What are the shapes of these compounds?

Or rather, my question is why are these the shapes for these compounds: ZnCl2 and CdCl4 2-

Apparently ZnCl2 has a molecular geometric shape that is linear and after working with the Lewis structure I can see why: One of the Chlorine atoms forms a double bond with the Zinc. But why can't two double bonds form in this structure leaving me with, say, a T-shape? Is it because the double bonded Cl atom has satisfied the octet rule?

Now for CdCl4 2-, this one was tricky. Apparently its shape is tetrahedral but I don't get why. Can someone explain why?

3 Answers

  • This is a poor question because you cannot apply VSEPR theory to these cmpds. ZnCl2 (partly ionic) forms three (four?) solid variations at RT that may be considered as close packed Cl^- ions with Zn^2+ ions occupying tetrahedral holes; the structure formed depends on trace moisture. "All four structures consist of tetrahedral ZnCl4 groups sharing vertices with 4 others."

    A. F. Wells, Structural Inorganic Chemistry 5th ed (1993) p 412

    ZnCl2 boils at 732°C (CRC Handbook 4-99) Zn [Ar] 3d^10 4s^2 so ZnCl2(g) is an AX2 system linear: "Electron diffraction studies show that ZnX2 (X = Cl, Br, I) have linear X-Zn-X structures in the gas phase."

    N. N. Greenwood, A. Earnshaw, Chemistry of the Elements 2nd ed. (1997) p 1211.

    You can't use VSEPR on cmplx ions. This is Cd^2+ [Kr] 4d^10 and it involves Cl^- forming dative covalent bonds to Cd^2+; Cl:→Cd^2+ The Cd^2+ is the spherically symmetric d^10 so yes [CdCl4]^2- is tetrahedral. CdCl3]^- and [CdCl5]3- are also known.

    VSEPR is best suited for covalent cmpds of the main group elements.

    N. N. Greenwood, A. Earnshaw, Chemistry of the Elements 2nd ed. (1997) p 1215.

  • ZnCl2 does not have any double bonds. Zinc is in the +2 oxidation state and the Chlorines are in the -1 oxidation state. The 2 Chlorines form normal single bonds to the Zinc which will give a linear molecule.

    In CdCl4 2-, the 4 Chlorines are in the -1 oxidation state and the Cadmium is in the +2 oxidation state (this is why there are 2 negative charges on the molecule). 2 of the Chlorines form normal single bonds with Cadium. The other 2 Chlorines share 2 of their electrons with the empty p orbitals on the Cadium. With 4 ligands and no lone pairs, the shape is a tetrahedron arrangement.

  • The Chloride ion being negative species wants to repel the adjacent chloride ion also negative. Remember like charges repel in a similar way to like poles on a bar magnet. Yet they all want to ligate to the Cd^2+ ion. .

    So they ligate onto the Cd ion but space themselves so that they are as far apart as possible. In 3-dimensional space they form the tetrahedron. In measuring any one Cl-Cd-Cl angle you will find it to be approximately 109.5 degrees.

    I hope that clarifies it for you.

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