How many electrons does  ${31}_{15}$P$^3-$ contain?

Analyze the question: We need the number of electrons in the phosphide ion, P³⁻, where the isotope is written as ³¹₁₅P.

Concept:

• The subscript (15) is the atomic number, which equals the number of protons.
• The superscript (31) is the mass number (protons + neutrons).
• In a neutral atom, electrons = protons.
• A charge of 3− means the atom has gained 3 electrons.

Solution:

1. Protons = atomic number = 15
2. Neutral P atom has 15 electrons
3. P³⁻ gains 3 electrons: 15 + 3 = 18 electrons

Check: 18 electrons is the same as argon's configuration (2, 8, 8), which is why phosphorus readily forms a 3− ion.

The mass number 31 is not needed for the electron count. It only tells us there are 31 − 15 = 16 neutrons, which is the trap behind option C.

Answer: D. 18.

Tip: For ions, start with the atomic number, then add electrons for negative charges and subtract electrons for positive charges. Remember: anions gain, cations lose.