Question

Give formulas for the following. a. potassium tetrachlorocobaltate(II) b. aquatricarbonylplatinum(II) bromide c. sodium dicyanobis(oxalato)ferrate(III) d. triamminechloroethylenediaminechromium(III) iodide

Short answer

The short answer for the formulas of the coordination compounds is: a. Potassium tetrachlorocobaltate(II): \(K2[CoCl4]\) b. Aquatricarbonylplatinum(II) bromide: \([Pt(H2O)(CO)3]Br2\) c. Sodium dicyanobis(oxalato)ferrate(III): \(Na3[Fe(CN)2(C2O4)2]\) d. Triamminechloroethylenediaminechromium(III) iodide: \([Cr(NH3)3Cl(en)]I3\)

Step-by-step solution

a. Potassium tetrachlorocobaltate(II)

In this coordination compound, the central metal atom is cobalt (Co) with an oxidation state of +2 (indicated by II in the name). The ligands are four chloride ions (Cl-), and the counterion is potassium (K+). So the formula for potassium tetrachlorocobaltate(II) is: K2[CoCl4]

b. Aquatricarbonylplatinum(II) bromide

In this compound, the central metal atom is platinum (Pt) with an oxidation state of +2 (indicated by II in the name). The ligands are one water molecule (H2O) and three carbonyl ligands (CO). The counterion is bromide (Br-). So the formula for aquatricarbonylplatinum(II) bromide is: [Pt(H2O)(CO)3]Br2

c. Sodium dicyanobis(oxalato)ferrate(III)

In this compound, the central metal atom is iron (Fe) with an oxidation state of +3 (indicated by III in the name). The ligands are two cyanide ions (CN-) and two oxalato ligands (C2O4²-). The counterion is sodium (Na+). So the formula for sodium dicyanobis(oxalato)ferrate(III) is: Na3[Fe(CN)2(C2O4)2]

d. Triamminechloroethylenediaminechromium(III) iodide

In this compound, the central metal atom is chromium (Cr) with an oxidation state of +3 (indicated by III in the name). The ligands are three ammonia molecules (NH3), one chloride ion (Cl-), and one ethylenediamine ligand (en, C2H4(NH2)2). The counterion is iodide (I-). So the formula for triamminechloroethylenediaminechromium(III) iodide is: [Cr(NH3)3Cl(en)]I3