Question

1. using the table below, determine how many valence electrons are in an atom of neon gas.

electron configurations and dot structures

element	atomic number	electron configuration
beryllium	4	$1s^{2}2s^{2}$
boron	5	$1s^{2}2s^{2}2p^{1}$
carbon	6	$1s^{2}2s^{2}2p^{2}$
nitrogen	7	$1s^{2}2s^{2}2p^{3}$
oxygen	8	$1s^{2}2s^{2}2p^{4}$
fluorine	9	$1s^{2}2s^{2}2p^{5}$
neon	10	$1s^{2}2s^{2}2p^{6}$

Explanation:

Step1: Identify outer - shell electrons

Valence electrons are in the outermost shell. For neon with electron configuration $1s^{2}2s^{2}2p^{6}$, the outer shell is the second shell ($n = 2$).

Step2: Count outer - shell electrons

In the $2s$ sub - shell, there are 2 electrons, and in the $2p$ sub - shell, there are 6 electrons. So, $2 + 6=8$.

Answer:

8