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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
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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
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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
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D
the attractive forces between the SiF4 molecules are strong whereas those between the AlF3 molecules are weak
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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 possible
Aluminum 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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