How many atoms are present in nodal plane of pie bond of ch2=sf4?
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4 atoms are present in the nodal plane of the double bond present in the compound- 2 hydrogen atoms and 2 fluorine atoms.
The hydrogen atoms bonded to the C atom are perpendicular to the plane of the pi bond and p-orbitals but lie in the plane of the node. Similarly, the 4 F atoms adopt a seesaw structure along with F, pushing 2 axial F atoms into the same plane as the 2 H atoms.
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4 atoms is your answer
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