Which orbital of the following is lower in energy in a many electron atom?

2 s

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3 d

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4 s

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5 f

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Solution

The correct option is B $2 s$
According to $(n + l)$ rule,

for $2 s$ orbital, $n = 2$ and $l = 0$
$∴$ $n + l$ = 2

Similarly for $3 d = (n + l) = 5$

$4 s = (4 + 0) = 4$

$5 f = (5 + 3) = 8$

$∴$ $2 s$ is lower in energy.

So, the correct option is $A$

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List - I	List - II
I) The electrons of same orbital differ in spin	a) Stability of half filled sublevel
II) Order of energy of orbitals is $2 s, 2 p, 3 s, 3 p, 4 s$	b) Stability of completely filled sublevel
III) Electronic configuration of $N$ is $1 s^{2} 2 s^{2} 2 p^{3}$	c) Pauli's principle
IV) Electric configuration of $C u$ is [Ar] $4 s^{1} 3 d^{10}$	d) Aufbau principle

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