Which of the following correctly represents the third ionization of aluminum?
a) al^2+ (g) + e^- > al^+ (g)
b) al^2+ (g) > al^3+(g) + e^-
c) al^2- (g) + e^- > a1^3- (g)
d) al (g) > al^+ (g) + e^-
e) al^2+ (g) + e^- > al^3+ (g)

Option B

Explanation:

Energy required to remove outermost electron from an neutral atom is called ionization energy or first ionization energy ()

Energy required to remove an electron from M+ ion is called second ionization energy ().

Energy required to remove an electron from M2+ ion is called third ionization energy ).

Therefore, among the given options:

Therefore, the correct option is b

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error