The following bimolecular elimination reaction $(E_2)$ is carried out with different halogen leaving groups. The per cent yield of the two products ($2$-hexene and $1$-hexene) for each leaving group is listed below.

Leaving group	Conj. Acid $pK_a$	$\%$-yield of $2$-hexene	$\%$-yield of $1$-hexene
$X = I$	$-10$	$81\%$	$19\%$
$X = Br$	$-9$	$72\%$	$28\%$
$X = Cl$	$-7$	$67\%$	$33\%$
$X = F$	$3.2$	$30\%$	$70\%$

Which of the following statement$(s)$ is/are true concerning this series of $E_2$ reactions?

• A
  Based on the $pK_a$ values of the conjugate acid,$I^{-}$ is the best leaving group and $F^{-}$ is the poorest leaving group.
• B
  When $I^{-}$,$Br^{-}$,and $Cl^{-}$ are used as leaving groups,Zaitsev's rule is followed.
• C
  $F^{-}$ is the strongest base (and therefore the poorest leaving group) and the transition state for reaction with fluoride as the leaving group has the least double bond character.
• D
  $a$,$b$,and $c$ are true.