- A88 to 101
- B89 to 102
- C90 to 103
- D91 to 104
Explanation:
Second inner transition elements are from thorium to lawrencium, i.e., from atomic numbers 90 to 103.
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$E_{{A^{3 + }}/A}^o = 1.50\,\,V\,,$ $E_{{B^{2 + }}/B}^o = 0.3\,\,V,$
$E_{{C^{3 + }}/C}^o = - \,0.74\,\,V,$ $E_{{D^{2 + }}/D}^o = - \,2.37\,\,V.$
The correct sequence in which the various metals are deposited at the cathode is
$(I)$ $\begin{array}{*{20}{c}} {C{H_3}\,\,\,\,\,\,Br} \\ {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,} \\ {C{H_3} - CH - CH - C{H_3}} \end{array}$
$(II)$ $[Figure\,2]$
$(III)$ $\begin{array}{*{20}{c}} {{C_6}{H_5} - CH - {C_6}{H_5}} \\ {|\,\,\,} \\ {Br} \end{array}$
$(IV)$ $[Figure\,4]$
Which one of the following is correct sequence of the halides given above in the decreasing order of their reactivity ?
$A{l_{\left( s \right)}}|A{l^{ + 3}}\left( {0.1\,M} \right)||F{e^{ + 2}}\left( {0.001\,M} \right)|Fe\left( s \right)$ If $E_{A{l^{ + 3}}/Al}^o = - 1.66\,V$ and $E_{Fe/F{e^{ + 2}}}^o = + 0.44\,V$
Reason : $Zn$ is deposited at anode, and $Cu$ is deposited at cathode.