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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?

rafi ur rashid
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  • related http://chemistry.stackexchange.com/questions/2547/why-are-some-salt-solutions-coloured – Mithoron May 26 '15 at 13:59
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    $\ce{HgCl2}$ and $\ce{CdCl2}$ are colourless; just for reference. – Jan May 26 '15 at 14:34
  • We see a similar situation with oxides of alkali metals: https://chemistry.stackexchange.com/questions/72243/why-color-of-alkali-metal-peroxides-superoxides-and-ozonides-deepens-down-the-g?r=SearchResults – Oscar Lanzi Feb 28 '19 at 02:46

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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.)

Melanie Shebel
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Aay jain
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