Which of the following ions has the smallest radius: Ca2+, K+, F-, or Cl-?
Would it be K+, because it is a cation, making it have a smaller radius?
Look at the electron configurations.
For example, between K^+ and Ca^+2.
K is 1s2 2s2 2p6 3s2 3p6 4s1 which makes
K^+ 1s2 2s2 2p6 3s2 3p6 and
Ca^+2 1s2 2s2 2p6 3s2 3p6
but K^+ has 19+ charges tugging.
Ca^+2 has 20+ charges tugging on the same number of electrons and in the same shells; therefore, Ca^+2 must be smaller than K^+.
Now look at F^- and Cl^- and go through the same, then compare with these + charged ions.
To determine which ion has the smallest radius among Ca2+, K+, F-, and Cl-, we need to consider the periodic trends.
As we move across a period in the periodic table from left to right, the atomic radius decreases. This is because the effective nuclear charge, or the attraction between the positively charged nucleus and negatively charged electrons, increases, causing the electrons to be pulled closer to the nucleus.
Comparing Ca2+ (calcium ion) and K+ (potassium ion), both are cations, meaning they have lost electrons. Since they have lost electrons, the net positive charge increases, resulting in a stronger attractive force between the nucleus and the remaining electrons. As a result, the ionic radius decreases. However, since calcium (Ca) is located to the left of potassium (K) in the periodic table, Ca2+ will have a smaller radius than K+.
Now comparing F- (fluoride ion) and Cl- (chloride ion), both are anions and have gained electrons. The additional electrons in the outermost shell result in increased electron-electron repulsions, leading to a larger ionic radius. As we move from F to Cl in the periodic table, the atomic radius generally increases. Therefore, Cl- will have a larger radius than F-.
In conclusion, out of the given ions (Ca2+, K+, F-, and Cl-), the ion with the smallest radius is Ca2+.