Question

Write four quantum numbers for 2s1 electron

Answer 1

n=2; l=0; m=0; s=+(1÷2)

Answer 2

Answer: The four quantum numbers are n = 2, l = 0, $m_l$ = 0 and $m_s=+\frac{1}{2}$

Explanation:

There are 4 quantum numbers:

Principle Quantum Number: It describes the size of the orbital and the energy level. It is represented by n. Where, 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 describes the electron shell or the energy level of an atom. The value of 'n' ranges from 1 to the shell containing outermost electron of that atom.

We are given:

Electronic configuration = $2s^1$

The four quantum numbers are:

• $n=2$       (Because it is in 2nd shell)
• $l=0$         (Because it is in 's' orbital)
• $m_l=0$     (Because l = 0)
• $m_s=+\frac{1}{2}$     (Because of the sign convention)

Hence, the four quantum numbers are n = 2, l = 0, $m_l$ = 0 and $m_s=+\frac{1}{2}$