Ncert Solutions Chemistry Class 11th
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6 months agoNew answer posted
6 months agoContributor-Level 10
4.45. (a) It should be noted that all the bond angles in trigonal bipyramidal geometry are not equivalent. In PCl5 the five sp3d orbitals of phosphorus overlap with the singly occupied
p orbitals of chlorine atoms to form five P–Cl sigma bonds. Three P–Cl bond lie in one plane and make an angle of 120° with each other; these bonds are termed as equatorial bonds.
The remaining two P–Cl bonds–one lying above and the other lying below the equatorial plane, make an angle of 90° with the plane.These bonds are called axial bonds. As the axial bond pairs suffer more repulsive interaction from the equatorial bond pairs, therefore
New answer posted
6 months agoContributor-Level 10
4.44. (c) R is an incorrect statement. Electrons placed in a bonding molecular orbital tend to hold the nuclei together and stabilise a molecule. Therefore, a bonding molecular orbital alwayspossesses lower energy than either of the atomic orbitals that have combined to form it.In contrast, the electrons placed in the antibonding molecular orbital destabilise the molecule.
This is because the mutual repulsion of the electrons in this orbital is more than the attraction between the electrons and the nuclei, which causes a net increase in energy.
New answer posted
6 months agoContributor-Level 10
4.43. (d) Basically the strength of a bond depends uponthe extent of overlapping. In case of sigma bond, the overlapping of orbitals takes place to alarger extent. Hence, it is stronger as comparedto the pi bond where the extent of overlappingoccurs to a smaller extent. Further, it isimportant to note that in the formation ofmultiple bonds between two atoms of amolecule, pi bond (s) is formed in addition to asigma bond.
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