Complete the hybridization and bonding scheme for xef4.

Xe = VIII = 8 valence electrons
F = VII = 4 (7 ve) = 28 valence electrons

total ve = 8 + 28 = 36 ve

36 - 4(2) = 28 ve
(there are 2 electrons in each bond x 4 bonds)

28 - 4(6) = 4 
(We assign the remaining electrons to F atoms)

4 - 2(2) = 0 
(Therefore 4 electrons left => we have 2 lone pairs)

The steric number = No. of σ bonds + #lone pairs

= 4 σ bonds + 2 lone pairs = 6 => d²sp³ (6 hybrid orbitals)

4 bonds + 2 lone pairs => square planar

substituting a hydrogen atom with a halogen in a hydrocarbon:

- the single bond remains single. a is wrong.

- the bond with hydrogen is as strong as with a halogen. c is wrong

- a hydrocarbon is saturated if there is no double bond. we do not know that so d is wrong.

- carbon will always have four valence electrons. e is wrong.

by elimination, the correct answer is b. the boiling point of the new compound increases.