
Consider the reaction,
$Pb{O_2}\xrightarrow{{}}PbO$ $\Delta {G_{298}} < 0$
$Sn{O_2}\xrightarrow{{}}SnO$ $\Delta {G_{298}} > 0$
Most probable oxidation state of $Pb \& Sn$ will be.
(1) $P{b^{4 + }}$,$S{n^{4 + }}$
(2) $P{b^{4 + }}$,$S{n^{2 + }}$
(3) $P{b^{2 + }}$,$S{n^{2 + }}$
(4) $P{b^{2 + }}$,$S{n^{4 + }}$
Answer
515.7k+ views
Hint:We know that the oxidation number of an atom is defined because the charge that an atom appears to possess on forming ionic bonds with other heteroatoms. An atom having higher electronegativity (even if it forms a covalent bond) is given a negative oxidation number.
Complete step by step answer:
As we move down the group II, oxidation number become more stable than IV so \[Pb\left( {II} \right)\] is more stable than \[Pb\left( {IV} \right)\] and \[Pb{O_2}\] has high tendency to urge electron and converts into \[PbO\]. So, the most probable oxidation number of lead and tin are going to be \[P{b^{2 + }}\],\[S{n^{4 + }}\]. Therefore, the option 4 is correct.
The sign and magnitude of Gibbs free energy may be a criterion of spontaneity for a process. When \[\Delta G{\text{ }} > 0\]or \[ + ve\] it means \[{G_{product}} > {G_{reactant}}\] as \[\Delta G{\text{ }} = {\text{ }}{G_{product}} - {\text{ }}{G_{reactants}}\] the reaction won't happen spontaneously, i.e. the reaction should be spontaneous in reverse direction.
If \[\Delta G > 0\]or \[ - ve\] the reaction occurs spontaneously.
Therefore, the correct option is 4. .
Note: We can discuss about the rules for oxidation numbers are,
1.A free element will be zero as its oxidation number.
2.Monatomic ions will have an oxidation number equal to charge of the ion.
3.In hydrogen, the oxidation number is ${\text{ + 1}}$ when combined with elements having less electronegativity; the oxidation number of hydrogen is -1.
4.In compounds of oxygen, the oxidation number of oxygen will be -2 and in peroxides it will be -1.
5.Group 1 elements will have +1 oxidation number.
6.Group 2 elements will have +2 oxidation number.
Group 17 elements will have -1 oxidation number.
7.Sum of oxidation numbers of all atoms in neutral compounds is zero.
8.In a polyatomic ion, the sum of the oxidation number is equal to the charge of the ion.
Complete step by step answer:
As we move down the group II, oxidation number become more stable than IV so \[Pb\left( {II} \right)\] is more stable than \[Pb\left( {IV} \right)\] and \[Pb{O_2}\] has high tendency to urge electron and converts into \[PbO\]. So, the most probable oxidation number of lead and tin are going to be \[P{b^{2 + }}\],\[S{n^{4 + }}\]. Therefore, the option 4 is correct.
The sign and magnitude of Gibbs free energy may be a criterion of spontaneity for a process. When \[\Delta G{\text{ }} > 0\]or \[ + ve\] it means \[{G_{product}} > {G_{reactant}}\] as \[\Delta G{\text{ }} = {\text{ }}{G_{product}} - {\text{ }}{G_{reactants}}\] the reaction won't happen spontaneously, i.e. the reaction should be spontaneous in reverse direction.
If \[\Delta G > 0\]or \[ - ve\] the reaction occurs spontaneously.
Therefore, the correct option is 4. .
Note: We can discuss about the rules for oxidation numbers are,
1.A free element will be zero as its oxidation number.
2.Monatomic ions will have an oxidation number equal to charge of the ion.
3.In hydrogen, the oxidation number is ${\text{ + 1}}$ when combined with elements having less electronegativity; the oxidation number of hydrogen is -1.
4.In compounds of oxygen, the oxidation number of oxygen will be -2 and in peroxides it will be -1.
5.Group 1 elements will have +1 oxidation number.
6.Group 2 elements will have +2 oxidation number.
Group 17 elements will have -1 oxidation number.
7.Sum of oxidation numbers of all atoms in neutral compounds is zero.
8.In a polyatomic ion, the sum of the oxidation number is equal to the charge of the ion.
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