All $Zn(II)$ compounds are white because

• A
  $Zn^{2+}$ has a $d^{10}$ configuration and the $d$-subshell is full
• B
  $Zn^{2+}$ shows $d-d$ transition
• C
  $Zn^{2+}$ has no electron in the $4s$-subshell
• D
  $Zn$ is not a transition element