Class 12th

Kindly consider the following figure

A > B > C > D

Lone pair is localized in (A) while all 3 have delocalized lone pair but they can be compared by 3° > 2° > 1° because methyl group increases the basicity.

S > O > Se > Te   

Leaching of Al₂O₃ with alkali;

$\begin{array}{l}A{l}_2{O}_3\left(s\right)+2NaOH\left(aq\right)+3H_{2}O\left(l\right)\to 2Na\left[Al{\left(OH\right)}_4\right]\left(aq\right)\\ X\end{array}$            

$\begin{array}{l}2Na\left[Al{\left(OH\right)}_4\right]\left(aq\right)+2C{O}_2\left(g\right)\to A{l}_2{O}_3.x{H}_{2}O\left(s\right)+2NaHC{O}_3\left(aq\right)\\ XYZ\end{array}$

Both (A) and (B) are correct but R is not correct explanation of A. In both $T\mathcal{l}{l}_{3}andCs{l}_3$ oxidation state of metal is +1, also both have similar lattice structure.   $T\mathcal{l}=4f^{14}5d^{10}6s^{2}6p^1$

(a) Sucrose -> α-D glucose and b-D fructose

(b) Lactose -> β-D- galactose and b-D glucose

(c) maltose -> α-D glucose and a-D-glucose

It does not contains 2^nd most abundant element by weight in earth crust because that is Si Calgon ->Na₂ [Na₄ (PO₃)₆] $\underset{}{\overset{Watersoluble}{\to }}2N{a}^{+}{\left[N{a}_4{\left(P{O}_3\right)}_6\right]}^{2-}$

$\underset{}{\overset{C{a}^{2+}}{\to }}2N{a}^{+}{\left[N{a}_{2}Ca{\left(P{O}_3\right)}_6\right]}^{2-}$