Answer to Question #22155 in Organic Chemistry for naeim vhora
right answer is B how?
F2O, Cl2O and H2O are all similar from the point of view that a single oxygen atom forms two single bonds with other atoms.
In all of these cases the molecule is a tetrahedral with oxygen in the center, 2 of its electron lone pairs on the two corners and the other atoms on the other two corners.
The angle is determined by the repulsion between the different electron pairs (electron pairs repell each other and try to be located as far away as possible from each other).
The strength of repulsion decreases in the manner;
lone pair-lone pair > lone pair-bond pair > bond pair-bond pair
The bond angle of X-O-X where X=F, Cl, H will be determined by the repulsion of the lone pairs and the bonding pairs. The more lone pair -bond pair repulsion increases the more the bonding angle increases.
The degree of this type of repulsion depends on the electronegativity of X. Hydrogen is the least electronegative. Thus the bonding pair of the O-H bond will be closer to O, resulting in a higher electron density near O at the site of the bond. This increases the repulsion with the lone pairs in the outer shell of oxygen and thus increases the bond angle.
For F it is the opposite; F is very electronegative and thus the electron density of the O-F bond near O is less than that in O-H.
Thus the repulsion with the lone pairs of O is lower and the 2 lone pairs of O squeeze the two bonding pairs closer together so that the lone pairs themselves can be further away from each other.
Thanks for your comment. Yes, the solution is incorrect. The correct order is Cl2O > H2O > OF2, so the correct answer is B. Take our apologies!
The O in H2 O has 2 bond pairs and 2 lone pairs ( 4 total pairs ). The electron pair orientation
around O is tetrahedral. Two corners of the tetrahedron are "missing" because they are
occupied by lone pairs, not atoms. The shape is called bent. The H-O-H bond angle is
F is very electronegative and thus the electron density of the O-F bond near O is less than
that in O-H.
In OF2, the bonding electrons are nearer of F atom because of higher electronegativity, so
the repulsion between the lone pairs exceeds that of bonding pairs, resulting in decrease of
The H-O-H and F-OF angles in H2O and F2 O are 104.5 o and 103.1o respectively