To determine which of the given ions is isoelectronic with \( P^{3-} \), we need to find the number of electrons in each ion and compare them with the number of electrons in \( P^{3-} \).
### Step-by-Step Solution:
1. Identify the atomic number of phosphorus (P):
- Phosphorus (P) has an atomic number of 15, which means a neutral phosphorus atom has 15 electrons.
2. Determine the number of electrons in \( P^{3-} \):
- The \( P^{3-} \) ion results when a neutral phosphorus atom gains 3 electrons. Therefore:
[tex]\[
15 \text{ (original electrons)} + 3 \text{ (gained electrons)} = 18 \text{ electrons}
\][/tex]
3. Calculate the number of electrons in each given ion:
- Ion: \( N^{3-} \)
- Nitrogen (N) has an atomic number of 7, so a neutral nitrogen atom has 7 electrons.
- As \( N^{3-} \), it gains 3 electrons:
[tex]\[
7 \text{ (original electrons)} + 3 \text{ (gained electrons)} = 10 \text{ electrons}
\][/tex]
- Ion: \( K^{+} \)
- Potassium (K) has an atomic number of 19, so a neutral potassium atom has 19 electrons.
- As \( K^{+} \), it loses 1 electron:
[tex]\[
19 \text{ (original electrons)} - 1 \text{ (lost electron)} = 18 \text{ electrons}
\][/tex]
- Ion: \( Na^{+} \)
- Sodium (Na) has an atomic number of 11, so a neutral sodium atom has 11 electrons.
- As \( Na^{+} \), it loses 1 electron:
[tex]\[
11 \text{ (original electrons)} - 1 \text{ (lost electron)} = 10 \text{ electrons}
\][/tex]
- Ion: \( O^{2-} \)
- Oxygen (O) has an atomic number of 8, so a neutral oxygen atom has 8 electrons.
- As \( O^{2-} \), it gains 2 electrons:
[tex]\[
8 \text{ (original electrons)} + 2 \text{ (gained electrons)} = 10 \text{ electrons}
\][/tex]
4. Compare the number of electrons:
- \( P^{3-} \) has 18 electrons.
- \( N^{3-} \) has 10 electrons.
- \( K^{+} \) has 18 electrons.
- \( Na^{+} \) has 10 electrons.
- \( O^{2-} \) has 10 electrons.
Only \( K^{+} \) has the same number of electrons (18) as \( P^{3-} \).
Therefore, the ion that is isoelectronic with [tex]\( P^{3-} \)[/tex] is [tex]\( K^{+} \)[/tex], which corresponds to option 2.
