Question

re - order each list in the table below, if necessary, so that the atoms or ions in it are listed in order of decreasing size.

Explanation:

Step1: Analyze the first row (Ne, O, Li)

• Atomic radius trends: In a period, atomic radius decreases from left to right. Li is in group 1, O in group 16, Ne in group 18 of the same period (period 2). So the order of decreasing size is \(Li>O > Ne\).

Step2: Analyze the second row (\(Cl^{-},I^{-},Br^{-}\))

• For ions of the same charge (all are - 1), as we go down a group, ionic radius increases. The order of the groups (from top to bottom) for these halide ions: \(Cl\) (group 17, period 3), \(Br\) (group 17, period 4), \(I\) (group 17, period 5). So the order of decreasing size is \(I^{-}>Br^{-}>Cl^{-}\).

Step3: Analyze the third row (\(Se,Se^{2 - },Kr\))

• For a neutral atom and its anion: \(Se^{2-}\) has more electrons (electron - electron repulsion increases the size), so \(Se^{2-}>Se\). \(Se\) and \(Kr\) are in the same period (period 4). As we move from left to right in a period, atomic radius decreases. \(Se\) is in group 16, \(Kr\) in group 18. But \(Se^{2-}\) has a larger size than \(Kr\) (because of the extra electrons in \(Se^{2-}\)). So the order of decreasing size is \(Se^{2-}>Se > Kr\).

Answer:

\(Li,O,Ne\); \(I^{-},Br^{-},Cl^{-}\); \(Se^{2-},Se,Kr\)