For a unielectronic species, compare the energy of the following orbitals (represented by S)
$S_{1} = a$ spherically symmetrical orbital having two spherical nodes
$S_{2} =$ an orbital which is double-dumb bell and has no radial nodes
$S_{3} =$ an orbital with orbital angular momentum zero and three radial nodes
$S 4 =$ an orbital having one planar and one radial node

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$S_{1} = S_{2} = S_{4} < S_{3}$

$S_{1} = S_{2} = S_{3} = S_{4}$

$S_{1} < S_{2} < S_{3} < S_{4}$

$S_{1} > S_{2} > S_{3} > S_{4}$

answer is B.

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Detailed Solution

$\begin{array}{l} S_{1} \to 3s \\ S_{2} \to 3d \\ S_{3} \to 4s \\ S_{4} \to 3p \end{array}$

For unielectronic species energy depends only on ‘n’

$∴ S_{1} = S_{2} = S_{4} < S_{3}$

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