Transition and InnerTransition Elements - Result Question 8

####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 $MnO _2$

Show Answer

Solution:

  1. 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”.



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