Question

there are 3 ways to show the electron configuration for an element or ion. first, we learned the complete electron configuration. then, we learned the noble gas shortened way. later, we will learn the box orbital diagram way.
in the noble gas shortened way we place the noble gas in brackets to describe the core electrons. the noble gas is the last one passed on the way to the element or ion. using li as an example, he 2s^1, indicates that li has all of the electrons as he and another in 2s^1. in this quiz we would give this as a2i
give the noble gas shortened electron configuration for f
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give the noble gas shortened electron configuration for the only f ion, f^-1
remember that ion formation usually results in completely full or completely empty subshells.
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a. he b. ne c. ar d. kr e. xe f. rn g. uuo h. s^0
i. s^1 j. s^2 k. s^3 l. p^0 m. p^1 n. p^2 o. p^3 p. p^4

Explanation:

Step1: Determine noble - gas core for F

Fluorine (F) has 9 electrons. The noble gas before F is He which has 2 electrons. So the noble - gas core is He.

Step2: Determine remaining electron configuration for F

After accounting for the 2 electrons from He, there are 7 remaining electrons. The next sub - shell is 2s and 2p. The 2s sub - shell can hold 2 electrons and the 2p sub - shell can hold 6 electrons. So the electron configuration is $2s^22p^5$.

Step3: Determine noble - gas core for $F^-$

The $F^-$ ion has 10 electrons. The noble gas with 10 electrons is Ne.

Step4: Determine electron configuration for $F^-$

Since $F^-$ has a full outer shell with 10 electrons and using the noble - gas Ne as the core, the remaining electron configuration is $2s^22p^6$.

Answer:

1. A. He
2. J. $s^2$
3. M. $p^5$
4. B. Ne
5. J. $s^2$
6. N. $p^6$