S – O bond length is maximum in :
(a) SOBr2 (b) SOCl2 (c) SOF2 (d) SO(CH3)2
Dear Student,
Larger the bond angle, shorter is the bond length.
Now, greater the size of halogen, greater will be the S-X bond angle and so less will be the S-O bond angle.
The order of size among halogens is Br>Cl>F
Therefore, order of S-X bond angle will be S-Br>S-Cl>S-F
And so, order of S-O bond angle will be SO(CH)3<SOBr 2<SOCl2<SOF2
In case of SO(CH3)3, due to steric hinderance its bond angle is least
Since S-O bond angle in SOBr2 is minimum, so its bond length will be maximum. Hence, the correct answer is SOBr2.
Larger the bond angle, shorter is the bond length.
Now, greater the size of halogen, greater will be the S-X bond angle and so less will be the S-O bond angle.
The order of size among halogens is Br>Cl>F
Therefore, order of S-X bond angle will be S-Br>S-Cl>S-F
And so, order of S-O bond angle will be SO(CH)3<SOBr 2<SOCl2<SOF2
In case of SO(CH3)3, due to steric hinderance its bond angle is least
Since S-O bond angle in SOBr2 is minimum, so its bond length will be maximum. Hence, the correct answer is SOBr2.