0
$\begingroup$

In $\ce{SiH2F2}$ four orbitals of silicon $(\mathrm{3s},$ $\mathrm{3p}_x,$ $\mathrm{3p}_y,$ $\mathrm{3p}_z)$ overlap with two hydrogen $(\mathrm{1s})$ and two fluorine $(\mathrm{2p}).$ I want to know how these orbitals are paired for overlapping. Is it based on strength of overlapping?

$\endgroup$
1
$\begingroup$

A simple model for bonding in this molecule is taking $\ce {Si}$ to adopt $\ce {sp^3}$ hybridisation, as suggested from the tetrahedral geometry of the molecule. Subsequently, we can refine our model by applying Bent's rule (see more about it here), which tells us that $\ce {p}$ orbital character concentrates towards more electronegative substituents. Since the electronegativity of the elements has the following order: $\ce {F}$ > $\ce {H}$ > $\ce {Si}$ with Pauling electronegativity values of $\ce {3.98}$, $\ce {2.20}$ and $\ce {1.90}$ respectively, the $\ce {3p}$ character of $\ce {Si}$ would concentrate towards $\ce {F}$ while $\ce {3s}$ character concentrates towards $\ce {H}$. Thus, we would have $\ce {Si}$ using $\ce {sp^{3+x}}$ orbitals to form the bonds with $\ce {F}$ and $\ce {sp^{3-x}}$ to form the bonds with $\ce {H}$, where $x$ is some small value between $\ce {0}$ and $\ce {1}$.

$\endgroup$
2
  • $\begingroup$ It means one hydrogen approach to S orbital to form 1s-3p overlapping and other hydrogen will approach to p orbital of Si to form 1s-3p overlapping and the rest orbitals for fluorine to for 2p-3p.Is it correct? $\endgroup$ – Private5661 Jul 13 '20 at 5:39
  • $\begingroup$ @Private5661 Are you familiar with the theory of hybridisation? Have a look here: en.wikipedia.org/wiki/Orbital_hybridisation $\endgroup$ – Tan Yong Boon Jul 13 '20 at 5:50

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