Number of electrons present in 4f orbital of Ho³⁺ ion is ____________. (Given Atomic No. of Ho = 67)

Solution

Chemistry ma’am · 3 steps

Write Ho as [Xe] 4f¹¹ 6s², then take the three electrons from 6s first and 4f last.

1. Neutral atom. Ho has 67 electrons. Xenon accounts for 54, which leaves 67 - 54 = 13 to place after the core. Lanthanides fill 4f, so these go as 4f¹¹ 6s².

   Ho: [Xe] 4f¹¹ 6s²

2. Remove three electrons. When an ion forms, electrons leave the outermost shell first. The two 6s electrons go first, and the third comes from 4f.

   Ho³⁺: [Xe] 4f¹⁰

3. So. The 4f subshell of Ho³⁺ holds 10 electrons.

The common slip: 8. Taking all three electrons from 4f gives 4f⁸, which is 8. The two 6s electrons must leave first.

The idea to keep. For any Ln³⁺ ion, remove the two 6s electrons first, then one electron from 4f or 5d. A quick check is that the number of 4f electrons equals the atomic number minus 57.

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