Using stock notation, represent the following compounds:
\[HAuC{l_2},{\text{ }}T{l_2}O,{\text{ }}FeO,{\text{ }}F{e_2}{O_3},{\text{ }}CuI,{\text{ }}CuO,{\text{ }}MnO,{\text{ }}and{\text{ }}Mn{O_2}\] .

Hint: Stock notation is the representation of the oxidation state of the central atom in roman numerals. It is a way of representation of the molecular formula.

Complete step by step answer:
The oxidation state of the central atom in different molecules has to be determined first. The stock notation is represented by writing the oxidation state of the central metal atom in roman numerals after the symbol of the element.
\[HAuC{l_2}\]: Let the oxidation state of \[Au\] be \[x\]. The \[HAuC{l_2}\] is a neutral compound in which \[Au\] is the central atom bonded to a hydrogen atom and a chlorine atom. The oxidation state of \[Au\] is determined as:
Valency of \[H\] + O.S. of \[Au\] + \[2{\text{ }} \times \] Valency of \[Cl\] = \[0\]
$ + 1 + x + 2( - 1) = 0$
$x = + 1$.
The O.S. of \[Au\] is \[ + 1\], so the stock notation is \[HAu\left( I \right)C{l_2}\].
\[T{l_2}O\]: Let the oxidation state of \[Tl\] be \[x\]. The \[T{l_2}O\] is a neutral compound in which two \[Tl\] atoms are bonded to one oxygen atom. The oxidation state of \[Tl\] is determined as:
\[2{\text{ }} \times \] O.S. of \[Tl\]+ Valency of \[O\]= \[0\]
$2(x) + ( - 2) = 0$
$x = + 1$.
The O.S. of \[Tl\] is \[ + 1\] , so the stock notation is \[T{l_2}\left( I \right)O\] .
\[FeO\]: Let the oxidation state of \[Fe\] be \[x\]. The \[FeO\] is a neutral compound in which the \[Fe\] atom is bonded to one oxygen atom. The oxidation state of \[Fe\] is determined as:
O.S. of \[Fe\] + Valency of \[O\] = \[0\]
$x + ( - 2) = 0$
$x = + 2$.
The O.S. of \[Fe\] is \[ + 2\], so the stock notation is \[Fe\left( {II} \right)O\].
\[F{e_2}{O_3}\] : Let the oxidation state of \[Fe\] be \[x\]. The \[F{e_2}{O_3}\] is a neutral compound in which two \[Fe\] atoms are bonded to three oxygen atoms. The oxidation state of \[Fe\] is determined as:
\[2{\text{ }} \times \] O.S. of \[Fe\] + \[3{\text{ }} \times \] Valency of \[O\] = \[0\]
$2x + 3( - 2) = 0$
$x = + 3$.
The O.S. of \[Fe\] is \[ + 3\], so the stock notation is \[F{e_2}\left( {III} \right){O_3}\] .
\[CuI\]: Let the oxidation state of \[Cu\] be \[x\]. The \[CuI\] is a neutral compound in which the \[Cu\] atom is bonded to one iodine atom. The oxidation state of \[Cu\] is determined as:
O.S. of \[Cu\] + Valency of \[I\] = \[0\]
$x + ( - 1) = 0$
$x = + 1$.
The O.S. of \[Cu\] is \[ + 1\], so the stock notation is \[CuI\].
\[CuO\]: Let the oxidation state of \[Cu\]be \[x\]. The \[CuO\] is a neutral compound in which the \[Cu\] atom is bonded to one oxygen atom. The oxidation state of \[Cu\] is determined as:
O.S. of \[Cu\] + Valency of \[O\] = \[0\]
$x + ( - 2) = 0$
$x = + 2$.
The O.S. of \[Cu\] is \[ + 2\], so the stock notation is \[Cu\left( {II} \right)O\].
\[MnO\]: Let the oxidation state of \[Mn\] be \[x\]. The \[MnO\] is a neutral compound in which the \[Mn\] atom is bonded to one oxygen atom. The oxidation state of \[Mn\] is determined as:
O.S. of \[Mn\] + Valency of \[O\] = \[0\]
$x + ( - 2) = 0$
$x = + 2$.
The O.S. of \[Mn\] is \[ + 2\], so the stock notation is \[Mn\left( {II} \right)O\].
\[Mn{O_2}\]: Let the oxidation state of \[Mn\] be \[x\]. The \[Mn{O_2}\] is a neutral compound in which the \[Mn\] atom is bonded to two oxygen atoms. The oxidation state of \[Mn\] is determined as:
O.S. of \[Mn\] + \[2{\text{ }} \times \] Valency of \[O\] = \[0\]
$x + 2( - 2) = 0$
$x = + 4$.
The O.S. of \[Mn\] is \[ + 4\], so the stock notation is \[Mn\left( {IV} \right)O\] .

Note:
The oxidation state of an atom is the charge on an atom which appears on forming ionic or covalent bonds. The oxidation state is very useful in determining redox reactions where change of oxidation state takes place.