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Annette [7]
3 years ago
13

Which metal atoms can form ionic bonds by losing electrons from both the outermost and next to outermost principal energy levels

?
Chemistry
1 answer:
RoseWind [281]3 years ago
5 0

Full question options;

(Fe, Pb, Mg, or Ca)

Answer:

Iron - Fe

Explanation:

We understand tht metals pretty much form bonds by losing their valence (outermost electrons). But this question specifically asks for metals that lose beyond their outermost electrons; next to outermost principal energy levels.

Pb, Mg, and Ca only lose their outermost electrons to form the following ions;

Pb2+, Mg2+, and Ca2+.

This is because their ions have achieved a stable octet configuration - the dreamland of atoms where they are satisfied and don't need to go into reactions again.

Iron on the other hand has the following electronic configurations;

Fe:  [Ar]4s2 3d6

Fe2+:  [Ar]4s0 3d6

Fe3+:  [Ar]4s0 3d5

This means ion can lose both the ooutermost electrons (4s) and next to outermost principal energy levels (3d). So correct option is Iron.

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Explanation:

The atomic number of an atom is the number of protons it has. If the atomic number is 20 then we know the atom has 20 protons.

•The mass number of an atom is the total number of protons and neutrons the atom contains. The mass number is 41 and the number of protons is 20, just subtract 20 from 41 and you will get the number of neutrons: 41 - 20= 21. The atoms has 21 neutrons.

•The number of electrons found in an atom is equal to the number of protons. The atoms has 20 protons which means it has 20 electrons.

So, the answer is:

20 protons, 20 electrons, and 21 neutrons

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