Question

Which of the following has been arranged in order of decreasing bond length ?
A. $P - O > Cl - O > S - O$
B. $P - O > S - O > Cl - O$
C. $S - O > Cl - O > P - O$
D. $Cl - O > S - O > P - O$

Answer 1

Hint: Bond length is defined as the equilibrium distance between the centres of the nuclei of the two bonded atoms is called its bond length . In other words it is the sum of the radii of the bonded atoms .

Complete step by step answer:
There are various factors on which bond length depends :
Size of the atom : The bond length increases with increase in size of atom , that means it increases as we go down a group and decreases across a period as we go from left to right .
Multiplicity of bond : The bond length decreases with multiplicity of the bond . This means doubly bonded atoms have bond length lesser than singly bonded atoms .
Type of hybridisation : AS we know that s - orbital is smaller therefore greater the s character , shorter is the hybrid orbital and hence shorter is the bond length .
We have to find the order of decreasing bond length in $P - O$ , $S - O$ and $Cl - O$ .
Now oxygen atom is common in all the three bonds , so we have to compare the radii, the phosphorus , sulphur and chlorine .
Phosphorus belongs to group 15 , sulphur belongs to group 16 and chlorine belongs to group 17 .
As we know that radii decreases as we move in a group from left to right , therefore the order of decreasing bond length is :
$P - O > S - O > Cl - O$
Hence , option B is correct .


Note:
The covalent radii of an atom is approximately equal to the radius of the core ( that is , leaving the valence shell ) whereas van der Waals radius represents the overall size of the atom including the valence shell .