Question

Pick out the correct statements-
A. ${\psi _{1s(H)}}$ and ${\psi _{2s(H)}}$ are orthogonal to each other.
B. The degeneracy of the orbitals of the H-atom having energy \[ - \dfrac{{Rydberg}}{{16}}\] is \[30\].
C. The average distance of the \[2s\] electron from the nucleus of H-atom is \[4{a_ \circ }\].
D. The most probable distance of an electron in the \[2p - \]orbital of an H-atom is \[4{a_ \circ }\].

Answer 1

Hint: We will analyze each statement accordingly. Wave functions are orthogonal to each other when their product is equal to zero. Then we will find the degeneracy of the orbital of the H-atom with the help of the ionization energy of the H-atom. We will use a formula for finding the average distance of an orbital electron from its nucleus. In the same manner we will calculate the most probable distance.
Formula Used:
\[(i){\text{ }}\int {{\psi _a}} {\text{ }}{\psi _b}{\text{ = }}\left\langle {\dfrac{a}{b}} \right\rangle \]
Where, \[{\psi _a}\] and \[{\psi _b}\] are wave functions of a and b orbital.
\[(ii)\] Ionization Energy \[{\text{ = - }}\dfrac{{Rydberg}}{{{n^2}}}\]
Where \[n\] is the Bohr orbit of a hydrogen atom.
\[(iii)\] Average Distance \[{\text{ = }}\dfrac{{{a_ \circ }}}{2}\left[ {3{n^2} - l\left( {l + 1} \right)} \right]\]
Where, \[n\] is the principal quantum number and \[l\] is azimuthal quantum number.

Complete Answer:
We will analyze each statement which are mention above and then we will be able to find the correct answer:
For A:
Let the wave function for a orbital and b orbital be represented by \[{\psi _a}\] and \[{\psi _b}\] then they are said to be orthogonal to other when the integration of their product is equal to zero. It can be represented as:
\[{\text{ }}\int {{\psi _a}} {\text{ }}{\psi _b}{\text{ = }}\left\langle {\dfrac{a}{b}} \right\rangle {\text{ = 0}}\]
According to the question we have ${\psi _{1s(H)}}$ and ${\psi _{2s(H)}}$ as our wave function. On comparing we get \[a = {\text{ }}1s\] and \[b = {\text{ }}2s\]. Thus on substituting the values in above equation we get the result as:
\[{\text{ }}\int {{\psi _a}} {\text{ }}{\psi _b}{\text{ = }}\left\langle {\dfrac{a}{b}} \right\rangle {\text{ }}\]
\[{\text{ }}\int {{\psi _{1s\left( H \right)}}} {\text{ }}{\psi _{2s\left( H \right)}}{\text{ = }}\left\langle {\dfrac{{1s}}{{2s}}} \right\rangle {\text{ }}\]
Since both belongs to same orbital we can say that,
\[{\text{ }}\int {{\psi _{1s\left( H \right)}}} {\text{ }}{\psi _{2s\left( H \right)}}{\text{ = }}\left\langle {\dfrac{{1s}}{{2s}}} \right\rangle {\text{ = 0}}\]
Hence they are orthogonal to each other.
For B:
We can find out the ionization energy of an electron by using relation:
Ionization Energy \[{\text{ = - }}\dfrac{{Rydberg}}{{{n^2}}}\]
According to the question it is given as \[ - \dfrac{{Rydberg}}{{16}}\]. Thus on comparing we get the value of n as,
\[{n^2}{\text{ }} = {\text{ 16}}\]
\[n{\text{ }} = {\text{ 4}}\]
The degeneracy of an orbital of an H- atom is equal to \[{n^2}\]. Since we got, \[n{\text{ }} = {\text{ 4}}\], therefore \[{n^2}{\text{ }} = {\text{ 16}}\].
But according to the question it is given as \[30\] , hence it is a wrong statement.
For C:
The average distance orbital electron can be found by using formula as;
Average Distance \[{\text{ = }}\dfrac{{{a_ \circ }}}{2}\left[ {3{n^2} - l\left( {l + 1} \right)} \right]\]
For \[2s\] electrons the value of \[n\] is two and the value of \[l\] is zero. Thus on substituting the value we get the average distance as,
Average Distance \[{\text{ = }}\dfrac{{{a_ \circ }}}{2}\left[ {3{{\left( 2 \right)}^2} - 0\left( {0 + 1} \right)} \right]\]
Average Distance \[{\text{ = }}\dfrac{{{a_ \circ }}}{2}{\text{ }} \times {\text{ 12}}\]
Average Distance \[{\text{ = 6}}{a_ \circ }{\text{ }}\]
But according to the question, the average distance is \[4{a_ \circ }\]. Hence it is also a wrong statement.
For D:
For finding most probable distance we have same formula for average distance which is equal to,
Most Probable Distance \[{\text{ = }}\dfrac{{{a_ \circ }}}{2}\left[ {3{n^2} - l\left( {l + 1} \right)} \right]\]
For \[2p - \]orbital we have \[n\] equal to two and \[l\] equals to one. On substituting values we get the most probable distance as,
Most Probable Distance \[{\text{ = }}\dfrac{{{a_ \circ }}}{2}\left[ {3{{\left( 2 \right)}^2} - 1\left( {1 + 1} \right)} \right]\]
Most Probable Distance \[{\text{ = }}\dfrac{{{a_ \circ }}}{2}{\text{ }} \times {\text{ 10}}\]
Most Probable Distance \[{\text{ = 5}}{a_ \circ }\]
But according to the question it is given as \[4{a_ \circ }\] , therefore it is also a wrong statement.

Hence on observing each statement we can say that only A. statement is correct.

Note:
The most probable and average distance is measured in terms of \[{a_ \circ }\] . Here \[{a_ \circ }\] is the radius of the orbit of the H- atom. The formula for calculating both distances is the same but their meanings are different. The unit of both distances is Angstrom \[({A_ \circ })\]. Wave function describes the motion of electrons in a particular orbit.