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What is the Oxidation number of Au in \[AuC{l_3}\]?
Answer
410.1k+ views
Hint: Gold(III) chloride, also known as auric chloride, is a gold-chlorine chemical compound. The term gold trichloride is a simplification of the molecular formula\[A{u_2}C{l_6}\], which refers to the analytical formula \[AuC{l_3}\]. The name's Roman numerals denote that the gold has an oxidation state of +3, which is typical of gold compounds.
Complete answer:
It's crucial to understand what oxidation numbers are before learning how to find the oxidation number of an atom in a given compound. The cumulative number of electrons lost or obtained by an atom is represented by its oxidation number.
The following principles may be used to assign an oxidation number to a given element or compound.
The oxidation number of any free element is zero.
The oxidation number of monatomic ions is often equal to the net charge associated with the molecule.
The oxidation state of the hydrogen atom (H) is +1. It has an oxidation number of -1 when bonded with an element that has less electronegativity than it.
A halogen (group 17 element) is given an oxidation number of -1 in compounds made up of two atoms.
The sum of all the oxidation numbers of the constituent atoms equals zero in neutral compounds.
When considering polyatomic ions, the total charge of the polyatomic ion is proportional to the total of all the oxidation numbers of the atoms that make it up.
So, now in \[AuC{l_3}\]using the condition
A halogen (group 17 element) is given an oxidation number of -1 in compounds made up of two atoms.
Let the oxidation number Au be x
Let the oxidation number Cl be -1
x + 3(-1)=0
x = 3
Oxidation state of Au in \[AuC{l_3}\] is +3.
Note:
The charge of an atom after ionic approximation of its heteronuclear bonds determines its oxidation state. The oxidation state is the same as the oxidation number. It's much better to figure out oxidation numbers from the Lewis structure than it is from the chemical formula. By subtracting the amount of lone pairs and electrons gained from bonds from the number of valence electrons, each atom's oxidation number can be determined. Homonuclear bonds (bonds between atoms of the same element) are often divided equally.
Complete answer:
It's crucial to understand what oxidation numbers are before learning how to find the oxidation number of an atom in a given compound. The cumulative number of electrons lost or obtained by an atom is represented by its oxidation number.
The following principles may be used to assign an oxidation number to a given element or compound.
The oxidation number of any free element is zero.
The oxidation number of monatomic ions is often equal to the net charge associated with the molecule.
The oxidation state of the hydrogen atom (H) is +1. It has an oxidation number of -1 when bonded with an element that has less electronegativity than it.
A halogen (group 17 element) is given an oxidation number of -1 in compounds made up of two atoms.
The sum of all the oxidation numbers of the constituent atoms equals zero in neutral compounds.
When considering polyatomic ions, the total charge of the polyatomic ion is proportional to the total of all the oxidation numbers of the atoms that make it up.
So, now in \[AuC{l_3}\]using the condition
A halogen (group 17 element) is given an oxidation number of -1 in compounds made up of two atoms.
Let the oxidation number Au be x
Let the oxidation number Cl be -1
x + 3(-1)=0
x = 3
Oxidation state of Au in \[AuC{l_3}\] is +3.
Note:
The charge of an atom after ionic approximation of its heteronuclear bonds determines its oxidation state. The oxidation state is the same as the oxidation number. It's much better to figure out oxidation numbers from the Lewis structure than it is from the chemical formula. By subtracting the amount of lone pairs and electrons gained from bonds from the number of valence electrons, each atom's oxidation number can be determined. Homonuclear bonds (bonds between atoms of the same element) are often divided equally.
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