- A$\mathrm{Mg}<\mathrm{Al}<\mathrm{S}<\mathrm{P}$
- ✓$\mathrm{Al}<\mathrm{Mg}<\mathrm{S}<\mathrm{P}$
- C$\mathrm{Mg}<\mathrm{Al}<\mathrm{P}<\mathrm{S}$
- D$\mathrm{Mg}<\mathrm{S}<\mathrm{Al}<\mathrm{P}$
$\quad\quad\quad\quad\quad\quad\mathrm{Mg}$ $\quad\quad\mathrm{Al}$ $\quad\quad\quad\mathrm{P}$ $\quad\quad\quad\mathrm{S}$
Valence $\left[\mathrm{N}_{\mathrm{e}}\right]: 3 \mathrm{~s}^{2}$ $\quad3 \mathrm{~s}^{2} 3 \mathrm{p}^{1}$ $\quad3 \mathrm{~s}^{2} 3 \mathrm{p}^{3} \quad3 \mathrm{~s}^{2} 3 \mathrm{p}^{4}$
$\quad\quad\quad\quad\quad\quad\uparrow$ $\quad\quad\quad\quad\quad\quad\uparrow$
$\quad\quad\quad\quad\quad$Full Filled Stable $\quad$Half Filled Stable
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$Me_2CHCH_2NHCH_2CH_2Me $ $\xrightarrow[\begin{smallmatrix}
(ii)\,A{{g}_{2}}O \\
(iii)\,\Delta
\end{smallmatrix}]{(i)\,C{{H}_{3}}-I(excess)}$ Product
$2KMn{O_4} + 2KOH \to 2{K_2}Mn{O_4} + {H_2}O + O$
Therefore its equivalent weight will be
$(I)$ $NO + Br_2 \rightleftharpoons NOBr_2 $ ........ Fast
$(II)$ $NOBr_2 + NO \rightarrow 2NOBr$ ......... Slow
The overall order of this reaction is
Identify $A$