Chemistry Phosphorous Halides and Oxoacids of Phosphorous
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Class 12 chapter7 p-block elements

Phosphorous Halides and Oxoacids of Phosphorous

Topics Covered :

● Preparation of Phosphorous Trichloride
● Properties of Phosphorous Trichloride
● Preparation of Phosphorous Pentachloride
● Properties of Phosphorous Pentachloride
● Oxoacids of Phosphorous

Phosphorus Halides :

Phosphorus forms two types of halides, `color{red}(PX_3 (X = F, Cl, Br, I))` and `color{red}(PX_5 (X = F, Cl, Br))`.

Phosphorus Trichloride :

Preparation and properties of phosphorous trichloride is given as follow :

Preparation :

`=>` It is obtained by passing dry chlorine over heated white phosphorus.

`color{red}(P_4+ 6Cl_2 → 4PCl_3)`

`=>` It is also obtained by the action of thionyl chloride with white phosphorus.

`color{red}(P_4 + 8SOCl_2 → 4PCl_3 +4SO_2+2S_2Cl_2)`

Properties :

`=>` It is a colourless oily liquid and hydrolyses in the presence of moisture.

`color{red}(PCl_3+3H_2O → H_3PO_3 +3HCl)`

`=>` It reacts with organic compounds containing `color{red}(–OH)` group such as `color{red}(CH_3COOH , C_2H_5OH)`

`color{red}(3CH_3COOH +PCl_3 → 3CH_3COCl+H_3PO_3)`

`color{red}(3C_2H_5 OH +PCl_3 → 3C_2H_5Cl +H_3PO_3)`

`=>` It has a pyramidal shape as shown, in which phosphorus is `color{red}(sp^3)` hybridised.
Q 3030491312

Why does `PCl_3` fume in moisture ?

Solution:

`PCl_3` hydrolyses in the presence of moisture giving fumes of `HCl`.
`PCl_3 + 3H_2O→H_3 PO_3 + 3HCl`

Phosphorus Pentachloride :

Preparation and properties of phosphorous pentachloride is given as follow :

Preparation :

`=>` Phosphorus pentachloride is prepared by the reaction of white phosphorus with excess of dry chlorine.

`color{red}(P_4+10 Cl_2 → 4PCl_5)`

`=>` It can also be prepared by the action of `color{red}(SO_2Cl_2)` on phosphorus.

`color{red}(P_4 +10 SO_2 Cl_2 → 4PCl_5+10 SO_2)`

Properties :

`=>` `color{red}(PCl_5)` is a yellowish white powder and in moist air, it hydrolyses to `color{red}(POCl_3)` and finally gets converted to phosphoric acid.

`color{red}(PCl_5+H_2O → POCl_3 +2 HCl)`

`color{red}(POCl_3+3H_2O → H_3PO_4 +3HCl)`

`=>` When heated, it sublimes but decomposes on stronger heating.

`color{red}(PCl_5 oversettext(Heat)→ PCl_3+Cl_2)`

`=>` It reacts with organic compounds containing `–OH` group converting them to chloro derivatives.

`color{red}(C_2H_5OH +PCl_5 → C_2H_5 Cl +POCl_3+HCl)`

`color{red}(CH_3COOH + PCl_5 → CH_3COCl + POCl_3+HCl)`

`=>` Finely divided metals on heating with `color{red}(PCl_5)` give corresponding chlorides.

`color{red}(2Ag + PCl_5 → 2Ag Cl +PCl_3)`

`color{red}(Sn +2PCl_5 → SnCl_4 +2PCl_3)`

`=>` It is used in the synthesis of some organic compounds, e.g., `color{red}(C_2H_5Cl, CH_3COCl).`

`=>` In gaseous and liquid phases, it has a trigonal bipyramidal structure as shown below.

● The three equatorial `color{red}(P–Cl)` bonds are equivalent, while the two axial bonds are longer than equatorial bonds.

● This is due to the fact that the axial bond pairs suffer more repulsion as compared to equatorial bond pairs.

● In the solid state it exists as an ionic solid, `color{red}([PCl_4]^+ [PCl_6]^–)` in which the cation, `color{red}([PCl_4]^+)` is tetrahedral and the anion, `color{red}([PCl_(6)]^–)`octahedral.
Q 3060491315

Are all the five bonds in `PCl_5` molecule equivalent? Justify your answer

Solution:

`PCl_5` has a trigonal bipyramidal structure and the three equatorial `P-Cl` bonds are equivalent, while the two axial bonds are different and
longer than equatorial bonds

Oxoacids of Phosphorus :

`=>` Phosphorus forms a number of oxoacids.

`=>` The important oxoacids of phosphorus with their formulas, methods of preparation and the presence of some characteristic bonds in their structures are given in Table 7.5.

● In oxoacids, phosphorus is tetrahedrally surrounded by other atoms.

● All these acids contain one `color{red}(P=O)` and at least one `color{red}(P–OH)` bond.

● The oxoacids in which phosphorus has lower oxidation state (less than `+5`) contain, in addition to `color{red}(P=O)` and `color{red}(P–OH)` bonds, either `color{red}(P–P)` (e.g., in `color{red}(H_4P_2O_6)`) or `color{red}(P–H)` (e.g., in `color{red}(H_3PO_2)`) bonds but not both.

● These acids in `+3` oxidation state of phosphorus tend to disproportionate to higher and lower oxidation states.

● `color{red}("Example")` : Orthophophorous acid (or phosphorous acid) on heating disproportionates to give orthophosphoric acid (or phosphoric acid) and phosphine.

`color{red}(4H_3PO_3 → 3H_3PO_4 +PH_3)`

● The acids which contain `P–H` bond have strong reducing properties. Thus, hypophosphorous acid is a good reducing agent as it contains two `color{red}(P–H)` bonds and reduces, for example, `color{red}(AgNO_3)` to metallic silver.

`color{red}(4AgNO_3+2H_2O +H_3PO_2 → 4Ag +4HNO_3+H_3PO_4)`

● The `color{red}(P–H)` bonds are not ionisable to give `color{red}(H^+)` and do not play any role in basicity.

● Only those `color{red}(H)` atoms which are attached with oxygen in `color{red}(P–OH)` form are ionisable and cause the basicity. Thus, `color{red}(H_3PO_3)` and `color{red}(H_3PO_4)` are dibasic and tribasic, respectively as the structure of `color{red}(H_3PO_3)` has two `color{red}(P–OH)` bonds and `color{red}(H_3PO_4)` three.
Q 3010591410

How do you account for the reducing behaviour of `H_3PO_2` on the basis of its structure ?

Solution:

In `H_3PO_2`, two `H` atoms are bonded directly to `P` atom which imparts reducing character to the acid.
 
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