If kb for nx3 is 1.5×10−6, what is the percent ionization of a 0.325 m aqueous solution of nx3?

Chemical reaction: NX₃ + H₂O ⇄ NX₃H⁺ + OH⁻.
c(NX₃) = 0,325 M.
Kb = 1,5·10⁻⁶.
[NX₃H⁺] = [OH⁻] = x.
[NX₃] = 0,325 M - x.
Kb = [NX₃H⁺] · [OH⁻] / [NX₃]. 
1,5·10⁻⁶ = x² / (0,325 M - x).
x = 0,0007 M.
percent of ionization:
α = 0,0007 M ÷ 0,325 M · 100% = 0,215%.

There is one s orbital, and there are three p orbitals, five d orbitals, and seven f orbitals. thus, the answer is b. do not confuse the number of orbitals in a subshell with the number of electrons the subshell can hold. each orbital can hold two electrons, so the capacity of an nf subshell is 14 electrons.