Explain why the enthalpy of vaporization of vanadium
(460 kJ/mol...

Question

Explain why the enthalpy of vaporization of vanadium 
(460 kJ/mol) is much larger than that of zinc (114 kJ/mol).

Accepted Answer

Hello everyone today we are being given the falling problem, account for the large difference in the entropy of vaporization of tungsten and mercury. So the first thing that we have to recall is what's known as band theory. So what is band theory? Well, band theory is as its name suggests a theory that states that transition metals or the metals in the middle of the periodic table that these medals contain both an S. And a D orbital that can overlap. And when they overlap, this forms in what's known as an S. D. Band, which is in simple terms composed of three bonding orbital's and three and high bonding orbital's. So now that we have this definition of bond theory out of the way we can go ahead and account for any trends with our entropy of vaporization. So the entropy of vaporization so happens to increase with more bonding electrons. So we have more electrons that are in the bonding orbital are entropy of vaporization will increase and as such, the same can be said will be decrease and when it decreases when our entropy of vaporization is going to decrease with anti bonding electrons. In other words, electrons that do not bond with each other or with one another another for three. Our third step we're going to account for the entropy of vaporization of transition metals specifically. And we can say that they reach A maximum of six valence electrons. This is due to the transition metals in this case being in group six B. And so if we have six valence electrons, we essentially have filled bonding orbital's and unfilled anti bonding orbital's and so now we have to account for which one of these will have the filled bonding and which will have the unfilled anti bonding out of mercury and tungsten. So for tungsten we look at its noble gas configuration, this is going to be starting with a noble gas that is directly before tungsten, which in this case is xenon and then we start with the regular electron configuration. So we have five D 4, six S, two and we simply have to add up their exponents to see how many valence electrons we have. So tungsten we have six valence electrons and with mercury, the noble gas that precedes it directly is also Xenon And its noble gas configuration is five D 10, six S 2. And so if we add up these exponents, this adds up to 12 valence electrons. So simply put, we can finally conclude that our tungsten here has filled bonding molecular orbital's and unfilled anti bonding orbital's since it directly matches up with what we stated before. And with that we have our answer overall, I hope that this helped and until next time

Watch next