Question

Compare both the numbers of protons and electrons present in each to rank the following ions in order of increasing radius: As3–, Br–, K+, Mg2+.

a) K+, Br–, As³–, Mg²+
b) Mg²+, K+, Br–, As³–
c) As³–, Br–, K+, Mg²+
d) Br–, As³–, Mg²+, K+

Answer 1

To rank the ions (As3−, Br−, K+, Mg2+) by increasing radius, we analyze their nuclear charge and electron count. As3− and Br− are isoelectronic, with As3− being larger due to fewer protons. The correct order is Mg2+, K+, Br−, As3−.

Step-by-step explanation:

To compare the numbers of protons and electrons present in each ion and rank them in order of increasing radius, you need to consider both the nuclear charge and the number of electrons. The general trend for ionic radius is that, for ions with the same number of electrons (isoelectronic species), the size decreases with an increasing number of protons because the greater positive charge pulls the electrons closer to the nucleus.

Conversely, for ions with the same nuclear charge, the size increases with more electrons due to increased electron-electron repulsion in the larger electron cloud.

Let's analyze the given ions: As3− has 33 protons and 36 electrons, Br− has 35 protons and 36 electrons, K+ has 19 protons and 18 electrons, and Mg2+ has 12 protons and 10 electrons.

Since As3− and Br− are isoelectronic, but As3− has fewer protons, it will have a larger radius than Br−. K+ and Mg2+ also have the same number of electrons but K+ has more protons, making Mg2+ the smallest.

Therefore, the correct order from smallest to largest radius is: Mg2+, K+, Br−, As3−.

So, the correct answer is b) Mg2+, K+, Br−, As3−.