The largest of the following is (a) an Ar atom; (b) a \(\mathrm{K}^{+}\) ion; \((c) a C a^{2+}\) ion; \((d) a C l^{-}\) ion.

Short Answer

Expert verified
\(Cl^{-}\) ion is the largest among the given atoms/ions.

Step by step solution

01

Analysis of Ar atom

Argon (Ar) is in Period 3 of the periodic table. It has 3 energy levels occupied by electrons. Since it is a neutral atom, there are no changes in the nuclear charge due to the loss or gain of electrons.
02

Analysis of \(K^{+}\) ion

Potassium (K) is also in Period 4 of the periodic table. However, the \(K^{+}\) ion indicates that it has lost one electron, which also means it has lost an energy level. Therefore, \(K^{+}\) has only 3 energy levels occupied by electrons. Additionally, with the loss of an electron, the nuclear charge increases which pulls the electrons closer to the nucleus, reducing the size of the ion.
03

Analysis of \(Ca^{2+}\) ion

Calcium (Ca) is in Period 4 of the periodic table. The \(Ca^{2+}\) ion means that Calcium has lost 2 electrons and therefore lost an energy level. Hence, \(Ca^{2+}\) has 3 energy levels occupied by electrons. Moreover, the increase in nuclear charge due to the loss of electrons decreases the size of the ion.
04

Analysis of \(Cl^{-}\) ion

Chlorine (Cl) is in Period 3 of the periodic table. The gaining of an electron to form \(Cl^{-}\) doesn't add an energy level. Thus, \(Cl^{-}\) has 3 energy levels. However, the nuclear charge decreases with the addition of an electron, increasing the size of the ion.
05

Comparison and conclusion

From the analysis of the 4 atoms/ions, we see that all have 3 energy levels occupied by electrons. However, the \(K^{+}\) and \(Ca^{2+}\) ions have an increased nuclear charge, which makes them smaller than Ar and \(Cl^{-}\). Comparing Ar and \(Cl^{-}\), the decreased nuclear charge of \(Cl^{-}\) makes it larger than Ar. Hence, among the given atoms/ions, \(Cl^{-}\) has the largest size.

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