Which solid in each pair has the higher melting point and why?
a. Fe(s) or CCl₄(s)
b. KCl(s) or HCl(s)
c. Ti(s) or Ne(s)
d. H₂O(s) or H₂S(s)

Question

Which solid in each pair has the higher melting point and why?

a. Fe(s) or CCl₄(s)

b. KCl(s) or HCl(s)

c. Ti(s) or Ne(s)

d. H₂O(s) or H₂S(s)

Accepted Answer

hey everyone in this example, we need to determine which of the molecules has a lower melting point value. We need to explain our answer. So in order to explain our answers, we want to recall upon the fact that the greater the strength of our inter molecular force present in the molecule will correspond to a higher value for melting point. So we're going to analyze our given molecules based on their inter molecular forces that are present that are keeping the atoms in the molecule held together. So we have our first molecule which is N. 02 nitrogen dioxide. So we have two oxygen atoms surrounding a central nitrogen atom. We want to recall that nitrogen is in group five a. And that corresponds to five valence electrons, meaning that two of its valence electrons are going to be in bonds. So we have three more valence electrons to fill in. We want to recall that nitrogen has a bonding preference of having a double bond. So three bonds and one lone pair. So we want to go ahead and place the last two valence electrons for nitrogen as a lone pair on our nitrogen. And so what that means is that now we're actually going to have a bent shape for our molecule. So we're going to redraw our molecule so that it's a bent shape And we can fill in the lone pairs on our oxygen atoms. Since we should recall that oxygen being in group six a. Should have six valence electrons. And right now we've only shown one of their six valence electrons. So we would have three sets of lone pairs, or sorry, two sets of lone pairs on each of our oxygen atoms. And actually this one should have only one lone pair since it would have 246 valence electrons filled in that way. So now that we have our molecule for nitrogen dioxide, we make note of the fact that it's a bent shape and we want to recognize that the inter molecular force present is going to be based on election negativity. So we want to recall that the election, sorry, the electro negativity trend on our periodic table increases as we go towards the top right of our periodic tables. And so there should be a negative, slightly negative charge on each of our oxygen atoms since our oxygen atoms would be more electro negative. So we'll have deep holes going in the direction of our oxygen atom. We also have a di pole in the direction of our lone pair electrons on the nitrogen atom which even though is less negative than oxygen has more negative character based on those electrons as a lone pair on the nitrogen. And so we have this third die pole which is disturbing the symmetry in this molecule. And so because it's an I bent shape, the disciples do not cancel. And so we can say that therefore this is a polar molecule meaning that our inter molecular force present is going to be the vander Waals dipole dipole force. And we should recall that this has medium strength. So let's make sure that everything is in view. So this has medium strength as an inter molecular force. So we would expect a higher value for melting point. So now we want to compare C. 02. So we would have carbon in the center surrounded by two oxygen atoms. Recall that carbon. And let's not draw a bent shape here because we don't know if carbon is going to have a lone pair yet. So we'll just draw a linear shape here. Recall that carbon on the periodic tables is in group four a corresponding to four valence electrons. So we'll have our base connections to the oxygen's representing four electrons. And to get to the bonding preference for carbon, we want to have a total of four bonds on carbons that we're going to make these double bonds and that would also complete the four valence electrons 123 and four for our carbon atom so that it has a neutral charge. So now we want to go ahead and add the lone pairs to our oxygen. Since we recognize oxygen being in group six A should have six valence electrons. Right now it has four in the bonds that shares with carbon and then we would have five and then six. So now our oxygen's are also neutral. And so even though we would say based on our election negativity trend for our or on our periodic tables, we would say that oxygen, since it's further to the right, has a slightly negative charge. And so we have di polls going in the direction of oxygen. However, we have symmetry, we have a linear molecule with symmetry. And so therefore the dye poles in the direction of oxygen cancel. And so we would say that C. 02 is a non polar Covalin molecule. And so we would say that the inter molecular force present is going to be our Vander wal's London dispersion force, which is a weaker force. And so therefore we would expect a higher value for its melting point. Or sorry, we would expect a lower value for its melting point since we have a weaker inter molecular force in this molecule. And so we can say that therefore based on what we've outlined above, our carbon dioxide has the weaker inter molecular force being again, as we stated, the London dispersion force Vander wal's Forest here. And so therefore it will have the lowest or we can say the lower melting point. And so this explanation here would be our final answer to complete this example. So I hope that everything I explained was clear. If you have any questions, please leave them down below. And I will see everyone in the next practice video

Which solid in each pair has the higher melting point and why?
a. Fe(s) or CCl₄(s)
b. KCl(s) or HCl(s)
c. Ti(s) or Ne(s)
d. H₂O(s) or H₂S(s)

Verified Solution

Key Concepts

Intermolecular Forces

Hydrogen Bonding

Melting Point

Which solid in each pair has the higher melting point and why? a. Fe(s) or CCl4(s)b. KCl(s) or HCl(s) c. Ti(s) or Ne(s)d. H2O(s) or H2S(s)

Verified Solution

Key Concepts

Intermolecular Forces

Hydrogen Bonding

Melting Point

Which solid in each pair has the higher melting point and why?
a. Fe(s) or CCl₄(s)
b. KCl(s) or HCl(s)
c. Ti(s) or Ne(s)
d. H₂O(s) or H₂S(s)