Question

A compound, XF4, is 52.44% X by mass. What is the molecular geometry of XF4?

A) tetrahedral B) square planar C) square pyramidal D) see-saw E) none of the above

Please explain to me why the answer is square planar. What needs to be calculate in order to do this problem?

Answer 1

100 g XF4 contain -------------------------> 52.44 g of X        and

                              ---------------------------> 47.56 g of F

F weight in XF4 = 47.56 g

F molar mass = 19 g / mol

F moles = weight / molar mass

              = 47.56 / 19

              = 2.503

XF4-----------------------------> 1 mole X + 4 moles F

so X moles = F moles / 4

    X moles = 2.503 /4

                  =0.626

X moles = weight / molar mass

0.626 = 52.44 / molar mass

molar mass = 52.44 / 0.626

                   = 83.8 g/ mol

molar mass 83.8 g /mol exactly equal to Kr . it is 18th group element .

so the XF4 compound is = KrF4

Kr configuration (36) = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰4s² 4p⁶

in the excited state configuration = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰4s² 4p²4p¹4p¹ 4d¹4d¹

one 4s orbital three 4p orbital s and two 4d orbitals overlap with each other to form 6 equivalent sp3d2

hybrid orbitals .

among 6 orbitals two orbitals having lone pairs. remaining 4 orbitals overlap with four fluorine pz orbital forms 4 sigma bonds.

molecule shape is square planar .