The coordination number and the oxidation state of element E in the complex [E(e n)2( C 2 O 4)] NO 2( where (e n) is ethylenediamine) are, respectively

Question

The coordination number and the oxidation state of element E in the complex [E(e n)2( C 2 O 4)] NO 2( where (e n) is ethylenediamine) are, respectively (A) 6 and 3 (B) 6 and 2 (C) 4 and 2 (D) 4 and 3

Tardigrade Admin · Accepted Answer

In the given complex [E(e n)2( C 2 O 4)] NO 2 ethylenediamine is a bidentate ligand and ( C 2 O 42-) oxalate ion is also bidentate ligand. Therefore, coordination number of the complex is 6 i.e., it is an octahedral complex. Oxidation number of E in the given complex is x+2 × 0+1 ×(-2)=+1 ⇒ x=+3

Q. The coordination number and the oxidation state of element $E$ in the complex $[E (e n)_{2} (C_{2} O_{4})] N O_{2} ($ where $(e n)$ is ethylenediamine) are, respectively

$6$ and $3$

$6$ and $2$

$4$ and $2$

$4$ and $3$

Q. The coordination number and the oxidation state of element E in the complex [E(en)2​(C2​O4​)]NO2​( where (en) is ethylenediamine) are, respectively

6 and 3

6 and 2

4 and 2

4 and 3

Q. The coordination number and the oxidation state of element $E$ in the complex $[E (e n)_{2} (C_{2} O_{4})] N O_{2} ($ where $(e n)$ is ethylenediamine) are, respectively

$6$ and $3$

$6$ and $2$

$4$ and $2$

$4$ and $3$