Answer:
Reduction
Explanation:
Reduction:
Reduction involve the gain of electron and oxidation number is decreased.
Mn⁺⁷ +3e⁻ → Mn⁴⁺
Mn gets three electrons , its oxidation state reduced from +7 to +4 so Mn gets reduced.
Examples:
Consider the following reactions.
4KI + 2CuCl₂ → 2CuI + I₂ + 4KCl
the oxidation state of copper is changed from +2 to +1 so copper get reduced.
CO + H₂O → CO₂ + H₂
the oxidation state of carbon is +2 on reactant side and on product side it becomes +4 so carbon get oxidized.
H₂S + 2NaOH → Na₂S + 2H₂O
The oxidation sate of sulfur is -2 on reactant side and in product side it is also -2 so it neither oxidized nor reduced.
We have to get the relationship between metallic character and atomic radius.
Metallic character increases with increase in atomic radius and decrease with decrease of atomic radius.
If electrons from outermost shell of an element can be removed easily, that atom can be considered to have more metallic character.
With increase in atomic radius, nuclear force of attraction towards outermost shell electron decreases which facilitates the release of electron.
With decrease in atomic radius, nuclear force of attraction towards outermost shell electrons increases, so electrons are hold tightly to nucleus. Hence, removal of electron from outermost shell becomes difficult making the atom less metallic in nature.
<h3><u>Answer</u>;</h3>
B.The rate of forward reaction increases.
<h3><u>Explanation;</u></h3>
- Le Chatelier's principle states that changing a factor such as concentration, temperature, or pressure of a reaction at equilibrium will cause the reaction to shift in the direction that counteracts the effect of that change.
- <em><u>Therefore, when reactants are added to a reaction at equilibrium shift when more reactants are added then the reaction shifts to the right to make more products.</u></em>
True. Nuclear fusion of hydrogen to form helium occurs naturallyin the sun and other stars. It takes place only at extremely high temperatures.
