Question

Write IUPAC names of the following :

(i) \( [\mathrm{Co(H_2O)(CN)(en)_2}]^{2+} \)

(ii) \( [\mathrm{PtCl_4}]^{2-} \)

(iii) \( [\mathrm{Cr(NH_3)_4Cl(ONO)}]^{+} \)

Hint:

Find each oxidation state from the charge, then list ligands alphabetically; an anionic complex ends in -ate and ONO is nitrito-O.

Answer 1

Step 1: Rules for naming a complex:
Name the ligands first in alphabetical order, then the metal, then its oxidation state in Roman numerals in brackets.
Anionic ligands end in -o (chlorido, cyanido, nitrito). Neutral ligands keep their name (aqua for \( \mathrm{H_2O} \), ammine for \( \mathrm{NH_3} \)).
A cationic complex keeps the plain metal name. An anionic complex gets the ending -ate.
A ligand with a long name, such as ethane-1,2-diamine (en), takes bis- and its name goes in brackets.

Step 2: Complex (i):
Let the oxidation state of Co be \( x \). Water and en are neutral, cyanide is \( -1 \) and the overall charge is \( +2 \).
\[ x + 0 + (-1) + 2(0) = +2 \Rightarrow x = +3 \]
Alphabetical order: aqua, cyanido, ethane-1,2-diamine. The complex is a cation, so the metal is called cobalt.
Name: aquacyanidobis(ethane-1,2-diamine)cobalt(III) ion.

Step 3: Complex (ii):
Let the oxidation state of Pt be \( x \). Four chloride ligands carry \( -4 \) in all and the charge is \( -2 \).
\[ x + 4(-1) = -2 \Rightarrow x = +2 \]
The complex is an anion, so platinum becomes platinate.
Name: tetrachloridoplatinate(II) ion.

Step 4: Complex (iii):
Let the oxidation state of Cr be \( x \). Ammonia is neutral, chloride is \( -1 \), nitrite is \( -1 \) and the charge is \( +1 \).
\[ x + 4(0) + (-1) + (-1) = +1 \Rightarrow x = +3 \]
The nitrite group is attached through its oxygen atom (written ONO), so it is nitrito-O. Alphabetical order: ammine, chlorido, nitrito.
Name: tetraamminechloridonitrito-O-chromium(III) ion.

Final Answer:
(i) aquacyanidobis(ethane-1,2-diamine)cobalt(III) ion
(ii) tetrachloridoplatinate(II) ion
(iii) tetraamminechloridonitrito-O-chromium(III) ion