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Suppose $\ce{Zn}$, $\ce{Cd}$ and $\ce{Hg}$ were to form a bond with the same element; say $\ce{ZnCl2}$, $\ce{CdCl2}$ and $\ce{HgCl2}$ Then which of the compounds would be showing most covalent character

I feel like there are two schools of thought with regards to the covalent nature of group 12 elements.

One says according to Fajans' rule covalent character is greatest for smaller cations. Therefore the covalent bond character should be $\ce{Zn}$ > $\ce{Cd}$ > $\ce{Hg}$

The other says (my theory), due to weaker shielding effect of the $\mathrm d$-block there would be more effective nuclear charge for $\ce{Hg}$ than $\ce{Zn}$. So it should be able to cause more distortion and be able have stronger covalent bond. Therefore the order should be $\ce{Zn}$< $\ce{Cd}$< $\ce{Hg}$.

Which one is correct and why?

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The ionic radii (VI coordination) increases from $\ce{Zn^2+}(\pu{0.74 Å)}$ to $\ce{Cd^2+}(\pu{0.95 Å)}$ to $\ce{Hg^2+}(\pu{1.02Å)}$. Applying Fajans' rule, the expected order of covalent character:

$$\ce{ZnCl2} > \ce{CdCl2} > \ce{HgCl2}$$

But consider the following properties:

  • Water solubility order (per $\pu{100 g}$ at $\pu{20 °C})$: $$\ce{ZnCl2}~(\pu{432 g}) > \ce{CdCl2}~(\pu{120 g}) > \ce{HgCl2}~(\pu{7.4g})$$(Source: Wikipedia)

  • Colour intensity comparison due to cations: $$\ce{ZnS}~\text{(white)} < \ce{CdS}~\text{(yellow)} < \ce{HgS}~\text{(black)}$$

Therefore, the correct order of covalent character must be

$$\ce{ZnCl2} < \ce{CdCl2} < \ce{HgCl2}$$

This may be due to increase of effective nuclear charge, for which Slater's rule also fails.

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    $\begingroup$ HgCl2 is a colourless/white solid, unless contaminated by metallic mercury. $\endgroup$
    – Poutnik
    Jul 26 at 10:06
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    $\begingroup$ They are all colorless crystals, not sure where yellow or black (!?) color comes from. It would be nice to see some references for physical properties and solubility data. $\endgroup$
    – andselisk
    Jul 26 at 10:12
  • $\begingroup$ @andselisk Edited, also added references. $\endgroup$
    – Apurvium
    Jul 26 at 10:53
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    $\begingroup$ "(Source: Wikipedia)" You are not serious, are you? $\endgroup$
    – andselisk
    Jul 26 at 10:54
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    $\begingroup$ @Apurvium Oh. I'm sorry you see this comment as sarcastic. Let's walk it through. First, there is no URL, just plain text which makes this addition useless. Second, Wikipedia is not a reliable source. Third, you ignored two comments regarding unrealistic colors and claimed you added references, which is not true. Fourth, "take this" part is what should've been added a reference in a first place. You may feel upset about critics, but others are equally upset about the quality and transparency of your answer. $\endgroup$
    – andselisk
    Jul 27 at 4:05

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