Chemistry

An atomic orbital is characterized by $n,\mathcal{l}$ $$ and m.

Xenon has 8 valence electrons. In ${\mathrm{X}\mathrm{e}\mathrm{F}}_2$ , it forms two bonds with fluorine, leaving three lone pairs. The steric number =2 bonds +3 lone pairs $=5$ .
As per the VSEPR Theory  the electron? pair geometry due to SN = 5 will be trigonal bipyramidal. The hybridization type will be ${\mathrm{s}\mathrm{p}}^3\mathrm{d}$ hybridisation. However, the three lone pairs occupy equatorial positions, resulting in a linear molecular geometry.

Total 3 streo- isomers

Mole of polyhydric alcohol =  $\frac{1.84*10^{-3}}{92}=2.0*10^{-5}mole.$

Mole of H₂ gas produced = $\frac{1.344}{22400}=6.0*10^{-5}mole.$  

No of -OH gp present =  $\frac{6.0*10^{-5}}{2.0*10^{-5}}=3$

Only H₂S₂O₈ has per-oxo bond.

Oxidation state of Co = 3

And co-ordination No = 6

Sum = 3 + 6 = 9

$\lambda t=\mathcal{l}n\frac{N_o}{N_t}$

$\frac{0.693}{t_{1/2}}*t=\mathcal{l}n\frac{1}{N_t}$

 Or  $\frac{0.693*100}{30}=\mathcal{l}n\frac{1}{N_t}$

  $\frac{1}{N_t}=10$

$\therefore N_t=\frac{1}{10}=0.1$              

Or N_t = 1 * 10^-1   $\mu g$

Cu_{(s) $\underset{}{\overset{}{\to }}$} Cu⁺⁺ + 2e^- (Oxidⁿ at cathode)

$\frac{2A{g}^{+}+2e\underset{}{\overset{}{\to }}2Ag\left(s\right)}{C{u}_{\left(s\right)}+2A{g}^{+}\underset{}{\overset{}{\to }}2A{g}_{\left(s\right)}+C{u}^{++}}\left(Re{d}^{n}atanode\right)$

$Q=\frac{\left[C{u}^{++}\right]}{{\left[A{g}^{+}\right]}^2}andn=2$

0.43 =   $E_{cell}^o=\frac{0.06}{2}log\frac{\left(0.001\right)}{{\left(0.01\right)}^2}$

$\therefore E_{cell}^o=0.46$

$E_{cell}^o=E_{A{g}^{+}/Ag}^o-E_{C{u}^{++}/Cu}^o$

0.46 = 0.80 - $E_{c{u}^{++}/Cu}^o$

$\therefore E_{c{u}^{++}/Cu}^o=0.34Volt=34*10^{-2}$  

$\Delta T_f=K_f*m$  

$\left(156-155.1\right)=2*\frac{1.8}{MW}*\frac{1000}{\left(62.5*0.8\right)}$                

  0.9 =   $\frac{2*1.8*1000}{MW*50}$

$\therefore MW=80gm/mol$      

Following have the sp³d² hybridisation