Chemistry NCERT Exemplar Solutions Class 12th Chapter Nine: Overview, Questions, Preparation

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= $\sqrt[]{n(n+2}=\mathrm{}\sqrt[]{5(5+2}$ =  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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Ans:

[Mn (CN)₆]³⁻

Electronic configuration is Mn³⁺= [Ar]3 d⁴ hence box electronic structure

(i) Type of hybridisation d²sp³

(ii) Inner orbital complex

(iii) paramagnetic, due to presence of three unpaired electrons.

(iv) Spin only magnetic moment is calculated using the formula : n=2 in this case, we get spin only magnetic moment in BMas = $\sqrt[]{2(2+2}$ = $\sqrt[]{8}$  = 2.87BM

[Co (NH₃)₆]³⁺

Electronic configuration of Co³⁺= [Ar]3 d⁶

(i) Hyb As shown in the above box electronic structure the type of hybridisation is . d²sp³

(ii) Inner orbital complex

(iii) Diamagnetic

(iv) Zero ( Since no unpaired electrons are present, as depicted above)

[Cr (H₂O)₆]³⁺

(i) type of hybridisation. d²sp³

(ii) Inner orbital complex

(iii) paramagnetic

(iv) 3.87BM

[Fe (Cl)₆]⁴⁻

electronic configuration is of Fe²⁺= [Ar]3 d⁶

(i) type of hybridisation will be sp³d²

(ii) Outer orbital complex

(iii) Paramagnetic

(iv) 4.9BM

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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

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Ans: When light wavelengths from a specific part of the spectrum are absorbed by a substance, the result is a complimentary colour. When a complex absorbs a wavelength of light, it reflects a complementary colour. If a violent colour is absorbed, for example, yellow is conveyed. The CFSE value, often known as the colour definer for any complex, comes next. To determine the wavelength value in order to determine which colour absorbs the most energy.

Δe=$\frac{\mathrm{h}\mathrm{c}}{\mathrm{\lambda }}$

As λ has shorter wavelength.

Low spin complexes absorb shorter wavelengths, while high spin complexes absorb longer ones. The colour theory, complex, and wavelength are all determined using this method.

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

Δt= ( $\frac{4}{9}$ )Δ₀

Δt < Δ₀

Δt= CFSE in tetrahedral field

Δ₀= CFSE in octahedral field

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

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 Short Answer Type Questions  as classified in NCERT Exemplar

Ans: The solution CrCl₃.4H₂O molar conductance shows that it contains one positive and one negative ion.

An ion is present outside the complex when silver chloride is treated with silver nitrate chloride. Outside the complex, there are two ions and one chloride, hence the name of the complex will be 

[Co (H₂O)4Cl₂]Cl, Tetraaquadichloridocobalt (III)chloride.

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Ans: [M (AA)₂X₂]n +  is a bidentate ligand and %X is a monodentate ligand. It is an example of octahedral geometry as well as an ion. There are cis and trans isomers of this ion. Trans isomers are symmetrically active, while cis isomers are optically active.

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Ans: The existence of five unpaired electrons in the ion's mid-orbits Mn²⁺  equates to a magnetic moment of 5.92 BM. The number of hybridizations in this case is sp³. As a result, the magnetic moment value of the tetrahedral structure complex is 5.92 BM.

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Ans: Weak field ligands, Electronic configuration of  Co (III) is t⁴_2g e⁰_g, has 4 paired electrons and it is paramagnetic.

Strong field ligands, Electronic configuration of is Co (III) t⁶2_ge⁰_g.

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

Ans: In a tetrahedral complex, the d-orbit splits too little compared to an octahedral complex. As a result, orbital energy alone are insufficient to couple. Low spin tetrahedral complexes do not form as a result.

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Ans: The ligands can be grouped in ascending order of increasing field intensity according to spectrochemical series.

[CoF6]³⁻, Co³⁺ → (d₆)→ t⁴_2g e⁰_g

[FE (CN)₆]⁴⁻, Fe²⁺→ (d₆)→ t⁶_2g e²_g

  [Cu (NH₃)₆]²⁺, Cu²⁺→ (d₉)→ t⁶_2g e²_g

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.

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: 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 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 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 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. 

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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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: D

When 1 mol [CrCl₃ (H₂O)₃]6H₂O is treated with excess of AgNO₃, 3 mol of AgCl is produced, i.e., [CrCl₃ (H₂O)₃]6H₂Ois dissociated in aqueous solution and all three chloride comes in solution.

[Cr (H₂O)₆]Cl₃→ [Cr (H₂O)₆]³⁺  + 3Cl^- 

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Ans: Correct option:  A

Diamminedichloridoplatinum (II) is also known as azane 

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Ans: Correct option: C

The chelate effect is when a chelating ligand coordinates the stabilisation of a molecule. Any ligand that binds to metal forms a ring with 5 or 6 members, which is the most stable.

Here, CO, CN, H₂O is monodentate ligand [Fe (C₂O₄)₃]^3-  is an oxalate ligand. 

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Ans: Correct option: A

This is due to the complex's octahedral structure and (cis-trans) isomerism. Two Cl ligands are next to each other in trans isomerism. Although these isomers have the same structure, they are not identical.

Possible isomers are:

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct option: A

CFSE for octahedral and tetrahedral complex is related as

Δt = $\frac{4}{9}$ Δ₀

Where Δ₀= CFSE for octahedral complex

Δ₀= CFSE for tetrahedral complex

Δ₀=1800 cm⁻¹

Δt =49×18000=8000 cm⁻¹

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Ans: Correct option: (A)

Linkage isomerism arises in a coordination compound containing ambidentate ligand.

A simple example is by complexes containing the thiocyanate ligand, NCS-which may bind through the nitrogen to give M-NCS or through sulphur to give M-SCN.

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Ans: Correct option: D

Two or more compounds with the same formula but distinct structures and characteristics are known as isomers.

So, in this case compounds 

[Co (SO₄) (NH₃)₅]Br and  [Co (SO₄) (NH₃)₅]Cl show no isomerism

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Ans: Correct option: A

A chelating agent binds to a single metal ion with two or more donor atoms. Oxaloto, Glycinato, and ethane-1,2-diamine all have two donor oxygen atoms. The gland Thiosulpato is ambidentate.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct option: B

Ligands are neutral ions that form a coordination complex with the central metal atom.

NH₄⁺  does not have a lone pair of electrons to give.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct option: B

Hydrate isomerism is isomerism in which water is used as a solvent.

Coordination compounds with the same composition but varying ligand connectivity are known as linkage isomers.

Solvate isomerism has the same composition as free solvent but distinct solvent ligand molecules.

Except for the ligand that swaps places with the anion, ionisation isomers are identical isomers.

Coordination isomers are coordination compounds with distinct metal and ligand compositions.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct option: C

[Pt (NH₃)₂Cl (NO₂)] is a neutral molecule. Ligands are arranged alphabetically.

So, NH₃ Diammine, chloride and the embedded linkage is nitrogen and last is the metal name.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct option: A and C

(i) Orbitals of Co³⁺ ion d²sp³ hybridised orbitals of Co³⁺ [Co (NH₃)₆]³⁺ (inner orbital or low spin complex)

No. of unpaired electron =0

Magnetic property = diamagnetic, due to absence of unpaired electrons

(ii) Electronic configuration is 3 d⁶ orbitals of Fe²⁺ ion:

As CN⁻is a strong field ligand, it causes the pairing of the unpaired 3d electrons. Since there are six ligands around the central metal ion, the most feasible hybridization d²sp³⋅d²sp³ hybridized oribtals of Fe²⁺ are:

6 electron pairs from CN⁻ ions occupy the six hybrid d²sp³ orbitals. Then,

Hence, the geometry of the complex is octahedral and the complex is diamagnetic (as there are no unpaired electrons.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct option: A and C

The outer octahedral complexes are weak ligands, number of pairing electrons and high spinning potential.

[MnCl⁶]^3-   - electronic configuration of Mn = 3d⁵,4s². Shifting the valence electrons become 

Mn³⁺ ,3d⁴ and number of unpaired electrons is 4.

[FeF₆]^3- - electronic configuration of Fe = 3d⁶,4s². Shifting the valence electrons become

Fe³⁺ ,3d⁵ and the number of unpaired electrons is 5.

[CoF₆]^{3 -} electronic configuration of Co  =  3d⁷,4s² shifting the valence electrons become

 Fe³⁺ ,3d⁶ and the number of unpaired electrons is 4.

[Ni (NH₃)₆]²⁺ - electronic configuration of Ni = 3d⁸,4s². Shuffling the valence electrons become 

Ni²⁺ ,3d⁶ and the number of unpaired electrons is 2.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct option: A and C

For answer at (A) Atomic number of Iron is 26 . In ferricyanide complex, iron is in oxidation state of 3+. The orbitals on the metal atom undergoes d₂sp₃ hybridizaion and hence the complex has octahedral shape.

Also it can be said that, [Fe (CN)₆]^3- has one unpaired electron which makes it weakiy paramagnetic. Therefore, option C is also correct.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct option: B and C

[Co (H₂O)₆]²⁺  is increased when excess of HCl is added. Tetrahedral complexes have smaller crystal field splitting than octahedral complexes because Δt = $\frac{4}{9}\mathrm{}$ Δ₀

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct options: A, B and C

A homoleptic complex is one that has only one species or group as a ligand.

Option A, B, and C each have only one ligand, but option D has two.

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Ans: Correct option: B and D

Heteroleptic is a compound that contains many ligands of various sorts. NH₃ and 

Cl as a ligand or also called as donor groups, is a heteroleptic complex.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct option: A and C

Optical compounds are highly imposable compounds that lack symmetry components. As a result, Option A and C are mirror reflections of each other.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct options: A, B and C

Option A, B, and C describe that ethane-1, 2-diamine is a neutral ligand due to its lack of charge, a bidentate ligand due to the presence of two donor sites on one nitrogen atom, and a chelating ligand due to its ability to chelate with metal.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Correct option: C

Coordination compounds with the same composition but varying ligand-metal connectivity are known as linkage isomerism.

[Co (NH₃)₅ (NO₂)]²⁺ , NO₂ has two donor sites N and O

[Cr (NH₃)₅SCN]²⁺ , S and N are two donor sites.

This is a Matching Type Questions as classified in NCERT Exemplar

Ans: Correct option: (ii)

This is a Matching Type Questions as classified in NCERT Exemplar

Ans: Correct option: (i)

This is a Matching Type Questions as classified in NCERT Exemplar

Ans: Correct option: (ii)

This is a Matching Type Questions as classified in NCERT Exemplar

Ans: Correct option: (iv)

This is a Matching Type Questions as classified in NCERT Exemplar

Ans: Correct option: (i)

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

Ans: Correct option: A

Toxic metal ions are removed by the chelating ligands. Chelate complexes tend to be more stable. When a solution of the chelating ligand is added to a solution containing toxic metals ligands chelates the metal ions by the formation of a stable complex.

Thus, EDT A (hexadentate ligand) forms a stable metal chelate and is used to remove toxic metal ions.

Both Assertion and Reason are correct and Reason is the correct  for Assertion.

Hence, option A is correct.

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

Ans: Correct option: B

[Cr (H₂O)₆]Cl₂ and [Fe (H₂O)₆]Cl₂ 

Due to the creation of a more stable complex ion after obtaining ion, [Cr (H₂O)₆]Cl₂ and [Fe (H₂O)₆]Cl₂ are reduced in nature. Both compounds have a coordination of 6 and so form an octahedral complex. The hybridisation state of is 3 and the oxidation state of is 2+. Hybridization is 3d⁴ and 3d⁶ is 2+. Both compounds have unpaired electrons and have weak field ligands. In both compounds, removing electrons is simple.

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

Ans: Correct option: A

The term "linkage isomerism" refers to two molecules that have similar ligands but differ in the donating site of the ligand. Ambidentate is made up of two different types of donor atoms.

Example [Fe (NO₂) (H₂O)₅]Cl₂. NO₂ is Ambidentate ligand as Nand O₂ are donors.

 As a result, linkage isomerism occurs in coordinations with ambidentate ligands. Reason and Assertion are both correct.

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

Ans: Correct option: B

From the Cis and Trans forms, geometrical isomerism forms compounds. All of the valences in the MX₆ ligand are the same, resulting in an identical molecule. L is a distinct ligand in MX₅L, and it can be replaced with any other valency in the structure to create an identical molecule; the molecule does not change. Because of the presence of symmetry, which is an essential prerequisite for revealing geometrical isomerism, complexes do not show geometrical isomerism.

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

Ans: Correct option: D

[Fe (CN)₆]^3-  ion shows magnetic moment corresponding to one unpaired electron. (CN) is a ligand with a strong field that couples electrons, resulting in hybridisation. d²sp³ and it has one unpaired electron, it has the magnetic moment of one paired electron.

  μ= $\sqrt[]{n(n+2}$   = 1 (1 + 2)

   = 3 = 1.73BM 

BCl₃ – electron deficient molecule

SF₆ – expanded octet molecule

NO – odd electron bond containing molecule

H₂SO₄ – expanded octet molecule

t = 0 $\frac{20}{28}$ $\frac{5}{2}$ 0

Here, N₂ is the limiting reagent

and No of moles of NH₃ = 2 × No of moles ofN₂  reacted

$2\times \frac{20}{28}=\frac{40}{28}=\frac{1.42}{}$

This is the method of preparation of lyophilic solution as condition is given.

Following is the trends of I.E. of given metals $\left(1^{st}I.P\right)Na<Al<Mg<Si$

In ores/mineral available earthy and undesired impurities are gangue

Following is the reaction of Zn with excess alkali,

$Zn+2NaOH+2H_{2}O\to N{a}_2{\left[Zn\left(OH\right)\right]}_4+H_2$

$N{a}_{2}Zn{O}_2+2H_{2}O$

$2N{a}^{+}Zn{O}_2^{--}$

 $4LiN{O}_3\underset{\Delta }{\overset{}{\to }}2L{i}_{2}O+4N{O}_2+O_2$

$NaN{O}_3\underset{\Delta }{\overset{}{\to }}NaN{O}_2+\frac{1}{2}{O}_2$

SCl₂ – 2 lone pair of central atom

O₃ – 1 lone pair of central atom

ClF₃ – 2 lone pair of central atom

SF₆ – 0 lone pair of central atom

  $S{c}^{3+}-3d^0$ electron in last subshell

$Z{n}^{2+}-3d^{10}$ electron in last subshell

$T{i}^{4+}-3d^0$ electron in last subshell

$C{u}^{2+}-3d^9$ electron in last subshell

$V^{2+}-3d^3$ electron in last subshell

$T{i}^{3+}-3d^1$ electron in last subshell

$M{n}^{2+}-3d^5$ electron in last subshell

Following is the reaction of KMnO4 in basic medium with thiosulphate anion.

$\stackrel{+7}{8Mn{O}_4^{-}}+3S_2{O}_3^{--}+H_{2}O\to \stackrel{+4}{8Mn{O}_{2|}}+6S{O}_4^{--}+2O{H}^{-}$

Change = 7 – 4 = 3

The pair of herbicides is sodium chlorate and sodium arsinite

Due the absence of inversion and non resonating nature

Drug used in the treatment of sleeping problems

Given cell constant $\frac{\mathcal{l}}{A}=129m^{-1}$

Ratio of ppm of two solution of KCl = Ratio of moles of per unit volume.

$\frac{{\left(ppm\right)}_{KC{l}_1}}{{\left(ppm\right)}_{KC{l}_2}}=\frac{M_1}{M_2}=\frac{74.5}{149.0}=\frac{1}{2}$

$\therefore \frac{{\wedge }_1}{{\wedge }_2}=1000\times 10^{-3}$

$\therefore$ x = 1000

Edge length of the unit cell AB = a = 2 $\left(r_{+}+r_{-}\right)$

$\therefore a=2\left(102+181\right)=2\left(283\right)=566$

According to Heisenberg’s uncertainly principle

$\Delta x\times \Delta p=\frac{h}{4\pi }$

$\therefore \Delta v=\frac{6.63\times 10^{-34}}{4\times 3.14\times 2\times 52.9\times 10^{-12}\times 9.1\times 10^{-31}}$

= 548273 ms^-1

HNO₃    +            NaOH    ®          NaNO₃      +        H₂O

t = 0,     (600 × 0.2)_meq    (400 × 0.1)_meq    40 meq               40 meq

120 meq             40 meq

t = $\infty$     80 meq               -

Total heat produced during neutralization ;

$=\frac{40}{1000}\times 57\times 1000J=2280J=m.s\theta$              

2280 = 1000 × 4.2 × q

$\therefore \theta =2280/4200=54.28\times 10^{-2}C$

According to Henry’s law

P = K_H × mole fraction of solute

$0.92=46.82\times 1000\times \frac{molof{O}_2}{55.5}$

$\therefore$ Mole of  O₂ = 1.09 × 10^-3

$\therefore Millimole=1.09\approx 1$

 $S=\sqrt[]{Ksp}=\sqrt[]{8\times 10^{-28}}=2.82\times 10^{-14}$

Or solubility of PbS = 282 × 10^-16

According to condition

Rate $\propto {\left[X\right]}^1{\left[Y\right]}^0$

By using experimental data I and II

$\frac{4\times 10^{-3}}{2\times 10^{-3}}=\frac{L}{0.1}\therefore L=0.2$

And by using experimental data I and II

$\frac{M\times 10^{-3}}{2\times 10^{-3}}=\frac{0.4}{0.1}$

$\therefore$ M = 8

Ratio of $$$\left(\frac{M}{L}\right)=\frac{8}{0.2}=40$

No. of Amino acid units are 4 and

No. of peptide bonds are 3

In tetrapeptide

Reaction is

Mass of product, 1, 2, 2- tetrobromopropane obtained

= $\frac{1\times 360}{160\times 2}\times \frac{27}{100}$

= 3.30375

= 3.0375 × 10^-1

In complex ${\left[Fe{\left(CN\right)}_6\right]}^{---}$

CN is the strong field ligand

$F{e}^{3+}\to 3d^5\left[t_{2g}^5{e}_g^0\right]$

CFSE value = 5 × $\left(-0.4{\Delta }_0\right)=-2.0{\Delta }_0$