Question

re - order each list in the table below, if necessary, so that the atoms or ions in it

atoms or ions	atoms or ions in order of decreasing size
$cl, s, s^{2 - }$	□, □, □
$mg, na, mg^+$	□, □, □

Explanation:

Step1: Analyze halide - ion size

For halide ions ($Cl^-, F^-, Br^-$), as we move down the group in the periodic table, the ionic radius increases. Bromine is in the 4th period, chlorine in the 3rd, and fluorine in the 2nd. So, $Br^-$ is the largest and $F^-$ is the smallest.

Step2: Analyze sulfur and chlorine species

For $Cl$, $S$, and $S^{2 - }$, sulfur has a larger atomic radius than chlorine because sulfur has one - less proton in its nucleus for the same number of electron shells (in the same period). Also, the $S^{2 - }$ ion is larger than the neutral $S$ atom because of the increased electron - electron repulsion due to the gain of two electrons.

Step3: Analyze magnesium and sodium species

For $Mg$, $Na$, and $Mg^+$, sodium has a larger atomic radius than magnesium in the same period because sodium has one - less proton, resulting in a weaker nuclear pull on the electrons. Also, the $Mg^+$ ion is smaller than the neutral $Mg$ atom because when a magnesium atom loses one electron to form $Mg^+$, the remaining electrons are pulled closer to the nucleus.

Answer:

$Br^-, Cl^-, F^-$
$S^{2 - }, S, Cl$
$Na, Mg, Mg^+$