Question

Consider the following statements
(i) Common oxidation state of Pr, Nd, Tb and Dy is +4 state.
(ii) $Z{r}^{4+}\text{and}H{f}^{4+}$ posses almost the same ionic radii.
(iii) $C{e}^{4+}$ can act as an oxidizing agent.
Which of the above is/are true?

• A. (i) and (iii)
• B. (ii) and (iii)
• C. (ii) only
• D. (i) and (ii)

Answer 1

The correct option is B (ii) and (iii)

As a result lanthanide contraction $Z{r}^{4+}\text{and}H{f}^{4+}$ possess almost the same ionic radii. $C{e}^{4+}$ is an oxidising agent.
$C{e}^{4+}$ gains electron acquire more stable $C{e}^{3+}$ state. Thus it act as good analytical agent.
$C{e}^{4+}+e^{-}\to C{e}^{3+};E^o=+1.74V$
Pr, Nd, Tb and Dy also ​exhibit +4 state but ​only in oxides $\left(M{O}_2\right)$