The correct order of ionic radii of Y3+, La3+, Eu3+ and Lu3+ is (Atomic nos. Y = 39, La = 57, Eu = 63, Lu = 71)
- Y3+ < La3+ < Eu3+ < Lu3+
- Y3+ < Lu3+ < Eu3+ < La3+
- Lu3+ < Eu3+ < La3+ < Y3+
- La3+ < Eu3+ < Lu3+ < Y3+
Answer
(2) Y3+ < Lu3+ < Eu3+ < La3+
Solution
Y (Z = 39) is a 4d element lying in period 5, just above La in group 3.
⇒ Y3+ has one shell less than La3+, Eu3+ and Lu3+ (period 6) ⇒ Y3+ is the smallest of the four.
As per NCERT, in the lanthanoid series the M3+ ionic radii decrease steadily from La3+ to Lu3+ (lanthanoid contraction, due to poor shielding by 4f electrons).
⇒ La3+ > Eu3+ > Lu3+
∴ Y3+ < Lu3+ < Eu3+ < La3+
Answer: (2)