Chemistry

B₂H₆ has 4 2 c -2e bonds and 2 3c-2e bonds.

Bridging (B-H) bonds have more value of bond- length then terminal (B – H) bonds

Bridging bonds are in one plane, but terminal bonds are in perpendicular plane.

Due to presence of (3c-2e) bonds, it behaves as electrons deficient and prone to get attached by lewis base.

$3NaB{H}_4+4B{F}_3\underset{}{\overset{\Delta }{\to }}3NaB{F}_4+2B_2{H}_6$                

Due to higher extent of polarization by Li⁺ and Mg²⁺, LiCl and Mgcl₂ have covalent character. Therefore they are soluble in ethanol.

Due to very high value of lattice energy, LiF is having very less solubility in water.

The highest industrial consumption of hydrogen gas is in the synthesis of ammonia gas (Having use in manufacturing of N-based fertizers)

Number of moles of C = Number of moles of CO₂ = $\frac{330}{44}$ moles

              Number of moles of H = 2 * no. of moles of H₂O = $\left(\frac{270}{18}*2\right)$ moles

              Mass of C = $\frac{330}{44}*12$ gm = 90 gm

              Mass of H = $\frac{270}{18}*2*1gm=30gm$

$\begin{array}{l}\mathrm{\%}ofC=\frac{90}{120}*100\mathrm{\%}=75\mathrm{\%}\\ \mathrm{\%}ofH=\frac{30}{120}*100\mathrm{\%}=25\mathrm{\%}\end{array}$  

Sphalarite           ZnS

Calamine            ZnCO₃

Galena                PbS

Siderite               FeCO₃

Acidic oxide => Cl₂O₇

Neutral oxide => N₂O, NO

Basic oxide => Na₂O

Amphoteric oxide => As₂O₃

Emulsions gets separated into two layers on standing.

For stabilization of emulsion. Emulsifying agents added into it but not electrolyte.

$\underset{\_}{\begin{array}{l}A\left(g\right)?B\left(g\right)+\frac{1}{2}C\left(g\right)\\ t=0amoles00\\ -a\alpha moles+a\alpha moles+\frac{a\alpha }{2}moles\end{array}}$

Eq.   a(1 - α)          aα           (aα/2)

Moles           moles       moles

Total no. of moles at equilibrium

= nA + nB + nC

$=a\left(1-\alpha \right)+a\alpha +\frac{a\alpha }{2}=a\left[1+\frac{\alpha }{2}\right]$

${\left(P_A\right)}_{eq}={\left(x_A\right)}_{eq}*P_{eq}=\frac{a\left(1-\alpha \right)}{\left(1+\alpha /2\right)}P=\frac{1-\alpha }{\left(1+\alpha /2\right)}P$

${\left(P_B\right)}_{eq}={\left(x_B\right)}_{eq}*P_{eq}=\frac{a\alpha }{a\left(1+\alpha /2\right)}P=\frac{\alpha }{\left(1+\alpha /2\right)}P$

${\left(P_C\right)}_{eq}={\left(x_C\right)}_{eq}*P_{eq}=\frac{a\alpha /2}{a\left(1+\alpha /2\right)}P=\frac{\alpha /2}{\left(1+\alpha /2\right)}P$

$K_P=\frac{{\left(P_B\right)}_{eq}*{\left(P_C\right)}_{eq}^{1/2}}{{\left(P_A\right)}_{eq}}=\frac{{\left(\alpha \right)}^{3/2}{P}^{1/2}}{\left(1-\alpha \right){\left(2+\alpha \right)}^{1/2}}$

${\left[PtC{l}_4\right]}^{2-}\to Pt$ has dsp² hybridization

BrF₅ -> Br has sp³d² hybridization

PCl₅ -> P has sp³d hybridization

[Co (NH₃)₆]³⁺ -> Co has d²sp³ hybridization.

Degenerate orbitals must have same value of energy

Orbitals with same n and   values are degenerate orbitals.