Question

The correct order of ionic radii of $Y^{3+},L{a}^{3+}$ $E{u}^{3+}$, and $L{u}^{3+}$ is : -
(Atomic nos. $Y=39,La=57,Eu=63,Lu=71$.)

• 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 1

Answer: (D)

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 lowei radii as comparison to $L{a}^{3+}$ because it lies immediately above it in the periodic table.