2
$\begingroup$

I'm wondering what is the exact reasoning to explain why in $\ce{GeX_2}$ the angle $\widehat{XGeX}$ is smaller when $\ce{X}=\ce{F}$ than when $\ce{X}=\ce{I}$.

I ask this question because I had an exams some days ago and I failed in this question and for me my reasoning is not false.

What I said is that

From $\ce{F}$ to $\ce{I}$ the difference of electronegativity between the germanium and the halogen is decreasing. Then the electrons in the $\ce{Ge-X}$ bond are more near $\ce{X}$ when $\ce{X}=\ce{F}$ than when $\ce{X}=\ce{I}$. Then the two bonds repulse themselves more for $\ce{GeI_2}$ than for $\ce{GeF_2}$ and so the angle increase from $\ce{F}$ to $\ce{I}$.

Why my reasoning is absolutely false, and what could be the perfect one?

$\endgroup$
3
  • 3
    $\begingroup$ chemistry.stackexchange.com/questions/15620/what-is-bents-rule $\endgroup$
    – Mithoron
    Commented Dec 9, 2015 at 21:31
  • 1
    $\begingroup$ This is not a duplicate of the question linked by @Mithoron. Yes, the short version of the answer is "Bent's rule", but just because the answers are similar, it does not mean the questions are the same. $\endgroup$ Commented Dec 10, 2015 at 8:10
  • $\begingroup$ For me bent' s rules don't explain it and i want to know why my reasonning is false. $\endgroup$
    – ParaH2
    Commented Dec 10, 2015 at 13:03

1 Answer 1

2
$\begingroup$

As for all atoms of period 3 and higher, we can consider germanium to be essentially unhybridised and the bonds being formed entirely by $\unicode[Times]{x3c3}_\mathrm{p-p}$ overlaps. Therefore, we expect a perfect $90^\circ$ angle for all three compounds.

However, while fluorine is small, the size of the halogens increases from fluorine to iodine. The larger the halogen the more space it requires. Since the central atom remains the same size (and, let’s approximate, is roughly the size of bromine) it cannot offer more space for a strict $90^\circ$ angle without lengthening (= weakening) the bonds which would be unfavourable. It is less unfavourable to widen the angle to generate more space. This requires the s-orbital to hybridise slightly (less favourable) but keeps the desired shorter $\ce{Ge-X}$ bond length (more favourable than anything else in this consideration). Not the bonding electrons repulse each other, but the steric stress of the halogen’s lone pairs causes them to move apart. The same effect can be observed in $\ce{NH3}$ versus $\ce{PH3}$ or $\ce{PH3}$ versus $\ce{PF3}$.

Note the following: If you have two bonds from a central atom which are each described by a bonding and an antibonding orbital, then these two sets of orbitals will always be orthogonal to each other (the integral of overlap is $0$). Thus they will be indifferent to each other, not repulsive.

$\endgroup$

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.