Chapter 3, Problem 3.123QP

What is the maximum number of electrons in an atom that can have the following quantum numbers? Specify the orbitals in which the electrons would be found. (a) n = 2, m_S = +; (b) n = 4, m_l = +1; (c) n = 3, l = 2; (d) n = 2, l = 0, m_S = −; (e) n = 4, l = 3, m_l = −2.

Interpretation Introduction

Interpretation:

The maximum number of electrons in an atom which can occupy in all orbitals having the given quantum numbers and the specification of the orbitals should be explained using the concept of quantum numbers.

Concept Introduction:

Quantum Numbers

The distribution of electron density in an atom is defined by Quantum numbers.  They are derived from the mathematical solution of Schrodinger’s equation in the hydrogen atom.  The four types of quantum numbers are the principal quantum number ($n$), the angular momentum quantum number ($l$), the magnetic quantum number ($m_l$) and the electron spin quantum number ($m_s$).  Each atomic orbital in an atom is characterized by a unique set of the quantum numbers.

Principal Quantum Number ($n$)

The size of an orbital and the energy of an electron are specified by the principal quantum number ( $n$).  If the value of $n$ is larger, then the average distance from the nucleus to the specified orbital of an electron will be greater.  Hence, the orbital’s size is large with the increasing energy. The principal quantum numbers get the integral values of $1,2,3$ and so on.  If the same value of ‘ $n$’ is present in the orbitals, then, all the electrons are occupied in the same shell (level).  The total number of the orbitals corresponding to a given $n$ value is found by $n^2$.

Angular Momentum Quantum Number ($l$)

The shape of the atomic orbital is given by the angular momentum quantum number ( $l$) which are integers and its values depend on the integral value of the principal quantum number, $n$.  The probable values of $l$ range are given from $0\text{ to }n-1$ for $n$ value.  There is one possible value of $l$ ($l\text{ = 0}$) for $n=1$.  There are two values of $l$ which are $0\text{ and }1$ for $n\text{ = 2}$.  There are three values of $l$ which are $\text{0, 1 and 2}$ for $n\text{ = 3}$.   The $l$ value gives the type of orbitals namely, $s,p,dandf$.  s orbital comes for $l\text{ = 0}$; p orbital for $l\text{ = 1}$; d orbital for $l\text{ = 2}$; f orbital  for $l\text{ = 3}$.  If the orbitals have the same $n$ and $l$ values, they are present in the same subshell (sublevel).  A smaller amount of energy is contributed by the $l$ values which increase with the subshell levels $\left(s<p<d<f\right)$.

Magnetic Quantum Number ($m_l$)

The orientation of the orbital in space is given the magnetic quantum number ( $m_l$).  The value of $m_l$ depends on the $l$ value in a subshell.  It divides the subshell into the individual orbitals which have the electrons.  There are $\left(2l+1\right)$ integral $m_l$ values for a $l$ value which is explained as follows:

${}_{ }{m}_l =-l\mathrm{...0...}+l$

There is one possible $m_l$ value which is $0$ for $l\text{ = 0}$.

There are three $m_l$ values which are $-1,0\text{ and }+1$ for $l\text{ = 1}$.

There are five $m_l$ values which are $-2,-1,0,+1\text{ and }+2$ for $l\text{ = 2}$.

There are seven $m_l$ values which are $-3,-2,-1,0,+1,+2\text{ and }+3$ for $l\text{ = 3}$ and so on.

For a particular $l$ value,  the number of $m_l$ values specifies the number of orbitals in a subshell.  Therefore, each $m_l$ value gets a different orbital.

Electron Spin Quantum Number ($m_s$)

For an electron, the orientation of the spin axis is given by it.  An electron can spin in two directions.  There are two possible ways to represent $m_s$ values.  They are +½ and ‒½.  One electron spins in the clockwise direction.  Another electron spins in the anticlockwise direction.  But, two electrons should not have the same spin quantum number.

Pauli exclusion principle

The two electrons in an atom should not have the four same quantum numbers.  Two electrons are occupied in an atomic orbital because there are two possible values of $m_s$.  If two electrons have the same values of $n,l{\text{ and }}_{ }{m}_l$, they should have different values of $m_s$.

To find: Count the maximum number of electrons in an atom that can have the quantum number $n=2$, $m_s=+$ (a) and specify the orbitals which have the electrons

Answer to Problem 3.123QP

The maximum number of electrons in an atom that can have the quantum number $n=2$, $m_s=+$ is 4.  $2s,2p_x,2p_{y}and2p_z$-orbitals are involved in which each orbital occupy a single electron.

Explanation of Solution

For a given value of $n$, the possible number of orbitals involved is $n^2$.  When $n=2$, $4\left(2^2\right)$ orbitals are involved.

For a given value of $n$, the possible values of $l$ range are from $0\text{ to }n-1$.  When $n=2$, the angular momentum quantum number ($l$) values are 0 and 1.  They correspond to 2s and 2p-subshells.

If $l\text{ = 0}$, the number of possible magnetic quantum number ($m_l$) values are calculated using the formula $\left(2l+1\right)$ for $n=2$.  Here, $\left(\text{2}\left(\text{0}\right)\text{ + 1}\right)=\text{ 1}$ results.  Therefore, there is only one orbital present when $l\text{ = 0}$.  It corresponds to 2s-atomic orbital.  If $l\text{ = 1}$, $\left(\text{2}\left(\text{1}\right)\text{ + 1}\right)=\text{ 3}$ results.  Therefore, there are three orbitals present when $l\text{ = 1}$.  They correspond to $2p_x,2p_{y}and2p_z$-atomic orbitals.  Totally, 4 atomic orbitals are present when $n=2$.

Here, $m_s=+$ is given.  Only one direction spin orbital is involved.  Hence, only one electron is occupied in each orbital.  Therefore, the maximum number of electrons in an atom that can have the quantum number $\text{n }=\text{ 2}$, $m_s=+$ (a) is 4.  $2s,2p_x,2p_{y}and2p_z$-orbitals are involved in which each orbital occupy a single electron.

Interpretation Introduction

Interpretation:

The maximum number of electrons in an atom which can occupy in all orbitals having the given quantum numbers and the specification of the orbitals should be explained using the concept of quantum numbers.

Concept Introduction:

Quantum Numbers

The distribution of electron density in an atom is defined by Quantum numbers.  They are derived from the mathematical solution of Schrodinger’s equation in the hydrogen atom.  The four types of quantum numbers are the principal quantum number ($n$), the angular momentum quantum number ($l$), the magnetic quantum number ($m_l$) and the electron spin quantum number ($m_s$).  Each atomic orbital in an atom is characterized by a unique set of the quantum numbers.

Principal Quantum Number ($n$)

The size of an orbital and the energy of an electron are specified by the principal quantum number ( $n$).  If the value of $n$ is larger, then the average distance from the nucleus to the specified orbital of an electron will be greater.  Hence, the orbital’s size is large with the increasing energy. The principal quantum numbers get the integral values of $1,2,3$ and so on.  If the same value of ‘ $n$’ is present in the orbitals, then, all the electrons are occupied in the same shell (level).  The total number of the orbitals corresponding to a given $n$ value is found by $n^2$.

Angular Momentum Quantum Number ($l$)

The shape of the atomic orbital is given by the angular momentum quantum number ( $l$) which are integers and its values depend on the integral value of the principal quantum number, $n$.  The probable values of $l$ range are given from $0\text{ to }n-1$ for $n$ value.  There is one possible value of $l$ ($l\text{ = 0}$) for $n=1$.  There are two values of $l$ which are $0\text{ and }1$ for $n\text{ = 2}$.  There are three values of $l$ which are $\text{0, 1 and 2}$ for $n\text{ = 3}$.   The $l$ value gives the type of orbitals namely, $s,p,dandf$.  s orbital comes for $l\text{ = 0}$; p orbital for $l\text{ = 1}$; d orbital for $l\text{ = 2}$; f orbital  for $l\text{ = 3}$.  If the orbitals have the same $n$ and $l$ values, they are present in the same subshell (sublevel).  A smaller amount of energy is contributed by the $l$ values which increase with the subshell levels $\left(s<p<d<f\right)$.

Magnetic Quantum Number ($m_l$)

The orientation of the orbital in space is given the magnetic quantum number ( $m_l$).  The value of $m_l$ depends on the $l$ value in a subshell.  It divides the subshell into the individual orbitals which have the electrons.  There are $\left(2l+1\right)$ integral $m_l$ values for a $l$ value which is explained as follows:

${}_{ }{m}_l =-l\mathrm{...0...}+l$

There is one possible $m_l$ value which is $0$ for $l\text{ = 0}$.

There are three $m_l$ values which are $-1,0\text{ and }+1$ for $l\text{ = 1}$.

There are five $m_l$ values which are $-2,-1,0,+1\text{ and }+2$ for $l\text{ = 2}$.

There are seven $m_l$ values which are $-3,-2,-1,0,+1,+2\text{ and }+3$ for $l\text{ = 3}$ and so on.

For a particular $l$ value,  the number of $m_l$ values specifies the number of orbitals in a subshell.  Therefore, each $m_l$ value gets a different orbital.

Electron Spin Quantum Number ($m_s$)

For an electron, the orientation of the spin axis is given by it.  An electron can spin in two directions.  There are two possible ways to represent $m_s$ values.  They are +½ and ‒½.  One electron spins in the clockwise direction.  Another electron spins in the anticlockwise direction.  But, two electrons should not have the same spin quantum number.

Pauli exclusion principle

The two electrons in an atom should not have the four same quantum numbers.  Two electrons are occupied in an atomic orbital because there are two possible values of $m_s$.  If two electrons have the same values of $n,l{\text{ and }}_{ }{m}_l$, they should have different values of $m_s$.

To find: Count the maximum number of electrons in an atom that can have the quantum number $n=4$, $m_s=+$ (b) and specify the orbitals which have the electrons

Answer to Problem 3.123QP

The maximum number of electrons in an atom that can have the quantum number $n=4$, $m_s=+$ is 6.  $4p,4dand4f$-orbitals are involved in which each orbital occupy two electrons.

Explanation of Solution

If $l\text{ = 0}$, the number of possible magnetic quantum number ($m_l$) values are calculated using the formula $\left(2l+1\right)$ for $n=4$.  Here, $\left(\text{2}\left(\text{0}\right)\text{ + 1}\right)=\text{ 1}$ results.  Therefore, there is only one orbital present when $l\text{ = 0}$ which get the $m_l$ value of 0.  It corresponds to 4s-atomic orbital which is not involved in (b).  If If $l\text{ = 1}$, $\left(\text{2}\left(\text{1}\right)\text{ + 1}\right)=\text{ 3}$ results.  Therefore, there are three orbitals present when $l\text{ = 1}$ which get the $m_l$ values of $-1,0\text{ and }+1$.  Here, one 4p-orbital which corresponds to $m_l$ values of +1 is involved.   If $l\text{ = 2}$, $\left(\text{2}\left(2\right)\text{ + 1}\right)=\text{ 5}$ results.  Therefore, there are five orbitals present when $l\text{ = 2}$ which get the $m_l$ values of $-2,-1,0,+1\text{ and }+2$.  Here, one 4d-orbital which corresponds to $m_l$ values of +1 is involved.  If $l\text{ = 3}$, $\left(\text{2}\left(3\right)\text{ + 1}\right)=\text{ 7}$ results.  Therefore, there are seven orbitals present when $l\text{ = 3}$ which get the $m_l$ values of $-3,-2,-1,0,+1,+2\text{ and }+3$.  Here, one 4f-orbital which corresponds to $m_l$ values of +1 is involved.  Therefore, one 4p, one 4d and one 4f-orbital are involved in (b).

Each of $4p,4dand4f$-orbitals occupy two electrons.  Hence, 6 electrons are resulted.  Therefore, the maximum number of electrons in an atom that can have the quantum number $n=2$, $m_s=+$ (b) is 6.  $4p,4dand4f$-orbitals are involved in which each orbital occupy two electrons.

Interpretation Introduction

Interpretation:

The maximum number of electrons in an atom which can occupy in all orbitals having the given quantum numbers and the specification of the orbitals should be explained using the concept of quantum numbers.

Concept Introduction:

Quantum Numbers

The distribution of electron density in an atom is defined by Quantum numbers.  They are derived from the mathematical solution of Schrodinger’s equation in the hydrogen atom.  The four types of quantum numbers are the principal quantum number ($n$), the angular momentum quantum number ($l$), the magnetic quantum number ($m_l$) and the electron spin quantum number ($m_s$).  Each atomic orbital in an atom is characterized by a unique set of the quantum numbers.

Principal Quantum Number ($n$)

The size of an orbital and the energy of an electron are specified by the principal quantum number ( $n$).  If the value of $n$ is larger, then the average distance from the nucleus to the specified orbital of an electron will be greater.  Hence, the orbital’s size is large with the increasing energy. The principal quantum numbers get the integral values of $1,2,3$ and so on.  If the same value of ‘ $n$’ is present in the orbitals, then, all the electrons are occupied in the same shell (level).  The total number of the orbitals corresponding to a given $n$ value is found by $n^2$.

Angular Momentum Quantum Number ($l$)

The shape of the atomic orbital is given by the angular momentum quantum number ( $l$) which are integers and its values depend on the integral value of the principal quantum number, $n$.  The probable values of $l$ range are given from $0\text{ to }n-1$ for $n$ value.  There is one possible value of $l$ ($l\text{ = 0}$) for $n=1$.  There are two values of $l$ which are $0\text{ and }1$ for $n\text{ = 2}$.  There are three values of $l$ which are $\text{0, 1 and 2}$ for $n\text{ = 3}$.   The $l$ value gives the type of orbitals namely, $s,p,dandf$.  s orbital comes for $l\text{ = 0}$; p orbital for $l\text{ = 1}$; d orbital for $l\text{ = 2}$; f orbital  for $l\text{ = 3}$.  If the orbitals have the same $n$ and $l$ values, they are present in the same subshell (sublevel).  A smaller amount of energy is contributed by the $l$ values which increase with the subshell levels $\left(s<p<d<f\right)$.

Magnetic Quantum Number ($m_l$)

The orientation of the orbital in space is given the magnetic quantum number ( $m_l$).  The value of $m_l$ depends on the $l$ value in a subshell.  It divides the subshell into the individual orbitals which have the electrons.  There are $\left(2l+1\right)$ integral $m_l$ values for a $l$ value which is explained as follows:

${}_{ }{m}_l =-l\mathrm{...0...}+l$

There is one possible $m_l$ value which is $0$ for $l\text{ = 0}$.

There are three $m_l$ values which are $-1,0\text{ and }+1$ for $l\text{ = 1}$.

There are five $m_l$ values which are $-2,-1,0,+1\text{ and }+2$ for $l\text{ = 2}$.

There are seven $m_l$ values which are $-3,-2,-1,0,+1,+2\text{ and }+3$ for $l\text{ = 3}$ and so on.

For a particular $l$ value,  the number of $m_l$ values specifies the number of orbitals in a subshell.  Therefore, each $m_l$ value gets a different orbital.

Electron Spin Quantum Number ($m_s$)

For an electron, the orientation of the spin axis is given by it.  An electron can spin in two directions.  There are two possible ways to represent $m_s$ values.  They are +½ and ‒½.  One electron spins in the clockwise direction.  Another electron spins in the anticlockwise direction.  But, two electrons should not have the same spin quantum number.

Pauli exclusion principle

The two electrons in an atom should not have the four same quantum numbers.  Two electrons are occupied in an atomic orbital because there are two possible values of $m_s$.  If two electrons have the same values of $n,l{\text{ and }}_{ }{m}_l$, they should have different values of $m_s$.

To find: Count the maximum number of electrons in an atom that can have the quantum number $n=3$, $l\text{ = 2}$ (c) and specify the orbitals which have the electrons

Answer to Problem 3.123QP

The maximum number of electrons in an atom that can have the quantum number $n=3$, $l\text{ = 2}$ is 10.  $3d_{xy},3d_{yz},3d_{zx},3d_{x^2-y^2}and3d_{z^2}$-orbitals are involved in which each orbital occupy two electrons.

Explanation of Solution

If $l\text{ = 2}$, the number of possible magnetic quantum number ($m_l$) values are calculated using the formula $\left(2l+1\right)$ for $n=3$.  Here, $\left(\text{2}\left(2\right)\text{ + 1}\right)=\text{ 5}$ results.  Therefore, there are five orbitals present when $l\text{ = 2}$ which get the $m_l$ values of $-2,-1,0,+1\text{ and }+2$.  Here, five 3d-orbitals are involved.  They are $3d_{xy},3d_{yz},3d_{zx},3d_{x^2-y^2}and3d_{z^2}$ orbitals involved in (c).

Each of 3d orbitals occupies two electrons.  Hence, 10 electrons are resulted.  Therefore, the maximum number of electrons in an atom that can have the quantum number $n=3$, $l\text{ = 2}$ (c) is 10.  $3d_{xy},3d_{yz},3d_{zx},3d_{x^2-y^2}and3d_{z^2}$-orbitals are involved in which each orbital occupy two electrons.

Interpretation Introduction

Interpretation:

The maximum number of electrons in an atom which can occupy in all orbitals having the given quantum numbers and the specification of the orbitals should be explained using the concept of quantum numbers.

Concept Introduction:

Quantum Numbers

The distribution of electron density in an atom is defined by Quantum numbers.  They are derived from the mathematical solution of Schrodinger’s equation in the hydrogen atom.  The four types of quantum numbers are the principal quantum number ($n$), the angular momentum quantum number ($l$), the magnetic quantum number ($m_l$) and the electron spin quantum number ($m_s$).  Each atomic orbital in an atom is characterized by a unique set of the quantum numbers.

Principal Quantum Number ($n$)

The size of an orbital and the energy of an electron are specified by the principal quantum number ( $n$).  If the value of $n$ is larger, then the average distance from the nucleus to the specified orbital of an electron will be greater.  Hence, the orbital’s size is large with the increasing energy. The principal quantum numbers get the integral values of $1,2,3$ and so on.  If the same value of ‘ $n$’ is present in the orbitals, then, all the electrons are occupied in the same shell (level).  The total number of the orbitals corresponding to a given $n$ value is found by $n^2$.

Angular Momentum Quantum Number ($l$)

The shape of the atomic orbital is given by the angular momentum quantum number ( $l$) which are integers and its values depend on the integral value of the principal quantum number, $n$.  The probable values of $l$ range are given from $0\text{ to }n-1$ for $n$ value.  There is one possible value of $l$ ($l\text{ = 0}$) for $n=1$.  There are two values of $l$ which are $0\text{ and }1$ for $n\text{ = 2}$.  There are three values of $l$ which are $\text{0, 1 and 2}$ for $n\text{ = 3}$.   The $l$ value gives the type of orbitals namely, $s,p,dandf$.  s orbital comes for $l\text{ = 0}$; p orbital for $l\text{ = 1}$; d orbital for $l\text{ = 2}$; f orbital  for $l\text{ = 3}$.  If the orbitals have the same $n$ and $l$ values, they are present in the same subshell (sublevel).  A smaller amount of energy is contributed by the $l$ values which increase with the subshell levels $\left(s<p<d<f\right)$.

Magnetic Quantum Number ($m_l$)

The orientation of the orbital in space is given the magnetic quantum number ( $m_l$).  The value of $m_l$ depends on the $l$ value in a subshell.  It divides the subshell into the individual orbitals which have the electrons.  There are $\left(2l+1\right)$ integral $m_l$ values for a $l$ value which is explained as follows:

${}_{ }{m}_l =-l\mathrm{...0...}+l$

There is one possible $m_l$ value which is $0$ for $l\text{ = 0}$.

There are three $m_l$ values which are $-1,0\text{ and }+1$ for $l\text{ = 1}$.

There are five $m_l$ values which are $-2,-1,0,+1\text{ and }+2$ for $l\text{ = 2}$.

There are seven $m_l$ values which are $-3,-2,-1,0,+1,+2\text{ and }+3$ for $l\text{ = 3}$ and so on.

For a particular $l$ value,  the number of $m_l$ values specifies the number of orbitals in a subshell.  Therefore, each $m_l$ value gets a different orbital.

Electron Spin Quantum Number ($m_s$)

For an electron, the orientation of the spin axis is given by it.  An electron can spin in two directions.  There are two possible ways to represent $m_s$ values.  They are +½ and ‒½.  One electron spins in the clockwise direction.  Another electron spins in the anticlockwise direction.  But, two electrons should not have the same spin quantum number.

Pauli exclusion principle

The two electrons in an atom should not have the four same quantum numbers.  Two electrons are occupied in an atomic orbital because there are two possible values of $m_s$.  If two electrons have the same values of $n,l{\text{ and }}_{ }{m}_l$, they should have different values of $m_s$.

To find: Count the maximum number of electrons in an atom that can have the quantum number $n=2$, $l\text{ = 0}$, $m_s=-$ (d) and specify the orbitals which have the electrons

Answer to Problem 3.123QP

The maximum number of electrons in an atom that can have the quantum number $n=2$, $l\text{ = 0}$, $m_s=-$1 is 1.  $2s$-orbital is involved in which it occupies a single electron.

Explanation of Solution

If $l\text{ = 0}$, the number of possible magnetic quantum number ($m_l$) values are calculated using the formula $\left(2l+1\right)$ for $n=2$.  Here, $\left(\text{2}\left(\text{0}\right)\text{ + 1}\right)=\text{ 1}$ results.  Therefore, there is only one orbital present when $l\text{ = 0}$ which get the $m_l$ value of zero only.  This corresponds to 2s orbital in (d).

Here, $m_s=-$ is given.  Only one direction spin orbital is involved.  Hence, only one electron is occupied in 2s orbital.  Therefore, the maximum number of electrons in an atom that can have the quantum number $n=2$, $l\text{ = 0}$, $m_s=-$ (d) is 1.  $2s$-orbital is involved in which it occupies a single electron.

Interpretation Introduction

Interpretation:

The maximum number of electrons in an atom which can occupy in all orbitals having the given quantum numbers and the specification of the orbitals should be explained using the concept of quantum numbers.

Concept Introduction:

Quantum Numbers

The distribution of electron density in an atom is defined by Quantum numbers.  They are derived from the mathematical solution of Schrodinger’s equation in the hydrogen atom.  The four types of quantum numbers are the principal quantum number ($n$), the angular momentum quantum number ($l$), the magnetic quantum number ($m_l$) and the electron spin quantum number ($m_s$).  Each atomic orbital in an atom is characterized by a unique set of the quantum numbers.

Principal Quantum Number ($n$)

The size of an orbital and the energy of an electron are specified by the principal quantum number ( $n$).  If the value of $n$ is larger, then the average distance from the nucleus to the specified orbital of an electron will be greater.  Hence, the orbital’s size is large with the increasing energy. The principal quantum numbers get the integral values of $1,2,3$ and so on.  If the same value of ‘ $n$’ is present in the orbitals, then, all the electrons are occupied in the same shell (level).  The total number of the orbitals corresponding to a given $n$ value is found by $n^2$.

Angular Momentum Quantum Number ($l$)

The shape of the atomic orbital is given by the angular momentum quantum number ( $l$) which are integers and its values depend on the integral value of the principal quantum number, $n$.  The probable values of $l$ range are given from $0\text{ to }n-1$ for $n$ value.  There is one possible value of $l$ ($l\text{ = 0}$) for $n=1$.  There are two values of $l$ which are $0\text{ and }1$ for $n\text{ = 2}$.  There are three values of $l$ which are $\text{0, 1 and 2}$ for $n\text{ = 3}$.   The $l$ value gives the type of orbitals namely, $s,p,dandf$.  s orbital comes for $l\text{ = 0}$; p orbital for $l\text{ = 1}$; d orbital for $l\text{ = 2}$; f orbital  for $l\text{ = 3}$.  If the orbitals have the same $n$ and $l$ values, they are present in the same subshell (sublevel).  A smaller amount of energy is contributed by the $l$ values which increase with the subshell levels $\left(s<p<d<f\right)$.

Magnetic Quantum Number ($m_l$)

The orientation of the orbital in space is given the magnetic quantum number ( $m_l$).  The value of $m_l$ depends on the $l$ value in a subshell.  It divides the subshell into the individual orbitals which have the electrons.  There are $\left(2l+1\right)$ integral $m_l$ values for a $l$ value which is explained as follows:

${}_{ }{m}_l =-l\mathrm{...0...}+l$

There is one possible $m_l$ value which is $0$ for $l\text{ = 0}$.

There are three $m_l$ values which are $-1,0\text{ and }+1$ for $l\text{ = 1}$.

There are five $m_l$ values which are $-2,-1,0,+1\text{ and }+2$ for $l\text{ = 2}$.

There are seven $m_l$ values which are $-3,-2,-1,0,+1,+2\text{ and }+3$ for $l\text{ = 3}$ and so on.

For a particular $l$ value,  the number of $m_l$ values specifies the number of orbitals in a subshell.  Therefore, each $m_l$ value gets a different orbital.

Electron Spin Quantum Number ($m_s$)

For an electron, the orientation of the spin axis is given by it.  An electron can spin in two directions.  There are two possible ways to represent $m_s$ values.  They are +½ and ‒½.  One electron spins in the clockwise direction.  Another electron spins in the anticlockwise direction.  But, two electrons should not have the same spin quantum number.

Pauli exclusion principle

The two electrons in an atom should not have the four same quantum numbers.  Two electrons are occupied in an atomic orbital because there are two possible values of $m_s$.  If two electrons have the same values of $n,l{\text{ and }}_{ }{m}_l$, they should have different values of $m_s$.

To find: Count the maximum number of electrons in an atom that can have the quantum number $n=4$, $l\text{ = 3}$, $m_l=-2$ (e) and specify the orbitals which have the electrons

Answer to Problem 3.123QP

The maximum number of electrons in an atom that can have the quantum number $n=4$, $l\text{ = 3}$, $m_l=-2$ is 2.  $4f$-orbital is involved in which two electrons are occupied.

Explanation of Solution

If $l\text{ = 3}$, the number of possible magnetic quantum number ($m_l$) values are calculated using the formula $\left(2l+1\right)$ for $n=4$.  Here, $\left(\text{2}\left(3\right)\text{ + 1}\right)=\text{ 7}$ results.  Therefore, there are seven orbitals present when $l\text{ = 3}$ which get the $m_l$ values of $-3,-2,-1,0,+1,+2\text{ and }+3$.  Here, one 4f-orbital which corresponds to $m_l$ values of −2 is involved.  Therefore, one 4f-orbital is involved in (e).

4f-orbital occupies two electrons.  Therefore, the maximum number of electrons in an atom that can have the quantum number $n=4$, $l\text{ = 3}$, $m_l=-2$ (e) is 2.  $4f$-orbital is involved in which two electrons are occupied.