# Which of these represents the correct electron configuration of rubidium? [Atomic number of rubidium is 37.] A. $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{6} 5 s^{2}$ B. $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{6} 5 s^{1}$ C. $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{8} 4 p^{6} 5 s^{1}$ D. $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{5} 5 s^{1}$
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Answer

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Step 1
: Write down the atomic number of rubidium and determine the number of electrons.

The atomic number of rubidium is 37, which means that a neutral rubidium atom has 37 electrons.

Step 2
: Write the electron configuration up to the noble gas configuration of the previous period.

[Ar] = 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6
So, the electron configuration of rubidium starts as:

Final Answer

Therefore, option B is the correct answer.