Two types of halides are available for
this group. One is MX3 and another is MX5 type.
Imp note:
Trisilyl amine, N(SiH3)3 is planar whereas trimethyl amines N(CH3)3 is pyramidal. Explain why?.
Why bond angle of hydrides of group 15, decreases down the group?
(A) MX3 Type or Trihalides:
MX3:
|
NCl3, PCl3,
AsCl3, SbCl3 and BiCl3
|
Characteristic of Tri
halides:
(1) All possible trihalides of all the
elements of nitrogen family are known except NCl3, NBr3,
and NI3.
(2) These are (NCl3, NBr3,
and NI3) unstable due to the low polarity of the (N – X) bond. And Weakness
of (N - X) bond due to large size difference.
(3) All
trihalides are covalent except BiF3
which is ionic.
(4) Like
hydrides these trihalides have pyramidal structure and a central atom is sp3
hybridized.
Structure of PCl3:
(5) These
trihalides can be easily hydrolyses by water except NX3.
SN
|
Halides
|
Hydrolysis
|
NF3
|
Do not Hydrolysed at
room temperature
|
|
1
|
NX3 (X= Cl,
Br, I)
|
NX3 +3H2O
à NH3 +
3HOX (At RT)
|
2
|
PX3 (X= F Cl, Br, I)
|
PCl3 + 3H2O
à H3PO3 + 3HCl
|
2
|
AsCl3
|
AsCl3 + H2O
à AsO3 + 6HCl or
AsCl3 + 3H2O
à H3AsO3 + 3HCl
|
4
|
SbCl3
|
SbCl3 + H2O à SbOCl (White
turbidity) + 2HCl
|
5
|
BiCl3
|
BiCl3 + H2O à BiOCl (White turbidity) + 2HCl
|
Imp note:
(i) NF3
Hydrolysed only at high temperature (300 0 C) and give N2O3
which is unstable and dissociate to produced NO and NO3 gas.
2NF3 + 3H2O à N2O3
+6HF à NO (gas) + NO2
(gas) + 6HF
(ii)When an aq. solution of BiCl3
is prepared, after some time turbidity appears which is milky in appearance and
finally a white ppt is formed due to formation of BiOCl.
(6) The
trihalides of P, As and Sb acts as Lewis acids and combine with Lewis bases.
PF3 +F2
|
à
|
PF5
|
SbF3 +2F
|
à
|
[SbF5]2-
|
(7) Lewis Base Order of trihalides follows as
NF3
< NCl3 <NBr3 < NI3
|
Imp note: As electronegativity of halides decrease
and availability of loan pair over nitrogen atom increases hence Lewis Base
character increase
(B) MX5 Type
or Pentahalides:
MX5:
|
PCl5, AsCl5, SbCl5
|
Characteristic of Pentahalides:
(1) Except N and Bi all form Pentahalides, N does not form due to
absence of d-orbital while Bi does not form due to inert pair effect.
(2) Phosphorous form Pentahalides of all halogens.
(3) Bismuth forms only BiF5.
(4) As and Sb form Pentafluorides and pentachlorides
only.
(5) The central atom of Pentahalides attains
sp3d
hybridization and forms five covalent bonds with five chlorine
atoms.
Structure of PCl5 :
(6) Penta halides have less thermal
stability as compared to trihalides. All Penta halides act as Lewis acids.
PF5 +F-
|
à
|
[PF6]-
|
SbCl3 +Cl-
|
à
|
[SbCl4]-
|
(7) PCl5 acts as an effective chlorinating
Agent so it decomposes into PCl5 à PCl3 + Cl2
(8) PCl5 exists as molecule in
gaseous state but in solid it exists as [PCl4]+ [PCl6]-
and is ionic in nature. PBr5, PI5 also exist in the ionic
form in solid state.
SN
|
Halides(Gaseous state)
|
Solid State
|
1
|
PCl5
|
[PCl4]+[PCl6]- [sp3
and sp3d]
|
2
|
PBr5
|
[PBr4]+[Br]-
|
3
|
PI5
|
[PI4]+[I]-
|
ILLUSTRATIVE EXAMPLE (1): Are all the
five bonds of PCl5 equivalent? Justify your answer.
SOLUTION: PCl5 has trigonal
bipiramidal structure. It has three equivalent equatorial and two equivalent
axial P – Cl bonds. However, due to greater bond pair – bond pair repulsions,
the axial P – Cl bonds are longer and hence different from the three equatorial
bonds.
ILLUSTRATIVE EXAMPLE (2): Phosphorus can form
PCl5 but nitrogen can not form NCl5 why?
SOLUTION: Phosphorus forms PCl5 due to
availability of vacant d – orbital though white P can extend its oxidation
state but this is not applicable in case of N due to unavailability of d –
orbitals.
ILLUSTRATIVE EXAMPLE (3): Why NCl3 can not be hydrolysed?
SOLUTION: Due to unavailability of vacant d – orbitals.
EXERCISE (1): Among the
trihalides of nitrogen, which one is the least basic?
(A) NF3 (B) NCl3
(C) NBr3 (D) NI3
EXERCISE (2): Which of
the following halides is most acidic?
(A) PCl3 (B) SbCl3
(C) BiCl3 (D) CCl4
EXERCISE (3): Which of
the following is not hydrolysed?
(A) AsCl3 (B) PF3
(C) SbCl3 (D) NF3
SN
|
EXERCISE
|
ANSWER KEY
|
1
|
1
|
(A)
|
2
|
2
|
(A)
|
3
|
3
|
(D)
|
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