P Block Elements

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.   $T\mathcal{l}=4f^{14}5d^{10}6s^{2}6p^1$

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

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

Blue cupric metaborate is reduced to cuprous metaborate in a luminous flame.

  $2Cu{\left(B{O}_2\right)}_2+2NaB{O}_2+C\underset{}{\overset{Luminous}{\to }}2CuB{O}_2+N{a}_2{B}_4{O}_7+CO$            

Cupric metaborate is obtained by heating boric anhydride & copper sulphate in a non-luminous flame as

$CuS{O}_4+B_2{O}_3\underset{}{\overset{Non-LuminiousFlame}{\to }}\underset{Blue}{Cu{\left(B{O}_2\right)}_2}+S{O}_3.$            

α - sulphur & β - sulphur – Diamagnetic, S₂ – form is paramagnetic due to presence of unpaired electron in π* orbital like O₂.

Blood – Negatively charged colloid.

According to Hardy – Schulze Rule

FeCl₃ is more efficient for blood clotting

$l^{-}+H_2{O}_2+2H^{+}\to l_2+2H_{2}O$

  (-1)                 oxidation       (0)

Here, I^- is reducing agent

$\therefore H_2{O}_2$   behaves like oxidizing agent

All given oxide have nitrogen – nitrogen bond except N₂O₅ as ;

Let a moles of SO₂Cl₂ is taken

Then no. of moles of H₂SO₄ = a moles

No. of moles of HCl = 2a moles

No. of moles of NaOH required = 2a + 2a = 4a = 16