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Determine the bond order angle-
$O{F_2},O{H_2},OB{r_2},OC{l_2}$

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Last updated date: 13th Jun 2024
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Answer
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HintTo determine the bond angle,we should know the structures of the molecules given,bonds and lone pairs. The bond angle depends on the electronegativity and lone pair of electrons on the central atom. These two factors combined will determine the bond angle of any compound.

Complete step by step answer:
 Bond angle is the angle between two bonds. There are a few points that we need to keep in mind while calculating the bond angle of any compound .
a) First of all ,we will find out the hybridisation of the central atom . The central atom is the one who is more electronegative or the atom with higher atomic number.
If it is $s{p^3}$, the bond angle is $109^\circ 28'$ .
If it is $s{p^2}$, the bond angle is $120^\circ $.
If it is $sp$, the bond angle is $180^\circ $.
b) If there are no lone pairs of electrons . However if there are lone pairs of electrons present the bond angles will be slightly different.
Now if the hybridisation of two central atoms are the same , the lone pair will determine the bond angle.
$L.P. = \dfrac{1}{{B.A.}}$
More the number of lone pairs of electrons, lesser will be the bond angle. This happens because of the repulsion between lone pairs –bond pairs.
c) If the hybridisation and lone pair of both the atoms are the same . Then ,check :
Case I: If the central atom is different and the surrounding atom same .
1- Then , check the electronegativity of the central atom.
2- More the electronegativity of the central atom more will be the bond angle.
Case II: If the central atom is the same and the surrounding atom is different .
1-Then , check the electronegativity of the surrounding atom.
2-More the electronegativity of the surrounding atom, lesser will be the bond angle.
Now , in the given set of series , the central atom is same in all the compounds i.e. $O$
The order of bond angle is : $OB{r_2} > OC{l_2} > O{H_2} > O{F_2}$

Note: However there is an exception , the bond angle of water molecule is supposed to be greater than $OC{l_2}$.But due to the larger size of chlorine molecule, the angle increases and hence the above order is obtained.