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what is exactly d shape of a hybride orbital??
Dear samidha,
An atomic orbital is the space around the nucleus in which the probability of finding the electron is maximum. These most probable regions can be diagrammatically represented by cloud density (dot) diagrams. The density of dots (or lack of them) in any region of the cloud diagram, indicates the degree of probability of finding the electron in that region. It is not always convenient to draw dot diagrams of orbitals, since the probability of finding an electron decreases with distance (but does not become zero), thus not giving it any definite shape. Drawing boundary surfaces, which enclose 95-99% of the probability of locating an electron, is the method generally used, to show the shape of an orbital.
As definite energies and angular movements characterize atomic orbitals, the permissible values of these parameters are expressed in terms of quantum numbers. The azimuthul quantum 'l', is related to orbital angular momentum of the electron through the number.
Thus different values of 'l' correspond to different orbital angular momentums. As a result the azimuthul quantum 'l' determines the shape of orbitals.
The orbital angular momentum of a 's' orbital is zero, as 'l' = 0. This means that the probability of finding an electron at a particular distance from the nucleus is the same in all directions/at all angles. Since the distribution of electron density is symmetrical, the shape representing the 's' orbital is a sphere.
It is to be primarily noted that the total number of concentric spheres at any given main energy level in an 's' orbital equals the principle quantum number of that level. Thus for example '1s' orbital consists of only one sphere while a '3s' orbital consists of three concentric spheres. Secondly, as the value of the principal quantum number 'n' increases, the 's' orbital becomes larger and the energy of the 's' orbital increases, while retaining the spherical symmetry. The energies of the various 's' orbitals follow the order 1s < 2s < 3s < 4s.
For 'p' orbitals 'l' = 1. So,
Thus the distribution around the nucleus is not spherical. The 'p' orbital probability diagram is dumb-bell shaped i.e., it consists of distorted spheres of high probability, one on each side of the nucleus, concentrated along a particular direction. The probability of finding the electron in a particular 'p' orbital is equal in both the lobes.
Since 'l' = 1, irrespective of the value of n, three values of magnetic quantum number 'm' exists i.e., +1, 0 and 1. Thus three 'p' orbitals exist in each 'p' subshell. These are oriented symmetrically around the three axes 'x', 'y'and 'z'.
Shapes of 2px, 2py and 2pz orbitals
For 'd' orbitals 'l' = 2. Therefore the angular momentum of an electron does not show a spherical symmetry. For 'l' = 2, five values of 'm' the magnetic quantum number exists i.e., -2, -1, 0, +1, and +1. Accordingly there are five space orientations for 'd' orbitals, which are not identical in shape. These are designated as:
m = -2 m = -1 m = 0 m = +1 m = +2
Shapes of five 'd' orbitals
We are all IITians and here to help you in your IIT JEE preparation. All the best. If you like this answer please approve it.... win exciting gifts by answering the questions on Discussion Forum Sagar Singh B.Tech IIT Delhi sagarsingh24.iitd@gmail.com
We are all IITians and here to help you in your IIT JEE preparation. All the best.
If you like this answer please approve it....
win exciting gifts by answering the questions on Discussion Forum
Sagar Singh
B.Tech IIT Delhi
sagarsingh24.iitd@gmail.com
The shape of d - orbital is dumbell with a dounaught in between of it.In case of hybridisation it depends over the shape of the hybrid orbitals as the shape of sp3d hybridisation is trigonal bipyramidal where as the shape of sp3d2 is octahedral shape. hence the conclusion is this that the shape of d - orbitals is dumbbell with a donought in between but show itself diffrently in case of hybridisation.
hope that you like my answer samidha
thank you
from best wishes
swapnil sudhir
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