JEE MAIN - Chemistry (2019 - 8th April Morning Slot - No. 14)
The lathanide ion that would show colour is :
Sm3+
Gd3+
Lu3+
La3+
Explanation
Electronic configuration of
Sm = [Xe] 4f6 6s2
$$ \therefore $$ Electronic configuration of
Sm+3 = [Xe] 4f5
Sm+3 shows yellow colour due to partially filled f orbital and f-f transition.
Electronic configuration of
La+3 = [Xe] 4f0
Lu+3 = [Xe] 4f14
Gd+3 = [Xe] 4f7
La+3, Lu+3 and Gd+3 are colourless due to absence of f-f transition.
Sm = [Xe] 4f6 6s2
$$ \therefore $$ Electronic configuration of
Sm+3 = [Xe] 4f5
Sm+3 shows yellow colour due to partially filled f orbital and f-f transition.
Electronic configuration of
La+3 = [Xe] 4f0
Lu+3 = [Xe] 4f14
Gd+3 = [Xe] 4f7
La+3, Lu+3 and Gd+3 are colourless due to absence of f-f transition.
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