Question

Consider the complex ion ${\left[ {Cr{{\left( {N{H_3}} \right)}_2}C{l_2}\left( {{C_2}{O_4}} \right)} \right]^ - }$. What is the
$\left( i \right)$ oxidation number of the metal atom?
$\left( {ii} \right)$ coordination number of the metal atom?
$\left( {iii} \right)$ charge on the complex if all ligands were chloride ions?
A. $- 1$, $6$, $- 3$
B. $+ 3$, $6$ ,$- 3$
C. $+ 3$, $6$, $- 1$
D. $- 1$, $5$ ,$- 3$

Answer 1

The internal central atom must have the vacant orbitals for the binding of the ligand with the structure resulting in the formation of a complex structure. The oxidation number can be calculated using the charge from the ligand and the total charge of the ions.

Complete step by step answer:
$\left( i \right)$ The ${\left[ {Cr{{\left( {N{H_3}} \right)}_2}C{l_2}\left( {{C_2}{O_4}} \right)} \right]^ - }$ is complex ion where the internal metal atom is $Cr$ which has a specific oxidation number based on the whole complex ion. Let us consider the oxidation of the $Cr$ atom to be $X$. The charge of the $N{H_3}$ ligand in the complex ion is $0$, that of Chlorine is $- 1$ and oxalate ion has the charge $- 2$. The equation for determining the oxidation number for $Cr$ in the complex ion is:
$X + ( - 2) + ( - 2) = ( - 1)$
$\Rightarrow X = + 4 - 1$
$\Rightarrow X = + 3$
The oxidation number of the $Cr$ residue is $+ 3$ for this complex ion.
$\left( {ii} \right)$ The coordination number of the metal atom in the complex ion is defined by the number of ligands involved. Hence there are two $N{H_3}$ ligands, there are two chlorides and an oxalate $\left( {{C_2}{O_4}} \right)$ residue. Hence the coordination number is $6$. This is because of the four linkages with ammonia and chloride and the oxalate associated with $Cr$ is a bidentate ligand. Thus, the resultant coordination number of the complex ion is $6$.
$\left( {iii} \right)$ If all the ligands were replaced by the chloride ions associated with the central metal atom as $Cr$ then the charge of the complex would differ. The charge of the $Cr$ residue is $+ 3$ while the charge of the Chloride is $- 1$ . There are six chloride residues which can be associated with the central $Cr$ atom. Therefore, the resultant charge of the complex is: $+ 3 - 6 = - 3$.
Hence the correct option is: B. $+ 3$,$6$,$- 3$.

Note:
The oxidation number and the charge of the complex ion is important for determining the metal ion which it can form the complex structure with. The coordination number can be used to determine the structure of the given complex as well based on the central atom and associated ligands.