Ncert Solutions Chemistry Class 12th

In sulphide ore, depressants selectively prevent impurity from coming to the fourth.

$Mn{O}_4^{-2}\to A+B$  

Oxidation state of Mn in B < A 

$Mn{O}_4^{-2}+H^{+}\to Mn{O}_4^{-}+Mn{O}_2$  

B is MnO₂

$\therefore$  Oxidation state of Mn = +4

${\therefore }_{25}M{n}^{+4}=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{0}3d^3$  

  $\therefore$ unpaired electron = 3

Spin only magnetic moment  $\left(\mu \right)$  

$=\sqrt[]{n\left(n+2\right)}=\sqrt[]{3\left(3+2\right)}=\sqrt[]{15}$  

$=3.87\approx 4$  

$C\mathcal{l}{F}_3$ ->T-shaped (sp³d)

IF₇ ->Pentagonal bipyramidal (sp³d³)

BrF₅ -> Square pyramidal (sp³d²)

BrF₃ ->T-Shaped (sp³d)

I₂Cl₆ -> Triangular bipyramidal (sp³d)

$I{F}_5\to$  Square Pyramidal (sp³d²)

ClF -> (sp³)

ClF₅ -> Square Pyramidal (sp³d²)

Br₅, IF₅ & ClF₅ ® Square Pyramidal

B.C.C structure

a = 300 pm = 300 * 10^-12 m

d = 6g/cm³

z = 2

$d=\frac{Z*M}{a^3}$

$6=\frac{2*A}{{\left(300*10^{-10}\right)}^3}=\frac{2*A}{27*10^{-24}}$  

$\therefore AtomsofM$ = 3.69 * 6.022 * 10²³

= 22.22 * 10²³

the nearest integer = 22

Sphalarite             ZnS

Calamine              ZnCO₃

Galena                  PbS

Siderite                FeCO₃

Density, $\rho =7.62gc{m}^{-3}$  

Cell edge, a = 0.4518nm = 0.458 * 10^-7 cm

Effective number of atoms, Z = 4 for CCP.

Using ; $\rho =\frac{Z*A}{N_A*a^3}$  

$\rho =\frac{4*A}{6.022*10^{23}*{\left(0.458*10^{-7}\right)}^3}$              

$7.62=\frac{4*A}{6.022*{\left(0.4518\right)}^3*10^2}$              

A = 105.79 g/mol

Ans. is 106 (the nearest integers)

Reducing power for group-15 hydrides increases down the group, so BiH₃ is the strongest reducing agent.

Sulphide ion (s^2-) form ores commonly with Pb & Ag as PbS and Ag₂S

Compound A is phenol since phenol gives dark green colour with FeCl₃.