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:
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.