
Concept explainers
Interpretation:
The difference in properties of sodium metal and sodium ions for example in sodium chloride should be explained
Concept Introduction:
- Octet rule: According to this rule, all atoms can lose, gain or share their electrons in order to complete their octet or get 8 valence electrons; the electron configuration of the nearest noble gas element.
- Valence electrons: The electrons present in the outer most energy levels of an atom are known as valence electrons. The number of valence electrons can be calculated by the group number of the element. Generally, the group number is same as the valance electrons of an atom of an element.
- There are two types of ions: positive and negative ions; which are formed when a neutral atom either loses or gains an electron.
Since a negative ion is formed by the addition of one or two electrons in a neutral atom, the resulting anion is larger than its parent atom.
Similarly, a positive ion is formed when an element, mostly a metal loses 1 or more electrons; the resulting cation is smaller than its parents.

Answer to Problem 1RQ
Solution:
Sodium metal is very reactive as compare to the sodium ions. The atomic radius of Na atom is more than the ionic radius of Na+ ion. The total number of electrons in sodium atom is 11 and that in sodium ion is 10. Sodium ion has complete octet, but sodium atom has 1 unpaired electron in its outermost shell. In the presence of more electronegative atom (with deficiency of 1 electron), sodium atom can readily convert into Na+ by losing 1 electron.
Explanation of Solution
The Lewis symbol for
is as follows:
The Lewis symbol for
is as follows:
Sodium metal loses one electron to from sodium ion. This sodium ion follows octet rule and get the electronic configuration of the nearest noble gas neon. Thus, sodium ion is chemically inactive due to complete of its octet while sodium metal is very reactive due to the presence of one unpaired electron.
Therefore; sodium metal is very reactive as compare to the sodium ions due to incomplete octet.
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Chapter 4 Solutions
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