QQuestionAnatomy and Physiology
QuestionAnatomy and Physiology
# 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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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:
Step 3:: Add the remaining electrons to the outermost shell.
Rubidium has 37 electrons in total, and after writing the noble gas configuration, we have accounted for 18 electrons. Therefore, we need to add 37 - 18 = 19 more electrons to the outermost shell (5s).
Step 4:: Write the correct electron configuration for rubidium.
[Ar] 5s^1
The correct electron configuration for rubidium is:
Final Answer
Therefore, option B is the correct answer.
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