We know that the electron configuration in $3d$ orbitals of $\ce{Zn^2+}$, in $4d$ orbitals of $\ce{Cd^2+}$ and in $5d$ orbitals of $\ce{Hg^2+}$ ions are completely same.

Then why the compounds of $\ce{Hg(II)}$ and $\ce{Cd}$, such as $\ce{HgO}$, $\ce{CdS}$ radiate versatile colour where $\ce{ZnO}$, $\ce{ZnS}$ are white?


1 Answer 1


The band gap decreases as either electronegativity of the anion decreases (electronegativity of anion in CdS < CdTe , ZnS < ZnSe) or the overlap decreases (overlap in ZnS > CdS).

Also, note that colored compounds only result when 1.8 < Band gap < 3.0. Larger gaps give white/transparent & smaller gaps give a black colour to compounds. (The band gap in ZnS is 3.54, the band gap in CdS is 2.45.)

  • $\begingroup$ The electronegativity decreases from S, over Se to Te. $\endgroup$
    – Mäßige
    Commented Oct 6, 2023 at 22:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.