A computational molecular modeling program was used to determine the energy for
ID: 1070482 • Letter: A
Question
A computational molecular modeling program was used to determine the energy for two pomble geometries of the molecule xeF_4 Geometry A was found to have a lower energy than Geometry B therefore, Geometry A is more stable geometry. A computational molecular modeling program was used to determine the HN-H bond angles for the compounds shown below. The program gave H N-H angle values of 10.69 degree, 107, 02 degree, and 109, 47 degree, write the H NH that to each molecule on the line under the molecule. Briefly discuss the H NIH angle trend based on VSEPR model concepts.Explanation / Answer
the geometry A in case of XeF4 is more stable and have lower energy. Because upon finding the shape of XEF4 with VSEPR theory, two lone pair of electron remanis on Xe. ie
XeF4- centarl metal atom=Xe have 8 valence eletrons.
hence total valence shell eletrn=12 i.e 6 pairs. 4 bond pairs and 2 lone pairs.
Four flourine will engage 4 electrons of Xe in bond formation and hence 4 unpaired electrons will be left
these two lone pair of electrons should be arranged in such a way so that it gives minimum repulsion. In Geometry A, the two pairs are in opposite sides of the molecule i.e in perpendicular plane, hence repulsion and less energy.
B)
According to electron pair repulsion theory, lone pair-lone pair repulsion > lone pair- bond pair > bond pair-bond pair repulsion.
In NH4+ there is bond pair-bond pair repulsion only, since no lone pairs present. so it will have largest bond angle=109.470
In NH3 THERE IS ONE LONE PAIR OF ELECTRON. HENCE THERE WILL BE lone pair-bond pair repulsion and lone pair will try to accomodate itself. hence bond angle will be slightly less than NH4 and will be 107.020
In NH2- there will be strong lone pair-lone pair repulsion due to presence of two lone pairs. hence bond angle will shrink to 104.690
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