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Although hexafluoroethane\[({{C}_{2}}{{F}_{6}},\text{ }b.p\text{ }-7.9{}^\circ C)\]and ethane \[({{C}_{2}}{{H}_{6}},\text{ }b.p\text{ }-8.9{}^\circ C)\] differ very much in their molecular weight, their boiling point differ only by\[10{}^\circ C\]. This is due to
(A) Low polarizability of F
(B) Nearly similar size to F and H
(C) Both (a) and (b)
(D) Neither of the two

Answer
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Hint: Electronegativity of fluorine is very high as compared to hydrogen because fluorine belongs to group \[{{17}^{th}}\], left side of the periodic table while hydrogen is the member of group \[{{1}^{st}}\], on the right side of the periodic table. As a long period in the periodic table an electron adds to the same orbital as before and the nuclear charge increases with every addition of an electron. Due to this effective nuclear charge on every electron increases and thus, fluorine becomes dense (high molecular weight) as compared to hydrogen.

Complete Step by Step Solution:
As a hint, it is clear that fluorine has a much greater molecular weight than hydrogen but still,they are slightly different in boiling point by \[10{}^\circ C\]. This is because of the comparable size of fluorine with hydrogen.

Fluorine’s atomic number is 9 whereas the atomic number of hydrogen is 1. So, fluorine should be large as compared to hydrogen but due to the high electronegativity value of fluorine (4), its size is compact and becomes comparable to hydrogen. As both have a similar size so both form similar bonding with carbon to form \[{{C}_{2}}{{F}_{6}}and\text{ }{{C}_{2}}{{H}_{6}}\]. Thus, their boiling point (energy needed to break bond) are quite similar and quite different because of the electronegativity of the fluorine atom.

Fluorine polarising power (not polarizability) is more thus, in bonding with carbon (\[2.5\text{ }electronegativity\]), fluorine tends to attract bond electrons towards itself. And the bond becomes partially polar and due to this, on some heating bond easily get broken because of polarity in between the bond. So, its boiling point is less \[({{C}_{2}}{{F}_{6}},\text{ }b.p\text{ }-7.9{}^\circ C)\]. Whereas carbon \[\left( 2.5 \right)\]and hydrogen \[\left( 2.2 \right)\]bond is non-polar because of less electronegativity difference between carbon and hydrogen \[({{C}_{2}}{{H}_{6}},\text{ }b.p\text{ }-8.9{}^\circ C)\].
Thus, the correct option is C.

Note: Polarizability is the tendency of any element to get polarised or favour charge separation in its electron cloud or can say the ability to get polarised. Whereas polarising power is termed as the power to make others polarise. Fluorine has great polarising power but not polarizability.