Question
From the ground state electronic configuration of the elements given below, pick up the one with highest value of second ionization energy
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$Xe{{F}_{6}}\,\xrightarrow{+\,Excess\,\,{{H}_{2}}O}\,'X'\,+\,HF$
$Xe{{F}_{6}}\,\xrightarrow{+\,2\,\,{{H}_{2}}O}\,'Y'\,+\,HF$
$\gamma_{1} A +\gamma_{2} B \rightarrow \gamma_{3} C +\gamma_{4} D$
Concentration of $C$ changes from $10\, mmol$ appearance of $D$ is $1.5$ times the rate of disappearance of $B$ which is twice the rate of disappearance $A$. The rate of appearance of $D$ has been experimentally determined to be $9 \,m\,mol$ $dm ^{-3} s ^{-1}$. Therefore the rate of reaction is $......\,m\,mol\, dm ^{-3} \,s ^{-1}$. (Nearest Integer)
$' A ^{\prime} \stackrel{ HNO _3}{\longrightarrow} X ^{\prime} \stackrel{ NaBH _4}{\longrightarrow} \underset{\text { Chiral compound }}{\text { } B ^{}}$ $X$ is $.......$.
(Given atomic number of $\mathrm{Cr}$ is 24 )
$(i)$ Enzymes lack in nucleophilic groups
$(ii)$ Enzymes are highly specific both in binding chiral substrates and in catalyzing their reactions
$(iii)$ Enzymes catalyse chemical reactions by lowering the activation energy
$(iv)$ Pepsin is a proteolytic enzyme