Question

write the electronic configuration of Ce3+ ion and calculate the magnetic moment on the basis of spin only formula

Answer 1

Answer:

Using Hund's rule to find the electronic configuration of Ce ion,

Ce : 1s² 2s² 2p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶ 4f¹ 5d¹ 6s²

Now, when the cesium atom looses three electrons to form Ce³⁺ ion, it will loose electrons from the 5th d-shell and 6th s-shell, hence,

Ce³⁺ :   1s² 2s² 2p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶ 4f¹ ;

Hence having only one unpaired electron.

Finding the magnetic spin (μ) only on the basis of spin-only formula,

μ = √n(n+2) where n = number of unpaired electrons.

Here n =2 (5s²)

μ = √2(2+2) = √8 = 2√2 = 2.828 BM (bohr magneton unit)

Answer 2

According to Hund's rule

Electronic configuration of Ce = $1s^2 2s^2 3s^2 3p^6 3d^{10} 4s^2 4p^6 4d^{10} 5s^2 5p^6 4f^1 5d^1 6s^2$

So two electrons from the 6s subshell and one electron from 5d subshell. Although 5d and 4f subshell are of the same energy level, however according to the second factor on which the energy level of an orbital depends lower the value of n(here n=4 for 4f and n=5 for 5d) lower will be the energy level. Hence as 5d is greater in energy level than 4f hence electron will come out from 5d subshell first.

Electronic configuration of $Ce^{+3}$ = $1s^2 2s^2 3s^2 3p^6 3d^{10} 4s^2 4p^6 4d^{10} 5s^2 5p^6 4f^1$

Next for the magnetic moment on the basis of spin only formula;

If unpaired electrons are present in the valence shell of an atom or ion then only the ion or atom is said to have magnetic moment.

The magnetic moment of an ion is calculated using the following formula

Magnetic Moment = $\sqrt{n(n+2)}$ B.M.(Bohr magnetum)

where n = Number of unpaired electrons

Therefore

Magnetic moment of $Ce^{+3}$ ion is = $\sqrt{1(1+2)} = \sqrt{3}$  = 1.732 B.M. (Bohr magnetum)