Question

Name the following coordination compounds. a. \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Br}\right] \mathrm{Br}_{2}\) b. \(\mathrm{Na}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]\) c. \(\left[\mathrm{Fe}\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\right)_{2}\left(\mathrm{NO}_{2}\right)_{2}\right] \mathrm{Cl}\) d. \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{2}\right]\left[\mathrm{PtI}_{4}\right]\)

Short answer

The short answer for the given coordination compounds are: a. Penta-aqua-bromo-chromium(III) bromide b. Hexacyano-cobalt(III) sodium c. Bis(ethylenediamine)dinitro-iron(II) chloride d. Tetraammine-diiodido-platinum(II) tetraiodido-platinum(IV)

Step-by-step solution

a. Naming the compound \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Br}\right] \mathrm{Br}_{2}\)

Step 1: Identify ligands and central metal The central metal here is \(\mathrm{Cr}\), which is Chromium. The ligands are \(\mathrm{H_{2}O}\) (water) and \(\mathrm{Br}\) (bromide). Step 2: Name ligands in alphabetical order \(\mathrm{H_{2}O}\) is named as "aqua," and \(\mathrm{Br}\) is named as "bromo". So we have Aqua and Bromo. Step 3: Name the central metal and its oxidation state Chromium in this coordination compound has an oxidation state of +3, so it will be written as "Chromium(III)". Step 4: Name the compound The complete name of the compound will be: "Penta-aqua-bromo-chromium(III) bromide".

b. Naming the compound \(\mathrm{Na}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]\)

Step 1: Identify ligands and central metal The central metal here is \(\mathrm{Co}\) (Cobalt). The ligands are \(\mathrm{CN}\) (cyanide) and the counter-ion is \(\mathrm{Na}\) (sodium). Step 2: Name ligands in alphabetical order \(\mathrm{CN}\) is named as "cyano". The counter-ion, \(\mathrm{Na}\), will be named "sodium" when used at the end of the name. Step 3: Name the central metal and its oxidation state Cobalt in this coordination compound has an oxidation state of +3, so it will be written as "Cobalt(III)". Step 4: Name the compound The complete name of the compound will be: "Hexacyano-cobalt(III) sodium".

c. Naming the compound \(\left[\mathrm{Fe}\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\right)_{2}\left(\mathrm{NO}_{2}\right)_{2}\right] \mathrm{Cl}\)

Step 1: Identify ligands and central metal The central metal here is \(\mathrm{Fe}\) (Iron). The ligands are \(\mathrm{NH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{NH}_{2}\) (ethylenediamine) and \(\mathrm{NO}_{2}\) (nitro), and the counter-ion is \(\mathrm{Cl}\) (chloride). Step 2: Name ligands in alphabetical order \(\mathrm{NH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{NH}_{2}\) is named as "ethylenediamine" and \(\mathrm{NO}_{2}\) is named as "nitro". So we have Ethylenediamine and Nitro. Step 3: Name the central metal and its oxidation state Iron in this coordination compound has an oxidation state of +2, so it will be written as "Iron(II)". Step 4: Name the compound The complete name of the compound will be: "Bis(ethylenediamine)dinitro-iron(II) chloride".

d. Naming the compound \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{2}\right]\left[\mathrm{PtI}_{4}\right]\)

Step 1: Identify ligands, central metals, and counter-ions There are two metal complexes in this compound: - \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{2}\right]\) with central metal \(\mathrm{Pt}\) (Platinum) and ligands \(\mathrm{NH}_{3}\) (ammine) and \(\mathrm{I}\) (iodido). - \(\left[\mathrm{PtI}_{4}\right]\) with central metal \(\mathrm{Pt}\) (Platinum) and ligands \(\mathrm{I}\) (iodido). Step 2: Name ligands in alphabetical order for each complex For the first complex: \(\mathrm{NH}_{3}\) is named as "ammine" and \(\mathrm{I}\) is named as "iodido". So we have Ammine and Iodido. For the second complex: \(\mathrm{I}\) is named as "iodido". Step 3: Name the central metal and its oxidation state for each complex For the first complex, platinum has an oxidation state of +2, so it will be written as "Platinum(II)". For the second complex, platinum has an oxidation state of +4, so it will be written as "Platinum(IV)". Step 4: Name the compound The complete name of the compound will be: "Tetraammine-diiodido-platinum(II) tetraiodido-platinum(IV)".