Question

(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 Fe²⁺ (z = 26), Mn²⁺ (z = 25) and argon (z=18)?

Answer 1

(I) An element X contains 35 electrons and 45 neutrons 

1. The number of protons must be equal to the number of electrons. So the number of protons = 35. 

2. Number of electrons = 35. So the electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁵.

3. The last electron i.e. 5^th electron in 4p orbital has the following quantum numbers. 

n = 4, 1 = 1, m= +1, s = -\(\frac{1}{2}\)

(II) Fe → Fe²⁺ + 3e^-

Fe (Z = 26) Fe³⁺ = number of electrons = 23

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² for Fe atom. 

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ for Fe³⁺ ion. 

So, it contain 5 unpaired electrons. Mn (Z = 25). Electronic configuration is

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵

Mn → Mn²⁺ + 2e^-

Number of unpaired electrons in Mn²⁺ = 5

Ar (Z = 18). Electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ .

All orbitals are completely filled. So, no unpaired electrons in it.