12
$\begingroup$

How do bond angles vary in molecules with a lone pair and central atom of different electronegativity, but in the same period so that electronegativity matters more than orbital size? Let's assume that the molecules I am comparing have the same substituents and one lone pair.

Will the molecule with the more electronegative atom in the middle have the lone pair electrons closer to the nucleus, so they take up more bonding space and the angles for the other substituents will be smaller?

$\endgroup$
  • 3
    $\begingroup$ see Why is the bond angle H-P-H smaller than H-N-H? $\endgroup$ – ron Oct 21 '14 at 19:18
  • $\begingroup$ @ron: I think the OP is comparing NH3 with H2O. $\endgroup$ – Abel Friedman Oct 21 '14 at 19:25
  • 1
    $\begingroup$ @AbelFriedman Would the arguments used in the above-referenced question\answer, especially those relating to electronegativity, change when discussing groups vs. periods? $\endgroup$ – ron Oct 21 '14 at 19:45
  • 1
    $\begingroup$ This seems really challenging to answer, since O vs. N involves different number of valence electrons and thus bonding patterns (e.g., comparing a lone pair to a bonding pair) $\endgroup$ – Geoff Hutchison Jun 9 '15 at 18:53
4
$\begingroup$

The electronegativity of the central atom does not really influence bond angles in any meaningful way. The size of atoms and the orbitals used and useable are much more important.

Sometimes, students may be taught concepts such as electronegativity or ‘lone pairs require space’ to explain why $\ce{H2O}$ has a smaller bond angle ($105^\circ$) than $\ce{NH3}\ (107^\circ)$ or similar. However, careful analysis of the molecular orbitals reveals that sterics are a much more important reason than ‘lone pairs requiring more space and thus making tetrahedral $\mathrm{sp^3}$ contract’. Indeed, as soon as a lone pair exists on a central atom, there is a tendency to have the lone pair occupy an orbital with an s-contribution as high as possible and to have bonding orbitals with a maximised p-contribution. p-orbitals, having a direction, can form stronger σ bonds due to better overlap.

This is why the bond angle drops quickly when comparing second-period to third-period central elements to close to $90^\circ$ but then hardly changes down the group. On the other hand, the reduced bond angles observed from ammonia to water are mainly due to more steric stress on nitrogen — which needs to accomodate an additional hydrogen and thus needs to space them out more.

Indeed, if you compare the bond angles of different $\ce{OX2}$ compounds, the only general trend is ‘the larger the second element’s atoms, the greater the bond angle.’ See the table below (all values taken from Wikipedia).

\begin{array}{ccc}\hline \ce{X} & \angle(\ce{X-O-X}) & \angle(\ce{X-N-X})\\ \hline \ce{H} & 104.5^\circ & 107.8^\circ \\ \ce{F} & 103^\circ & 102.5^\circ \\ \ce{Cl} & 110.9^\circ & -/- \\ \ce{Br} & 112^\circ & -/- \\ \hline \end{array}

The drop in bond angles from hydrogen to fluorine is sufficiently explained by the much longer bond lengths of the $\ce{OF2}/\ce{NF3}$ species when compared to the corresponding hydrogen species due to fluorine using p-orbitals for bonding where hydrogen uses its 1s orbital.

While the bond lengths also increase when going from fluorine to chlorine and bromine, the angle also does. Note especially $\ce{OBr2}$ whose bond angle is larger than the tetrahedral bond angle. Any argument involving electronegativity should assume a smaller angle here since the electronegativity of the partners drops from fluorine to bromine.

$\endgroup$

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.