Ncert Exercises & Solutions

Match the property given in Column I with the element given in Column II.

	Column I (Property)		Column II (Element)
A.	Lanthanoid which shows + 4 oxidation state	1.	Pm
B.	Radioactive lanthanoid	2.	Ce
C.	Lanthanoid which has 4 $f^{7}$ electronic configurations in +3 oxidation state	3.	Lu
D.	Lanthanoid which has 4 $f^{14}$ electronic configurations in +3 oxidation state	4.	Gd

Codes:

Hint: Cerium shows +3 and +4 oxidation state

Lanthanoid which shows +4 oxidation state is cerium.

{}_{58}C e = [X e] 4 f^{2} 5 d^{0} 6 s^{2};

Oxidation state = +3 , +4

Lanthnoid which has

4 f^{7}

electronic configuration in +3 oxidation state is gadolinium.

{}_{64}G d = [X e] 4 f^{7} 5 d^{1} 6 s^{2};

Oxidation state = +3

Radioactive lanthanoid is promethium. It is the only synthetic (man-made) radioactive lanthanoid.

Lanthanoid which has

4 f^{14}

electronic configuration in +3 oxidation state is lutetium

{}_{71}L u = [X e] 4 f^{14} 5 d^{1} 6 s^{2};

Oxidation state = +3

Hence, option third is the correct answer.