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

a. \[K_2CoCl_4\] b. \[[Pt(CO)_3(H_2O)]Br_2\] c. \[Na_3[Fe(CN)_2(C_2O_4)_2]\] d. \[[Cr(NH_3)_3Cl(en)]I_3\]

Step-by-step solution

a. potassium tetrachlorocobaltate(II)

First, we identify the cations and anions in the compound. Here, potassium is the cation (K^+) and tetrachlorocobaltate(II) is the anion. Cobalt has an oxidation state of +2, so the formula for the tetrachlorocobaltate(II) anion is CoCl4^2-. Now, we determine the number of each ion to balance their charges: one potassium ion (K^+) and one tetrachlorocobaltate(II) anion (CoCl4^2-). So, the chemical formula of potassium tetrachlorocobaltate(II) is \[K_2CoCl_4\].

b. aquatricarbonylplatinum(II) bromide

Here, we have aquatricarbonylplatinum(II) as the cation and bromide as the anion. The cation has platinum with an oxidation state of +2 along with three carbonyl ligands (CO) and one water ligand (H2O). Therefore, the formula for aquatricarbonylplatinum(II) cation is [Pt(CO)3(H2O)]^2+. Bromide anion has the formula Br^-. Now, we determine the number of each ion to balance their charges: one aquatricarbonylplatinum(II) cation [Pt(CO)3(H2O)]^2+ and two bromide anions Br^-. So, the chemical formula of aquatricarbonylplatinum(II) bromide is \[[Pt(CO)_3(H_2O)]Br_2\].

c. sodium dicyanobis(oxalato)ferrate(III)

In this compound, sodium is the cation (Na^+) and dicyanobis(oxalato)ferrate(III) is the anion. Ferrate(III) has an oxidation state of +3, so we have Fe(CN)2(C2O4)2^3-. Now, we determine the number of each ion to balance their charges: three sodium ions (Na^+) and one dicyanobis(oxalato)ferrate(III) anion (Fe(CN)2(C2O4)2^3-). So, the chemical formula of sodium dicyanobis(oxalato)ferrate(III) is \[Na_3[Fe(CN)_2(C_2O_4)_2]\].

d. triamminechloroethylenediaminechromium(III) iodide

Here, triamminechloroethylenediaminechromium(III) is the cation and iodide is the anion. The cation consists of chromium with an oxidation state of +3, three ammonia ligands (NH3), one chloro ligand (Cl), and one ethylenediamine ligand (en, which is C2H4(NH2)2). Therefore, the formula for triamminechloroethylenediaminechromium(III) cation is [Cr(NH3)3Cl(en)]^3+. Iodide anion has the formula I^-. Now, we determine the number of each ion to balance their charges: one triamminechloroethylenediaminechromium(III) cation [Cr(NH3)3Cl(en)]^3+ and three iodide anions I^-. So, the chemical formula of triamminechloroethylenediaminechromium(III) iodide is \[[Cr(NH_3)_3Cl(en)]I_3\].