Questions: Which represents the oxidation number of each of the elements in BaNaPO4?
+2; +4; +5; +2
+2; +1; +5; -2
+1; +2; +5; -2
+2; +1; +4; -2
+1; +1; -5; -2
Transcript text: 18. Which represents the oxidation number of each of the elements in $\mathrm{BaNaPO}_{4}$ ?
$+2 ;+4 ;+5 ;+2$
$+2 ;+1 ;+5 ;-2$
$+1 ;+2 ;+5 ;-2$
$+2 ;+1 ;+4 ;-2$
$+1 ;+1 ;-5 ;-2$
Solution
Solution Steps
Step 1: Identify the Elements and Their Common Oxidation States
In the compound $\mathrm{BaNaPO}_{4}$, the elements present are Barium (Ba), Sodium (Na), Phosphorus (P), and Oxygen (O). We need to determine the oxidation number of each element.
Step 2: Assign Known Oxidation Numbers
Barium (Ba) typically has an oxidation number of \(+2\).
Sodium (Na) typically has an oxidation number of \(+1\).
Oxygen (O) typically has an oxidation number of \(-2\).
Step 3: Determine the Oxidation Number of Phosphorus (P)
The compound $\mathrm{BaNaPO}_{4}$ is neutral, so the sum of the oxidation numbers of all the atoms must equal zero.
Let the oxidation number of Phosphorus (P) be \(x\).
The total oxidation number can be calculated as:
\[
(+2) + (+1) + x + 4(-2) = 0
\]
Step 4: Solve for the Oxidation Number of Phosphorus (P)
\[
2 + 1 + x - 8 = 0
\]
\[
3 + x - 8 = 0
\]
\[
x - 5 = 0
\]
\[
x = +5
\]
Final Answer
The oxidation numbers of the elements in $\mathrm{BaNaPO}_{4}$ are:
Barium (Ba): \(+2\)
Sodium (Na): \(+1\)
Phosphorus (P): \(+5\)
Oxygen (O): \(-2\)
Thus, the correct answer is:
\[
\boxed{+2 ;+1 ;+5 ;-2}
\]