Which of the following sets of molecules contains ONLY those...

The octet rule states that atoms tend to form compounds in ways that give them eight valence electrons. Molecules that do not have eight electrons around the central atom violate this rule.

In $BF_3$, the central boron atom has $3$ valence electrons and forms $3$ single bonds with fluorine atoms. The total number of electrons around boron is $6$. This is an electron-deficient (hypovalent) molecule and violates the octet rule.

In $SF_6$, the central sulfur atom has $6$ valence electrons and forms $6$ single bonds with fluorine atoms. The total number of electrons around sulfur is $12$. This is an expanded octet (hypervalent) molecule and violates the octet rule.

In $IF_7$, the central iodine atom has $7$ valence electrons and forms $7$ single bonds with fluorine atoms. The total number of electrons around iodine is $14$. This is an expanded octet (hypervalent) molecule and violates the octet rule.

In option (B), $H_2O$, $CH_4$, and $CO_2$ all have exactly $8$ electrons around their central atoms, following the octet rule.

In option (C), $CS_2$ has $8$ electrons around the central carbon atom, following the octet rule.

In option (D), $NF_3$ has $8$ electrons around the central nitrogen atom ($3$ bond pairs and $1$ lone pair), following the octet rule.

Therefore, only the set in option (A) contains molecules that all violate the octet rule.

Answer: $BF_3,\ SF_6,\ IF_7$