How compounds exist in solid state ?
Compounds exist as ions in solid state. Consider the following example –
PCl5 molecule – This molecule has trigonal bipyramidal (TBP) geometry, which is a little less stable as compared to octahedral or tetrahedral geometry. As there are 5 substituents , this molecule is sp3d hybridized.Thus, in solid state , it splits into two ions as follows –
[PCl6] – ion has six substituents/ligands and so it has an octahedral geometry.
The central atom phosphorous is pentavalent i.e it has 5 valence electrons. In this ion it is attached to six chlorine atoms.This is possible as phosphorous atom can have an expanded octet i.e it has valence vacant d-orbitals where it can accommodate extra electrons from another chlorine atom. One extra electron , which is accommodated in P atom’s d-orbital (which comes from another PCl5 molecule) gives this ion a negative charge.
[PCl4] + ion has four substituents/ligands and so it has a tetrahedral geometry. It is sp3 hybridised.
Although phosphorous is pentavalent , in [PCl4] + ion it is attached to only 4 substituents in this ion. The fifth valence electron is transferred to another PCl5 molecule to form [PCl6] – ion. As one electron is transferred to another species, the net charge on this ion is +1.
The transfer can be shown as follows –
Many compounds show this phenomenon.Some examples are –
- PBr5 →PBr4++ Br –…. Note that this molecule forms ions differently than PCl5 as size of bromine atom is bigger than chlorine and so six bigger Br atoms do not sterically fit around P atom.
- PI5 →PI4++ I –
-
Cl206 → Cl202+ + Cl204–
In our next post we shall study another rule which is important in the hybridization series.Till then ,
Be a perpetual student of life and keep learning ..
Good day !
References and Further Reading –
PI5 doesn’t show solid state hybridisation
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