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

<p>We need to determine the number of electrons each element has in its <strong>4d orbitals</strong> and then add them together.</p> <hr /> <h3>1. <strong>Niobium $\mathrm{(Nb, Z = 41)}$</strong></h3> <p><p>The ground‐state electronic configuration of niobium is:</p> <p>$ [\mathrm{Kr}]\,4d^4\,5s^1. $</p> <p>Hence, <strong>Niobium has 4 electrons in its 4d orbitals</strong>.</p></p> <hr /> <h3>2. <strong>Ruthenium $\mathrm{(Ru, Z = 44)}$</strong></h3> <p><p>The ground‐state electronic configuration of ruthenium is:</p> <p>$ [\mathrm{Kr}]\,4d^7\,5s^1. $</p> <p>Hence, <strong>Ruthenium has 7 electrons in its 4d orbitals</strong>.</p></p> <hr /> <h3>3. <strong>Sum of 4d Electrons</strong></h3> <p>$ x \;=\; 4 \quad (\text{for Nb}), \quad y \;=\; 7 \quad (\text{for Ru}). $</p> <p>Therefore,</p> <p>$ x + y \;=\; 4 \;+\; 7 \;=\; 11. $</p> <hr /> <h4><strong>Answer: 11</strong></h4>