The electronic configuration of Helium is $1 s^{2}$ and of oxygen is:

1 s^{2}, 2 s^{2}, 2 p^{5}

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1 s^{2}, 2 s^{2}, 2 p^{4}

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1 s^{2}, 2 s^{2}, 2 p^{3}

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1 s^{2}, 2 s^{2}, 2 p^{6}

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Solution

The correct option is B $1 s^{2}, 2 s^{2}, 2 p^{4}$
Atomic number of Helium is 2, So electronic configuration is $1 s^{2}$ .
Atomic number of Oxygen is 8, So electronic configuration is $1 s^{2} 2 s^{2} 2 p^{4}$ .
So answer is B.

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