Question

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

Hint:

The number of electrons in the d-orbitals for transition metals can be determined by their electron configuration, which can be identified from the periodic table.

Answer 1

Correct Answer: 11

To determine \( x + y \), where \( x \) represents the number of electrons in niobium's (Nb) 4d orbitals and \( y \) represents the number of electrons in ruthenium's (Ru) 4d orbitals, we examine their electron configurations:

1. Niobium (Nb): Atomic number 41. Electron configuration: \([Kr] 4d^4 5s^1\). This indicates \( x = 4 \) electrons in the 4d orbital.
2. Ruthenium (Ru): Atomic number 44. Electron configuration: \([Kr] 4d^7 5s^1\). This indicates \( y = 7 \) electrons in the 4d orbital.

The sum is calculated as:

\( x + y = 4 + 7 = 11 \)

The sum of 11 is within the expected range of \([11, 11]\), validating the result.

The value of \( x + y \) is 11.