Class 12th

In aniline –NH₂ is strong activating group due to presence of lone pair in nitrogen.

Aniline does not show Friedel-Craft alkylation reaction, because anhydrous AlCl₃ and aniline form salt together

[Ni (H₂O)₆]²⁺ Þ Ni²⁺ is octahedral and [Ni (CN)₄]^2– is square planar.

In [Ni (H₂O)₆]²⁺ Þ Ni²⁺ has two unpaired electrons and in [Ni (CN)₄]^2– Þ Ni²⁺ has no unpaired electrons. [Ni (H₂O)₆]²⁺ is coloured as it absorbs red light due to suitable d-d transition and complementary light emitted is green.

[Ni (CN)₄]^2– has strong field ligand so the electrons of Ni²⁺ pair up and it is colourless as it cannot absorb light from visible region.

KCN   + CN^–

            R – X + KCN  R – CN + KX

            R – X + AgCN  R – NC + AgX

            KCN is ionic therefore ionised and attack occurs through carbon.

            AgCN is covalent therefore attack starts with Nitrogen.

$r=R_0{\left(192\right)}^{\frac{1}{3}}$

$\frac{r}{2}=R_0{\left(m\right)}^{\frac{1}{3}}$

$m=\frac{192}{8}=24$

As NaCl is strong electrolyte, its degree of dissociation (a) will remain same.

            I = 2

            For each sample,

            i₁ = i₂ = i₃

Kindly go through the solution

R_eq = r

$P=\frac{V^2}{r}=\frac{4}{2}=2W$

The balanced reaction is.

            Cr₂  + 14H⁺ + 6e^– ® 2Cr³⁺ + 7H₂O

            x = 14

            y = 6

            A = 7

$\frac{1}{2}{\epsilon }_0{E}^2=\frac{B^2}{2{\mu }_0}$

$?E=CBandC=\frac{1}{{\mu }_0{\epsilon }_0}$          

$?=\frac{h}{p}\left[??\frac{1}{p}\right]$

Þ        lp = h (constant)

            So, the plot is a rectangular hyperbola.