Niobium ( Nb) and ruthenium $(\mathrm{Ru})$ have " $x$ " and " $y$ " number of electrons in their respective…
Niobium ( Nb) and ruthenium $(\mathrm{Ru})$ have " $x$ " and " $y$ " number of electrons in their respective 4 d orbitals. The value of $x+y$ is ______.
Solution
$\begin{aligned} & \mathrm{Z}=41 \rightarrow \mathrm{Nb} \text { (Niobium) : }[\mathrm{Kr}]_{36} 4 \mathrm{~d}^4 5 \mathrm{~s}^1 \\ & \text { Number of electron in } 4 \mathrm{~d}=4=\mathrm{x} \\ & \mathrm{Z}=44 \rightarrow \mathrm{Ru} \text { (Ruthenium) }[\mathrm{Kr}]_{36} 4 \mathrm{~d}^7 5 \mathrm{~s}^1 \\ & \text { Number of electron in } 4 \mathrm{~d}=7=\mathrm{y} \\ & \mathrm{x}+\mathrm{y}=11\end{aligned}$