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Bismuth(III), if I am not mistaken, has only two valence electrons in the 6s orbital. However it is able to form 3 bonds with hydroxide. This goes against my intuition. How is it able to form 3 bonds with 3 hydroxide molecules with only 2 valence electrons?

my best guess is that i am mistaken in thinking that the bismuth(III) is forming 3 covalent bonds when in actually it is forming ionic bonds. As a result, bismuth does not need to offer/share 3 electrons to each hydroxide, but simply take 3. Is this correct?

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There isn't a unique sturcture for bismuth hydroxide, but ions such as

$$\ce{[Bi6O4(OH)4]^{6+}}$$

are involved, where the 6 Bi atoms are at the corners of an octahedron, none of the Bi atoms are directly bonded to each other, and O atoms bridge pairs of Bi atoms.

See Solvation of the Bismuth(III) Ion by Water... Inorg. Chem. 2000, 39, 4012-4021 as a starting point.

But supposing the structure was $\ce{Bi(OH)3}$, Bi3+ has two electrons, each OH- has 8 electrons, so Bi could share 2 electrons from each OH- to form an octet.

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  • $\begingroup$ Bismuth hydroxide has a trigonal planar geometry, meaning it has no lone electron pairs. It would only be borrowing 2 electrons from each hydroxide, giving it only 6 valence electrons, not an octet. $\endgroup$ – Sam D20 Dec 4 '14 at 19:43
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    $\begingroup$ @ByronS If Bi3+ has two valence electrons, and borrows 2 from each of 3 hydroxides, isn't that 8 altogether? $\endgroup$ – DavePhD Dec 4 '14 at 19:58
  • $\begingroup$ Yessir, but in doing so, it would take on a "pyramidal" shape in terms of geometry (3 covalent bonds, 1 lone electron pair). However, my resource is telling me that bismuth(III) hydroxide takes on a "trigonal planar" geometry, meaning it forms 3 covalent bonds and has no leftover lone electron pairs. Is my resource incorrect? $\endgroup$ – Sam D20 Dec 4 '14 at 20:59
  • $\begingroup$ @ByronS I don't have any resources that even say the such a molecule exists, are you sure it isn't boron instead of bismuth in your resource? $\endgroup$ – DavePhD Dec 4 '14 at 21:08
  • $\begingroup$ No, it is bismuth. My resource is most likely wrong, considering it is simply a question on an old exam made by my professor. Question asks for the molecular geometry of "Bismuth(III) Hydroxide". According to your logic, which also makes sense to me, it should be forming a pyramidal shape, but my professor claims that it takes on a trigonal planar shape. I'll have to speak with him about it soon. Thank you for the help. $\endgroup$ – Sam D20 Dec 4 '14 at 21:34

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