Tag: p- block elements-ii

Questions Related to p- block elements-ii

How is $ {P}{Cl} _{3} $ produced industrially?

  1. Reacting chlorine with white phosphorus

  2. Reacting chlorine with phosphorus acid

  3. Reacting chlorine with black phophorous

  4. All of the above


Correct Option: A
Explanation:
Industrially Phosphorous trichloride is prepared by the reaction of Chlorine with a solution of white Phosphorous in phosphorous trichloride.
The reaction is given by:
$P _4$ $+$ $6Cl _2$ $\longrightarrow $ $4PCl _3$

What is the structure of phosphorus trichloride?

  1. Trigonal Pyramidal

  2. Trigonal Planar

  3. Tetrahedral

  4. Square Planar


Correct Option: A
Explanation:

In ${ PCl } _{ 3 }$, $P$ is ${ sp }^{ 3 }$ hybridized with a lone pair. This makes its structure trigonal pyramidal. So, the correct answer is option $A$.

What is the physical state of ${PCl} _{3}$ under normal condition?

  1. Solid

  2. Liquid

  3. Gaseous

  4. None of these


Correct Option: B
Explanation:

${PCl} _{3}$ is liquid under normal conditions. ${PCl} _{3}$ is a covalent compound. Phosphorus and each of the chlorine atoms fulfil their octet by mutually sharing their valence electrons. In general, covalent compounds have weak bond energy and hence they form liquid.

Dehydrated phosphorus trichloride in water gives:

  1. $H _3PO _3$

  2. $H _3PO _4$

  3. $H _3PO _2$

  4. none


Correct Option: A
Explanation:

Dehydrated phosphorus trichloride in water gives phosphoric acid.

$PCl _3 + 3H _2O\rightarrow  H _3PO _3 + 3 HCl$

Hence option A is correct.

Which of the following is not correctly matched?

  1. $PCl _{5}$ - $sp^{3}d$ hybridisation.

  2. $PCl _{3}$ - $sp^3$ hybridisation.

  3. $PCl _{5}$ (solid) - $[PtCl _{4}]^{+}$+ $[PtCl _{6}]^{-}$.

  4. $PCl _{5}$ - brownish powder.


Correct Option: D
Explanation:


 Colour of $PCl _5$ is not brownish but a greenish yellow crystalline solid with an irritating odour.

Hence option D is correct.

On reaction with $Cl _{2}$, phosphorus forms two types of halides 'A' and 'B'. Halide 'A' is yellowish-white powder but halide 'B' is colourless oily liquid. What would be the hydrolysis products of 'A' and 'B' respectively?

  1. $H _{3}PO _{4}$,$H _{3}PO _{3}$

  2. $HOPO _{3}$, $H _{2}PO _{2}$

  3. $H _{3}PO _{3}$,$H _{3}PO _{4}$

  4. $HPO _{3}$,$H _{3}PO _{3}$


Correct Option: A
Explanation:

$P+Cl _2\longrightarrow \underset {(B)}{PCl _3} +\underset {(A)}{PCl _5}$

$PCl _5+4H _2O\longrightarrow H _3PO _4+5HCl$
$PCl _3+3H _2O\longrightarrow H _3PO _3+3HCl$

In solid state $PCl _{5}$ is a :

  1. covalent solid

  2. octahedral structure

  3. ionic solid with $[PCl _{6}]^{+}$ octahedral and $[PCl _{4}]^{-}$ tetrahedral

  4. ionic solid with $[PCl _{4}]^{+}$ tetrahedral and $[PCl _{6}]^{-}$ octahedral


Correct Option: D
Explanation:

In solid state $PCl _5$ tries  to exist as oppositely charged ions like

(1) $PCl _4^{+}$ and (2) $PCl _6^{-}$ as the ionic bonding enhances the crystalline nature .

also $PCl _4^{+}$  is tetrahedral , while $PCl _6^{-}$ is octahedral . these structure fit well into each other which gives more stability to solid structure .

Hence option D is correct.

Among the following, the number of compounds that can react with $PCl _{5}$ to give $POCl _{3}$ is: 


$O _{2},\ CO _{2},\ SO _{2},\ H _{2}O,\ H _{2}SO _{4},\ P _{4}O _{10}$

  1. 4

  2. 5

  3. 3

  4. 6


Correct Option: A
Explanation:

$PCl$ on reaction with these $4$ compounds produces $POCl _3$:


$PCl _5+SO _2\rightarrow POCl _3+SOCl _2$
$PCl _5+H _2O\rightarrow POCl _3+2HCl$
$PCl _5+H _2SO _4\rightarrow SO _2Cl _2+2POCl _3+2HCl$
$6PCl _5+P _4O _{10}\rightarrow 10POCl _3$.

The most powerful chlorinating agent is :

  1. $\mathrm{P}\mathrm{Cl} _{3}$

  2. $\mathrm{Cl} _{2}\mathrm{O}$

  3. $\mathrm{P}\mathrm{Cl} _{5}$

  4. $\mathrm{Cl}\mathrm{O} _{2}$


Correct Option: C
Explanation:

$PCl _{5}$ readily dissociates into $PCl _{3}$ and chlorine gas and therefore acts as a good chlorinating agent.

Which of the following pentahalides does not exist?

  1. $\mathrm{P}\mathrm{F} _{5}$

  2. $\mathrm{P}\mathrm{Cl} _{5}$

  3. $\mathrm{P}\mathrm{B}\mathrm{r} _{5}$

  4. $\mathrm{PI} _{5}$


Correct Option: D
Explanation:

SI is too big for small P.
SI atoms cannot fil around small P. So $PI _{s}$ does not exist.