Match List I with List II LIST I LIST II A. Lanthanum I. +3, +4 B. Pr…

The correct answer is option 3. A-II, B-I, C-III, D-IV.

LIST I	LIST II
A. Lanthanum	II. +3
B. Praseodymium	I. +3, +4
C. Neodymium	III. +2, +3, +4
D. Samarium	IV. +2, +3

The given elements belong to the lanthanide series. Lanthanides refers to 14 metallic elements present in f-block with atomic numbers ranging from 58-71.

The general electronic configuration of lanthanides is

\[4f^{1−14}5d^{0−1}6s^2\)

The last electron always enters into the 4f orbital (Aufbau principle).

All lanthanides show an oxidation state of \(+3\). It also shows an oxidation state of \(+2\) and \(+4\) in complexes, but \(+3\) is the most stable. During oxidation, electrons are removed from 6s (2 electrons) and \(4f\) orbital (1 electron). A big energy difference lies between \(6s\) and \(4f\) orbital. Being closer to the nucleus, 4f doesn’t lose electrons as easily.

Element		Electronic Configuration		Electronic Configuration of (M3+)		Oxidation States
Lanthanum	La	[Xe]          5d1	6s2	[Xe]			+3
Cerium	Ce	[Xe]4f15d1	6s2	[Xe]	4f1		+3	(+4)
Praseodymium	Pr	[Xe]4f3	6s2	[Xe]	4f2		+3	(+4)
Neodymium	Nd	[Xe]4f4	6s2	[Xe]	4f3	(+2)	+3
Promethium	Pm	[Xe]4f	6s2	[Xe]	4f4	(+2)	+3
Samarium	Sm	[Xe]4f6	6s2	[Xe]	4f5	(+2)	+3
Europium	Eu	[Xe]4f7	6s2	[Xe]	4f6	(+2)	+3
Gadolinium	Gd	[Xe]4f75d1	6s2	[Xe]	4f7		+3
Terbium	Tb	[Xe]4f9	6s2	[Xe]	4f8		+3	(+4)
Dysprosium	Dy	[Xe]4f10	6s2	[Xe]	4f9		+3	(+4)
Holmium	Ho	[Xe]4f11	6s2	[Xe]	4f10		+3
Erbium	Er	[Xe]4f12	6s2	[Xe]	4f11		+3
Thulium	Tm	[Xe]4f13	6s2	[Xe]	4f12	(+2)	+3
Ytterbium	Yb	[Xe]4f14	6s2	[Xe]	4f13	(+2)	+3
Lutetium	Lu	[Xe]4f145d1	6s2	[Xe]	4f14		+3