Bond Angles


by gravenewworld
Tags: angles, bond
gravenewworld
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Dec16-04, 11:15 PM
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Which compound has the smallest bond angle in the series


NH3 PH3 AsH3
PI3 AsI3 SbI3



I would think that the molecule with the smallest center atom would have the smallest bond angles because the lone electron pair is held closer to the bonds which means it can push them closer easer, so I would pick NH3, and PI3. Is this correct reasoning?
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Gokul43201
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Dec17-04, 10:48 AM
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Actually, it's the other way round.

Keep in mind that there are two different repulsive forces in play here. There's the lone pair-bonded pair repulsion, and the bonded pair-bonded pair repulsion, of which the former is greater in magnitude, largely because the lone pair-bonded pair distance is typically smaller due to the lone pair being closer to the central atom. At large separations (large compared to the size of the outer atoms), the first repulsion is nearly twice as large as the second...but not when the separation starts to get small.

Now, N is a smaller central atom, and additionally, it is more electronegative. So, as you correctly pointed out, it attracts all the electrons closer to itself. While this increases the LP-BP repulsion, it clearly also increases the BP-BP repulsion. So which one dominates ?

At smaller distances, the increase in repulsion between BPs is greater (note: the increase is greater; not the repulsion itself), because you can only put two H atoms (or any other atom for that matter) so close to each other. When the central atom gets smaller and smaller, steric (repulsive) effects between the outer atoms prevent them from "touching" each other. So, smaller central atoms (N in NH3) give rise to larger bond angles. So, the H-N-H bond angle should be significantly larger than the H-P-H angle.

When the central atom is big enough that the H-H distance is large compared to the size of the H atom, the steric effects grow vanishingly small. Hence, the decrease in bond angle from say, AsH3 to SbH3 will be negligible.

To summarize : (in terms of the bond angles) NH3 >> PH3 > AsH3 ~ SbH3 (and all angles are less than 109.5 deg)

The same is true for the iodide series. And maybe all the more so, because I is a fairly large atom. In fact a likely reason why NI3 does not exist is that the N atom being so small in comparison, the steric forces between the I atoms would be really enormous, making the molecule pretty unstable.

As a cautionary note, keep in mind that it's harder to apply this analysis when varying the outer atom, because then, the effect of changing electronegativity counters the effect of changing atomic size. So, in the series PF3, PCl3, PBr3, PI3, while the outer atom gets bigger it also gets less electronegative, and so it's hard to tell if the bonded electron pairs are farther away or closer. In fact, I think (except for PF3) the bond angles in this series are almost identical.


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