Identify the correct statements from the following : I. The ionic radius of \(\mathrm{Pr}^{3+},…
Identify the correct statements from the following :
I. The ionic radius of \(\mathrm{Pr}^{3+}, \mathrm{Dy}^{3+}\) and \(\mathrm{Sm}^{3+}\) follow the order, \(\mathrm{Sm}^{3+} > \mathrm{Pr}^{3+} > \mathrm{Dy}^{3+}\).
II. \(\mathrm{Eu}^{2+}\) acts as strong reducing reagent.
III. Pu exhibits +7 oxidation state.
I, II only
I, III only
I, II, III
II, III only
Solution
(i) The ionic radius of lanthanoids follow the following relation:
Ionic radius \(\propto \frac{1}{\text { atomic number }}\)
Atomic number \((Z)\) for \(\mathrm{Pr}=59\)
Atomic number \((Z)\) for \(\mathrm{Sm}=62\)
Atomic number \((Z)\) for \(D y=66\)
and has ionic radii for positive ions as given below:
\(\begin{aligned}
\mathrm{Pr}^{3+} & =1.013 Å \\
\mathrm{Sm}^{3+} & =0.964 Å \\
\mathrm{Dy}^{3+} & =0.908 Å
\end{aligned}\)
Thus, statement (i) is not correct.
(ii) Element Eu \((Z=63)\) can exist in \((+) 2\) and \((+) 3\) oxidation state.
\(\therefore(+)\) 3oxidation state is the most common oxidation state for lanthanoids.
Thus, (+) 2oxidation state oxides oxidises to (+) 3 easily and \(\mathrm{Eu}^{2+}\) can act as strong reducing agent.
Hence, the statement (II) is correct.
(iii) The electronic configuation for
\(\text {(Pu), }(Z=94)=5 f^6 6 d^0 7 s^2\)
Therefore, it can use maximum of \(6+2=8\) electrons.
Thus, can show +7 oxidation state.
Thus, statement (III) is also correct.
Hence, option (d) is the correct answer.