How many valence electrons are in Na, O and He, respectively? Group of answer choices a) 1, 6, 2 b)11, 8, 8 c)2, 4, 2 d)1, 6, 8 e)11, 8, 2
The Correct Answer and Explanation is:
Correct Answer: a) 1, 6, 2
To determine the number of valence electrons in sodium (Na), oxygen (O), and helium (He), we need to examine their positions on the periodic table and understand how valence electrons are determined.
Sodium (Na) is in Group 1 of the periodic table and has an atomic number of 11. This means it has 11 electrons in total. The electron configuration is 1s² 2s² 2p⁶ 3s¹. The outermost shell is the third shell, which contains one electron (3s¹). Therefore, sodium has 1 valence electron.
Oxygen (O) is in Group 16 and has an atomic number of 8. Its electron configuration is 1s² 2s² 2p⁴. The outermost shell is the second shell (2s and 2p), which together contain six electrons (2s² 2p⁴). Hence, oxygen has 6 valence electrons.
Helium (He) is a noble gas with an atomic number of 2. Its electron configuration is 1s². It only has one shell, and it is completely filled with two electrons. While Group 18 elements typically have 8 valence electrons, helium is an exception because its outer shell can hold a maximum of two electrons. Therefore, helium has 2 valence electrons.
Understanding valence electrons is essential in predicting the chemical behavior of elements. Valence electrons are the electrons in the outermost shell of an atom and are primarily responsible for the element’s ability to bond with other elements. Elements with similar valence electron configurations tend to exhibit similar chemical properties. For example, alkali metals like sodium are highly reactive due to having only one valence electron, which they readily lose to form positive ions. In contrast, oxygen seeks to gain two electrons to fill its outer shell, making it a good oxidizing agent. Helium is stable and inert because its only shell is full, which makes it non-reactive.
