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in water there are two bond pairs and two lone pairs where as in CH4 there are are four bond pairs nad no lone pair. in ch4 there is only bond pair to bond pair repulsion but in water there are three types of repulsions, lone to lone (greatest repulsion), lone to bond ( lesser repulsion ) and bond to bond ( the least repulsion) , therefore due to the presence of two lone pairs in water the bond pairs are repelled with greater force and they get compressed, reducing the ideal bond angle from 109.5 to 104.5 on the other hand, ch4 has only bond pairs and they dont repel each other that strongly so its angle is greater n its 109.5..
No, the bond angle for linear structure is 180 degrees.
ClO3 has the smaller bond angle than ClO4
Yes, water is a bend molecule with a bond angle of about 105 degrees. They are described as bent planar (or V shaped)
bond angle
Water: 104.45° Ozone: 116.8° Difference: 12.4°
The water molecule's bond angle is about 104.45 degrees.
Santa clause crawled into my bed and told me because he said so
The lone pair - OH bond repulsion in water is greater than the OH bond- OH bond repulsion. In methane all of the bonds are the same so it has perfect tetrahedral symmetry. This is VSEPR theory
If the question is "What is the likely bond angle for selenium hydride," then the answer is 105 degrees, just like a water molecule.
104.5 degrees
oxygen difluoride
in water there are two bond pairs and two lone pairs where as in CH4 there are are four bond pairs nad no lone pair. in ch4 there is only bond pair to bond pair repulsion but in water there are three types of repulsions, lone to lone (greatest repulsion), lone to bond ( lesser repulsion ) and bond to bond ( the least repulsion) , therefore due to the presence of two lone pairs in water the bond pairs are repelled with greater force and they get compressed, reducing the ideal bond angle from 109.5 to 104.5 on the other hand, ch4 has only bond pairs and they dont repel each other that strongly so its angle is greater n its 109.5..
Increases
Bond angle is 109.5 degrees.It is equal in every bond
The bond angle of the H-O-H is equivalent to 105 degrees.
The larger the central atom is the less the hydrogens have to spread out (because the electron repulsions are smaller) and the smaller the resulting angle.