NEET UG 2003 — Chemistry Inorganic Chemistry

2003mcqeasy

The correct order of ionic radii of $\mathrm{Y}^{3+}$ , $\mathrm{La}^{3+}$ and $\mathrm{Lu}^{3+}$ is: (Atomic nos $\mathrm{Y}=39, \mathrm{La}=57, \mathrm{Eu}=63 \mathrm{Lu}$ $=71$ )

Official previous-year question

Held on 30 Apr 2003 · Verified 9 Jul 2026.

Options

A
$\mathrm{Y}^{3+} < \mathrm{La}^{3+} < \mathrm{Eu}^{3+} < \mathrm{Lu}^{3+}$
B
$\mathrm{Y}^{3+} < \mathrm{Lu}^{3+} < \mathrm{Eu}^{3+} < \mathrm{La}^{3+}$
C
$\mathrm{Lu}^{3+} < \mathrm{Eu}^{3+} < \mathrm{La}^{3+} < \mathrm{Y}^{3+}$
D
$\mathrm{La}^{3+} < \mathrm{Eu}^{3+} < \mathrm{Lu}^{3+} < \mathrm{Y}^{3+}$

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Solution

Eu and Lu are the members of lanthanide series (so they show lanthanide contraction) and $\mathrm{La}$ is the representative element of all elements of such series and $\mathrm{Y}^{3+}$ ion has lower radii as comparison to $\mathrm{La}^{3+}$ because it lies immediately above it in the periodic table. $\mathrm{Y}^{3+} < \mathrm{Lu}^{3+} < \mathrm{Eu}^{3+} < \mathrm{La}^{3+}$

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Options

$\mathrm{Y}^{3+} < \mathrm{La}^{3+} < \mathrm{Eu}^{3+} < \mathrm{Lu}^{3+}$

$\mathrm{Y}^{3+} < \mathrm{Lu}^{3+} < \mathrm{Eu}^{3+} < \mathrm{La}^{3+}$

$\mathrm{Lu}^{3+} < \mathrm{Eu}^{3+} < \mathrm{La}^{3+} < \mathrm{Y}^{3+}$

$\mathrm{La}^{3+} < \mathrm{Eu}^{3+} < \mathrm{Lu}^{3+} < \mathrm{Y}^{3+}$

Solution

NEET UG 2003 — Chemistry Inorganic Chemistry

Options

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NEET UG 2003 — Chemistry Inorganic Chemistry

Options

Did you get this right?

Solution

Your note

NEET UG Inorganic Chemistry in other years

More NEET UG Inorganic Chemistry questions

Other Chemistry topics for NEET UG

Physical Chemistry

Organic Chemistry

Keep practicing Inorganic Chemistry