Consider an imaginary ion ₂₂⁴⁸ X³⁻. The nucleus contains 'a'% more neutrons than the number of electrons in the ion. The value of 'a' is _______________. [nearest integer]

Solution

Chemistry ma’am · 3 steps

Get neutrons from the nucleus and electrons from the charge, then take the percentage on the electron count.

1. Count neutrons. The mass number is 48 and the atomic number is 22. So the nucleus has 48 - 22 = 26 neutrons.
2. Count electrons. A neutral X atom has 22 electrons. The 3- charge means it has gained 3 electrons, so the ion has 22 + 3 = 25 electrons.

   a = (26 - 25)/25 × 100 = 4

3. So. There are 4% more neutrons than electrons, so a = 4.

The common slip: 37. Subtracting 3 for the 3- charge gives 19 electrons and a = 7/19 × 100 ≈ 37. A negative ion has extra electrons, not fewer.

The idea to keep. Neutrons depend only on the nucleus (A - Z). Electrons depend on the charge: add for a negative ion, subtract for a positive one. "More than B" means divide by B.

Answer from the NTA final answer key. Solution by PrepWiser.