Answer
(a) In order of increasing atomic radii:
$F \lt Cl \lt Br \lt I \lt At$
(b) In order of increasing ionic radii:
$F^- \lt Cl^- \lt Br^- \lt I^- \lt At^- $
(c) In order of increasing electronegativity:
$At \lt I \lt Br \lt Cl \lt F$
(d) In order of increasing melting points and boiling points:
$F_2 \lt Cl_2 \lt Br_2 \lt I_2 \lt At_2$
(e) See (d).
(f) In order of increasing standard reduction potentials:
At$_2$ $\lt$ $I_2 \lt Br_2 \lt Cl_2 \lt F_2$
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Work Step by Step
(a) In order of increasing atomic radii:
$F \lt Cl \lt Br \lt I \lt At$
(b) In order of increasing ionic radii:
$F^- \lt Cl^- \lt Br^- \lt I^- \lt At^- $
(c) In order of increasing electronegativity:
$At \lt I \lt Br \lt Cl \lt F$
(d) In order of increasing melting points and boiling points:
$F_2 \lt Cl_2 \lt Br_2 \lt I_2 \lt At_2$
In nature, halogens exist as nonpolar diatomic molecules. London dispersion forces are the only forces of attraction acting between the molecules. These forces increase with increasing molecular size.
(e) See (d).
(f) In order of increasing standard reduction potentials:
At$_2$ $\lt$ $I_2 \lt Br_2 \lt Cl_2 \lt F_2$
F$_2$ has the most positive standard reduction potential and therefore is the strongest of all common oxidizing agents. Oxidizing strengths of the diatomic halogen molecules decrease down Group 7A.
