Chemistry

$l^{-}+H_2{O}_2+2H^{+}\to l_2+2H_{2}O$

  (-1)                 oxidation       (0)

Here, I^- is reducing agent

$\therefore H_2{O}_2$   behaves like oxidizing agent

Cu⁺ = 3d⁹ 4s⁰

n = 1

$\mu =\sqrt[]{1*\left(1+2\right)}=\sqrt[]{3}=1.73BM$

the nearest integer is 2.

In Ce & Eu (+3) oxidation state is more stable

$?C{e}^{+4}+e^{?}?C{e}^{+3}\left(Reduction\right)$     so CeO₂ is oxidising agent

$E{u}^{+2}?E{u}^{+3}+e?\left(oxidation\right)$ so EuSO₄ is reducing agent

(1) $H{e}_2^{-}{\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^{1}B.O=\frac{3-2}{2}=\frac{1}{2}$  

(2)    $H{e}_2^{+}{\sigma }_{1s}^2{\sigma }_{1s}^{*1}B.O=\frac{2-1}{2}=\frac{1}{2}$

(3) $B{e}_2{\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}B.O=\frac{4-4}{2}=0$  

So Be₂ does not exits

$E=\frac{hc}{\lambda }=\left\{\frac{6.63*10^{-34}*3*10^8}{663*10^{-9}}\right\}$

= 0.03 * 10^-17 = 3.0 * 10^-19 J/atom

= 18.06 * 10¹ kJ/mole = 180.6 181 KJ/mole

Please find the below image

Angle of H_t – B – H_t > angle of H_b – B - H_b

Bond angle $\propto \mathrm{\%}S-Character\propto \frac{1}{\mathrm{\%}ofP-character}$  

$\angle H_t$ B H_t is more so % P-Character is less.

M_eq of NaOH = M_eq H₂C₂O₄

4.44 * N = 1.25 * 2 * 10

$\therefore N=\frac{1.25*2*10}{4.44}=5.63$

$\therefore N=M=5.63$

$AB?A^{+}+B^{-}$

1-x x x   $\therefore i=1+x$

$\begin{array}{l}\Delta T_b=i*k_b*m\\ 2.5=\left(1+x\right)*0.52m\end{array}$

$\therefore molality=\frac{2.5}{1.75*0.52}=2.74$

the nearest integer is 3.