Question

Give a possible set of values of the four quantum numbers for all the electrons in a boron atom and a nitrogen atom is each in the ground state.

Short answer

Boron:$\mathrm{n}=2,\mathrm{l}=1,{\mathrm{m}}_{\mathrm{l}}=-1,{\mathrm{m}}_{\mathrm{s}}=-\frac{1}{2}$

Nitrogen:$\mathrm{n}=3,\mathrm{l}=1,{\mathrm{m}}_{\mathrm{l}}=-1,{\mathrm{m}}_{\mathrm{s}}=-\frac{1}{2}$

Step-by-step solution

Step 1:

$\mathrm{B}:1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^1$Boron ‘B’ is located in period 2, group 13 of the periodic table and has an atomic number equal to 5.

the energy level is given by the principal quantum number, n, which in the case of Boron is $2.$

Azimuthal quantum numbers can take the values from 0 to n-1. In the case of Boron , $\mathrm{l}=0,1$

• The actual orbital in which the electron can be found is given by the magnetic quantum number ${\mathrm{m}}_{\mathrm{l}}$
• For the p-subshell, in the case of boron, you have a total number of three possible orbitals.

${\mathrm{m}}_{\mathrm{l}}=-1\to \mathrm{the}{\mathrm{P}}_{\mathrm{x}}$orbital

${\mathrm{m}}_{\mathrm{l}}=0\to \mathrm{the}{\mathrm{P}}_{\mathrm{y}}$orbital

${\mathrm{m}}_{\mathrm{l}}=+1\to \mathrm{the}{\mathrm{P}}_{\mathrm{z}}$the orbital

• ${\mathrm{m}}_{\mathrm{s}}=\pm \frac{1}{2}$for spin-up $\mathrm{\xi }-\frac{1}{2}$for spin down.
• $1^{\mathrm{st}}$electron : $\mathrm{n}=2,\mathrm{l}=1,{\mathrm{m}}_{\mathrm{l}}=-1,{\mathrm{m}}_{\mathrm{s}}=+\frac{1}{2}$($2\mathrm{p}$orbital)
• $2^{\mathrm{nd}}$electron: $\mathrm{n}=2,\mathrm{l}=0,{\mathrm{m}}_{\mathrm{l}}=0,{\mathrm{m}}_{\mathrm{s}}=-\frac{1}{2}$($2\mathrm{s}$orbital)
• $3^{\mathrm{rd}}$electron: $\mathrm{n}=2,\mathrm{l}=0,{\mathrm{m}}_{\mathrm{l}}=0,{\mathrm{m}}_{\mathrm{s}}=+\frac{1}{2}$($2\mathrm{s}$orbital)
• $4^{\mathrm{th}}$electron: $\mathrm{n}=2,\mathrm{l}=0,{\mathrm{m}}_{\mathrm{l}}=0,{\mathrm{m}}_{\mathrm{s}}=-\frac{1}{2}$ ($1\mathrm{s}$orbital)
• $5^{\mathrm{th}}$electron: $\mathrm{n}=1,\mathrm{l}=0,{\mathrm{m}}_{\mathrm{l}}=0,{\mathrm{m}}_{\mathrm{s}}=+\frac{1}{2}$ ($1\mathrm{s}$orbital)

Step 2:

The atomic number of nitrogen is $7$.

Number of electrons = $7$

Electronic configuration = $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^3$ or $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}_{\mathrm{x}}^{1}2{\mathrm{p}}_{\mathrm{y}}^{1}2{\mathrm{p}}_{\mathrm{z}}^1$

When $\mathrm{l}=0$, it S-subshell.

$\mathrm{l}=1$, its P-subshell.

For electrons in $1\mathrm{s}$, both the two electrons will have $\mathrm{n}=1,\mathrm{l}=0,{\mathrm{m}}_{\mathrm{l}}=0,{\mathrm{m}}_{\mathrm{s}}=+\frac{1}{2}$for one electron $\mathrm{\xi s}=-\frac{1}{2}$ for the other electron.

For electrons in $1\mathrm{s}$, both the electrons will have $\mathrm{n}=2,\mathrm{l}=0,{\mathrm{m}}_{\mathrm{l}}=0,{\mathrm{m}}_{\mathrm{s}}=+\frac{1}{2}$ for one electron $\mathrm{\xi s}=-\frac{1}{2}$for the other electron.

For an electron in $2{\mathrm{p}}_{\mathrm{x}},\mathrm{n}=2,\mathrm{l}=1,\mathrm{m}=-1\mathrm{and}\mathrm{s}=+\frac{1}{2}\mathrm{or}-\frac{1}{2}$ .

For an electron in $2{\mathrm{p}}_{\mathrm{y}},\mathrm{n}=2,\mathrm{l}=1,\mathrm{m}=0\mathrm{and}\mathrm{s}=+\frac{1}{2}\mathrm{or}-\frac{1}{2}$.

For an electron in .