IMAT 2018 Q44 [Energy States]

Which of the following species have the same number and arrangement of electrons in their lowest energy states?

_{19}K^+
_{11}Na^+
_8O^–
_8O^{2–}
_{16}S^–

A. _8O^– and _8O^{2–}
B. _{19}K^+ and _{16}S^–
C. _{11}Na^+ and _8O^{2–}
D. _{19}K^+ and _{11}Na^+
E. _8O^– and _{16}S^–

_{19}K^+
To find the number of protons, find the atomic number (in this case it’s 19, so we have 19 protons here). In a neutral atom, protons = electrons. Now we know that a neutral K atom should have 19 electrons, however, we want to find K^+, so we are going to remove one electron (we are looking to write out the structure for K^+ so we will need to fill in 18 electrons.

1s^2 2s^2 2p^6 3s^2 3p^6

_{11}Na^+
10 electrons.

1s^2 2s^2 2p^6

_8O^{-}
9 electrons.

1s^2 2s^2 2p^5

_8O^{2-}
10 electrons.

1s^2 2s^2 2p^6

_{16}S^-
17 electrons.

1s^2 2s^2 2p^6 3s^2 3p^5

_{11}Na^+ and _8O^{2-} both have the same number of electrons and therefore the same electron structure.

\fcolorbox{red}{grey!30}{Therefore the answer is C.}