Ncert Solutions Chemistry Class 12th

This is a Short Answer Type Questions  as classified in NCERT Exemplar

Ans: Complexes with more number of ions show more conductivity.

[Co (NH₃)₃Cl₃],   [Co (NH₃)₄Cl₂] <  [Cr (NH₃)₅Cl]Cl₂3)6]Cl3 ,  

This is a Long Answer Type Questions  as classified in NCERT Exemplar

Ans: The octahedral and tetrahedral splitting up of d orbitals is different.

Δt= ( $\frac{4}{9}$ )Δ₀

Δt < ₀

Δt= CFSE in tetrahedral field

Δ₀= CFSE in octahedral field

Comparing CFSE with energy= $\frac{\mathrm{h}\mathrm{c}}{\mathrm{\lambda }}$

This is a Long Answer Type Questions  as classified in NCERT Exemplar

Ans: (i) 'A' is [Co (NH₃)₅SO₄]Cl

'B' is [Co (NH₃)₅Cl]SO₄

_{(ii) Type of isomerism is ionisation isomerism}

_(iii) IUPAC name of isomer 'A' is pentaaminesulphatocobalt (III)chloride

IUPAC name of isomer 'B' is pentaaminechlorocobalt (III)sulphate

This is a Long Answer Type Questions as classified in NCERT Exemplar

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetI={\displaystyle \int \frac{x^2}{\left(x^2+a^2\right)\left(x^2+b^2\right)}}dx\\ Put{x}^2=tforthepurposeofpartialfraction.\\ Weget\frac{t}{\left(t+a^2\right)\left(t+b^2\right)}\\ Put\frac{t}{\left(t+a^2\right)\left(t+b^2\right)}=\frac{A}{t+a^2}+\frac{B}{t+b^2}\left[whereAandBarearbitrar\mathrm{yconstants}.\right]\\ \frac{t}{\left(t+a^2\right)\left(t+b^2\right)}=\frac{A\left(t+b^2\right)+B\left(t+a^2\right)}{\left(t+a^2\right)\left(t+b^2\right)}\\ \Rightarrow t=At+A{b}^2+Bt+B{a}^2\\ Comparingtheliketerms,weget\\ A+B=1andA{b}^2+B{a}^2=0\\ \Rightarrow A=\frac{-a^2}{b^2}B\\ \therefore \frac{-a^2}{b^2}B+B=1\\ B\left(\frac{-a^2}{b^2}+1\right)=1\Rightarrow B\left(\frac{-a^2+b^2}{b^2}\right)=1\\ \Rightarrow B=\frac{b^2}{b^2-a^2}andA=\frac{-a^2}{b^2}*\frac{b^2}{b^2-a^2}=\frac{a^2}{a^2-b^2}\\ So,A=\frac{a^2}{a^2-b^2}andB=\frac{-b^2}{a^2-b^2}\\ \therefore {\displaystyle \int \frac{x^2}{\left(x^2+a^2\right)\left(x^2+b^2\right)}}dx=\frac{a^2}{a^2-b^2}{\displaystyle \int \frac{1}{x^2+a^2}dx-}\frac{b^2}{a^2-b^2}{\displaystyle \int \frac{1}{x^2+b^2}dx}\\ =\frac{a^2}{a^2-b^2}*\frac{1}{a}ta{n}^{-1}\frac{x}{a}-\frac{b^2}{a^2-b^2}.\frac{1}{b}.ta{n}^{-1}\frac{x}{b}\\ =\frac{a}{a^2-b^2}ta{n}^{-1}\frac{x}{a}-\frac{b}{a^2-b^2}ta{n}^{-1}\frac{x}{b}+C\\ Hence,I=\frac{1}{a^2-b^2}\left[ata{n}^{-1}\frac{x}{a}-bta{n}^{-1}\frac{x}{b}\right]+C.\end{array}$

This is a Fill in the blanks Type Question as classified in NCERT Exemplar

Ans: Correct option B

If A→B then the concentration of both reactants and the products vary exponentially with time. But, in option B graph the reactant concentration decreases exponentially and the product concentration increases.

This is a Fill in the blanks Type Question as classified in NCERT Exemplar

Ans: Correct option C

Let's start with what a pseudo-first-order response is.

Although the pseudo-first-order reaction looks to be an order, it belongs to another order. It's a second-order reaction because it involves two reactants.

Let's have a look at a reaction.

CH₃Br + OH→CH₃OH + Br^- 

So, the rate law for the reaction is

Rate = k [OH] [CH₃Br]

Rate = k [OH^- ] [CH₃Br] = k (constant) [CH₃Br] = K' [CH₃Br]

Only the concentration of CH₃Br will change during the reaction, and the rate will be determined by the reaction's modifications.