
Electronic transition from $\text{ 4s }$ to$\text{ 3s }$ orbital is forbidden meaning that it cannot occur because:
A) There will be no change in the orbital angular momentum of an electron, although the emitted photon has angular momentum
B) There will be a change in the orbital angular momentum whereas the emitted photon has no momentum
C) $\text{ }\Delta {{\text{m}}_{\text{1}}}\text{ }$ values between $\text{ 4s }$ and $\text{ 3s }$orbital is not zero, which is an important selection rule for the allowed transition.
D) In $\text{ 4s }$ an $\text{ 3s }$ orbital, the wavelength of the electron wave is the same
Answer
570.9k+ views
Hint: According to the selection rule, the transition is allowed when the overall change in the orbital angular momentum is by 1 unit. $\text{ }\Delta l\text{ = }\pm \text{ 1 }$. If the change in the angular momentum is zero$\text{ }\Delta l\text{ = 0 }$, then such transitions are called the forbidden electronic transition. This is a Laporte selection rule.
Complete step by step answer:
The transition of an electron from the higher energy level to the lower energy level takes place by the emission of radiation. As there is an energy difference between the high and low energy level orbital, the electronic transition takes place only when the excess of the energy is liberated in the form of radiations.
According to the Laporte selection rule, the centrosymmetric environment transitions between the s-s, p-p, d-d, or f-f transitions are said to be forbidden. This is not allowed transition. Transition that is allowed should have the overall change in the orbital angular momentum or azimuthal quantum number ‘l’ by one unit. That is,
$\text{ }\Delta l\text{ = }\pm \text{ 1 or }-1\text{ }$
Let's have a look at the quantum numbers of the $\text{ 4s }$ and $\text{ 3s }$orbital. The principal quantum number that is shell number for $\text{ 4s }$orbital is $\text{ n = 4 }$ and for $\text{ 3s }$orbital it is found to be$\text{ n = 3 }$.
Here, both the orbitals under consideration are ‘s’ orbitals, thus despite the different shell numbers, the $\text{ 4s }$ and $\text{ 3s }$orbital have the same azimuthal quantum number. That is,
$\text{ }{{l}_{4s}}\text{ = }{{l}_{3s}}\text{ = 0 }$
Here, change is the azimuthal quantum number is equal to,
$\text{ }\Delta l={{l}_{4s}}\text{ }-\text{ }{{l}_{3s}}\text{ = 0 }$
The transition of an electron from the $\text{ 4s }$(higher energy orbital) to the $\text{ 3s }$(lower energy orbital ) is accomplished by the emission of radiation or emission of a photon having the energy equal to the energy difference between the high and low energy orbitals.
During the transition, the change in the spin quantum number should be equal to zero. Such a transition is allowed.
$\text{ }\Delta \text{S = 0 }$
Here, the photon is emitted during the transition of electrons from $\text{ 4s }$ to $\text{ 3s }$orbital. The photon has a spin angular moment of 1. Thus, the spin angular momentum is no longer equal to zero. Similarly the angular momentum should have changed by a 1 unit. But this change is taken away by the emitting photon.
$\text{ }\Delta \text{S }\ne \text{ 0 }$
The transition of electrons from $\text{ 4s }$ to $\text{ 3s }$orbital is a forbidden transition because there will be no change in the orbital angular momentum of the electron$\text{ }\Delta l\text{ = 0 }$, although the emitted photon has angular momentum.So, the correct answer is “Option A”.
Note: Note that, apart from angular momentum and spin momentum the selection rule is also given based on the symmetry of the orbital. The parity (symmetry) of the orbitals should change during the transition. That is if the symmetry of the orbital is ‘g’ must change to ‘u’ or vice versa. The transition between orbitals having the same symmetry is said to be a forbidden transition. For example, the transition between two s have the symmetry of ‘g’ (gerade, even) is forbidden.
$\begin{align}
& \text{ Allowed : g}\leftrightarrow \text{u , u}\leftrightarrow \text{g} \\
& \text{ Forbidden : u}\leftrightarrow \text{u , g}\leftrightarrow \text{g } \\
\end{align}$
Complete step by step answer:
The transition of an electron from the higher energy level to the lower energy level takes place by the emission of radiation. As there is an energy difference between the high and low energy level orbital, the electronic transition takes place only when the excess of the energy is liberated in the form of radiations.
According to the Laporte selection rule, the centrosymmetric environment transitions between the s-s, p-p, d-d, or f-f transitions are said to be forbidden. This is not allowed transition. Transition that is allowed should have the overall change in the orbital angular momentum or azimuthal quantum number ‘l’ by one unit. That is,
$\text{ }\Delta l\text{ = }\pm \text{ 1 or }-1\text{ }$
Let's have a look at the quantum numbers of the $\text{ 4s }$ and $\text{ 3s }$orbital. The principal quantum number that is shell number for $\text{ 4s }$orbital is $\text{ n = 4 }$ and for $\text{ 3s }$orbital it is found to be$\text{ n = 3 }$.
Here, both the orbitals under consideration are ‘s’ orbitals, thus despite the different shell numbers, the $\text{ 4s }$ and $\text{ 3s }$orbital have the same azimuthal quantum number. That is,
$\text{ }{{l}_{4s}}\text{ = }{{l}_{3s}}\text{ = 0 }$
Here, change is the azimuthal quantum number is equal to,
$\text{ }\Delta l={{l}_{4s}}\text{ }-\text{ }{{l}_{3s}}\text{ = 0 }$
The transition of an electron from the $\text{ 4s }$(higher energy orbital) to the $\text{ 3s }$(lower energy orbital ) is accomplished by the emission of radiation or emission of a photon having the energy equal to the energy difference between the high and low energy orbitals.
During the transition, the change in the spin quantum number should be equal to zero. Such a transition is allowed.
$\text{ }\Delta \text{S = 0 }$
Here, the photon is emitted during the transition of electrons from $\text{ 4s }$ to $\text{ 3s }$orbital. The photon has a spin angular moment of 1. Thus, the spin angular momentum is no longer equal to zero. Similarly the angular momentum should have changed by a 1 unit. But this change is taken away by the emitting photon.
$\text{ }\Delta \text{S }\ne \text{ 0 }$
The transition of electrons from $\text{ 4s }$ to $\text{ 3s }$orbital is a forbidden transition because there will be no change in the orbital angular momentum of the electron$\text{ }\Delta l\text{ = 0 }$, although the emitted photon has angular momentum.So, the correct answer is “Option A”.
Note: Note that, apart from angular momentum and spin momentum the selection rule is also given based on the symmetry of the orbital. The parity (symmetry) of the orbitals should change during the transition. That is if the symmetry of the orbital is ‘g’ must change to ‘u’ or vice versa. The transition between orbitals having the same symmetry is said to be a forbidden transition. For example, the transition between two s have the symmetry of ‘g’ (gerade, even) is forbidden.
$\begin{align}
& \text{ Allowed : g}\leftrightarrow \text{u , u}\leftrightarrow \text{g} \\
& \text{ Forbidden : u}\leftrightarrow \text{u , g}\leftrightarrow \text{g } \\
\end{align}$
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