Class 11th

Combined equation of pair of lines OP and OQ is

$x^2+2y^2=2{\left(x+y\right)}^2$

$\Rightarrow x^2+4xy=0\Rightarrow x\left(x+4y\right)=0\Rightarrow \{\begin{array}{l}x=0\left(lineOP\right)\\ y=-\frac{x}{4}\left(lineOQ\right)\end{array}$

$tan\left(90°+\theta \right)=-\frac{1}{4}$

$-cot\theta =-\frac{1}{4}\Rightarrow tan\theta =4$

$\theta =ta{n}^{-1}4=co{t}^{-1}\frac{1}{4}=\frac{\pi }{2}-ta{n}^{-1}\frac{1}{4}$

$\angle POQ=\pi -\theta =\frac{\pi }{2}+ta{n}^{-1}\frac{1}{4}$

${\alpha }^2-6\alpha -2=0,{\beta }^2-6\beta -2=0$

${\alpha }^2-2=6\alpha ,{\beta }^2-2=6\beta$

$\frac{a_{10}-2a_8}{3a_9}=\frac{{\alpha }^{10}-{\beta }^{10}-2\left({\alpha }^8-{\beta }^8\right)}{3\left({\alpha }^9-{\beta }^9\right)}=\frac{{\alpha }^8\left({\alpha }^2-2\right)-{\beta }^8\left({\beta }^2-2\right)}{3\left({\alpha }^9-{\beta }^9\right)}$

$\frac{{\alpha }^8.6\alpha -{\beta }^8.6\beta }{3\left({\alpha }^9-{\beta }^9\right)}=2$

(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

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

$\therefore N=M=5.63$

$\underset{SS}{AgC{N}_{\left(S\right)}?A{g^{+}}_{\left(aq\right)}+C{N^{-}}_{\left(aq\right)}}$            

$C{N}^{-}+H^{+}?HCN$

Before reaction     S            10^-3       0

After reaction       0            10^-3       S

$\therefore HCN?H^{+}+C{N}^{-}$

$S^2=\frac{2.2*10^{-16}}{6.2*10^{-7}}=\frac{2.2}{6.2}*10^{-9}$

$S=\sqrt[]{\frac{2.2}{6.2}*10^{-9}}=\sqrt[]{\frac{22}{6.2}*10^{-10}}$

$=\sqrt[]{3.54*10^{-10}}=1.88*10^{-5}=1.9*10^{-5}$           

If $\Delta E=0$

Q = W

W = -P_ext $\left(\Delta V\right)=-4.3nRT\left[\frac{1}{P_2}-\frac{1}{P_1}\right]$

= $-4.3*5*8.314*293\left[\frac{1}{1.3}-\frac{1}{2.1}\right]$

= 15347.70 K = 15.3 kJ

Q = 15 kJ