \\ =\sqrt{5(5+2} \\ \mu =5.92BM \end{gathered}$$

1. For ${\left[\mathrm{Co}\left(\mathrm{en}\right){\left({\mathrm{NH}}_3\right)}_2{\mathrm{Cl}}_2\right]}^{+}$ , ${\mathrm{NH}}_3$ is a strong field ligand for ${\mathrm{Co}}^{+2}$ .
2. Hence all electrons become paired.

From the above diagram, it is clear that ${\left[\mathrm{Co}\left(\mathrm{en}\right){\left({\mathrm{NH}}_3\right)}_2{\mathrm{Cl}}_2\right]}^{+}$ have ${\mathrm{d}}^2{\mathrm{sp}}^3$ hybridization and geometry are octahedral and the number of unpaired electrons is $0$

$$\begin{gathered} \mathrm{\mu }=\sqrt{\mathrm{n}(\mathrm{n}+2} \\ =\sqrt{0(0+2)} \\ \mathrm{\mu }=0 \end{gathered}$$

Therefore, ${\left[{\mathrm{FeCl}}_4\right]}^{-}$ have tetrahedral geometry and has a higher spin only magnetic moment than ${\left[\mathrm{Co}\left(\mathrm{en}\right){\left({\mathrm{NH}}_3\right)}_2{\mathrm{Cl}}_2\right]}^{+}$

The explanation for the incorrect option:

(B) There are three geometrical isomers possible for ${\left[\mathrm{Co}\left(\mathrm{en}\right){\left({\mathrm{NH}}_3\right)}_2{\mathrm{Cl}}_2\right]}^{+}$ as seen in the below diagram:

Hence, option (B) is incorrect.

(D) The above option (A) explanation shows that ${\left[\mathrm{Co}\left(\mathrm{en}\right){\left({\mathrm{NH}}_3\right)}_2{\mathrm{Cl}}_2\right]}^{+}$ have ${\mathrm{d}}^2{\mathrm{sp}}^3$ hybridization. Hence option (D) is incorrect.

Therefore, options (A) and (C) are correct.