Step 1: Restate the octet rule.
The octet rule says atoms tend to have $8$ electrons around the central atom after bonding. We must find the option where the central atom does not obey this rule. Both electron deficient and electron rich central atoms violate it.
Step 2: Examine option with $SF_6$, $PCl_5$, $XeF_2$.
In $SF_6$ sulphur forms six bonds, giving $12$ electrons around it, an expanded octet.
Step 3: Continue with $PCl_5$ and $XeF_2$.
In $PCl_5$ phosphorus forms five bonds, giving $10$ electrons around it. In $XeF_2$ xenon has two bonding pairs and three lone pairs, also an expanded octet beyond $8$. So all three break the octet rule.
Step 4: Check $CO_2$, $SiH_4$, $BeCl_2$.
In $CO_2$ carbon completes its octet, in $SiH_4$ silicon completes its octet, but $BeCl_2$ has only $4$ electrons on beryllium. This set is mixed, not all violating in the same expanded way the question targets.
Step 5: Check the remaining options.
$H_2O$, $Cl_2O$, $CO_2$ all have central atoms obeying the octet, and $CH_4$, $NH_3$, $OF_2$ also obey the octet on their central atoms.
Step 6: Choose the answer.
Only the set $SF_6$, $PCl_5$, $XeF_2$ has every central atom exceeding the octet, so it is the correct answer.
\[ \boxed{SF_6,\; PCl_5,\; XeF_2} \]