Question

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:

Boron ‘B’ is located in period 2, group 13 of the periodic table and has an atomic number equal to 5.

$\mathrm{B}:1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^2$

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

• The azimuthal quantum number can take the values for 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.

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

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

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

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

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

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

For electrons in $1\mathrm{s}$, both the electrons will have$\mathrm{n}=2,1=0,{\mathrm{m}}_{\mathrm{l}}=0,{\mathrm{m}}_{\mathrm{s}}=+\frac{1}{2}$

for one electron$\mathrm{\xi s}=-\frac{1}{2}$ for the other electron.

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

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

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