Class 11th

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

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

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

So Be2 does not exits

Kindly go throuigh the solution

Given   ${\displaystyle \sum _{i=1}^{18}\left(x_i-\alpha \right)=36i.e}{\displaystyle \sum _{i=1}^{18}{x}_i-18\alpha =36..........\left(i\right)}$

&       ${\displaystyle \sum _{i=1}^{18}{\left(x_i-\beta \right)}^2=90i.e{\displaystyle \sum _{i=1}^{18}{x_i}^2}}-2\beta {\displaystyle \sum _{}^{}{x}_i+18{\beta }^2=90.............\left(ii\right)}$      

(i) & (ii)   ${\displaystyle \sum _{i=1}^{18}{x_i}^2}=90-18\beta +36\beta \left(\alpha +2\right).............\left(iii\right)$

Now variance ${\sigma }^2=\frac{\sum {x_i}^2}{n}-{\left(\frac{\sum x_i}{n}\right)}^2$ = 1 given

->(a - b) (a - b + 4) = 0

Since $\alpha \ne \beta so\left|\alpha -\beta \right|=4$  

Angle of Ht – B – Ht > angle of Hb – B - Hb

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

${}_{}$  $\angle H_t$ B Ht is more so % P-Character is less.

Kindly consider the following figure

$Given\left|z+5\right|\le 4..........\left(i\right)$

->Represent a circle ${\left(x+5\right)}^2+y^2\le 16$  

$=z\left(1+i\right)+\overline{z}\left(1-i\right)\ge -10..........\left(ii\right)$           

->Represent a line X – y $\ge -5$  

So max |z + 1|² = AQ²

$={\left(-4-2\sqrt[]{2}\right)}^2+8$         

$\begin{array}{l}=32+16\sqrt[]{2}\\ =\alpha +\beta \sqrt[]{2}\end{array}$        

Hence a + b = 48

Kindly consider the following figure

Kindly consider the following figure

18 = 3² * 2

For G.C.D to be 3. no. of four digits should be only multiple of 3, but not multiple of 9 & also should not be even.

As we know no. of the form                         

9 k -> 1000

9 k + 1 -> 1000

9 k + 2 -> 1000

9 k + 3 -> 1000  -> Total no. = 2000

9 k + 4 -> 1000

9 k + 5 -> 1000

9 k + 6 ->1000

9 k + 7 -> 1000

9 k + 8 -> 1000

In which half will be even & half be odd so Required no. = 1000

Given sequence is -16, 8, -4, 2, .

are in G.P. with first term a = -16 & common ratio r =   $\frac{-1}{2}$

Now $t_p=a{r}^{P-1}=-16{\left(-\frac{1}{2}\right)}^{p-1}\&t_q=a{r}^{q-1}=-16{\left(-\frac{1}{2}\right)}^{q-1}$  

So A.M. = -8 $\left[{\left(-\frac{1}{2}\right)}^{p-1}+{\left(\frac{-1}{2}\right)}^{q-1}\right]=\alpha \left(let\right)$

Given equation is 4x² – 9x + 5 = 0 gives $x=1,\frac{5}{4}$  

From roots we get possible value of b = 1 so

$16{\left(-\frac{1}{2}\right)}^{\frac{P+q-2}{2}}=1OR{\left(-\frac{1}{2}\right)}^{\frac{p+q-2}{2}}=\frac{1}{16}-{\left(-\frac{1}{2}\right)}^4$

$\Rightarrow \frac{p+q-2}{2}=4\Rightarrow p+q=10$       

Blood – Negatively charged colloid.

According to Hardy – Schulze Rule

FeCl₃ is more efficient for blood clotting