Is it more difficult to remove a valence electron from a Magnesium (Mg) atom or a Chlorine (Cl) atom

Chlorine(Cl) is the more difficult to remove a valence electron from which is

because it needs one electron to achieve a stable octet configuration.

Magnesium (Mg) is an element which has an atomic number of 12. The

electronic configuration is 2, 8, 2. This means it has to lose its  2 valence

electrons needed to attain a stable octet configuration.

Chlorine (Cl) is an element which has an atomic number of 17. The electronic

configuration is 2, 8, 7. This means it has to gain one valence electrons

needed to attain a stable octet configuration.

Thus it is more difficult to remove a valence electron from a Chlorine (Cl)

atom than Magnesium atom.

Read more on link

 It is more difficult to remove a valence electron from a Chlorine (Cl) atom

Explanation:

Ionization energy is the energy required to remove the valence electron from an isolated gaseous atom.

Magnesium (Mg) is the 12th element of periodic table and has an atomic number of 12. The electronic configuration is 2, 8, 2. It can easily lose its valence 2 electrons to attain stable configuration.

Chlorine (Cl) is the 17th element of periodic table and has an atomic number of 17. The electronic configuration is 2, 8, 7. It has a tendency to gain electron to attain stable configuration. It cannot lose its valence electron easily as the valence electrons experience more nuclear charge.

Thus it is more difficult to remove a valence electron from a Chlorine (Cl) atom

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.

a compound is a molecule made of atoms from different elements.

a. by the combination of an unlimited number of elements