Question

The coordination number and the oxidations state of the element E in the complex.

$\lbrack E(en)_{2}(C_{2}O_{4})\rbrack NO_{2}$ (where (en) is ethylene diamine are, repectively)

• A. 6 and 3
• B. 6 and 2
• C. 4 and 2
• D. 4 and 3

Answer 1

Answer: (A)

6 and 3

Answer 2

In the given complex $\lbrack E(en)_{2}(C_{2}O_{4})\rbrack^{+}NO_{2}^{-}$ethylene diamine is a bidentate ligand and $(C_{2}O_{4}^{2 -})$ oxalate ion is also bidentate ligand. Therefore co-ordinations number of the complex is 6 i.e., it is an octahedral complex. Oxidations number of E in the given complex is

x + 2 × 0+ 1 × (-2) = +1

$\therefore x = 3$