
Consider the following statements:
1. Electron density in the $xy$ -plane in \[3{d_{{x^2} - {y^2}}}\] orbital is zero.
2. Electron density in the $xy$ -plane in \[3{d_{{z^2}}}\] orbital is zero
3. $2s$ orbital has one nodal surface
4. For $2px$ orbital $yz$ is the nodal plane
Which are the correct statements?
A. 3 and 4
B. 2 and 3
C. Only 2
D. All are correct
Answer
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Hint: To know this answer we need to know about s, p, d and f orbitals and each orbitals having their own orbitals. Also, electron density is the measure of the probability of an electron being present at an infinitesimal element of space surrounding any given point.
Complete Step by step answer: Electron density also known as electronic density in quantum chemistry is the measure of the probability of an electron being present at an infinitesimal element of space surrounding any given point. There are four major types of orbitals and they are s, p, d and f.
S orbital is a spherically symmetrical orbital around the atomic nucleus. The energy level increases as we move away from the nucleus and therefore the orbitals get bigger. The order of size is 1s<2s<3s and so on.
Nodal point is a point at which there is a zero probability of finding the electron. Further it is divided into radial and angular nodes. Nodal planes are the planes of zero probability region to find the electron. The number of planes is equal to I.
P orbital are dumbbell shaped. The node in the p orbital occurs at the centre of the nucleus. The p orbital can occupy a maximum of six electrons due to the presence of three orbitals. The three p orbitals are oriented right angles to each other. Its size depends on the principal quantum number. Its three p orbitals are \[xy\] , \[yz\] and \[zx\] .
D orbital is a cloverleaf or two dumbbells in a plane. The values of magnetic quantum numbers corresponding to d orbitals are -2, -1, 0, 1 and 2. Therefore there are five d orbitals. They are \[{d_{xy}}\] , \[{d_{yz}}\] , \[{d_{xz}}\] , \[{d_{{x^2} - {y^2}}}\] and \[{d_{{z^2}}}\] .
F orbitals have a diffused shape. For f orbital the values of ml corresponding to f orbital are -3, -2, -2, 0, 1, 2 and 3. The seven f orbitals are \[{f_{x({x^2} - {y^2})}}\] , \[{f_{y({x^2} - {y^2})}}\] , \[{f_{xyz}}\] , \[{f_{{z^3}}}\] , \[{f_{yz}}\] , \[{f_{x{z^2}}}\] , \[{f_{z({x^2} - {y^2})}}\] .
Now, according to the question, options 3 and 4 are correct as electron density in \[3{d_{{x^2} - {y^2}}}\] orbital is non zero and in \[3{d_{{z^2}}}\] orbital is also non zero.
Therefore, 1 and 2 are incorrect and 3 and 4 are correct. Therefore, our answer is option A.
Note: Electron density in \[3{d_{{x^2} - {y^2}}}\] orbital is non zero and in \[3{d_{{z^2}}}\] orbital is also non zero. $2s$ orbital has one nodal surface and for $2px$ orbital $yz$ is the nodal plane. It should be kept in mind that the energy of an orbital in higher shells can be less than energy of orbitals in lower shells.
Complete Step by step answer: Electron density also known as electronic density in quantum chemistry is the measure of the probability of an electron being present at an infinitesimal element of space surrounding any given point. There are four major types of orbitals and they are s, p, d and f.
S orbital is a spherically symmetrical orbital around the atomic nucleus. The energy level increases as we move away from the nucleus and therefore the orbitals get bigger. The order of size is 1s<2s<3s and so on.
Nodal point is a point at which there is a zero probability of finding the electron. Further it is divided into radial and angular nodes. Nodal planes are the planes of zero probability region to find the electron. The number of planes is equal to I.
P orbital are dumbbell shaped. The node in the p orbital occurs at the centre of the nucleus. The p orbital can occupy a maximum of six electrons due to the presence of three orbitals. The three p orbitals are oriented right angles to each other. Its size depends on the principal quantum number. Its three p orbitals are \[xy\] , \[yz\] and \[zx\] .
D orbital is a cloverleaf or two dumbbells in a plane. The values of magnetic quantum numbers corresponding to d orbitals are -2, -1, 0, 1 and 2. Therefore there are five d orbitals. They are \[{d_{xy}}\] , \[{d_{yz}}\] , \[{d_{xz}}\] , \[{d_{{x^2} - {y^2}}}\] and \[{d_{{z^2}}}\] .
F orbitals have a diffused shape. For f orbital the values of ml corresponding to f orbital are -3, -2, -2, 0, 1, 2 and 3. The seven f orbitals are \[{f_{x({x^2} - {y^2})}}\] , \[{f_{y({x^2} - {y^2})}}\] , \[{f_{xyz}}\] , \[{f_{{z^3}}}\] , \[{f_{yz}}\] , \[{f_{x{z^2}}}\] , \[{f_{z({x^2} - {y^2})}}\] .
Now, according to the question, options 3 and 4 are correct as electron density in \[3{d_{{x^2} - {y^2}}}\] orbital is non zero and in \[3{d_{{z^2}}}\] orbital is also non zero.
Therefore, 1 and 2 are incorrect and 3 and 4 are correct. Therefore, our answer is option A.
Note: Electron density in \[3{d_{{x^2} - {y^2}}}\] orbital is non zero and in \[3{d_{{z^2}}}\] orbital is also non zero. $2s$ orbital has one nodal surface and for $2px$ orbital $yz$ is the nodal plane. It should be kept in mind that the energy of an orbital in higher shells can be less than energy of orbitals in lower shells.
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