Question

lewis symbols part 1
using the electron configuration,
determine the number of valence
electrons for all the period 2
elements.
electron configuration 1s²2s¹ 1s²2s² 1s²2s²2p¹ 1s²2s²2p² 1s²2s²2p³ 1s²2s²2p⁴ 1s²2s²2p⁵ 1s²2s²2p⁶

Explanation:

Step1: Recall the concept of valence electrons

Valence electrons are the electrons in the outermost shell of an atom. For period 2 elements, the outermost shell is the second shell (\(n = 2\)).

Step2: Analyze each electron configuration

• For \(1s^{2}2s^{1}\): The second - shell (\(n = 2\)) has \(1\) electron (\(2s^{1}\)), so the number of valence electrons is \(1\).
• For \(1s^{2}2s^{2}\): The second - shell (\(n = 2\)) has \(2\) electrons (\(2s^{2}\)), so the number of valence electrons is \(2\).
• For \(1s^{2}2s^{2}2p^{1}\): The second - shell (\(n = 2\)) has \(2 + 1=3\) electrons (\(2s^{2}2p^{1}\)), so the number of valence electrons is \(3\).
• For \(1s^{2}2s^{2}2p^{2}\): The second - shell (\(n = 2\)) has \(2+2 = 4\) electrons (\(2s^{2}2p^{2}\)), so the number of valence electrons is \(4\).
• For \(1s^{2}2s^{2}2p^{3}\): The second - shell (\(n = 2\)) has \(2 + 3=5\) electrons (\(2s^{2}2p^{3}\)), so the number of valence electrons is \(5\).
• For \(1s^{2}2s^{2}2p^{5}\): The second - shell (\(n = 2\)) has \(2+5 = 7\) electrons (\(2s^{2}2p^{5}\)), so the number of valence electrons is \(7\).
• For \(1s^{2}2s^{2}2p^{6}\): The second - shell (\(n = 2\)) has \(2+6 = 8\) electrons (\(2s^{2}2p^{6}\)), so the number of valence electrons is \(8\).

Answer:

The number of valence electrons for the electron configurations \(1s^{2}2s^{1}\), \(1s^{2}2s^{2}\), \(1s^{2}2s^{2}2p^{1}\), \(1s^{2}2s^{2}2p^{2}\), \(1s^{2}2s^{2}2p^{3}\), \(1s^{2}2s^{2}2p^{5}\), \(1s^{2}2s^{2}2p^{6}\) are \(1\), \(2\), \(3\), \(4\), \(5\), \(7\), \(8\) respectively.