5
$\begingroup$

Many books state that $R_\ce{N} > R_\ce{O}$ which is in accordance with the general trend. However, some books say that $R_\ce{O} > R_\ce{N}$ because of repulsion caused by pairing of electrons.

Which one is correct and should be followed?

Older version of NCERT, cengage, arihant and JD Lee state the latter, while the newer version of NCERT, Wikipedia and many websites say the former. Why was the theory discarded?

Note: van der Waal's radii follow the general trend.

$\endgroup$
1
  • $\begingroup$ The radius of atoms decreases from left to right across the table. As the smallest radius per row will always be the noble gas, then opposite the noble has having the largest radius in the row. $\endgroup$ Commented Sep 6, 2020 at 10:48

1 Answer 1

8
$\begingroup$

The van der Waals radius of nitrogen is larger than that of oxygen, and has been calculated as such for quite a long time.

$$ \begin{array}{lll} \hline \text{Reference} & R_\ce{O} & R_\ce{N} \\ \hline \text{Pauling, 1939} & 1.40 & 1.5 \\ \text{Bondi, 1964} & 1.52 & 1.55 \\ \text{Zefirov, 1974} & 1.29 & 1.50 \\ \text{Gavezzotti, 1983–1999} & 1.40 & 1.50 \\ \text{Batsanov, 1995} & 1.51 & \\ \text{Wieberg, 1995} & 1.5 & 1.6 \\ \text{Rowland, 1996} & 1.58 & 1.64 \\ \hline \end{array} $$

As is evident from the table, $R_\ce{N} > R_\ce{O}.$ An interactive webpage of this reference can be seen here, which has the van der Waals radii of quite a few other elements as well.

Edit:

After OP mentioned the abnormality in Covalent Radii, I checked the radii in Concise Inorganic Chemistry, by J.D. Lee. The Covalent radius of Nitrogen $R_N = \pu{0.77 Å}$, whereas that of Oxygen is the same; $R_O = \pu{0.77 Å}$. Lee mentions that the radii are taken from Tom Moeller's Inorganic Chemistry book (which is from 1952)

OP should also be aware of the method used for calculating covalent radii: The Covalent Radius is defined as half the length of a single bond between two uncharged atoms. According to a more recent reference2 (which is used by Wikipedia), $R_N = \pu{73.4 pm}$ and $R_O = \pu{70.2 pm}$

Even more recent references, such as (3) use a different method of calculating covalent radius:

The covalent radius for nitrogen was obtained from N–N bond distances in substituted hydrazines (only H or C atoms bonded to N) with three-coordinated nitrogen atoms, excluding all structures with an R factor larger than 10% or presenting disorder or errors. The resulting value, $\pu{0.706(13) Å}$, is the average of 2200 crystallographically independent data and was used throughout with three decimal figures for calculating other radii from N– element bond distances to minimize rounding-off errors.

The covalent radius for $\ce{O}$ was derived from a sample of 10,000 acyclic $\ce{C–O}$ bond distances, in a search limited to purely organic compounds with two-coordinate oxygen and four-coordinate carbon atoms with R ≤ 5%, and the value obtained was $\pu{0.676(28) Å}$

A comparision of covalent radii has also been provided by the authors:

$$ \begin{array}{lll} \hline \text{Author} & R_O\ (Å) & R_N\ (Å) \\ \hline \text{Cordero et al.} & 0.706(13) & 0.661(19) \\ \text{Alcock} & 0.702 & 0.659 \\ \text{Mingos} & 0.74 & 0.72 \\ \text{Butler and Harrod} & 0.75 & 0.73 \\ \text{Wells} & 0.74 & 0.74 \\ \hline \end{array} $$

References

  1. Batsanov, S. S. Van Der Waals Radii of Elements. Inorganic Materials 2001, 37 (9), 871–885. DOI: 10.1023/A:1011625728803. (PDF)
  2. Sanderson, R. T. “Electronegativity and Bond Energy.” Journal of the American Chemical Society, vol. 105, no. 8, Apr. 1983, pp. 2259–61. doi:10.1021/ja00346a026.
  3. Cordero, Beatriz, et al. “Covalent Radii Revisited.” Dalton Transactions, no. 21, 2008, p. 2832. doi:10.1039/b801115j.
$\endgroup$
7
  • $\begingroup$ Indeed. But here we are not talking about van der Waal's radii. $\endgroup$
    – DatBoi
    Commented Sep 6, 2020 at 9:01
  • $\begingroup$ Then what is it that we are talking about? $\endgroup$ Commented Sep 6, 2020 at 10:03
  • $\begingroup$ @Safdar the covalent radii! $\endgroup$
    – DatBoi
    Commented Sep 6, 2020 at 12:27
  • $\begingroup$ My bad. I'll do it right away $\endgroup$
    – DatBoi
    Commented Sep 6, 2020 at 12:34
  • $\begingroup$ Nice one! Can you also provide some info on the effect of repulsion of $e^{−}$s? $\endgroup$
    – DatBoi
    Commented Sep 6, 2020 at 14:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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