Here is my question
Niobium has a density of $8.57 \pu{g/cm^3}$ and crystallizes with the body-centered cubic unit cell. Calculate the radius of a niobium atom.
Here is my solution
Asked
Active
Viewed 7,998 times
1
Mascular
- 21
- 1
- 2
-
The thing I cannot get is there are too many different answer for this atomic radius questions. I mean like I've seen three different answer on websites. I've rechecked my question but couldn't find anymistake. Take a look here: http://www.gordonengland.co.uk/elements/nb.htm and here https://www.periodni.com/nb.html Here is anohter different answer too: http://www.elementsdatabase.com/Niobium-Nb-41-element/ Are those all radius correct? – Mascular Dec 04 '17 at 19:32
-
5That answer is clearly wrong because you do not give units! Please do so, or else we may think that your answer is $100$ times as large as you likely intended (assuming length is in S.I. units). – Oscar Lanzi Dec 04 '17 at 19:56
-
The answer is not certain. As I said take a look that sites. You will see how right I am. Everyone finds different answers. According to Wikipedia, it seems $1.46 \text {pm}$. – Mascular Dec 04 '17 at 20:22
-
According to Wikipedia, answer is $1.46$, on other sites, $1.45$ and $1.44$. on my textbook, $1.43$. Moreover, I rechecked my solutions twice. I'm sure about I didn't do mistake. – Mascular Dec 04 '17 at 20:33
-
using crystallographic density does not account for free space in the material due to vacancies or voids at the grain boundary. Any answer, while close will be off from the true value by a small amount. – A.K. Dec 04 '17 at 23:41
-
1Uh ... how is the radius of an atom roughly $1$ picometer? I seem to recall that the Bohr radius of the hydrogen atom is $51$ picometers and that is just hydrogen. I would guess that the decimal points in the answers quoted here are not really there! – Oscar Lanzi Dec 04 '17 at 23:49
-
@A.K. What do you mean? – Mascular Dec 05 '17 at 05:03
-
@Mascular your equation assumes that you are working with a single crystal with all atomic sites occupied. real materials have many crystals and vacancies in the lattice where atoms should be. – A.K. Dec 05 '17 at 23:22
-
@A.K. - Huh?!? the only way to do the problem is to assume a perfect crystal. Also I highly doubt that vacancies are cause of variations in tabulated values for the radius. Frankly anyone who believes that an atom has one "real" radius is kidding themselves. The diameter of an orange depends on how hard you squeeze. So it is with atoms. – MaxW Dec 06 '17 at 00:46
-
@Mascular - You did make two minor mistakes in your calculations. First you used 93 for the atomic weight instead of 92.906. Second in the next to the last step you truncated to $r^3 = 0.29\times10^{-23}$. You should have carried more significant figures and only rounded to 3 when you got to $r$. – MaxW Dec 06 '17 at 00:54
-
@MaxW I'm sure that the tabulated values are from gaseous species or TEM observations, I was talking in regard to the OP's value. I know it may seem trivial but the results are real. If you do the arithmetic for the density of iridium and osmium, based on atomic radius you find that iridium should be denser, though empirically this is not observed; due to a high valency concentration in iridium, osmium actually ends up being a denser bulk material than iridium. – A.K. Dec 06 '17 at 05:15
-
Do atomic physicist know this? – Mascular Dec 06 '17 at 14:06
-
@A.K. - The assumption that atoms are hard spheres is a bad one which was my point about the orange. So I would expect values calculated from the bulk density, gases species, and TEM to be different. Also if you want to look at trends iridium and osmium crystallize differently. – MaxW Dec 06 '17 at 15:05
-
@MaxW yes they do crystallize differently but FCC and HCP have the same atomic packing factor so that is moot. – A.K. Dec 06 '17 at 18:08
-
@A.K. - That is the point. It is not moot. There is a reason that some metallic atoms prefer FCC or HCP. How the atoms pack effects how they interact (bond) which in turn effects the distance between them. Atoms are not ball bearings which would have absolutely defined packing fractions. – MaxW Dec 06 '17 at 18:15
-
All answers are correct. They are quite smilar. – Mascular Dec 06 '17 at 18:36
1 Answers
4
This is how I would do the calculation...
Given data:
$$\text {Weight of crystal} = \pu{92.90638 g/mole} \ce{->[Round] \pu{92.906} g/mole}$$ $$\text {Density} = \pu{8.57 g/cm^3}$$
Ok, the density is good to only 3 significant figures so the answer shouldn't have any more than that. But doing the whole problem, I'll carry 5 significant figures throughout all the intermediate calculations to try avoid rounding errors within the multiple calculations. I'll round the final result to 3 significant figures.
$$\text{Volume of mole} =\text{V}_{mole} = \frac {92.906}{8.57} = \pu{10.841 cm^3/mol}$$
In BCC, there are 2 atoms in one unit cell. Also note that the accepted figure for Avagadro's constant to five significant figures is now $6.0221\times10^{23}$
$$\text{Volume of unit cell} = \text{V}_{cell} = 10.841 \times \frac {2}{6.0221 \times 10^{23}} = 3.6004 \times 10^{-23}\text{ cm}^3$$ A unit cell is a cube with each side being $a$ $$\text{V}_{cell} = a^3$$ $$\therefore a = \sqrt[3]{3.6004 \times 10^{-23}} = 3.3021\times 10^{-8}\text{ cm}$$
But the BCC crystal, spheres of radius, $r$, packed inside a cube with side, $a$, the geometric relationship between $a$ and $r$ is: $$a = \dfrac{4}{\sqrt{3}}r$$ so
$$r = \dfrac{\sqrt{3}\times(3.3021\times 10^{-8})}{4} = 1.4299\times 10^{-8}\text{ cm}$$
Now rounding $r$ to 3 significant figures gives
$$r = 1.43\times 10^{-8}\text{ cm} = 143\text{ pm}$$
You shouldn't chase values through websites and literature. You can only calculate given whatever data you're suppose to take as given. For instance no temperature was given for the density which certainly changes with temperature. Also, to make the point again, with a calculator you should carry extra significant figures in the intermediate calculations to try to avoid round off errors in the final result(s).
