Question
Identify the correct statements from the following:
A) $\left\lbrack Fe\left( C_{2}O_{4} \right)_{3} \right\rbrack^{3 -}$ is the most stable complex among $\left\lbrack Fe(OH)_{6} \right\rbrack^{3 -},\left\lbrack Fe\left( C_{2}O_{4} \right)_{3} \right\rbrack^{3 -}$ and $\left\lbrack Fe(SCN)_{6} \right\rbrack^{3 -}$
B) The stability of $\left\lbrack Cu\left( {NH}_{3} \right)_{4} \right\rbrack^{2 +}$ is greater than that of $\left\lbrack Cu(en)_{2} \right\rbrack^{2 +}$
C) The hybridization of Fe in $K_{4}\left\lbrack Fe(CN)_{6} \right\rbrack$ is $d^{2}{sp}^{3}$
D) $\left\lbrack Fe\left( {NO}_{2} \right)_{3}{Cl}_{3} \right\rbrack^{3 -}$ exhibits linkage isomerism
E) ${NO}_{2}^{-}$and ${SCN}^{-}$ligands are NOT ambidentate ligands
Choose the correct answer from the options given below:
Options
Solution
Sol. [Key Concept: Chelation makes $\lbrack Fe(C_{2}O_{4})_{3}\rbrack^{3 -}$ most stable; $K_{4}\lbrack Fe(CN)_{6}\rbrack$ has $d^{2}{sp}^{3}$ Fe; $\lbrack Fe({NO}_{2})_{3}{Cl}_{3}\rbrack^{3 -}$ shows linkage isomerism because ${NO}_{2}^{-}$ is ambidentate.]
A) $\left\lbrack Fe\left( C_{2}O_{4} \right)_{3} \right\rbrack^{3 -}$ has oxalate ion as ligand which leads to chelation thus maximum stability among given complexes.
B) Stability : $\left\lbrack Cu\left( {NH}_{3} \right)_{4} \right\rbrack^{2 +} < \left\lbrack Cu(en)_{2} \right\rbrack^{2 +}$; due to higher CFSE and chelation.
C) $K_{4}\left\lbrack Fe(CN)_{6} \right\rbrack\ \Rightarrow {Fe}^{2 +} = \lbrack Ar\rbrack 3{\text{ }d}^{6} $In presence of SFL the electronic configuration of ${Fe}^{+ 2}\left\lbrack t_{2\text{ }g}^{6},e_{g}^{0} \right\rbrack$, thus $d^{2}{sp}^{3}$ hybridized.
D) $\left\lbrack Fe\left( {NO}_{2} \right)_{3}{Cl}_{3} \right\rbrack^{3 -}$ has ${NO}_{2}^{-}$as ambidentate ligand thus can show linakge isomerism.
E) ${NO}_{2}^{-}$and SCN both are ambidentate ligands.
Create a free account to view solution
View Solution Free