1
$\begingroup$

When the ionic character of a bond is above 50% should I use ionic radius instead of covalent one when calculating the dipole moment?

The formula I have for dipole moment is $\mu = \text{(% ionic character)} \cdot e \cdot d$, where $e$ is the elementary charge and $d$ the sum of the two radii.

$\endgroup$
  • $\begingroup$ just added it. sorry for that $\endgroup$ – user1790813 Oct 16 '15 at 21:06
  • $\begingroup$ No need to apologise. I was just wondering. But now I wonder even more... 1) How do you calculate percentage ionic character? 2) What is $e$? 3) I assume $d$ is the sum of the radii of the atoms, and you are asking whether this should be taken as the sum of the ionic radii or the covalent radii? $\endgroup$ – orthocresol Oct 16 '15 at 21:28
  • $\begingroup$ percentage ionic character can be obtained from the periodic table taking into account the difference between electronegativities. e is the elementary charge. The problem with d is the one you just mentioned. $\endgroup$ – user1790813 Oct 16 '15 at 21:51
  • $\begingroup$ Related: Dipole moment - calculation of percentage ionic character $\endgroup$ – Loong Oct 16 '15 at 22:46
2
$\begingroup$

Personally, I think "50%" as a cut-off is very arbitrary to begin with. It's not like I can suggest anything better, since proper dipole moment calculations rely on quantum mechanics.

But, to answer the question:

The whole premise of my answer is that you want to get the sum of your radii to match the actual bond length in the compound. And to do that, you have to look at the context. For example, for $\ce{BF3}$, you might want to use covalent radii since the atoms are covalently bonded. And for a ionic compound like $\ce{KF}$, you'd better use ionic radii or you might run into some problems.

Now, note that the formula you're using is already extremely inaccurate in itself - if you take the covalent radii for B and F, you get $d = (82+71) \text{ pm} = 153 \text{ pm}$. And... the actual bond length is very short - it's $130 \text{ pm}$. Clearly, there is something about that compound that the crude model of adding covalent radii cannot account for. Still, it's better than if you were to use the ionic radii: $(41 + 119)\text{ pm} = 160 \text{ pm} > 153 \text{ pm}$ (sum of covalent radii) $> 130 \text{ pm}$ (actual bond length).


"Hey @orthocresol, using $\ce{BF3}$ isn't fair! That compound has crazy strong $\pi$ bonding between B and F!"

Okay. Let's look at $\ce{BCl3}$, $\ce{BBr3}$ and $\ce{BI3}$, where the $\pi$ bonding is weaker.

$$\begin{array}{|c|c|c|c|} \hline \ce{X} & \text{Actual B-X bond length / pm} & \text{Sum of covalent radii / pm} & \text{Sum of ionic radii / pm} \\ \hline \ce{Cl} & 175 & 82 + 99 = 181 & 41 + 167 = 208\\ \ce{Br} & 187 & 82 + 114 = 196 & 41 + 182 = 223 \\ \ce{I} & 210 & 82 + 133 = 215 & 41 + 206 = 247\\ \hline \end{array}$$

The covalent radii are closer every time.


"That's still not fair... Can't we just use a compound without $\pi$ bonding?"

Fine. Let's talk about the tetrahedral $\ce{BF4-}$ anion. There's no $\pi$ bonding here and as a result, the B-F bond length is increases to $145 \text{ pm}$. But the covalent radii still win.

You can extend the discussion to something like $\ce{SiX4}$. But, honestly, I think that representing $\ce{SiX4}$ as $\ce{Si^4+}$ and $\ce{X-}$ is not that a great an idea, to say the least.

Also, I'm lazy to dig up bond lengths in $\ce{AlCl3}$, so I'll stop here.


"Okay, now what about ionic compounds?"

You'd better use those ionic radii, because guess how those ionic radii are tabulated? They're calculated by measuring the distance between cations and anions in an ionic lattice and assigning a certain part of that to the cation and another certain part to the anion. So if you add them back up, they'll of course give you the internuclear separation that you want.

Let me use the simple example of $\ce{NaCl}$. You'll have to trust that I'm doing the right things here; if you're interested, you can read up on the rock salt structure. The unit cell parameter of $\ce{NaCl}$, denoted by $a$, is $564 \text{ pm}$. This corresponds to two times the distance between $\ce{Na+}$ and $\ce{Cl-}$ ions, so we could say that the actual "bond length" is $564 \text{ pm} / 2 = 282 \text{ pm}$.

Now, the ionic radii of 6-coordinate $\ce{Na+}$ and $\ce{Cl-}$ are $102\text{ pm}$ and $181\text{ pm}$ respectively. That sums up to $283\text{ pm}$ - a beautiful match. And the atomic radii? They're $191\text{ pm}$ and $199\text{ pm}$ for $\ce{Na}$ and $\ce{Cl}$ respectively. That sums up to $290\text{ pm}$ - pretty close but not quite as close as the sum of the ionic radii.


Data is taken from this website, Wikipedia:Ionic radius, Shriver & Atkins, Inorganic Chemistry, 6th ed., and Greenwood & Earnshaw, Chemistry of the Elements, 2nd ed. Apparently, the numbers in that Wikipedia figure are taken from R. D. Shannon, Acta Cryst., 1976, A32, 751, which is in fact one of the most widely-used sources of ionic radii (all inorganic textbooks reference it) - so it shouldn't be that bad.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.