Question

What are the quantum numbers of the five electrons of Boron?

Answer 1

Hint: \[ \bullet \] Boron, B, has an atomic number of 5. This implies that around a neutral Boron atom there will be 5 electrons surrounding its nucleus.
 \[ \bullet \] Electronic configuration of Boron is: \[1{{s}^{2}}2{{s}^{2}}2{{p}^{1}}\]

Complete answer: \[ \bullet \] Quantum numbers are basically a set of numbers which are used to describe the position and energy of the electron in an atom.
 \[ \bullet \] As we have 5 electrons here that means we will have 5 sets of quantum numbers.

Number	Symbol	Possible values
Principle Quantum number	$n$	$1,2,3,.....$
Angular momentum Quantum number	$i$	$0,1,2,.....,(n-1)$
Magnetic Quantum number	${{m}_{l}}$	$-l,....,-1,0,1,....,l$
Spin Quantum number	${{m}_{s}}$	$+\dfrac{1}{2},-\dfrac{1}{2}$

\[ \bullet \] Here, the Electronic configuration of Boron is: \[1{{s}^{2}}2{{s}^{2}}2{{p}^{1}}\] so,
 \[ \bullet \] The ${{1}^{st}}$ electron from the orbital which is highest in energy will come. Because of which, this electron will come from a $2p$ -orbital.
 \[ \bullet \] Principal Quantum number, $n$, gives the energy level and here it is equal to $2$.
 \[ \bullet \] Angular Momentum Quantum number, $l$, gives the subshell in which electrons can be found. Here the electron is located in ${{2}^{nd}}$ energy level so,
\[ \bullet \] $l=0\to $ the $2s$ -subshell
\[ \bullet \] $l=1\to $ the $2p$ -subshell
 \[ \bullet \] Magnetic Quantum Number, ${{m}_{l}}$, gives the actual orbital in which the electrons can be found. There are three possible orbitals for p-subshells
\[ \bullet \] ${{m}_{{{l}_{{}}}}}=-1\to $the ${{p}_{x}}$orbital
 \[ \bullet \] ${{m}_{l}}=0\to $the ${{p}_{y}}$ orbital
\[ \bullet \] ${{m}_{l}}=+1\to $the ${{p}_{z}}$ orbital
\[ \bullet \] Spin Quantum Number, ${{m}_{s}}$, gives the spin of electrons
 \[ \bullet \] ${{m}_{s}}=+\dfrac{1}{2}$$\to $Spin-up
\[ \bullet \] ${{m}_{s}}=-\dfrac{1}{2}\to $Spin-down
 \[ \bullet \] So, the quantum numbers for the First electron:
First Electron: $n=2$, $l=1$, ${{m}_{l}}=-1$, ${{m}_{s}}=+\dfrac{1}{2}$
 \[ \bullet \] The Second and Third electrons will come in $2s$ orbital so they will have:
Second Electron: $n=2$, $l=0$, ${{m}_{l}}=0$, ${{m}_{s}}-\dfrac{1}{2}$
Third Electron: $n=2$, $l=0$, ${{m}_{l}}=0$, ${{m}_{s}}=+\dfrac{1}{2}$
 \[ \bullet \] The Fourth and Fifth electron will come in $1p$ orbital so they will have:
Fourth Electron: $n=1$,$l=0$, ${{m}_{l}}=0$, ${{m}_{s}}-\dfrac{1}{2}$
Fifth Electron: $n=1$, $l=0$, ${{m}_{l}}=0$, ${{m}_{s}}=+\dfrac{1}{2}$

Note:
It is really interesting to note that only three quantum numbers are required to specify an orbital while four quantum numbers are required to specify an electron. So, for orbital, three and for electrons, we need four quantum numbers.