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# How many valence electrons are transferred from the calcium atom to iodine in the formation of the compound calcium iodide ?

Last updated date: 22nd Feb 2024
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Hint: Calcium ($Ca$) is the element of the $2nd$ group and $\;4th$period. The electronic configuration of Calcium is -$2,{\text{ }}8,{\text{ }}8,{\text{ }}2$.There are two valence electrons of Calcium.
Iodine $(I)\;$ is the element of the $17th$ group and the $5th$ period. The electronic configuration of Iodine is -$2,{\text{ }}8,{\text{ }}18,{\text{ }}18,{\text{ }}7$.There are seven valence electrons of iodine.

Calcium iodide $(Ca{I_2})$ is an ionic bond, which defines that electrons are transferred.
In order for $Ca$ to become the ion $C{a^{2 + }}$, the calcium atom needs to lose two electrons. (Electrons have a negative charge, so when an atom loses two electrons, its ion turns into more positive.)
In order for Iodine $(I)\;$to become the ion ${I^{1 - }}$, the iodine atom needs to gain one electron. (When an atom gains an electron, its ion becomes more negative.)
However, the formula for calcium iodide is $Ca{I^{2 - }}$ there are two iodine ions existing. This makes sense because the iodine ion has a charge of$- 1$, so two iodine ions have to be present to contradict the $+ 2$ charge of the calcium ion.
Calcium iodide, $(Ca{I_2})$ is formed when two valence electrons of calcium are loosed and they are gained by two atoms of iodine.