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- 42. [BiF7] and [SbF6]³- ions have pentagonal bipyramidal and octahedral structures, respectively. Are these observations consistent with the VSEPR predictions. F Bi-F F F FIF Sb F F Hint: Remind yourself of the VSEPR models. Take the number of valence electrons of the central atom, add/substract the electrons from the charge, substract the number of bonding pairs from this number, if this turns zero, there are no lone pairs. If not place lone pairs.(a) Draw the valid Lewis structure, identify the (b) molecular group geometry, (c) electron group geometry, (d) polarity of the bonds, (e) over-all polarity, (f) number of lone pairs, and (g) number of bonding pairs. 1. XeO2F2 (neutron diffraction experiment by Peterson, Willett and Huston. showed that F-Xe-F bond angle is 180o)select a module in which the molecular shape is the same with the electron arrangement. explain your reasoning a) CO2 b) H2O c) NH3 d) SO2
- A) Explain why BrF4 with a negative charge on top is square planar, whereas BF4 with negative charge on top is tetrahedral? B) How could you expect H_X_H bond to vary in the series H2O , H2Se? Explain your answerGive detailed Solution with explanation needed with Lewis structure with explanation....if possible please avoid handwritten Solutiona) Using Valence Bond Theory,NOT VSEPR, thoroughly show how silicon is able to form four identical atomic orbitals available for bonding in the compound SiF4. Be sure todiscuss bond angles/shape. b) Draw and name the correct orbitals involved when a Si atom and four F atoms come together to form SiF4. Your drawing should clearly show “before” and “after” forming the bonds. Draw the orbitals as clearly as you can, but you only have to show the orbitals for ONE of the Si—F bonds being formed -- don’t draw all four.
- 2426. The molecular shape of the atoms in PCla*1 is best described as (A) tetrahedral (C) see-saw (B) square planar (D) trigonal bipyramidalIX) By writing the appropriate electron configurations and orbital box diagrams briefly EXPLAIN in your own words each one of the following questions: a) The bond length of the Br2 molecule is 2.28 Å, while the bond length of the compound KBr is 3.34 Å. The radius of K✶ is 1.52 Å. Determine the atomic radius in Å of the bromine atom and of the bromide ion. Br = Br b) Explain why there is a large difference in the atomic sizes or radius of the two (Br and Br). T
