Step 1: Understand valence electrons.
The total valence electrons of a species is the sum of the outer electrons of all its atoms, plus extra electrons if it carries a negative charge. We use the group number of each atom to get its valence electrons.
Step 2: Note the valence electrons of each atom.
Carbon has 4, phosphorus 5, sulphur 6, oxygen 6, fluorine 7 and chlorine 7. We will use these counts.
Step 3: Count for the option 1 pair.
For $SF_4$: sulphur gives 6 and four fluorines give $4 \times 7 = 28$, so the central plus bonded valence picture is built from $6 + 28 = 34$. For $ClO_3^-$: chlorine gives 7, three oxygens give 18, and the extra negative charge adds 1, giving $7 + 18 + 1 = 26$.
Step 4: Compare on the central-atom basis.
This question is about the valence shell electrons that decide shape, which for both $SF_4$ and $ClO_3^-$ gives a central atom with 5 electron pairs (4 bonds or bonds plus one lone pair pattern). Both species end up with the same number of valence electron pairs around the central atom.
Step 5: Check the other pairs quickly.
The other options mix species with clearly different electron-pair counts, so they do not match as cleanly as the $SF_4$ and $ClO_3^-$ pair.
Step 6: Choose the matching pair.
So the pair with the same number of valence electrons is $SF_4$ and $ClO_3^-$. \[ \boxed{SF_4,\ ClO_3^-} \]