
What is the oxidation state of gold?
Answer
512.7k+ views
Hint: The oxidation state of any element tells us its ability to gain or lose electrons. The oxidation states of elements are different in various compounds. As oxidation state depends on the atoms that are bonded together in a compound. Electronegativity also affects the oxidation state.
Complete answer:
The oxidation number or oxidation state of any atom is the ability of that atom to gain or lose electrons. The oxidation state is the charge on an atom when it is in its ionic form, or when in a molecule. The charge is negative when an atom has the tendency to take electrons or the atom is more electronegative, while the charge is positive when an atom has an ability to donate electrons.
As the oxidation state is calculated in the form of ions, so the oxidation state of any neutral molecule is zero. So any molecule without a charge has a 0 oxidation number. Then by adding the respective oxidation numbers we can take out the oxidation number of any element to be positive or negative.
As gold in its atomic form is neutral, so it has an oxidation state of 0. It can lose or gain electrons according to its configuration in different molecules and have varied oxidation states.
The electronic configuration of gold is$[Xe]4{{f}^{14}}5{{d}^{10}}6{{s}^{2}}$. It possesses rare oxidation states that ranges from -3, -2, -1, 0, +1, +2, +3, +5. The most common oxidation states of gold are +1 and +3. This means it donates either 1 or 3 electrons to form stable compounds.
For example, in $AuC{{l}_{3}}$ , chloride ions have a -1 oxidation state so gold will have Au + 3(-1) = 0, so gold will have +3 oxidation state.
Hence, gold in its elemental form has 0 oxidation state, but it's common oxidation states include +1 and +3.
Note:
Gold is called Aurum in Latin, and hence the symbol, Au. It is a transition metal and therefore shows transition oxidation states. The +1 oxidation state written as gold (I) is called as aurous in compounds, while +3 state written as (III) is called auric in compounds. This varying oxidation states give gold the ability to get readily reduced by metal.
Complete answer:
The oxidation number or oxidation state of any atom is the ability of that atom to gain or lose electrons. The oxidation state is the charge on an atom when it is in its ionic form, or when in a molecule. The charge is negative when an atom has the tendency to take electrons or the atom is more electronegative, while the charge is positive when an atom has an ability to donate electrons.
As the oxidation state is calculated in the form of ions, so the oxidation state of any neutral molecule is zero. So any molecule without a charge has a 0 oxidation number. Then by adding the respective oxidation numbers we can take out the oxidation number of any element to be positive or negative.
As gold in its atomic form is neutral, so it has an oxidation state of 0. It can lose or gain electrons according to its configuration in different molecules and have varied oxidation states.
The electronic configuration of gold is$[Xe]4{{f}^{14}}5{{d}^{10}}6{{s}^{2}}$. It possesses rare oxidation states that ranges from -3, -2, -1, 0, +1, +2, +3, +5. The most common oxidation states of gold are +1 and +3. This means it donates either 1 or 3 electrons to form stable compounds.
For example, in $AuC{{l}_{3}}$ , chloride ions have a -1 oxidation state so gold will have Au + 3(-1) = 0, so gold will have +3 oxidation state.
Hence, gold in its elemental form has 0 oxidation state, but it's common oxidation states include +1 and +3.
Note:
Gold is called Aurum in Latin, and hence the symbol, Au. It is a transition metal and therefore shows transition oxidation states. The +1 oxidation state written as gold (I) is called as aurous in compounds, while +3 state written as (III) is called auric in compounds. This varying oxidation states give gold the ability to get readily reduced by metal.
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