Question

1. determine the oxidation numbers for each element in the following species? a. br₂ b. fe³⁺ c. so₃ d. so₃²⁻ e. ca₃(po₄)₂

Explanation:

Step1: Oxidation number of an element in its elemental form

For $Br_2$, which is an elemental form of bromine, the oxidation number of each $Br$ atom is 0.

Step2: Oxidation number of a monatomic ion

For $Fe^{3 + }$, the oxidation number is equal to its charge, so it is +3.

Step3: Oxidation number rules for compounds

In $SO_3$, oxygen usually has an oxidation number of -2. Let the oxidation number of sulfur be $x$. Using the rule that the sum of oxidation numbers in a neutral compound is 0, we have $x+3\times(- 2)=0$. Solving for $x$ gives $x = +6$.

Step4: Oxidation number rules for poly - atomic ions

In $SO_{3}^{2 - }$, oxygen has an oxidation number of -2. Let the oxidation number of sulfur be $y$. Using the rule that the sum of oxidation numbers in a poly - atomic ion is equal to its charge, we have $y + 3\times(-2)=-2$. Solving for $y$ gives $y=+4$.

Step5: Oxidation number rules for compounds with multiple elements

In $Ca_3(PO_4)_2$, calcium ($Ca$) has an oxidation number of +2. Oxygen has an oxidation number of -2. Let the oxidation number of phosphorus be $z$. For the whole compound, the sum of oxidation numbers is 0. First, considering the $Ca$ atoms: $3\times(+2)=+6$. For the $PO_4$ groups, since there are 2 $PO_4$ groups, and the sum of oxidation numbers for the whole compound is 0, the total negative charge from oxygen in 2 $PO_4$ groups is $2\times4\times(-2)=-16$. Then for the 2 phosphorus atoms, we have $+6 + 2z-16 = 0$. Solving for $z$ gives $z = +5$.

Answer:

a. $Br$ in $Br_2$: 0
b. $Fe$ in $Fe^{3 + }$: +3
c. $S$ in $SO_3$: +6, $O$ in $SO_3$: -2
d. $S$ in $SO_{3}^{2 - }$: +4, $O$ in $SO_{3}^{2 - }$: -2
e. $Ca$ in $Ca_3(PO_4)_2$: +2, $P$ in $Ca_3(PO_4)_2$: +5, $O$ in $Ca_3(PO_4)_2$: -2