Let be the number of nodal planes in a orbital, and be the number of nodal surfaces (radial nodes) i — Atomic Structure Chemistry Question

💡 Solution & Explanation

The number of nodal planes (angular nodes) equals $l$. For a $d$ orbital ($l=2$), $a = 2$. Specifically, $d_{x^2-y^2}$ has two nodal planes at a $45^\circ$ angle to the $x$ and $y$ axes. The number of radial nodes is $(n-l-1)$. For a $4s$ orbital ($n=4, l=0$), $b = 4 - 0 - 1 = 3$. The product $(a \times b) = 2 \times 3 = 6$.