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

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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.

Complete answer:
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.
Therefore, the calcium atom transfers two valence electrons, one to each iodine atom, to form the ionic bond.
Calcium iodide, \[(Ca{I_2})\] is formed when two valence electrons of calcium are loosed and they are gained by two atoms of iodine.

Note:
-The tendency to form classes that have eight electrons in the valence shell is called the octet rule.
-The attraction of oppositely charged ions affected by electron transfer is called an ionic bond.
-The strength of ionic bonding is determined by the magnitude of the charges and the sizes of the ions.
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