Which ion is the smallest of the set: Br-, Rb+, Sr2+, Se-2
Sr2+
Br-
Se-2
Rb+
To determine which ion is the smallest, we need to compare their respective atomic radii. The atomic radius is a measure of the size of an atom or ion.
One way to determine the relative sizes is to consider the position of the ions in the periodic table. As we move across a period from left to right, the atomic radius generally decreases due to an increase in the effective nuclear charge, which attracts the electrons more strongly. Similarly, as we move down a group, or column, of the periodic table, the atomic radius generally increases due to the addition of new electron shells.
Now, let's compare the given ions:
- Br- (Bromide ion): Bromine is located in the halogen group (Group 17) of the periodic table. Since it is in the same group as chlorine and iodine, both of which have larger atomic radii than bromine, we can infer that the bromide ion (Br-) is larger than bromine.
- Rb+ (Rubidium ion): Rubidium is a metal located in Group 1, also known as the alkali metal group. As we move down Group 1, the atomic radius generally increases. Therefore, the rubidium ion (Rb+) is larger than rubidium itself.
- Sr2+ (Strontium ion): Strontium is an alkaline earth metal located in Group 2. Similar to the trend in Group 1, as we move down Group 2, the atomic radius generally increases. Hence, the strontium ion (Sr2+) is larger than strontium.
- Se-2 (Selenium ion): Selenium is a nonmetal located just below sulfur in Group 16. In this group, the atomic radius generally increases as we move down. Therefore, the selenium ion (Se-2) is larger than selenium.
Based on the trends mentioned above, we can conclude that the Sr2+ ion is the smallest among the given ions.