Explanation

(a)
The $d$ -block elements form coloured compounds. These compounds have ions with unpaired electron in $d$-subshell.
i) Na and Mg belong to $s$-block , so NaCl and ${MgCl}_{2}$ are colourless compounds.
ii) ${CuF}_{2}$
Oxidation state of Cu in ${CuF}_{2}$ is +2
${Cu}^{2 +}$=$1s^{2},2s^{2},2p^{6},3s^{2},3p^{6},4s^{0},3d^{9}$
[table] ⥮ | ⥮ | ⥮ | ⥮ | $\upharpoonleft$ [/table]
$\therefore\ {CuF}_{2}$ in which Cu has one unpaired electron is coloured.
iii )$CuI$
Oxidation state of Cu in $CuI$ =+1
${Cu}^{+}$=$1s^{2},2s^{2},2p^{6},3s^{2},3p^{6},4s^{0},3d^{10}$
It has no unpaired electron. So, $CuI$ is colourless.
$\therefore\$Only ${CuF}_{2}$ is coloured among given choices.