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This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct option: B

When 0.1 mol CoCl₃ (NH₃)₅ is treated with excess of AgNO₃, In the given reaction [CoCl3 (NH₃)₅]Cl₂

then electrolytic solution must contain 

[CoCl₃ (NH₃)₅]Cl²⁺  and two Cl^-  ions. 

Hence, it is 1:2 electrolyte.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct option: A

Strong field ligands have five degenerate energy levels, which means they have more energy separation than weak field ligands.

Here, ΔE= $\frac{\mathrm{h}\mathrm{c}}{\mathrm{\lambda }}$

ΔE α $\frac{1}{\mathrm{\lambda }}$

$\mathrm{\lambda }$ α $\frac{1}{\mathrm{\Delta }\mathrm{E}}$

The wavelength decreases as the energy separation rises.

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18. MET

This is a short answer type question as classified in NCERT Exemplar

By considering the diagram

_{$\theta$ 1}= ${\theta }_{o}sin(wt+{\varnothing }_1)$

_{$\theta$ 2}= ${\theta }_o\mathrm{s}\mathrm{i}\mathrm{n}?\left(wt+{\varnothing }_2\right)$

As it is clear given that amplitude time period being equal but phases being different. Now for first pendulum at any time t

_{$\theta$ 1}= $\theta$ ₂

So sin $\frac{\pi }{2}$ = sin(wt+ ${\varnothing }_1$ )

wt+ ${\varnothing }_1=\frac{\pi }{2}$

where $\theta$ _o=2^o is the angular amplitude of first pendulum . for the second pendulum , the angular displacement is one degree , therefore $\theta$ ₂= $\frac{{\theta }_0}{2}$  and negative sign is taken to show for being left to mean position.

- $\frac{{\theta }_o}{2}={\theta }_{0}sin(wt+{\theta }_2)$

Sin(wt+ ${\varnothing }_2$ )=-1/2

So (wt+ $\theta$ ₂)=- $\frac{\pi }{6}or\frac{7\pi }{6}$

So by making their difference

(wt+ $\theta$ ₂)-( $wt+\theta$ ₁)=7 $\frac{\pi }{6}-\frac{\pi }{2}$ =4 $\frac{\pi }{6}$

( $\theta$ _{2- $\theta$ 1})= 120⁰

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This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct options: B

The bigger the value of constant, the better will be the stability.

Here, has the highest value of logk which corresponds to the highest value of k. 

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This is a short answer type question as classified in NCERT Exemplar

Consider the diagram of the spring block system. It is a SHM with amplitude of 5cm about the mean position

Spring constant k=50N/m

Mass =2kg

Angular frequency w= $\sqrt[]{\frac{k}{w}}=\sqrt[]{\frac{50}{2}}=5rad/s$

Y (t)= Asin (wt+ $\varnothing$ )

Y (0)=Asin (w $\times 0+\varnothing$ )

sin $\varnothing$ =1 $\varnothing =\frac{\pi }{2}$

y (t)=Asin (wt+ $\frac{\pi }{2})$ = Acoswt

A=5cm w=5rad/s

Y=5sin5t

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

Ans: Ambidentate ligands are ligands which have two donating sites. Coordinating compounds containing ambidentate ligands show linkage isomerism due to two different binding positions. Linkage isomerism have same ligand and geometry attached to a central metal ion by different donating sites

Examples:

(i) 

(ii) M←SCN

thiocyanato

M←NCS

isothiocyanato

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

Ans: In CuSO₄⋅5H₂O, water acts as ligand and causes crystal field splitting. This makes d - d transitions possible. On the other hand, in CuSO₄, splitting is not possible due to the lack of ligand in the crystal field. As a result, no colour is visible.

This is a short answer type question as classified in NCERT Exemplar

given potential energy associated with the field

U (x)=U_o (1-cos $\alpha x$ )

F=-dU (x)/dx

F=-d (U_o-U_ocos $\alpha x$ )=-U_o $\alpha$ sin $\alpha x$

F=-U_{o ${\alpha }^{2}x$}

F $\propto -x$

Motion is SHM for small oscillatons

F=-mw²X

Mw²=U_{o ${\alpha }^2$}

w²= $\frac{U_o{\alpha }^2}{m}$  , w= $\sqrt[]{\frac{U_o{\alpha }^2}{m}}$

T= $\frac{2\pi }{w}$ = $2\pi \sqrt[]{\frac{m}{U_o{\alpha }^2}}$

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

Ans: Due to the presence of weak and strong field ligands in complexes, compounds with comparable geometry have distinct magnetic moments. The magnetic moment decreases when the CFSE increases, and vice versa.

Example: [Fe (CN)₆]^3-  - Fe³⁺ ,3d⁵, CN^-  (strong field ligand, pairs electron)

  [Fe (H2O)₆]^3-  - Fe³⁺ ,3d5, H₂O (weak field ligand, does not pair)

This is a short answer type question as classified in NCERT Exemplar

Let us assume that the required displacement be x

Potential energy of the simple harmonic oscillator =1/2 kx²

k= force constant=mw²

PE= ½ mw²x²

Maximum energy of oscillator

TE= ½ mw²A²

PE=1/2 TE

½ mw²x²= $\frac{1}{2}\left[\frac{1}{2}m{w}^2{A}^2\right]$

So x= $\sqrt[]{\frac{A^2}{2}}$ = $\pm \frac{A}{\sqrt[]{2}}$

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

Ans: CFSE is higher when the complex contains strong field ligand. Thus, crystal field splitting energy increases in the order

[Cr (Cl)₆]³⁻< [Cr (NH₃)₆]³⁺< [Cr (CN)₆]^3−.

Because according to spectrochemical series the order of field strength is

Cl⁻3−

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

Ans: CN^-  is a strong ligand and [Fe (CN)₆]^3-  has one paired electron

 H₂O is a weak ligand and [Fe (H₂O)₆]³⁺  has five paired electrons.