Q10. Calcium fluoride can be made from the reaction of calcium metal with fluorine gas. The image shows this reaction. Explain how the product is formed in terms of electron movement and what the final electron configurations are

The answer to your question is given below

Explanation:

In this reaction, the calcium metal, Ca loses two electrons to form calcium ion, Ca^2+ and two fluorine atoms, F receives the two electrons to form two fluoride ion, F^-.

This is illustrated by the following equation:

Ca —> Ca^2+ + 2e

2F + 2e –> 2F^-

Ca + 2F + 2e —> Ca^2+ + 2F^- + 2e

Cancel out 2e

Ca + 2F —> Ca^2+ 2F^-

Thus, the electronic configuration can be written as follow:

Before combination:

Calcium, Ca (20) => 2, 8, 8, 2

Two fluorine atom, F (9) => 2, 7 and 2, 7

After combination:

Calcium ion, Ca^2+ => 2, 8, 8

Two fluoride ion, F^- => 2,8 and 2,8

43.452

step-by-step explanation:

284 times .153 = 43.452