
What is the oxidation number of $AgN{{O}_{3}}$?
Answer
519.3k+ views
Hint: Oxidation state of any element is their ability to gain or lose electrons. This is the property that an atom exhibits in a molecule due to the other bonded atoms. Oxidation states are positive and negative depending on their ability to gain or lose electrons and the electronegativities.
Complete answer:
An oxidation state is the ability of atoms to gain or lose electrons to complete their octet. Oxidation state of any atom depends on the nature of bonded atoms in that molecule, whether they are electronegative or electropositive, this gives the atom, the sign of positive and negative, depending on its electronegativity and the ability to gain or lose electrons.
The oxidation state of any atom can be calculated by the valence of that atom. In the given compound $AgN{{O}_{3}}$which is silver nitrate. The individual oxidation states will be calculated for the anion and the cation.
In $AgN{{O}_{3}}$the cation is silver, Ag that has a configuration of $[Kr]4{{d}^{10}}5{{s}^{1}}$ , this means that Ag can lose one electron to form $A{{g}^{1+}}$, this shows that it has an oxidation state of +1.
Now, for the anion which is nitrate, $N{{O}_{3}}^{-}$ , the individual oxidation states of oxygen and nitrogen are added which is equal to the charge on the ion, so, it will be ${{N}_{oxidation\,number}}+3\times {{O}_{oxidation\,number}}=-1$ .
The oxidation state of oxygen is -2, due to its configuration, $[He]2{{s}^{2}}2{{p}^{4}}$ that has a lack of 2 electrons. So from calculating, ${{N}_{oxidation\,number}}+3\times (-2)=-1$, the oxidation number of nitrogen is +5.
Hence, the oxidation number of Ag is +1 and $N{{O}_{3}}^{-}$has, N with +5 and O with -2 oxidation states in $AgN{{O}_{3}}$.
Note:
The oxidation state of any atom is calculated in its ion form and not for the molecule, hence individual oxidation numbers are calculated. Here nitrogen is in its highest group oxidation number which is +5. The formula to calculate the oxidation number of any polyatomic species, is the addition of their individual oxidation number equal to the charge upon that molecule.
Complete answer:
An oxidation state is the ability of atoms to gain or lose electrons to complete their octet. Oxidation state of any atom depends on the nature of bonded atoms in that molecule, whether they are electronegative or electropositive, this gives the atom, the sign of positive and negative, depending on its electronegativity and the ability to gain or lose electrons.
The oxidation state of any atom can be calculated by the valence of that atom. In the given compound $AgN{{O}_{3}}$which is silver nitrate. The individual oxidation states will be calculated for the anion and the cation.
In $AgN{{O}_{3}}$the cation is silver, Ag that has a configuration of $[Kr]4{{d}^{10}}5{{s}^{1}}$ , this means that Ag can lose one electron to form $A{{g}^{1+}}$, this shows that it has an oxidation state of +1.
Now, for the anion which is nitrate, $N{{O}_{3}}^{-}$ , the individual oxidation states of oxygen and nitrogen are added which is equal to the charge on the ion, so, it will be ${{N}_{oxidation\,number}}+3\times {{O}_{oxidation\,number}}=-1$ .
The oxidation state of oxygen is -2, due to its configuration, $[He]2{{s}^{2}}2{{p}^{4}}$ that has a lack of 2 electrons. So from calculating, ${{N}_{oxidation\,number}}+3\times (-2)=-1$, the oxidation number of nitrogen is +5.
Hence, the oxidation number of Ag is +1 and $N{{O}_{3}}^{-}$has, N with +5 and O with -2 oxidation states in $AgN{{O}_{3}}$.
Note:
The oxidation state of any atom is calculated in its ion form and not for the molecule, hence individual oxidation numbers are calculated. Here nitrogen is in its highest group oxidation number which is +5. The formula to calculate the oxidation number of any polyatomic species, is the addition of their individual oxidation number equal to the charge upon that molecule.
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