Atomic Radius & Density of Transition Metals - Online Tutor, Practice Problems & Exam Prep

• Atomic radius decreases from left to right across a period and increases down a group for main group elements, while transition metals show a more gradual change due to constant electron count in the outermost s orbital.
• In transition metals, atomic radius changes slightly across periods 5 and 6 due to the lanthanide contraction, which increases effective nuclear charge without increasing shell size, leading to poor shielding by f orbital electrons.
• Density of transition metals increases both across a period and down a group, primarily because mass increases while volume remains relatively constant due to the addition of electrons to inner d or f orbitals and effects of lanthanide contraction.

Atomic Radius

Now recall that when it comes to atomic radius of main group elements, it's going to decrease from left to right across a period and going up a group. And when it comes to transition metals though, it's going to follow the same general trend, but change in size is more gradual. Now what's the reason for this? Well, across a period, the number of outermost electrons, which is some shell number which we're going to use the principal quantum number N.

And then these transition metals have an outer shell which is in the S orbitals and they can have one or two electrons involved. You will put two electrons and that's going to be constant. So for example, if we're taking a look here, we say that the general trend again is it decreases as we head towards the right top portion of the periodic table. Here we have our transition metals that we're paying attention to. We're only paying attention to rows 4-5 and six. 7 or row seven is more unpredictable because it's more synthetically made elements.

Heavy elements, their behavior is not easy to describe, so we're going to leave out the 7th row. If we're looking at rows 5 and six in particular, we can say that we know that the atomic radius decreases as we head up a group. But if we're comparing their atomic radiuses here, we can see that in some cases they stay the same and in other cases they only decrease slightly. If we're to take a look at TC and RE here, TC if we're to write its electron configuration, we'd have Krypton 455 5S22 and then Re rhenium would be Xenon 4F1414 5D556S22.

We see that their outer shells for TC is in the 5th shell has two electrons in the outer shell. Here is 2 electrons in the 6th shell. The electrons that we're adding in are getting added to either D or F orbitals, which are in the inner shells. The outer shells stay in the same. We're just packing in more electrons in the center or in the shells that are closer to the nucleus, because they are shells staying constant in size. We just see small variable changes in our atomic radius between the transition metals.

Again, this change in atomic radius is more prominent amongst main group elements, and it's only a gradual change within our transition metals. And again, that's having to do with us having a constant number of electrons in the outer shell. And later on, we'll learn about other phenomenons that result in this gradual change in our atomic radius.

Atomic Radius and Density of Transition Metals Example

Which element from each pair would you expect to have the biggest atomic size? So remember, atomic size, atomic radius. As we head towards the top right corner of our periodic table, we should expect it to decrease.

So if we take a look at the first one, we have NI versus TI, nickel versus titanium. Titanium. If you look at your periodic table, it's more to the left than nickel, so titanium would be larger. For the next one we have TC and Ru. Again, take a look at the periodic table. Look to see where do you see them. TC is to the left more so than Ru, so TC would be larger.

Next RH and MB. Again, take a look at your periodic table. Look and see where you find each of these elements. NB is more to the left again than RH. So NB would win. And then we have Y and AG, Yttrium versus silver again, which one is more to the left? Yttrium would be more to the left. So yitrium will be the larger out of these pair.

So here we've identified which element in each pair would have a larger atomic radius. Again, as we head towards the top right corner, we expect our atomic radius to decrease. So if you want the larger element, look which one is more to the left and lower down. These would result in larger atomic radii.

Lanthanide Contraction

Now when we look down a group, we can say that periods 5 to 6 of transition metals have relatively constant size. Now this is due to what we call the lanthanide contraction. Now according to the lanthanide contraction, we're going to have an increase in our effective nuclear charge which is ZEFF and that's due to 4 F4 sub shells, these 4F electron shield poorly.

Remember, effective nuclear charge itself is just the attractive force between our nuclear and the surrounding electrons. We're going to say that. Recall that if we increase the number of electrons in a shell, this is going to increase our attractive force. So just imagine that you are adding additional electrons as we go from 5 to 6. What are we adding or adding electrons in F orbitals? The shell number is not increasing. Those F orbital electrons are going into our inner shells, which are closer to the nucleus.

You're packing in more electrons. You're adding more protons to your nucleus. There's going to be a greater attraction between those protons in the nucleus and those surrounding electrons, those F orbital electrons. This causes an increase in our effective nuclear charge, pulling them closer and thereby causing your overall atomic size of your element to contract just slightly. So here, this is going to cause a decrease in expected atomic size of 6 transition metals.

So basically, once you introduce your F orbital electrons, without increasing the shell size, you're just causing more of an attractive force between your nucleus protons and your surrounding electrons. Now here, if you want a refresher in terms of this, make sure you take a look at our periodic trend under effective nuclear charge for more information. Here we just have an example of our nucleus, which has our protons and our neutrons, and then we have these different shells.

We're starting off by looking at the 4th shell because that's where we first have the introduction of our F orbital electrons packing them in, causing a greater attraction between my nucleus here and the F orbital electrons that I'm adding. So just remember, while we're looking down a group, we have this lanthanide contraction which is supported by the introduction of F orbital electrons.

Atomic Radius and Density of Transition Metals Example

In this example question it says which of the following transition metals would you expect to be larger but are actually same or nearly same size as technetium? Well, if we look on our periodic table we have technidium which is in. 5. So if we look just one row down below it in. 6:00, we'd have rhenium.

Remember we saw that Rhenium and Tecnidium have similar atomic radius. That's because of the lanthanide contraction. When we go into the sixth period of the periodic table, we have the introduction of F orbital electrons. The introduction of more of these electrons into one of our inner shells causes a greater attraction between those electrons and are more positive nucleus.

This is termed our effective nuclear charge. So here option D would be our final answer.

Density

Now when it comes to the density of our transition metals, just remember that increases as the mass of the metal increases. We're going to say an increase in density down a group is more significant then across a period. And we're going to say here that the general trend is as we head from the left to the right side of our periodic table in relation to our transition metals, in this case our density is going to increase and that's because our mass is increasing.

And then we're going to say as we head down a group our mass also still increases. Now remember as we're heading across a group our size is staying relatively constant because we're just adding additional electrons to either our D or F orbitals. So our volume is staying more or less the same. d=mV. So your volume is staying more or less the same, but your mass is increasing. This causes an increase in your density.

If we're looking at going down a group, well, as we're going down a group, our mass is still increasing because we're going from lighter elements up top to heavier elements on the bottom. And then we have lanthanide contractions that are possible. So that kind of keeps your volume more or less close to the same as you're heading down a group. Again, your volume is staying relatively the same. Your mass is increasing greatly. This also causes an increase in your density.

So just remember the formula for density. Remember with our transition metals, because we're being in the same row or because of the lanthanide contraction, our volume is staying more or less the same, but our mass is increasing. This causes an overall increase in our density.

Atomic Radius and Density of Transition Metals Example

Hey everyone. So in this example question, it says identify transition metal with the highest density O. Remember our density is going to increase as our mass increases. And also remember that there's going to be a greater spike in density as we go down a group than going across a period. So going down a group, there's more significant increase in our density.

If we take a look here at our options, we'd say here that A&B represent transition metals that are going to be in four. And if we look at C&D, well, they represent elements that are in eriod 6. Ince C&D are lower down in terms of groups, they're in row six. Their densities are going to be more impactful. They're going to have a more significant higher value. So that means A&B are out.

Now here if we look at C versus D, remember the general trend is as we had towards the right across a period, our density increases and as we go down to group, our density increases. Osmium C is more to the right than option D So osmium would have a larger density. So here my answer would be option C.

So just remember we look and see where they are in relation to each other on the periodic table. The ones lowered down in the group would have a more significant higher density than the ones higher up a group. That's what eliminate A&B. And then to break the tie between C&D, just remember more towards the right of the periodic table equals a higher density, giving us option C as our final answer.