Question

Number of lone pairs around phosphorus in $PC{l_5}$ , $PC{l_4}^ + $ and $PC{l_6}^ - $ are respectively
(A) $0,1,2$
(B) $0,0,0$
(C) $1,2,3$
(D) $0,2,1$

Answer 1

Hint: As we know, the valency of Phosphorus is $5$ . By using the valency and charges on the present molecules we can easily find the number of lone pairs present on the following compounds given above.
Complete step by step solution:
We know that, In the above question we have to find the number of lone pairs of three compounds which are $PC{l_5}$ , $PC{l_4}^ + $ and $PC{l_6}^ - $ .





So, let we take all one by one by which we can find the solution easily,
Firstly, we take $PC{l_5}$ as given above that the valency of phosphorus is $5$ . As from all the five electrons are bonded to chlorine atoms. By which we can say that there is no available lone pair on the $PC{l_5}$ compound.
Now in the $PC{l_4}^ + $ compound there the four electrons are bonded with the chlorine atom and the left one is donated to any other group of atoms from which a positive charge is applied on the top of the compound. As a result, in this compound there are no lone pairs present.
In $PC{l_6}^ - $ compound, as we know that all the five phosphorus atom are bonded to chlorine atom and left one chlorine atom is bonded to the received electron from any another compound By which a negative charge is applied on the top of phosphorus atom. As a result, there is also no lone pair available on this compound.
Therefore, from getting the whole conclusion from above the correct answer is $0,0,0$ .

Hence, the correct option is (B)
Note: In chemistry, a lone pair is an unshared pair or non-bonding pair of valence electrons that do not exchange places with another atom during the formation of a covalent bond. Lone pairs are present in the outermost electron shell of atoms. They could be specified through the use of a Lewis structure.