MCQ
The number of orbitals in the fourth principal quantum number will be
  • A
    $4$
  • B
    $8$
  • C
    $12$
  • $16$

Answer

Correct option: D.
$16$
d
(d) No. of electrons$ = 2{n^2}$ hence no. of orbital $ = \frac{{2{n^2}}}{2} = {n^2}$.

$ \Rightarrow {4^2} = 16$

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