Chemistry

$B_2={\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\pi }_{2px}^1={\pi }_{2py}^1\to$ Paramagnetic

$\begin{array}{l}L{i}_2={\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2\to Diamagnetic\\ C_2={\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\pi }_{2px}^2\equiv {\pi }_{2py}^2\to Diamagnetic\\ C_2^{-}={\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\pi }_{2px}^2\equiv {\pi }_{2py}^2{\sigma }_{2pz}^1\to Paramagnetic\end{array}$  

$O_2^{-2}={\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\sigma }_{2pz}^2{\pi }_{2px}^2\equiv {\pi }_{2py}^2{\pi }_{2px}^{*2}\equiv {\pi }_{2py}^{*2}$ ® Diamagnetic

$O_2^{+}={\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\sigma }_{2pz}^2{\pi }_{2px}^2\equiv {\pi }_{2py}^2{\pi }_{2px}^{*1}\equiv {\pi }_{2py}^0\to$ Paramagnetic

$H{e}_2^{+}={\sigma }_{1s}^2{\sigma }_{1s}^{*1}\to$ Paramagnetic

Paramagnetic molecules are  $=B_2,C_2^{-},O_2^{+},H{e}_2^{+}$  

Ans 4.

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

x      +      y    +   3z     =     xyz₃

1 mole 1 mole 0.05 mole

$?n_x=1$

$n_y=1$

$n_z=\frac{0.05}{3}=0.0167$ here z is limiting reagent.

$?\frac{0.05}{3}$ mole z gives 1 mole xyz₃

$?$ mass of xyz₃ = n * molecular mass

 = $\frac{0.05}{3}*\left(10+20+3*30\right)a.m.u.$  

= 0.5 * 4 = 2g

Sphalarite             ZnS

Calamine              ZnCO₃

Galena                  PbS

Siderite                FeCO₃

Acidic oxide -> Cl₂O₇

Neutral oxide -> N₂O, NO

Basic oxide ->Na₂O

Amphoteric oxide -> As₂O₃

$T_{90\mathrm{\%}}=\frac{2.303}{k}log\frac{100}{10}$

$T_{50\mathrm{\%}}=\frac{2.303}{k}log\frac{100}{50}$

$\frac{T_{90\mathrm{\%}}}{T_{50\mathrm{\%}}}=\frac{log10}{log2}=\frac{1}{0.301}=3.32$

$2C\left(s\right)+2H_2\left(g\right)\text{?}\text{?}\to C_2{H}_6\left(g\right)\Delta H=\Delta H_f\left(C_2{H}_6\right)$  

$\Delta H_f\left(C_2{H}_6\right)=\left[2*\left(-394\right)\right]+\left[3*\left(-286\right)\right]-\left[1*\left(-1560\right)\right]KJ/mole$  

= (-788 – 858 + 1560) KJ/mole = -86 kJ/mole

Bond order of $C_2^{2-}=\frac{N_B-N_A}{2}=\frac{10-4}{2}=3$  

Bond order of  $N_2^{2-}=\frac{N_B-N_A}{2}=\frac{10-6}{2}=2$   

Bond order of  $O_2^{2-}=\frac{N_B-N_A}{2}=\frac{10-8}{2}=1$  

N_B = No. of electrons in bonding molecular orbitals.

N_A = No. of electron is Anti bonding molecular orbitals

Energy of 1 photon = $\frac{hc}{\lambda }$  

Energy of 1 mole of photon =  $N_A*\frac{hc}{\lambda }$  

$=\frac{6.022*10^{23}*6.63*10^{-34}*3*10^8}{300*10^{-9}}$  

= 0.399 * 10⁶ J = 399 KJ

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 =  $\frac{270}{18}*2*1gm=30gm$

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