Number of sub energy levels (orbitals) in a given shell is given by:

2 n - 1

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n^{2}

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n^{2} - 1

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2 n + 1

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Solution

The correct option is B $n^{2}$
In an orbital, one of the electrons has up spin and the other down spin and the number of orbitals in a shell is given by $n^{2}$ where n is the shell number.

Total number of orbitals in a shell with principal quantum number 'n' is $n^{2}$ .

Thus, for n=1, the number of orbitals = $1^{2}$ = 1. It includes 1s orbital.
For n=2, the number of orbitals = $2^{2}$ = 4. It includes 2s, $2 p_{x}$ , $2 p_{y}$ and $2 p_{z}$ orbitals.

Hence, option B is correct.

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Similar questions

\frac{C_{0}}{1} - \frac{C_{1}}{4} + \frac{C_{2}}{9} - . . . + \frac{(- 1)^{n} C_{n}}{n^{2} + 2 n + 1}

= \frac{1}{n + 1} + \frac{1}{2 n + 2} + . . . + \frac{1}{n^{2} + 2 n + 1} .

${lim}_{n \to \infty} [\frac{1 - 2 + 3 - 4 + 5 - 6 + \dots (2 n - 1) - 2 n}{\sqrt{n^{2} + 1} + \sqrt{n^{2} - 1}}]$ is equal to

2 n + 1

1^{2}, 2^{2}, 3^{2}, . . . . n^{2}

2 n + 1

lim n \to \infty [\frac{1}{\sqrt{n^{2}}} + \frac{1}{\sqrt{n^{2} + 1}} + \frac{1}{\sqrt{n^{2} + 2}} + . . . . . . . . . + \frac{1}{\sqrt{n^{2} + 2 n}}]

$lim n \to \infty \frac{1}{\sqrt{n^{2} - 1}} + \frac{1}{\sqrt{n^{2} - 4}} + \frac{1}{\sqrt{n^{2} - 9}} + . . . . . \frac{1}{\sqrt{2 n - 1}} =$

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