Question
The correct order of ionic radii of $${Y^{3 + }},L{a^{3 + }},E{u^{3 + }}$$ and $$L{u^{3 + }}$$ is $$\left( {{\text{At}}{\text{. no}}{\text{.}}} \right.\,Y = 39,La = 57,$$ $$\left. {Eu = 63,Lu = 71} \right)$$
A.
$$L{u^{3 + }} < E{u^{3 + }} < L{a^{3 + }} < {Y^{3 + }}$$
B.
$$L{a^{3 + }} < E{u^{3 + }} < L{u^{3 + }} < {Y^{3 + }}$$
C.
$${Y^{3 + }} < L{a^{3 + }} < E{u^{3 + }} < L{u^{3 + }}$$
D.
$${Y^{3 + }} < L{u^{3 + }} < E{u^{3 + }} < L{a^{3 + }}$$
Answer :
$${Y^{3 + }} < L{u^{3 + }} < E{u^{3 + }} < L{a^{3 + }}$$
Solution :
The correct order of ionic radii of $${Y^{3 + }},L{a^{3 + }},E{u^{3 + }}$$ and $$L{u^{3 + }},$$ is $${Y^{3 + }} < L{u^{3 + }} < E{u^{3 + }} < L{a^{3 + }}$$ because $$Eu$$ and $$Lu$$ are the members of lanthanide series (so they show lanthanide contraction) and $$La$$ is the representative element of all elements of such series and $${Y^{3 + }}$$ $$ion$$ has lower radii as comparison to $$L{a^{3 + }}$$ because it lies immediately above it in the periodic table.