Transition and InnerTransition Elements - Result Question 82
####8. The correct order of atomic radii is
(2019 Main, 12 Jan II)
(a) $Ho>N>Eu>Ce$
(b) $N>Ce>Eu>Ho$
(c) $Eu>Ce>Ho>N$
(d) $Ce>Eu>Ho>N$
$9 \underline{A} \xrightarrow{4 KOH, O _2} \underset{(\text { Green })}{2 B}+2 H _2 O$
$3 \underline{B} \xrightarrow{4 HCl} \underset{\text { (Purple) }}{2 \underline{C}}+MnO _2+2 H _2 O$
$2 \underline{C} \xrightarrow{H _2 O, KI} 2 \underline{A}+2 KOH+\underline{D}$
In the above sequence of reactions, $\underline{A}$ and $\underline{D}$, respectively, are
(2019 Main, 11 Jan II)
(a) $KI$ and $KMnO _4$
(c) $KI$ and $K _2 MnO _4$
(b) $MnO _2$ and $KIO _3$
(d) $KIO _3$ and $MnO _2$
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Solution:
- The correct order of atomic radii is
Note
(i) $N$ being the member of $p$-block and second period, have the smallest radii.
(ii) Rest of all the 3 members are lanthanides with Eu having stable half-filled configuration thus with bigger size than rest two.
(iii) Among $Ce$ and $Ho$, $Ce$ has larger size which can be explained on the basis of “Lanthanoid contraction”.
