Question

what is the oxidation state of p in ca3(po4)2?
a. +5
b. +3
c. -5
d. -3

Explanation:

Step1: Determine oxidation state of Ca

Calcium (Ca) is in Group 2 of the periodic - table and has an oxidation state of +2 in most compounds. In $Ca_3(PO_4)_2$, there are 3 Ca atoms, so the total contribution from Ca is $3\times(+2)= + 6$.

Step2: Determine oxidation state of O

Oxygen (O) usually has an oxidation state of -2 in compounds. In $Ca_3(PO_4)_2$, there are $4\times2 = 8$ O atoms, so the total contribution from O is $8\times(-2)=-16$.

Step3: Let oxidation state of P be x

The compound $Ca_3(PO_4)_2$ is neutral, so the sum of the oxidation states of all atoms is 0. We can set up the equation: $3\times(+2)+2\times(x)+8\times(-2)=0$.

Step4: Solve the equation for x

First, simplify the left - hand side of the equation: $6 + 2x-16 = 0$. Then, combine like terms: $2x-10 = 0$. Add 10 to both sides: $2x=10$. Divide both sides by 2: $x = + 5$.

Answer:

A. +5