Question

The coordination number and the oxidation state of the element E in the complex $[E{(en)_2}({C_2}{O_4})]N{O_2}$ (where (en) is ethylene diamine) are respectively:
A) 4 and 3
B) 6 and 3
C) 6 and 2
D) 4 and 2

Answer 1

Hint:. The total number of coordinate bonds formed by the ligands in the complex $[E{(en)_2}({C_2}{O_4})]N{O_2}$ with the metal ion (here, element E) will be the coordination number of E. Oxidation state of element E in the given complex can be calculated by adding the charges of all the ligands and the counter ion present in the coordination entity.

Complete step by step answer:
We are given a complex, $[E{(en)_2}({C_2}{O_4})]N{O_2}$, where (en) is ethylene diamine. E is the metal ion in the complex, en and ${C_2}{O_4}^{2 - }$ (oxalate ion) are ligands joined to metal ion E with the coordinate covalent bonds. $N{O_2}$ is the counter ion.
- Coordination number: The total number of coordinate bonds formed by the ligands in the complex with the metal ion is called coordination number or it can also be defined as the number of ligand donor atoms to which the metal ion is directly bonded.
Now, in the given complex, ligands are en (ethylene diamine) and ${C_2}{O_4}^{2 - }$ (oxalate ion) and their structure is as follows:
                          

Both ligands en and ${C_2}{O_4}^{2 - }$ are bidentate ligands because they both have two donor atoms that can bind to the metal E. Donor atoms in en are two nitrogen atoms while donor atoms in ${C_2}{O_4}^{2 - }$ are two oxygen atoms.
Thus in complex $[E{(en)_2}({C_2}{O_4})]N{O_2}$, two en ligands will bind to the metal E with four coordinate bonds and one ${C_2}{O_4}^{2 - }$ ligand will bind to the metal E with two coordinate bonds. Therefore, the total number of coordinate bonds formed by the ligands in the complex with the metal ion are $4 + 2 = 6$ . Hence, 6 is called coordination number of the element E in the complex.

- We know that oxidation states of an element can be calculated by adding the charges of all the ligands and the counter ions (i.e., ions present outside the central atom).
Ligand en is a neutral ligand, therefore charge on en is 0. Charge on ligand ${C_2}{O_4}^{2 - }$ is -2 and on the counter ion i.e. $N{O_2}$ is -1. Therefore, the oxidation state of element E (let $x$) in the complex, $[E{(en)_2}({C_2}{O_4})]N{O_2}$ is:
$x + (2 \times 0) + ( - 1) + ( - 1) = 0$
$x - 2 - 1 = 0$
$x = + 3$
Thus, the oxidation state of E in the complex is +3.
So, the correct answer is “Option B”.

Note: It should be known that when a ligand binds to the central metal ion through two donor atoms as in en and ${C_2}{O_4}^{2 - }$, the ligand is said to be bidentate or tridentate ligand. The oxidation number is represented by a roman numeral in the naming of the coordination complex. For example, the oxidation number of E in $[E{(en)_2}({C_2}{O_4})]N{O_2}$ is +3 and it is written as E(III).