Let be the total number of nodes for a orbital, be the number of angular nodes for a orbital, and be — Atomic Structure Chemistry Question

💡 Solution & Explanation

The total number of nodes is given by $(n-1)$. For $5f$ ($n=5$), $x = 5 - 1 = 4$. The number of angular nodes is given by $l$. For $4d$ ($l=2$), $y = 2$. The number of radial nodes is given by $(n-l-1)$. For $6p$ ($n=6, l=1$), $z = 6 - 1 - 1 = 4$. The sum $(x + y + z) = 4 + 2 + 4 = 10$.