Question

$C{o}^{2+}(aq.)+SC{N}^{-}\left(aq\right)\to$ Complex$\left(X\right)$
$N{i}^{2+}\left(aq\right)+$ dimethylglyoxime $\stackrel{N{H}_{4}OH}{\to }$ Complex$\left(Y\right)$
The coordination number of cobalt and nickel in both the complexes (X) and (Y) is four. The IUPAC names of the complexes (X) and (Y) are respectively :

• A. tetrathiocyanato-S-cobalt(II) and bis(dimethylglyoximate)nickel(II)
• B. tetrathiocyanato-S-cobaltate(II) and bis(dimethylglyoximato)nickel(II)
• C. tetrathiocyanato-S-cobaltate(II) and bis(dimethylglyoximato)nickelate(II)
• D. tetrathiocyanato-S-cobaltate(III) and bis(dimethylglyoximato)nickel(II)

Answer 1

The correct option is B tetrathiocyanato-S-cobaltate(II) and bis(dimethylglyoximato)nickel(II)

The coordination number of metal in complexes (X) and (Y) is four. $SC{N}^{-}$ is monodentate ligand and coordinates through S atom. Dimethylglyoxime is bidentate ligand. Thus, for each metal ion, four $SC{N}^{-}$ and two DMG ligands are required.

Thus, the complexes are $\left(X\right)=\left[Co\right(SCN{)}_4{]}^2;\left(Y\right)=\left[Co\right(dmg{)}_2].$ Their IUPAC names are tetrathiocyanato-S-cobaltate(II) and bis(dimethylglyoximato)nickel(II). Cobalt is written as cobaltate as it is part of complex anion. Nickel is written as nickel as it is part of neutral complex. Both cobalt and nickel have +2 oxidation state. Hence, roman numeral II are used.