JEE Main 2025 · 22 January · evening · Numerical answer · +4 / −1

Niobium (Nb) and ruthenium (Ru) have " x " and " y " number of electrons in their respective 4 d orbitals. The value of x+y is __________.

Answer and solution

Answer 11

Solution

Chemistry ma’am · 4 steps

Niobium and ruthenium both break the aufbau order, so use their real configurations: 4d⁴ 5s¹ and 4d⁷ 5s¹.

  1. Start from krypton. Both elements come after krypton, which has 36 electrons. Niobium (Z = 41) has 5 electrons beyond [Kr], and ruthenium (Z = 44) has 8. These go into the 5s and 4d subshells.
  2. Niobium is anomalous. The aufbau order predicts [Kr] 4d³ 5s². The observed configuration is [Kr] 4d⁴ 5s¹, because in the 4d series the 4d and 5s levels are very close in energy. So x = 4.
  3. Ruthenium is anomalous. The aufbau order predicts [Kr] 4d⁶ 5s². Again one 5s electron moves into 4d, giving [Kr] 4d⁷ 5s¹. So y = 7.
  4. So. x + y = 4 + 7 = 11.

The common slip: 9. Filling 5s² first by the aufbau rule gives 4d³ for Nb and 4d⁶ for Ru, so the sum comes to 9 instead of 11.

The idea to keep. In the 4d series, 4d and 5s are so close that many elements (Nb, Mo, Ru, Rh, Pd, Ag) do not follow aufbau. Learn their actual configurations; do not just fill 5s² first.

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

Study Atomic Structure

Asked about 170 times in JEE Main since 2020. This question is from 2025.