Answer
The formula and name of the compounds are as follows:
a. Tetraaquadinitritoiron (III)
\[{{\left[ Fe{{\left( {{H}_{2}}O \right)}_{4}}{{\left( ONO \right)}_{2}} \right]}^{+1}}\]
b. bis (ethylenediamine)dichloridvanadium(III) tetrachloronickelate
\[{{\left[ V{{\left( en \right)}_{2}}C{{l}_{2}} \right]}_{2}}\left[ NiC{{l}_{4}} \right]\]
Work Step by Step
a. It is given that in a complex ion, metal iron cation is present as \[F{{e}^{3+}}\]which is written as \[Fe\left( III \right)\]. This cation is surrounded by four water molecules and two nitrito, \[ON{{O}^{-}}\]ions which is written as \[{{\left( {{H}_{2}}O \right)}_{4}}\]and \[{{\left( ONO \right)}_{2}}\]. The cation, iron has +3 charge, water carries no charge and nitro ion has -1 charge. The charge on the complex ion is calculated as:
\[\begin{align}
& Ch\arg e=+3+2\left( -1 \right) \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\,+1 \\
\end{align}\]
Thus, the given complex ion is written as: \[{{\left[ Fe{{\left( {{H}_{2}}O \right)}_{4}}{{\left( ONO \right)}_{2}} \right]}^{+1}}\]. It can be named as: water is written as aqua as there are four water molecules, so name would be written along with prefix tetra as tetraaqua. Two nitrito molecules is written as, dinitrito and iron is written as\[iron\left( III \right)\]. Thus, the name of the given complex ion is tetraaquadinitritoiron (III).
b. It is given that there are two complex ions are present. In one complex ion, vanadium cation in +3 oxidation state is present which is surrounded by two ethylenediamine molecules and two chloride ions. Ethylenediamine molecule is a bidentate ligand and it is represented as (en)2. The cation, vanadium has +3 charge, ethylenediamine carries no charge and chloride ion has -1 charge. The charge on the complex ion is calculated as:
\[\begin{align}
& Ch\arg e=+3+2\left( -1 \right) \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\,+1 \\
\end{align}\]
Thus, the formula of the complex ion, having two ethylene molecules and two \[C{{I}^{-}}\]ions as ligands, is \[{{\left[ V{{\left( en \right)}_{2}}C{{l}_{2}} \right]}^{+1}}\]. As it carries a positive charge, it should be a cation Thus, cation can be named as: for two ethylenediamine molecules, the prefix ‘-bis’ is used. Thus, it would be written as bisethylenediamine , for two chloride ions, prefix di is used and vanadium is written as\[vanadium\left( III \right)\]. Thus, the name of the given complex cation is bis (ethylenediamine)dichloridvanadium(III).
Similarly, the formula and name for anion is derived as:
In complex anion, nickel cation in +2 oxidation state is present which is surrounded by four chloride ions as a coordination number of nickel cation is four. Chloride ion has -1 charge. The charge on the complex anion is calculated as:
\[\begin{align}
& Ch\arg e=+2+4\left( -1 \right) \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\,-2 \\
\end{align}\]
Thus, the formula of the complex anion having Ni (II) with four \[C{{l}^{-}}\]as ligands, is\[{{\left[ Ni{{\left( Cl \right)}_{4}} \right]}^{2-}}\]. Thus, anion can be named as: for four chloride ions, prefix tetra is used and the word ‘-ate’ is suffixed at the end of the name of nickel as its complex is present as an anion. Thus, the name of the given complex anion is tetrachloronickelate(II).
Hence, the given anions and cations are: \[{{\left[ V{{\left( en \right)}_{2}}C{{l}_{2}} \right]}^{+1}}\]\[{{\left[ Ni{{\left( Cl \right)}_{4}} \right]}^{2-}}\]. Thus, the formula of the coordination compound is \[{{\left[ V{{\left( en \right)}_{2}}C{{l}_{2}} \right]}_{2}}\left[ Ni{{\left( Cl \right)}_{4}} \right]\]and name would be bis (ethylenediamine)dichloridvanadium(III) tetrachloronickelate(II).
