Zinc phosphide, Zn3P2, is often used as a rat poison. Phosphorus has 5 valence electrons. Assuming that phosphorus is following the octet rule, how many valence electrons does each zinc atom lose

Two electrons each!

Explanation:

The question pretty much requires us to find the oxidation number of Zinc in the compound.

Zn3P2

Following oxidation number rules;

O.N of Zn3P2 = 0

.Since phosphorus has valence of 5, it needs three more electrons to achieve its octet state. Hence;

Oxidation number of P = -3

Let oxidation number of Zn = x

We have;

3x  +  2 (-3) = 0

3x + (-6) = 0

3x = 6

x = 6/3 = 2

This means each zn electrons lost 2 electrons.

answer: i would say b.

explanation:

increased stability in the completion of fluorine's outer shell.