Question

Choose the correct statement(s) among the following:

• A. ${\left[FeC{l}_4\right]}^{-}$has higher spin-only magnetic moment than ${\left[Co\right(en\left){\left(N{H}_3\right)}_{2}C{l}_2\right]}^{+}$.
• B. ${\left[Co\right(en\left){\left(N{H}_3\right)}_{2}C{l}_2\right]}^{+}$ has $2$ geometrical isomers
• C. The cobalt ion in ${\left[Co\right(en\left){\left(N{H}_3\right)}_{2}C{l}_2\right]}^{+}$ has $s{p}^3{d}^2$ hybridization.
• D. ${\left[FeC{l}_4\right]}^{-}$ has tetrahedral geometry

Answer 1

Answer: (A), (D)

${\left[FeC{l}_4\right]}^{-}$ has tetrahedral geometry

(a) ${\left[FeC{l}_4\right]}^{-}$ : $C{l}^{-}$ is weak field ligand So, electrons cannot be paired (unpaired electrons)
${\left[FeC{l}_4\right]}^{-}$ is $s{p}^3$ hybridized and has tetrahedral geometry with number of unpaired electrons $\left(n\right)=5$.

(b) ${\left[Co\right(en\left){\left(N{H}_3\right)}_{2}C{l}_2\right]}^{+}$ has three geometrical isomers.
(c) Number of unpaired electron (n)=5
So, spin-only magnetic moment$\mu \left(s\right)=\sqrt{n(n+2)}$

$\mu \left(s\right)=\sqrt{5(5+2)}=\sqrt{35}=5.92B.M.{\left[Co\right(en\left){\left(N{H}_3\right)}_{2}C{l}_2\right]}^{+}$ hs zero unpaired electrons. So,
$\mu \left(s\right)=\sqrt{0(0+2)}=\sqrt{0}=0B.M.$
(d) ${\left[Co\right(en\left){\left(N{H}_3\right)}_{2}C{l}_2\right]}^{+}$; $N{H}_3$ and $en$ are strong field ligand so all unpaired electrons get paired.



$d^{2}s{p}^3$hybridization (octahedral geometry).