Class 12th

BOD is the amount of oxygen required by bacteria to break down the organic matter present in a certain volume of sample of water.

Polymer Monomers
Buna-N CH? = CH – CH = CH?
and
CH? = CH – CN
Neoprene H? C = C (Cl) – CH = CH?
Nylon-6, 6 Hexamethylenediamine + Adipic Acid
PHBV CH? – CH (OH) – CH – COOH
and
CH? – CH? – CH (OH) – CH? – COOH

Inversion + Inversion = Retention

(H? S + dil HCl) is 2nd group reagent so Cu²? get precipitated as CuS.

[Mn (CN)? ]³? → Mn³? i.e 3d? → t? g? , eg? (d²sp³), CN? is strong field ligand.
[Fe (CN)? ]³? → Fe³? i.e 3d? → t? g? , eg? (d²sp³), CN? is strong field ligand.
[Co (C? O? )? ]³? → Co³? i.e, 3d? → t? g? , eg? (d²sp³), C? O? ²? is strong field ligand.
[MnCl? ]? → Mn³? = 3d? → t? g²¹¹¹, eg¹¹? (4 unpaired e? ), Cl? is weak field ligand.
[FeF? ]³? → Fe³? = 3d? → t? g³¹¹, eg²¹¹ (5 unpaired e? ) F? is strong field ligand.
∴ S? and S? are true.

In hexagonal unit cell, the shortest distance between two identical layers is
(2√2x) / √3 [x = edge length]

h = (2√2x) / √3

${\left[Co{\left(H_{2}O\right)}_6\right]}^{2+}+N{H}_{3\left(excess\right)}\to \underset{Diamagneticnature}{{\left[Co{\left(N{H}_3\right)}_6\right]}^{3+}+6H_{2}O}$

$C{o}^{3+}\to 3d^{6}4s^0$

$\Rightarrow {t_{2g}}^{6}e{g}^0$

$\Rightarrow {t_{2g}}^{6}e{g}^0$ (form of paired electron)

              No. of electron in t_2g = 6

$K_{cat}=A{e}^{-}\frac{{\left(Ea\right)}_{cat}}{RT}and{K}_{uncat}=A{e}^{-}\frac{{\left(Ea\right)}_{uncat}}{RT}$

 $\frac{K_{cat}}{K_{uncat}}=e^{\frac{{\left(Ea\right)}_{uncat}-{\left(Ea\right)}_{cat}}{RT}}=e^{\frac{10*1000}{8.314*300}}=e^{4.009}=e^x$

$\therefore x=4$

FeO (Impurities) + SiO? (Flux) → FeSiO? (Iron silicate) Slag