Question

Give the quantum number set for one electron in the 2p sublevel of a nitrogen (N) atom.

Answer 1

Answer : The  values of all the quantum numbers for one electron in the 2p sublevel of a nitrogen (N) atom are, $(n=2,l=1,m_l=-1,0,+1,m_s=+\frac{1}{2}\text{ or }-\frac{1}{2})$

Explanation :

There are 4 quantum numbers :

Principle Quantum Numbers : It describes the size of the orbital. It is represented by n. n = 1,2,3,4....

Azimuthal Quantum Number : It describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...

Magnetic Quantum Number : It describes the orientation of the orbitals. It is represented as $m_l$. The value of this quantum number ranges from $(-l\text{ to }+l)$. When l = 2, the value of $m_l$ will be -2, -1, 0, +1, +2.

Spin Quantum number : It describes the direction of electron spin. This is represented as $m_s$ The value of this is $+\frac{1}{2}$ for upward spin and $-\frac{1}{2}$ for downward spin.

The electronic configuration of nitrogen atom is, $1s^22s^22p^3$

Now we have to calculate the values of all the quantum numbers for one electron in the 2p sublevel of a nitrogen (N) atom.

For $2p^1$ :

$n=2$       (Because it is in 2nd shell)

$l=1$         (Because it is in 'p' orbital)

$m_l=-1,0,+1$     (Because l = 1)

$m_s=+\frac{1}{2}\text{ or }-\frac{1}{2}$   (Because of the sign convention)

Therefore, the values of all the quantum numbers for one electron in the 2p sublevel of a nitrogen (N) atom are, $(n=2,l=1,m_l=-1,0,+1,m_s=+\frac{1}{2}\text{ or }-\frac{1}{2})$

Answer 2

 it can form 3 bonds because 1 electron is in each of the 3 2p oribitals. 

The electron configuration of nitrogen (atomic number 7) is 1s2 2s2 2px1 2py1 2pz1 

so specifically the 2p oribital is : 

2px1 2py1 2pz1

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