Question

Briefly explain why electronic configurations of lanthanoids are not known with certainty.

Answer 1

In the lanthanoids, 4f and 5d subshells are very close in energy. The outermost 6s-orbital remains filled with two electrons (6s²). The electrons can easily jump from 4f to 5d or vice-versa. Further, irregularities in electronic configurations are also related to the stabilities of f⁰ , f⁷ and f¹⁴ occupancy of f-orbitals. Hence, their electronic configurations are not known with certainty.

Electronic configuration of lanthanum and 4f-series of f-block elements:

i. The 4f-series includes elements from cerium (Ce) to lutetium (Lu). The electronic configuration of these elements can be expressed in terms of its nearest inert gas Xe (Z = 54).

ii. Electronic configuration of Xe (Z = 54) = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹0 5s² 5p⁶ Therefore, general electronic configuration of 4f-series is [Xe] 4f^1-14 5d^0-1 6s² .

iii. Lanthanum has electronic configuration [Xe]4f⁰ 5d¹ 6s² . It does not have any 4f electrons.

Element	Symbol	Atomic Number	Expected electronic configuration	Observed electronic configuration
Lanthanum	La	57	[Xe] 4f0 5d1 6s2	[Xe] 4f0 5d1 6s2
Cerium	Ce	58	[Xe] 4f1 5d1 6s2	[Xe] 4f2 5d0 6s2
Praseodymium	Pr	59	[Xe] 4f2 5d1 6s2	[Xe] 4f3 5d0 6s2
Neodymium	Nd	60	[Xe] 4f3 5d1 6s2	[Xe] 4f4 5d0 6s2
Promethium	Pm	61	[Xe] 4f4 5d1 6s2	[Xe] 4f5 5d0 6s2
Samarium	Sm	62	[Xe] 4f5 5d1 6s2	[Xe] 4f6 5d0 6s2
Europium	Eu	63	[Xe] 4f6 5d1 6s2	[Xe] 4f7 5d0 6s2
Gadolinium	Gd	64	[Xe] 4f7 5d1 6s2	[Xe] 4f7 5d1 6s2
Terbium	Tb	65	[Xe] 4f8 5d1 6s2	[Xe] 4f9 5d0 6s2
Dysprosium	Dy	66	[Xe] 4f9 5d1 6s2	[Xe] 4f10 5d0 6s2
Holmium	Ho	67	[Xe] 4f10 5d1 6s2	[Xe] 4f11 5d0 6s2
Erbium	Er	68	[Xe] 4f11 5d1 6s2	[Xe] 4f12 5d0 6s2
Thulium	Tm	69	[Xe] 4f12 5d1 6s2	[Xe] 4f13 5d0 6s2
Ytterbium	Yb	70	[Xe] 4f13 5d1 6s2	[Xe] 4f14 5d0 6s2
Lutetium	Lu	71	[Xe] 4f14 5d1 6s2	[Xe] 4f14 5d1 6s2