We need to determine the number of electrons each element has in its 4d orbitals and then add them together.

1. Niobium $\mathrm{(Nb, Z = 41)}$

The ground‐state electronic configuration of niobium is:

$ [\mathrm{Kr}]\,4d^4\,5s^1. $

Hence, Niobium has 4 electrons in its 4d orbitals.

2. Ruthenium $\mathrm{(Ru, Z = 44)}$

The ground‐state electronic configuration of ruthenium is:

$ [\mathrm{Kr}]\,4d^7\,5s^1. $

Hence, Ruthenium has 7 electrons in its 4d orbitals.

3. Sum of 4d Electrons

$ x \;=\; 4 \quad (\text{for Nb}), \quad y \;=\; 7 \quad (\text{for Ru}). $

Therefore,

$ x + y \;=\; 4 \;+\; 7 \;=\; 11. $

Answer: 11