How many $p$ electrons are there in an atom of rubidium (Atomic number: $37$ )?

12

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18

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6

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9

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Solution

The correct option is B $18$
Electronic configuration of rubidium $37 : 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{6} 5 s^{1}$ .
Number of $p$ electrons = $6 + 6 + 6 = 18$ .

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How many p electrons are there in an atom of rubidium (Atomic number: 37)?

The correct option is B 18Electronic configuration of rubidium 37:1s22s22p63s23p64s23d104p65s1. Number of p electrons =6+6+6=18.

How many $p$ electrons are there in an atom of rubidium (Atomic number: $37$ )?

The correct option is B $18$
Electronic configuration of rubidium $37 : 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{6} 5 s^{1}$ .
Number of $p$ electrons = $6 + 6 + 6 = 18$ .