Question

# The melting point of $$\displaystyle \:AlF_{3}$$ is $$1291^{\circ}C$$ and that $$\displaystyle \:SiF_{4}$$ is $$77^{\circ}C$$ (it sublimes ) because :

A
there is a very large difference in the ionic character of the AlF and SiF bonds
B
in AlF3, Al3+ interacts very strongly with the neighbouring F ions to give a three dimensional structure but in SiF4 no such interaction is possible
C
the silicon ion in the tetrahedralSiF4 is not shielded effectively from the fluoride ions whereas inAlF3, the Al3+ ion is shielded on all sides
D
the attractive forces between the SiF4 molecules are strong whereas those between the AlF3 molecules are weak

Solution

## The correct option is B in $$\displaystyle \:AlF_{3}$$, $$\displaystyle \:Al^{3+}$$ interacts very strongly with the neighbouring $$F^{-}$$ ions to give a three dimensional structure but in $$SiF_{4}$$ no such interaction is possibleAluminum fluoride is ionic and has high melting point. Large aluminum cation has vacant d orbitals. It attains a coordination number of six towards the relatively small fluorine atom. Due to this, a three dimensional structure is obtained. This is not possible in silicon tetrafluoride. Hence, the melting point of silicon tetra fluoride is much lower than the melting point of aluminum trifluoride.Chemistry

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