Which of the following element has least number of electrons in its M-shell?

The correct option is K.

To determine which element has the least number of electrons in its M-shell (the third energy level, where $n=3$), we need to write the electronic configurations of each of the given elements.

The M-shell consists of the $3s$, $3p$, and $3d$ subshells. Therefore, the number of electrons in the M-shell is the sum of the electrons in these subshells.

Let's analyze each option:

1. Potassium (K):
Atomic number ($Z$) = 19
Electronic configuration: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^1$
Electrons in the M-shell ($n=3$): $3s^{2}3p^6$ contains $2+6=8$ electrons.

2. Scandium (Sc):
Atomic number ($Z$) = 21
Electronic configuration: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{1}4s^2$
Electrons in the M-shell ($n=3$): $3s^{2}3p^{6}3d^1$ contains $2+6+1=9$ electrons.

3. Manganese (Mn):
Atomic number ($Z$) = 25
Electronic configuration: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{5}4s^2$
Electrons in the M-shell ($n=3$): $3s^{2}3p^{6}3d^5$ contains $2+6+5=13$ electrons.

4. Nickel (Ni):
Atomic number ($Z$) = 28
Electronic configuration: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{8}4s^2$
Electrons in the M-shell ($n=3$): $3s^{2}3p^{6}3d^8$ contains $2+6+8=16$ electrons.

Comparing the number of M-shell electrons:
- K: 8 electrons
- Sc: 9 electrons
- Mn: 13 electrons
- Ni: 16 electrons

Therefore, Potassium (K) has the least number of electrons in its M-shell (8 electrons).