(i) An atom of an element contains 35 electrons and 45 neutrons. Deduce
1. the number of protons
2. the electronic configuration for the element
3. All the four quantum numbers for the last electron
(ii) How many unpaired electrons are present in the ground state of `Fe^(3+) (z = 26), Mn^(2+) (z = 25)` and argon (z = 18)?

Explanation

(a) (i) An element X contains 35 electrons and 45 neutrons 2. Number of electrons= 35. So the electronic configuration is `1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(10) 4p^(5)`
3. The last electron i.e., `5^(th)` electron in 4p orbital has the following quantum numbers. n=4, l=1, `m= +1, s= -½`
(ii) `Fe rarr Fe^(3+) + 3e^(-)`
Fe (Z=26) `Fe^(3+)`= number of electrons =23
`1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(6) 4s^(2)` for F atom.
`1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(5)` for `Fe^(3+)` ion.
So, it contains 5 unpaired electrons.
Mn (Z=25). Electronic configuration is `1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 4s^(2) 3d^(5)`
`Mn rarr Mn^(2+) + 2e^(-)`
Number of unpaired electrons in `Mn^(2+) =5`
Ar(Z=18) . Electronic configuration is `1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6)`.
All orbials are completely filled. So, no unpaired electrons in it.