Question

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

Answer 1

Hint: Calcium ($$Ca$$) is the element of the $$2nd$$ group and $$4th$$period. The electronic configuration of Calcium is -$$2,8,8,2$$.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 -$$2,8,18,18,7$$.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.