Question

What quantum numbers specify a 6s orbital?

Answer 1

Hint: To solve this question we first need to know what quantum numbers are. The quantum numbers govern the position and the energy of an electron in an atom containing various shells, subshells, and orbitals.

Complete answer:
There are four quantum numbers.
1) Principal quantum number (n): This corresponds to the shells of an atom
It has integer values n=1,2,3,4…
The shell closest to the nucleus, K, has principal quantum number 1. The shell next to K is L and has principal quantum number 2, and so on.
In a shell, there can be a maximum of $2{{n}^{2}}$ electrons.
2) Azimuthal quantum number ($\ell $): This gives the number of subshells in a shell.
It ranges from 0 to (n-1).
It has values $\ell $=0,1,2,3 corresponding to s, p, d, and f.
3) Magnetic quantum number (m): This gives the number of orbitals in a subshell.
The number of orbitals in a subshell range from -$\ell $ to +$\ell $, including 0.
4) Spin quantum number (s): This states that an electron can have either a positive spin-up $(+\dfrac{1}{2})$ or a negative spin-down $(-\dfrac{1}{2})$.
Since Pauli's exclusion principle states that in an atom, 2 electrons can't have the same values for all four quantum numbers, hence there can only be 2 electrons of opposite spin in an orbital.
Now, we have to determine which quantum number specifies orbital 6s.
So, its principal quantum number n = 6 corresponds to shell 6 or shell P.
And its azimuthal quantum number is $\ell $=0 corresponding to the s orbital.

Note:
It should be noted that since $\ell $=0 for s orbital, the value of magnetic quantum number m = 0, i.e., the s subshell has 1 orbital which can accommodate 2 electrons of opposite spins.
All atoms having atomic numbers equal to or greater than 55 have electrons in their 6s orbital. The atoms cesium (atomic number- 55) and barium (atomic number- 56) have their valence electrons in the 6s orbital.