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Class 12th Chemistry NCERT Exemplar Solutions Class 12th Chapter Nine

Using crystal field theory, draw energy level diagram, write electronic configuration of the central metal atom/ion and determine the magnetic moment value in the following:

(i) [CoF₆]^3- ,[Co(H₂O)₆]²⁺ ,[Co(CN)₆]^3-

^{(ii) [FeF₆]3- ,[Fe(H₂O)₆]2 + ,[Fe(CN)₆]4 -}

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Vishal Baghel

Contributor-Level 10

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

Ans: (i) Electronic cnfiguration: Co³⁺ =[Ar]3d⁶

Energy level diagram:

Magnetic moment:

Number of unpaired electrons (n)=4

Magnetic moment = μ= $\sqrt[]{n (n + 2}$ = $\sqrt[]{4 (4 + 2}$

= $\sqrt[]{24}$ = 4.9 BM

[Co(H₂O)₆]²⁺

Electronic cnfiguration: Co²⁺=[Ar]3 d⁷

Energy level diagram:

Magnetic moment: Since ,number of unpaired electrons (n)=3, therefore magnetic moment = $\sqrt[]{3 (3 + 2}$ = $\sqrt[]{15}$ = 3.87BM

[Co(CN)₆]³⁻

Electronic configuration: [Ar]Co³⁺=3 d⁶

Energy level diagram:

(ii)

Ans: [FeF₆]³⁻

Electronic configuration: Fe³⁺=[Ar]3 d⁵

Energy level d

...more

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

Ans: (i) Electronic cnfiguration: Co³⁺ =[Ar]3d⁶

Energy level diagram:

Magnetic moment:

Number of unpaired electrons (n)=4

Magnetic moment = μ= $\sqrt[]{n (n + 2}$ = $\sqrt[]{4 (4 + 2}$

= $\sqrt[]{24}$ = 4.9 BM

[Co(H₂O)₆]²⁺

Electronic cnfiguration: Co²⁺=[Ar]3 d⁷

Energy level diagram:

Magnetic moment: Since ,number of unpaired electrons (n)=3, therefore magnetic moment = $\sqrt[]{3 (3 + 2}$ = $\sqrt[]{15}$ = 3.87BM

[Co(CN)₆]³⁻

Electronic configuration: [Ar]Co³⁺=3 d⁶

Energy level diagram:

(ii)

Ans: [FeF₆]³⁻

Electronic configuration: Fe³⁺=[Ar]3 d⁵

Energy level diagram:

Number of unpaired electrons = μ= $\sqrt[]{n (n + 2} = \sqrt[]{5 (5 + 2}$ = μ= $\sqrt[]{35}$ = 5.95BM [Fe(H₂O)₆]²⁺

Fe²⁺=[Ar]3 d⁶

Energy level diagram:

Magnetic moment

Since, number of unpaired electrons (n)=4, therefore magnetic moment is-

$\sqrt[]{4 (4 + 2}$ = $\sqrt[]{24}$ = 4.9BM

[Fe(CN)₆]⁴⁻

Electronic configuration: Fe²⁺=[Ar]3 d⁶

Energy level diagram

Since CN⁻is strong field ligand, it forces all the electrons to get paired up Magnetic moment.

In the absence of unpaired electrons it behaves as diamagnetic ( ie. no magnetic behaviour ).

So magnetic moment = zero

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9 months ago

Class 12th Maths Integrals

Kindly consider the following

$\int e^{- 3 x} c o s^{3} x d x$

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} L e t ? ? ? ? I = ? \underset{I I}{e^{? 3 x}} \underset{I}{c o s^{3} x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = c o s^{3} x . ? e^{? 3 x} ? d x ? ? (D (c o s^{3} x) . ? e^{? 3 x} ? d x) d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = c o s^{3} x . \frac{e^{? 3 x}}{? 3} ? ? (3 c o s^{2} x (? s i n x) . \frac{e^{? 3 x}}{? 3}) d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? c o s^{2} x s i n x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? (1 ? s i n^{2} x) s i n x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x + ? \underset{I}{s i n^{3} x} . \underset{I I}{e^{? 3 x}} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x + s i n^{3} x ? e^{? 3 x} d x ? ? (D (s i n^{3} x) . ? e^{? 3 x} ? d x) d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x + s i n^{3} x . \frac{e^{? 3 x}}{? 3} ? ? (3 s i n^{2} x . c o s x . \frac{e^{? 3 x}}{? 3}) d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x ? \frac{1}{3} e^{? 3 x} s i n^{3} x + ? s i n^{2} x c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x ? \frac{1}{3} e^{? 3 x} s i n^{3} x + ? (1 ? c o s^{2} x) c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? I = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? [s i n x . \frac{e^{? 3 x}}{? 3} ? ? c o s x . \frac{e^{? 3 x}}{? 3} d x] ? \frac{1}{3} e^{? 3 x} s i n^{3} x + ? c o s x . e^{? 3 x} d x ? ? c o s^{3} x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x + s i n x . \frac{e^{? 3 x}}{3} ? ? c o s x . \frac{e^{? 3 x}}{? 3} d x ? \frac{1}{3} e^{? 3 x} s i n^{3} x + ? c o s x . e^{? 3 x} d x ? I \\ ? ? ? ? ? ? ? ? ? 2 I = \frac{e^{? 3 x}}{? 3} [c o s^{3} x + s i n^{3} x] ? [s i n x . \frac{e^{? 3 x}}{3} ? ? c o s x . \frac{e^{? 3 x}}{? 3} d x] + ? c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{e^{? 3 x}}{? 3} [c o s^{3} x + s i n^{3} x] + \frac{1}{3} s i n x . e^{? 3 x} ? \frac{1}{3} ? c o s x . e^{? 3 x} d x + ? c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? 2 I = \frac{e^{? 3 x}}{? 3} [c o s^{3} x + s i n^{3} x] + \frac{1}{3} s i n x . e^{? 3 x} + \frac{2}{3} ? c o s x . e^{? 3 x} d x \\ N o w, ? I_{1} = \frac{2}{3} ? \underset{I}{c o s x} . \underset{I I}{e^{? 3 x}} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{2}{3} [c o s x . ? e^{? 3 x} d x ? ? (D (c o s x) . ? e^{? 3 x} ? d x) d x] \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{2}{3} [c o s x . \frac{e^{? 3 x}}{? 3} ? ? ? s i n x . \frac{e^{? 3 x}}{? 3} d x] \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{2}{3} [c o s x . \frac{e^{? 3 x}}{? 3} ? \frac{1}{3} ? s i n x . e^{? 3 x} d x] \end{array}$

$\begin{array}{l} = \frac{2}{? 9} c o s x . e^{? 3 x} ? \frac{2}{9} ? s i n x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = \frac{2}{? 9} c o s x . e^{? 3 x} ? \frac{2}{9} [s i n x . \frac{e^{? 3 x}}{? 3} ? ? c o s x . \frac{e^{? 3 x}}{? 3} d x] \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} ? \frac{2}{27} ? c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} ? \frac{1}{9} . \frac{2}{3} ? c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} ? \frac{1}{9} . I_{1} \\ I_{1} + \frac{1}{9} I_{1} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} \\ ? ? ? ? ? ? ? \frac{10 I_{1}}{9} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = ? \frac{1}{10} c o s x . e^{? 3 x} + \frac{1}{15} s i n x . e^{? 3 x} \\ S o, ? ? ? ? ? 2 I = ? \frac{1}{3} e^{? 3 x} [s i n^{3} x + c o s^{3} x] + \frac{1}{3} s i n x . e^{? 3 x} ? \frac{1}{10} c o s x . e^{? 3 x} + \frac{1}{15} s i n x . e^{? 3 x} \\ ? ? ? ? ? ? ? ? ? ? ? ? ? I = ? \frac{1}{6} e^{? 3 x} [s i n^{3} x + c o s^{3} x] + \frac{1}{6} s i n x . e^{? 3 x} ? \frac{1}{20} c o s x . e^{? 3 x} + \frac{1}{30} s i n x . e^{? 3 x} \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{6} e^{? 3 x} [s i n^{3} x + c o s^{3} x] + \frac{1}{5} s i n x . e^{? 3 x} ? \frac{1}{20} c o s x . e^{? 3 x} \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{e^{? 3 x}}{24} [s i n 3 x ? c o s 3 x] + \frac{3 e^{? 3 x}}{40} [s i n x \end{array}$

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New answer posted 9 months ago

Class 12th Maths Integrals

Kindly consider the following

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alok kumar singh Contributor-Level 10

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

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New answer posted 9 months ago

Class 12th Maths Integrals

Kindly consider the following

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alok kumar singh Contributor-Level 10

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

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New answer posted 9 months ago

Class 12th Maths Integrals

$\int e^{t a n^{- 1} x} (\frac{1 + x + x^{2}}{1 + x^{2}}) d x$

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} L e t I = \int e^{t a n^{- 1} x} (\frac{1 + x + x^{2}}{1 + x^{2}}) d x \\ P u t t a n^{- 1} x = t \Rightarrow \frac{1}{1 + x^{2}} . d x = d t \\ = \int e^{t} (1 + t a n t + t a n^{2} t) d t = \int e^{t} (s e c^{2} t + t a n t) d t \\ H e r e, f (t) = t a n t \\ ∴ f^{'} (t) = s e c^{2} t \\ = e^{t} . f (t) = e^{t} t a n t = e^{t a n^{- 1} x} . x + C \\ [? \int e^{t} [f (x) + f^{'} (x)] d x = e^{x} f (x) + C] \\ H e n c e, I = e^{t a n^{- 1} x} . x + C \end{array}$

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New answer posted

9 months ago

Class 12th Maths Integrals

$\int \frac{2 x - 1}{(x - 1) (x - 2) (x - 3)} d x$

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alok kumar singh

Contributor-Level 10

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

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New answer posted

9 months ago

Class 12th Maths Integrals

Kindly consider the following

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alok kumar singh

Contributor-Level 10

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

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New answer posted

9 months ago

Class 12th Maths Integrals

$\int \frac{x^{2} d x}{(x^{2} + a^{2}) (x^{2} + b^{2})}$

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} L e t I = \int \frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x \\ P u t x^{2} = t f o r t h e p u r p o s e o f p a r t i a l f r a c t i o n . \\ W e g e t \frac{t}{(t + a^{2}) (t + b^{2})} \\ P u t \frac{t}{(t + a^{2}) (t + b^{2})} = \frac{A}{t + a^{2}} + \frac{B}{t + b^{2}} [w h e r e A a n d B a r e a r b i t r a r yconstants .] \\ \frac{t}{(t + a^{2}) (t + b^{2})} = \frac{A (t + b^{2}) + B (t + a^{2})}{(t + a^{2}) (t + b^{2})} \\ \Rightarrow t = A t + A b^{2} + B t + B a^{2} \\ C o m p a r i n g t h e l i k e t e r m s, w e g e t \\ A + B = 1 a n d A b^{2} + B a^{2} = 0 \\ \Rightarrow A = \frac{- a^{2}}{b^{2}} B \\ ∴ \frac{- a^{2}}{b^{2}} B + B = 1 \\ B (\frac{- a^{2}}{b^{2}} + 1) = 1 \Rightarrow B (\frac{- a^{2} + b^{2}}{b^{2}}) = 1 \\ \Rightarrow B = \frac{b^{2}}{b^{2} - a^{2}} a n d A = \frac{- a^{2}}{b^{2}} * \frac{b^{2}}{b^{2} - a^{2}} = \frac{a^{2}}{a^{2} - b^{2}} \\ S o, A = \frac{a^{2}}{a^{2} - b^{2}} a n d B = \frac{- b^{2}}{a^{2} - b^{2}} \\ ∴ \int \frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x = \frac{a^{2}}{a^{2} - b^{2}} \int \frac{1}{x^{2} + a^{2}} d x - \frac{b^{2}}{a^{2} - b^{2}} \int \frac{1}{x^{2} + b^{2}} d x \\ = \frac{a^{2}}{a^{2} - b^{2}} * \frac{1}{a} t a n^{- 1} \frac{x}{a} - \frac{b^{2}}{a^{2} - b^{2}} . \frac{1}{b} . t a n^{- 1} \frac{x}{b} \\ = \frac{a}{a^{2} - b^{2}} t a n^{- 1} \frac{x}{a} - \frac{b}{a^{2} - b^{2}} t a n^{- 1} \frac{x}{b} + C \\ H e n c e, I = \frac{1}{a^{2} - b^{2}} [a t a n^{- 1} \frac{x}{a} - b t a n^{- 1} \frac{x}{b}] + C . \end{array}$

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New answer posted

9 months ago

Class 12th Maths Integrals

Kindly consider the following

$\int^{} \frac{x^{2} d x}{x^{4} - x^{2} - 12}$

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alok kumar singh

Contributor-Level 10

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

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New answer posted

9 months ago

Class 12th Chemical Kinetics

Consider the reaction A⇌B. The concentration of both the reactants and the products varies exponentially with time. Which of the following figures correctly describes the change in concentration of reactants and products with time?

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alok kumar singh

Contributor-Level 10

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.

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