Answer
In diagram, the reaction would be:
Work Step by Step
1) The electron configuration of $Ca$ is $$[Ar]4s^2$$
The electron configuration of $F$ is $$1s^22s^22p^5$$
2) We see that $Ca$ has 2 valence electrons and $F$ has 7 valence electrons.
According to octet rule, atoms tend to lose, gain or share electrons until they have 8 valence electrons.
We find that if $Ca$ loses both of its valence electrons, it would reach an octet $$1s^22s^22p^63s^23p^6$$
which is also the electron configuration of $Ar$, with 8 valence electrons.
In the case of $F$, if $F$ gains 1 electron, it would reach an octet $$1s^22s^22p^6$$
with 8 valence electrons in the n=2 shell.
While $Ca$ atom is willing to lose both valence electrons, a $F$ atom is only willing to accept 1 electron so that it can achieve an octet. So, only one is satisfied here.
To deal with this, another $F$ atom is involved, which means there are 2 $F$ atoms in the reaction. $Ca$ atom would lose 2 electrons, and each $F$ atom gains 1 electron. Eventually, all reach an octet.
In diagram, the reaction would be: