Class 11th

$\mathrm{}{}_{26}\mathrm{F}\mathrm{e}=\left[\mathrm{A}\mathrm{r}\right]3{\mathrm{d}}^{6}4{\mathrm{S}}^2$

Among Ni, Co, Mn, Fe

Fe having minimum third ionization energy

In a hydrogen atom, which we know has a single electron, the orbital energy will only depend on the principal quantum number (n). Here, the orbitals like 2s and 2p have the same energy (degenerate).

It's a little different with multielectron atoms. The energy depends on both n and the azimuthal quantum number (l). This causes splitting. And, the energies increase in the order: s < p < d < f.

$3+4+8+9+13+14+\dots .$ . upto 40 terms

$\Rightarrow 7+17+27+\dots \dots ..20$ terms

$\mathrm{S}=\frac{20}{2}[2*7+19*10]$

$=102*20=102\mathrm{m}$

$\therefore \mathrm{m}=20$

$y=-\frac{3x}{4}+3\sqrt[]{2}$ line is tangent to ellipse

$\therefore {\mathrm{c}}^2={\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2$

$18=\frac{9{\mathrm{a}}^2}{16}+9$

${\mathrm{a}}^2=16$

$\therefore {\mathrm{e}}^2=1-\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}$

$\mathrm{e}=\frac{\sqrt[]{7}}{4}$

Distance between focii $=2\mathrm{a}\mathrm{e}$

$=2\sqrt[]{7}$

$\mathrm{}6\cdot {}^{35}{\mathrm{C}}_{\mathrm{r}}=\left({\mathrm{k}}^2-3\right)\cdot \frac{36}{\mathrm{r}+1}{}^{35}{\mathrm{C}}_{\mathrm{r}}$

$\Rightarrow {\mathrm{k}}^2-3=\frac{\mathrm{r}+1}{6},{\mathrm{k}}^2-3>0$

(i) $k=\pm 2$ gives $r=5$

(ii) $k=\pm 3$ gives $r=35$

4 ordered pairs

$500\mathrm{m}\mathrm{L}$ has $\mathrm{H}\mathrm{C}\mathrm{l}=25*{10}^{-3}\mathrm{m}\mathrm{o}\mathrm{l}=25\mathrm{m}\cdot \mathrm{m}\mathrm{o}\mathrm{l}$

$120\mathrm{m}\mathrm{L}$ has $=1\mathrm{m}\cdot \mathrm{m}\mathrm{o}\mathrm{l}$

and $500\mathrm{m}\mathrm{L}$ has ${\mathrm{C}\mathrm{H}}_3\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H}=\frac{1}{20}*{10}^3\mathrm{m}\cdot \mathrm{m}\mathrm{o}\mathrm{l}$

so $20\mathrm{m}\mathrm{L}$ has $\frac{{10}^3}{20}*\frac{20}{500}=2\mathrm{m}\cdot \mathrm{m}\mathrm{o}\mathrm{l}$

$\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}$ added in $20\mathrm{m}\mathrm{L}$ is $5*\frac{1}{2}=2.5\mathrm{m}\cdot \mathrm{m}\mathrm{o}\mathrm{l}$

So, $\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}$ (left) $+{\mathrm{C}\mathrm{H}}_3\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H}?{\mathrm{C}\mathrm{H}}_3\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H}+{\mathrm{H}}_2\mathrm{O}$
1.5
2
1.5

$\Rightarrow \mathrm{p}\mathrm{H}={\mathrm{p}}^{{\mathrm{K}}_{\mathrm{a}}}+\mathrm{l}\mathrm{o}\mathrm{g}?\frac{\text{salt}}{\text{acid}}=4.75+\mathrm{l}\mathrm{o}\mathrm{g}?0.4771=5.2271$

192.5

$\mathrm{}2\mathrm{C}\left(\mathrm{g}\mathrm{r}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)?{\mathrm{C}}_2{\mathrm{H}}_6\left(\mathrm{g}\right)$

$\mathrm{\Delta }{\mathrm{H}}_{\mathrm{f}}=2*\mathrm{\Delta }{\mathrm{H}}_{\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{b}}\left(\mathrm{C}\right)+3*\mathrm{\Delta }\sigma {\mathrm{H}}_{\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{b}}\left({\mathrm{H}}_2\right)-\mathrm{\Delta }{\mathrm{H}}_{\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{b}}\left({\mathrm{C}}_2{\mathrm{H}}_6\right)$

$\Rightarrow \sigma {\mathrm{H}}_{\mathrm{f}}=2(-286)+3(-393.5)-(-1560)=192.5$

$\mathrm{}0.3675$

Moles of $\mathrm{H}\mathrm{C}\mathrm{l}=$ moles of ${\mathrm{N}\mathrm{H}}_3$

$=2*$ moles of urea $=2*\frac{0.6}{60}=0.02$

$\Rightarrow \mathrm{N}.\mathrm{V}=0.02$

Kindly go through the solution

Conceptual.