JEE Main 2025 Jan 22 Shift 2, Chemistry Q22: Quantum Numbers

JEE Main2025Jan 22, Shift 2Chemistry

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

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Solution

Both elements show anomalous configurations due to stabilisation associated with half-filled or particular d-subshell occupancies.

Niobium, $Z = 41$ : $[K r] 4 d^{4} 5 s^{1}$ . Hence $x = 4$ .

Ruthenium, $Z = 44$ : $[K r] 4 d^{7} 5 s^{1}$ . Hence $y = 7$ .

$∴ x + y = 4 + 7 = 11$ .

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