Question

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

Answer 1

Answer: (A)

${Zn}^{2+}$ has a $d^{10}$ configuration and the $d$-subshell is full

The electronic configuration of $Z{n}^{2+}$ is $3d^{10}4s^0$. All the electrons are paired.
Hence, all $Z{n}^{2+}$ compounds are colorless or white.