Class 12th

Due the absence of inversion and non resonating nature

$A^2=\left(\begin{array}{ll}\mathrm{c}\mathrm{o}\mathrm{s}?2\theta & i\mathrm{s}\mathrm{i}\mathrm{n}?2\theta \\ i\mathrm{s}\mathrm{i}\mathrm{n}?2\theta & \mathrm{c}\mathrm{o}\mathrm{s}?2\theta \end{array}\right)$

Similarly, $A^5=\left(\begin{array}{cc}\mathrm{c}\mathrm{o}\mathrm{s}?5\theta & i\mathrm{s}\mathrm{i}\mathrm{n}?5\theta \\ i\mathrm{s}\mathrm{i}\mathrm{n}?5\theta & \mathrm{c}\mathrm{o}\mathrm{s}?5\theta \end{array}\right)=\left(\begin{array}{ll}a& b\\ c& d\end{array}\right)$

(1) $a^2+b^2={\mathrm{c}\mathrm{o}\mathrm{s}}^2?5\theta -{\mathrm{s}\mathrm{i}\mathrm{n}}^2?5\theta =\mathrm{c}\mathrm{o}\mathrm{s}?10\theta =\mathrm{c}\mathrm{o}\mathrm{s}?{75}^{?}$

(2) $a^2-d^2={\mathrm{c}\mathrm{o}\mathrm{s}}^2?5\theta -{\mathrm{c}\mathrm{o}\mathrm{s}}^2?5\theta =0$

(3) $a^2-b^2={\mathrm{c}\mathrm{o}\mathrm{s}}^2?5\theta +{\mathrm{s}\mathrm{i}\mathrm{n}}^2?5\theta =1$

(4) $a^2-c^2={\mathrm{c}\mathrm{o}\mathrm{s}}^2?5\theta +{\mathrm{s}\mathrm{i}\mathrm{n}}^2?5\theta =1$

$\int {\left(\frac{x}{x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x}\right)}^{2}dx=\int \left(\frac{x}{\mathrm{c}\mathrm{o}\mathrm{s}?x}\right)\cdot \frac{x\mathrm{c}\mathrm{o}\mathrm{s}?xdx}{(x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x{)}^2}$

$\begin{array}{rr}& =\frac{x}{\mathrm{c}\mathrm{o}\mathrm{s}?x}\left(-\frac{1}{x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x}\right)+\int \left(\frac{\mathrm{c}\mathrm{o}\mathrm{s}?x+x\mathrm{s}\mathrm{i}\mathrm{n}?x}{{\mathrm{c}\mathrm{o}\mathrm{s}}^2?x}\right)\left(\frac{1}{x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x}\right)dx\\ & =-\frac{x\mathrm{s}\mathrm{e}\mathrm{c}?x}{x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x}+\int {\mathrm{s}\mathrm{e}\mathrm{c}}^2?xdx\\ & =-\frac{x\mathrm{s}\mathrm{e}\mathrm{c}?x}{x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x}+\mathrm{t}\mathrm{a}\mathrm{n}?x+C\end{array}$

The pair of herbicides is sodium chlorate and sodium arsinite

Since $\left(\mathrm{3,3}\right)$ lies on $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$\frac{9}{a^2}-\frac{9}{b^2}=1$

Now, normal at $\left(\mathrm{3,3}\right)$ is $y-3=-\frac{a^2}{b^2}(x-3)$ ,

which passes through $\left(\mathrm{9,0}\right)\Rightarrow b^2=2a^2$

So,  ${\mathrm{e}}^2=1+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}=3$

Also,  ${\mathrm{a}}^2=\frac{9}{2}$

(From (i) and (ii)

Thus,  $\left({\mathrm{a}}^2,{\mathrm{e}}^2\right)=\left(\frac{9}{2},3\right)$

Following is the reaction of KMnO4 in basic medium with thiosulphate anion.

$\stackrel{+7}{8Mn{O}_4^{-}}+3S_2{O}_3^{--}+H_{2}O\to \stackrel{+4}{8Mn{O}_{2|}}+6S{O}_4^{--}+2O{H}^{-}$

Change = 7 – 4 = 3

$\left(\frac{\mathrm{K}-2}{\mathrm{h}-1}\right)\left(\frac{1-2}{3-1}\right)=-1\Rightarrow \mathrm{K}=2\mathrm{h}$

$?[?\mathrm{A}\mathrm{B}\mathrm{C}]=5\sqrt[]{5}$

$\Rightarrow \frac{1}{2}\left(\sqrt[]{5}\right)\sqrt[]{(\mathrm{h}-1{)}^2+(\mathrm{K}-2{)}^2}=5\sqrt[]{5}$

  $S{c}^{3+}-3d^0$ electron in last subshell

$Z{n}^{2+}-3d^{10}$ electron in last subshell

$T{i}^{4+}-3d^0$ electron in last subshell

$C{u}^{2+}-3d^9$ electron in last subshell

$V^{2+}-3d^3$ electron in last subshell

$T{i}^{3+}-3d^1$ electron in last subshell

$M{n}^{2+}-3d^5$ electron in last subshell

$\mathrm{u}=\frac{2\mathrm{z}+\mathrm{i}}{\mathrm{z}-\mathrm{k}\mathrm{i}}$

$=\frac{2x^2+(2y+1)(y-k)}{x^2+(y-k{)}^2}+i\frac{\left(x\right(2y+1)-2x(y-k\left)\right)}{x^2+(y-k{)}^2}$

Since $\mathrm{R}\mathrm{e}?\left(u\right)+\mathrm{I}\mathrm{m}?\left(u\right)=1$

$\Rightarrow 2{\mathrm{x}}^2+(2\mathrm{y}+1)(\mathrm{y}-\mathrm{k})+\mathrm{x}(2\mathrm{y}+1)-2\mathrm{x}(\mathrm{y}-\mathrm{k})={\mathrm{x}}^2+(\mathrm{y}-\mathrm{k}{)}^2$

$\begin{array}{c}\mathrm{P}\left(0,{\mathrm{y}}_1\right)\\ \mathrm{Q}\left(0,{\mathrm{y}}_2\right)\end{array}?\Rightarrow {\mathrm{y}}^2+\mathrm{y}-\mathrm{k}-{\mathrm{k}}^2=0?\begin{array}{c}{\mathrm{y}}_1+{\mathrm{y}}_2=-1\\ {\mathrm{y}}_1{\mathrm{y}}_2=-\mathrm{k}-{\mathrm{k}}^2\end{array}$

$?\mathrm{P}\mathrm{Q}=5$

$\Rightarrow \left|{\mathrm{y}}_1-{\mathrm{y}}_2\right|=5\Rightarrow {\mathrm{k}}^2+\mathrm{k}-6=0$

$\Rightarrow \mathrm{k}=-\mathrm{3,2}$

So, $k=2(k>0)$

SCl₂ – 2 lone pair of central atom

O₃ – 1 lone pair of central atom

ClF₃ – 2 lone pair of central atom

SF₆ – 0 lone pair of central atom