For the compounds CCl4, Br2, HF, LiBr there are four possible boiling points -161.4 degree C -34 degree C 102.3 degree C 1465 degree C Please assign one boiling point to each one of the compounds....

For the compounds CCl4, Br2, HF, LiBr there are four possible boiling points

-161.4 degree C

-34 degree C

102.3 degree C

1465 degree C

Please assign one boiling point to each one of the compounds. Please give the reason?

Asked on by islnds

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valentin68 | College Teacher | (Level 3) Associate Educator

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Let us first specify the type of chemical bonding in each of the above compounds. CCl4 has a covalent bonding (the hybridization of C atom is sp3), LiBr has an ionic bonding, Br2 has also a covalent bonding and HF presents an weak hydrogen bonding (also named as dipole-dipole bonding).

 Since the strongest bonding is the ionic one, one expect that LiBr has the highest boiling point. Also, since the weaker bonding is the dipole-dipole interaction, one expects that HF has the lowest boiling point.

Now to distinguish between CCl4 and Br2 (both having covalent bonding) we need to observe that the Br2 molecule is totally symmetric and because of this the bonding is 100% covalent, whereas in the molecule of CCl4 there are 2 types of different atoms each of these having different electronegativity. It means that in CCl4 the bonding will have a certain (small) percentage of ionic bonding also.

Therefore the boiling point of CCl4 should be a bit higher than the boiling point of Br2.

Now we can assign a boiling point to each of the above compounds

LiBr : 1465 degree C (almost perfect ionic bond)

CCl4 : 102.3 degree C (covalent bond)

Br2 : -34 degree C (almost perfect covalent bond)

HF : -161.4 degree C (weak hydrogen bond)

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