Question

Which of the following represent incorrect order of the related properties:

• A. $B{F}_3<BC{l}_3<BB{r}_3<B{I}_3$ (Acidic character)
• B. $BaS{O}_4<SrS{O}_4<CaS{O}_4<MgS{O}_4$
  (Solubility in water)
• C. $CHC{l}_3<CH{F}_3$
  (Acidic strength)
• D. $(C{F}_3{)}_{3}N<(C{H}_3{)}_{3}N$
  (Basic strength )

Answer 1

The correct option is C $CHC{l}_3<CH{F}_3$

(Acidic strength)
a) Boron atom in boron halides has 6 electrons in its valence shell. Thus it is electron defficient and has acidic properties. Since the effective orbital overlap decreases down the group, thus boron become more electron defficient down the group, accordingly acidic property also increases.

b) The sulphate become less soluble as we go down the group.

c) $CHC{l}_3$ is more acidic than $CH{F}_3$ because if we compare the conjugate base of both of them , $\stackrel{-}{C}C{l}_3$ and $\stackrel{-}{C}{F}_3$, $Cl$ has a vacant 3d orbital whereas $F$ doesn't. Therefore $Cl$ can exert -M effect due to availability of vacant d orbital but $F$ cannot. -M effect stabilises the negative charge and increases acidic strength. So $CHC{l}_3$ is more acidic than $CH{F}_3$.

d) In $(C{H}_3{)}_{3}N$ the lone pair in $N$ is concentrated on $N$, so it can easily donate its electron but in $(C{F}_3{)}_{3}N$ due to high electronegativity of $F$ atom, electron density is reduced on $N$ atom. Thus $(C{H}_3{)}_{3}N$ is more basic in nature.