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

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.