Tin (II) sulfide, a mainly covalent compound, bonds using its p orbitals only (the 5s orbitals are considered to be part of the core in divalent tin). Its structure is made up of Sn–S–Sn–S chains, which are further stabilized by Sn–S bonding across the chains:

Tin (II) sulfide structure

How would VB or MO theory explain the fact that the tin molecule bonds three times, when only two of its p orbitals are half filled (Electron configuration: [Kr]5s24d105p2), while the third is empty? The observed bond angles of ca. 90° (96°/88°) would indeed suggest plain p orbital bonding, as suggested by this paper, but how is this possible?

And how could sulphur bond three times (Electron configuration: [Ne]3s23p4) with similar bond angles, when one p orbital is full?

The only thing that comes up to my mind is a dative covalent (co-ordinate) bond, in which sulphur provides both p electrons, thus permitting coordination. But then, the tin atoms would have a negative charge attached to them (due to having more electrons around than protons) and the sulphur atoms a positive one. Wouldn’t this be energetically unconvenient, considering that the compound has ca. 9% of ionic character, which depends on the tin ions being positive and the sulphur ions negative?


The dative bond assumption clashes against the bond lengths observed in tin (II) sulfide. The paper I cited above states that the Sn–S bond length on the chains are 2.68 Å long, while those between the chains are 6 pm shorter (2.62 Å). The IUPAC Gold Book says that dative bonds

are distinguished by their significant polarity, lesser strength, and greater length

when compared to covalent bonds. This is exactly the opposite of what happens in tin (II) sulfide.


1 Answer 1


I think your conclusion of a dative bond between sulphide anions and tin cations (assuming pure ionic bonding, which is clearly not the case if you study the image closely) is correct.

Note that the charges you are observing are nothing but formal charges. In a macroscopic picture, tin is still less electronegative than sulphur and the electrons would be considered sulphur’s e.g. when determining oxidation states. The mere fact that an electron pair with correct symmetry is adjacent to an atom is enough to stabilise said atom’s empty orbital.

  • $\begingroup$ And yet, the charge on the tin atoms is more negative with the dative bonds than it would be without them... So how is the bond convenient? Wouldn't single polymer chains be more thermodynamically stable? $\endgroup$
    – Corundum
    Commented Dec 29, 2015 at 10:25
  • $\begingroup$ @Corundum Think octet rule. That way, all three equivalent p-orbitals take part in some binding interaction; none is left alone in a non-bonding state. At least, that is how I would explain it quickly. $\endgroup$
    – Jan
    Commented Dec 29, 2015 at 12:30
  • $\begingroup$ and what about the problem with bond length (explained in the addendum)? $\endgroup$
    – Corundum
    Commented Dec 30, 2015 at 20:21
  • $\begingroup$ @Corundum That requires me to do more thinking. I may be back later (or delete this answer; I’ll have to see). $\endgroup$
    – Jan
    Commented Dec 30, 2015 at 21:29

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