Class XII Chemistry SET 2 2014

8/10/2019 Class XII Chemistry SET 2 2014

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Class XII Chemistry, Board Paper 2014, Set-2 DELHIGeneral Instructions: (i) All questions are compulsory(ii) Question numbers 1 to 8 are very short-answer questions and carry 1 mark each.(iii) Question numbers 9 to 18 are short-answer questions and carry 2 marks each.(iv) Question numbers 19 to 27 are also short-answer questions and carry 3 marks.(v) Question numbers 28 to 30 are long-answer questions and carry 5 marks each.(vi) Use Log Tables, if necessary. Use of calculators is not allowed.

Question 1

Questions Q1

Give one example each of sol and gel . Solution:

Dispersed phase Dispersedmedium Examples

Sol Solid Liquid Paints, Milk of magnesia, MudwaterGel Liquid Solid Cheese, Butter, Jellies

Q2 Which reducing agent is employed to get copper from the leached low-grade copper ore?

Solution:

Copper can be obtained from low-grade ore through the process of leaching using acid or bacteria (leaching is a process in which ore is treated with suitable reagent that dissolves ore butnot the impurities).

The solution containing copper can be reduced with the help of reducing agents such as scrapiron or H 2 to get copper metal.

Cu 2+(aq) + Fe Cu(s) + Fe 2+(aq)Cu 2+ (aq) + H 2 (g) Cu(s) + 2 H +(aq)

Q3 Write the IUPAC name of the compound

Solution:
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The IUPAC nomenclature of the given organic compound is 3-aminobutanal.

Q4 Which of the following isomers is more volatile:o-nitrophenol or p-nitrophenol?

Solution:

o-nitrophenol forms intra-molecular hydrogen bond. On the other hand, p-nitrophenol isinvolved in stronger inter-molecular hydrogen bonding and hence, it has higher boiling pointthan o-nitrophenol. So, o-nitrophenol is more volatile than p-nitrophenol.

Q5 Some liquids on mixing form 'azeotropes'. What are 'azeotropes'? Solution:

Azeotropic solutions are those solutions whose boiling points remain constant. As aresult, both components boil at the same temperature, regardless of difference in theirrespective boiling points.Thus, binary mixtures that have the same composition in liquid and vapour phase andhave constant boiling points are called azeotropes. It is not possible to separate thecomponents of azeotropes by fractional distillation.

Q6 Arrange the following in increasing order of basic strength:

C6H5NH2, C 6H5NHCH 3, C 6H5CH 2NH2 Solution:C6H5 NHCH 3 is more basic than C6H5NH2 due to the presence of electron-donating CH 3 groupinC6H5NHCH 3. Again, in C6H5NHCH 3, C 6H5 group is directly attached to the N-atom .However, it is not so in C6H5CH 2NH2. Thus, in C6H5NHCH 3, the R effect of C 6H5 groupdecreases the electron density over the N-atom . Therefore, C6H5CH 2NH2 is more basicthan C6H5NHCH 3.Hence, the increasing order of the basic strengths of the given compounds is as follows:C6H5NH2 < C 6H5NHCH 3 < C 6H5CH 2NH2

Q7 Which component of starch is a branched polymer of -glucose and insoluble in water?

Solution:Starch is a polymer of -glucose that has two components, namely amylose andamylopectin. Out of these, amylopectin is insoluble in water and constitutes about 80-85% of starch.

Q8
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Which of the following is a more stable complex and why ?

[Co(NH 3)6]3+ and [Co(en) 3]3+ Solution:

Chelating ligands form more stable complexes as compared to non-chelating ligands. Sinceethylene diammine is a bidentate ligand so it forms stable chelate, [Co(en) 3]3+ will be a morestable complex than [Co(NH 3)6]

3+. Q9

State Henry's Law. What is the effect of temperature on the solubility of a gas in aliquid? Solution:

Henry's Law states that the solubility of a gas in a liquid is directly proportional to thepressure of the gas. If we use the mole fraction of a gas in the solution as a measure ofits solubility, then it can be said that the mole fraction of gas in the solution isproportional to the partial pressure of the gas over the solution. The most commonlyused for m of Henrys Law states that the partial pressure of a gas in vapour phase (p)is directly proportional to the mole fraction of the gas (x) in the solution, i.e.p xp = K H x

where K H = Henrys Law Constant

The solubility of gases is dependent on temperature. An increase in temperature resultsin a decrease in gas solubility in liquid, while a decrease in temperature results in anincrease of gas solubility in liquid.

An increase in temperature causes an increase in kinetic energy, resulting in a morerapid motion of molecules, breaking intermolecular bonds which enable molecules toescape from the solution allowing the gases dissolved to evaporate more readily. Hencecausing decrease in the solubility of gases in liquids.

Q10 Define the following terms:

(i) Pseudo first-order reaction(ii) Half-life period of reaction (t 1/2 ). Solution:

i) Pseudo first-order reaction Pseudo first-order reaction: The reaction which is bimolecular but whose order is one is

called pseudo first-order reaction. This happens when one of the reactants is present ina large amount. For example, in acidic hydrolysis of ester (ethyl acetate), water ispresent in a large quantity.

CH 3COOC 2H5 + H 2O CH 3COOH + C 2H5OH

Similarly, inversion of cane sugar or hydrolysis of cane sugar is an example of pseudofirst-order reaction.
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ii) Half-life period of a reaction Half-life period of a reaction is the time taken for half of the reaction to be completed. Inother words, the time in which the concentration of a reactant is reduced to half of its

original value is called half-life period of the reaction.For zero-order reaction,

t 1/2=[ Ro]2k

For first-order reaction,

t 1/2=0.693 k

Q11 Write the principle behind the following methods of refining:

(i) Hydraulic washing(ii) Vapour-phase refining Solution:

The principle behind the given methods of refining are:

i) Hydraulic Washing: This method is based on difference of gravity between ores and gangue. An upwardstream of water is used to wash the powdered ore. The lighter gangue particles are

washed away, leaving behind the heavier ones.

ii) Vapour-phase refining :Vapour-phase refining is the process of refining metal by converting it into its volatilecompound and then, decomposing it to obtain a pure metal. To carry out this process,(i) the metal should form a volatile compound with an available reagent.(ii) the volatile compound should be easily decomposable, so that the metal can beeasily recovered. This method is used for refining nickel, zirconium and titanium.

Q12 Draw the structure of major monohalo product in each of the following reactions :

(i)

(ii) Solution:
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(i)

(ii) Addition in presence of peroxide yields product according to anti-Markovnikov rule ofaddition.

Q13 (i) Which alkyl halide from the following pair is chiral and undergoes faster S N2 reaction?

(ii) Out of S N1 and S N2, which reaction occurs with

(a) Inversion of configuration

(b) Racemisation Solution:

(i) Among the given pair of compounds, alkyl halide (b) has a chiral centre.

The alkyl halide (a) does not contain a chiral centre and and it also gives fasterS N2 reaction as S N2 is more favourable in primary alkyl halides.

(ii)(a) Inversion of configuration takes place in S N2 reaction.
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(b) Racemisation takes place in S N1 reaction.

Q14 Complete the following chemical equations :(i) Ca 3P 2 + H 2O (ii) Cu + H 2SO 4(conc.)

OR

Arrange the following in the order of the property indicated against each set :(i) HF, HCl, HBr, HI increasing bond -dissociation enthalpy.(ii) H2O, H 2S, H 2Se, H 2Te increasing acidic character . Solution:

The balanced reactions are given below:

i) Ca 3P 2 + 6H 2O 3Ca(OH) 2 + 2PH 3 ii) Cu + 2H 2SO 4(conc.) CuSO 4 + SO 2 + 2H 2O

OR

i)The arrangement of the given hydrogen halides in increasing order of bond-dissociationenthalpy is given below:HI < HBr < HCl < HF

ii)
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The increasing order of acidic character of the given hydrides of Group 16 elements isgiven below:H2O < H 2S < H 2Se < H 2Te

Q15 Write the IUPAC name of the complex [Cr(NH 3)4Cl2]+. What type of isomerism does it

exhibit? Solution:

The IUPAC name of the complex [Cr(NH 3)4Cl2]+ is Tetraamminedichlorochromium(III)ion.This complex exhibits geometrical isomerism. [Cr(NH 3)4Cl2]+ is a [MA 4B2] type ofcomplex, in which the two chloride ligands may be oriented cis and trans to each other.

Q16 An element with density 11.2 g cm 3 forms a f.c.c. lattice with edge length of 4 10 8 cm.Calculate the atomic mass of the element.(Given : N A = 6.022 10 23 mol 1) Solution:

Given,Density, d = 11.2 g cm -3 Edge length, a = 4 x 10 -8 cmAvogadro number, N A = 6.022 X 10 23

Number of atoms present per unit cell, Z (fcc) = 4

We know for a crystal system ,

Thus, atomic mass of the element is 107.91 g. Q17
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Examine the given defective crystal:

Answer the following questions :(i) What type of stoichiometric defect is shown by the crystal?(ii) How is the density of the crystal affected by this defect?(iii) What type of ionic substances show such defect? Solution:

(i) Schottky defect is shown by the mentioned crystal, as equal number of cations andanions are missing in the crystal lattice.

(ii) This defect leads to decrease in density, as equal number of the cations and anionsare missing from the crystal lattice.

(iii) This kind of defect is shown by those ionic substance in which the cations andanions are of almost similar sizes.Examples: NaCl, KCl and CsCl.

Q18 Calculate the mass of a compound (molar mass = 256 g mol 1 ) to be dissolved in 75 gof benzene to lower its freezing point by 0.48 K (K f = 5.12 K kg mol 1). Solution:

Q19 Give the structures of A, B and C in the following reactions :

(i)

(ii)

OR

How will you convert the following?(i) Nitrobenzene into aniline

(ii) Ethanoic acid into methanamine
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(iii) Aniline into Nphenylethanamide

(Write the chemical equations involved. ) Solution:

(i)

(ii)

OR (i) Nitrobenzene into aniline

(ii) Ethanoic acid into methanamine

(iii) Aniline into Nphenylethanamide

Q20 (a) Write the mechanism of the following reaction :

(b) Write the equation involved in Reimer-Tiemann reaction . Solution:
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a. The reaction proceeds through nucleophilic substitution bimolecular (S N2) mechanism, asshown:

Inversion of configuration takes place during the reaction.

b.Reimer-Tiemann reaction:

Q21 After the ban on plastic bags, students of a school decided to make people aware of the

harmful effects of plastic bags on the environment and Yamuna River. To make theawareness more impactful, they organised a rally by partnering with other schools anddistributed paper bags to vegetable vendors, shopkeepers and departmental stores. Allthe students pledged not to use polythene bags in the future to save the Yamuna River.

After reading the above passage, answer the following questions:(i) What values are shown by the students?(ii) What are bio-degradable polymers? Give one example.(iii) Is polythene a condensation or an addition polymer? Solution:

(i) From the given passage, we can conclude that the students show awareness about theenvironment.

(ii) A polymer that can be decomposed by microorganisms within a definite period of time, sothat the polymer or its degraded product does not cause any harm to the environment, is called a

bio-degradable polymer. For example, poly- -hydroxybutyrate-CO- - hydroxy valerate (PHBV) is a bio-degradablealiphatic polyester.


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(iii) Polythene is an addition polymer that is formed by addition of ethene molecules.

Q22 (a) Draw the structures of the following:

(i) XeF 2 (ii) BrF 3

(b) Write the structural difference between white phosphorus and red phosphorus . Solution:

a.

(b) Structural difference between White P and Red P:

White P Red P It consists of four P atoms,linked with one another togive rise to a tetrahedralshape.

It has a polymeric structure, consisting of chains of P 4 tetrahedralunits that are linked together.

Structure:Structure:

Q23 (a) In reference to Freundlich adsorption isotherm, write the expression for adsorption ofgases on solids in the form of an equation.(b) Write an important characteristic of lyophilic sols.


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(c) Based on the type of particles of dispersed phase, give one example each ofassociated colloid and multimolecular colloid . Solution:

(a)Freundlich adsorption isotherm for adsorption of gases on solids:

where x is the mass of the adsorbate, m is the mass of the absorbent and p is the pressure of thegas and n is a constant which is greater than 1.

(b) Lyophilic sols are sols that are solvent-attracting. An important characteristic of these sols is thatif the dispersion medium is separated from the dispersion phase by any method, the sol can bereconstituted by simply remixing the two again. That is why these sols are also knownas reversible sols.(c)Example of associated colloid: Soap solutionExample of multimolecular colloid: Gold sol

Q24 Account for the following :

(i) Bi(V) is a stronger oxidizing agent than Sb(V).(ii) N N single bond is weaker than P P single bond.

(ii) Noble gases have very low boiling points . Solution:

i)Because of inert pair effect +5 oxidation state is less stable in Bi than in Sb. So the Biin +5 state will immediately get reduced to its +3 state and thus will be a betteroxidizing agent than +5 state of Sb.

ii)N-N single bond is weaker than P-P bond due to smaller size of N as compared to P.Smaller size of N leads to smaller N-N bond length. As a result, the lone pair ofelectrons on the both the N atoms repel each other leading to unstability or weakeningof N-N bond.

iii)Noble gases have very weak interatomic forces between them and thus have a verylow boiling point.

Q25 (i) Name sweetening agent used in the preparation of sweets for a diabetic patient.

(ii) What are antibiotics? Give an example.


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(iii) Give two examples of macromolecules that are chosen as drug targets . Solution:

(i) Saccharin is the sweetening agent used in the preparation of sweets for a diabeticpatient.

(ii) Antibiotics: These drugs are used to treat infections because of their less toxicityfor humans and animals. These are produced wholly or partly by chemical processesand they inhibit the growth or even destroy microorganisms by intervening in theirmetabolic processes.Example: Salvarsan

(iii) Enzymes and receptors are the two examples of macromolecules that are chosenas drug targets.

Q26 The following data were obtained during the first-order thermal decomposition ofSO 2C l 2 at a constant volume:SO 2C l 2(g) SO 2(g) + C l 2(g)

Experiment Time/s Total pressure/atm

12

0100

0.40.7

Calculate the rate constant.(Given : log 4 = 0.6021, log 2 = 0.3010 ) Solution:

The thermal decomposition of SO 2Cl2 at a constant volume is represented by thefollowing equation:

After time t , total pressure,P t = P 0 + p

which on rearrangement gives:

p = P t - P 0

Therefore,= 2 P 0 P t For a first-order reaction,


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When t = 100 s,

Q27 (i) Deficiency of which vitamin causes rickets?(ii) Give an example each for fibrous protein and globular protein.(iii) Write the product formed on reaction of D-glucose with Br 2 water . Solution:

(i) Deficiency of vitamin D causes rickets.

(ii)Fibrous protein: Collagen, keratin (any one)Globular protein: Hemoglobin, carboxypeptidase (any one)

(iii)Br 2 water oxidises D- glucose into D- gluconic acid.

Q28 (a) Write the products of the following reactions:

(i)

(ii)

(iii)(b) Give simple chemical tests to distinguish between the following pairs of compounds:

(i) Benzaldehyde and Benzoic acid(ii) Propanal and Propanone

OR

(a) Account for the following:(i) CH 3CHO is more reactive than CH 3COCH 3 towards reaction with HCN.(ii) Carboxylic acid is a stronger acid than phenol.


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(b) Write the chemical equations to illustrate the following name reactions:(i) Wolff-Kishner reduction(ii) Aldol condensation(iii) Cannizzaro reaction

Solution:

(a)(i)

(ii)

(iii)

(b) (i) Benzaldehyde and Benzoic acidTest-1 Through sodium bicarbonate Benzaldehyde does not react with sodium bicarbonate. However, benzoic acid will produce briskeffervescence on reaction with sodium bicarbonate as shown in the given reaction:C6H5COOH + NaHCO 3 C6H5COONa + H 2O + CO 2

Test-2 Through Tollen's reagent Benzaldehyde reacts with ammoniacal solution of silver nitrate to form a silver mirror. C6H5CHO + 2[Ag(NH 3)2]+ + 3OH C6H5COO + 2Ag + 2H 2O + 4NH 3 However, no such reaction is given by benzoic acid.

(ii)Test-1 Iodoform Test Propanone gives positive iodoform test, as it contains CH 3CO group, whereas propanal does notgive iodoform test. The reaction is as follows:

CH 3CH 2CHO + 3I 2 + 4NaOH No reaction Heat Propanal

OR


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(a)(i) CH 3COCH 3 is sterically hindered than CH 3CHO due to the presence of alkyl group on both sidesof the carbonyl carbon, making them less reactive towards nucleophilic attack because bothmethyl groups have electron releasing tendency due to -I effect. These alkyl groups

make ketone less reactive by donating electron to carbonyl group.Therefore, acetaldehyde ismore reactive towards reaction with HCN.

(ii)Carboxylic acids are acidic due to resonance stabilisation of carboxylate anion and in phenols,acidic character is present due to resonance stabilisation of phenoxide anion. Carboxylic acidsare more acidic than phenols because the negative charge in carboxylate anion is more spread outcompared to the phenoxide ion, as there are two electronegative O-atoms in carboxylate anioncompared to one in phenoxide ion. In the resonance structures of carboxylate anion, the negativecharge is present on the O-atoms, while in resonance of phenoxide ion, negative charge is also

present on electropositive carbon atom, which leads to less stability of phenoxide ion

than carboxylate anion.(b)


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Q29 (a) Define the following terms :

(i) Limiting molar conductivity(ii) Fuel cell

(b) Resistance of a conductivity cell filled with 0.1 mol L 1 KCl solution is 100 . If theresistance of the same cell when filled with 0.02 mol L 1 KCl solution is 520 , calculatethe conductivity and molar conductivity of 0.02 mol L 1 KCl solution. The conductivity of0.1 mol L 1 KCl solution is 1.29 10 2 1 cm 1 .

OR

(a) State Faraday's first law of electrolysis. How much charge in terms of Faraday isrequired for the reduction of 1 mol of Cu 2+ to Cu.

(b) Calculate emf of the following cell at 298 K:Mg(s) | Mg 2+(0.1 M) || Cu 2+ (0.01) | Cu(s)[Given E ocell = +2.71 V, 1 F = 96500 C mol 1] Solution:

(a)(i) When concentration of an electrolyte approaches zero, then its molar conductivity isknown as limiting molar conductivity.(ii)Fuel cells are the galvanic cells in which the energy of combustion of the fuels likehydrogen, methanol. etc is directly converted into electrical energy.


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(b)Given that:Concentration of the KCl solution = 0.1 mol L 1 Resistance of cell filled with 0.1 mol L 1 KCl solution = 100 ohmCell constant = G * = conductivity resistance

1.29102

ohm1

cm1

100 ohm = 1.29 cm1

= 129 m1

Cell constant for a particular conductivity cell is a consant.

Conductivity of 0.02 mol L 1 KCl solution = =

Concentration = 0.02 mol L 1

= 1000 0.02 mol m 3 = 20 mol m 3 Now,

Molar conductivity =

Therefore, the molar conductivity of 0.02 mol L 1 KCl solution is .

OR

(a)Faraday's first law of electrolysis states that "the amount of chemical reaction whichoccurs at any electrode during electrolysis by a current is proportional to the quantity ofelectricity passed through the electrolytic solution or melt".

The reduction of one mol of Cu 2+ to Cu can be represented as:

Since, in this reaction there are two moles of electrons involved, so the amount ofcharge required is 2F.

(b) The cell reaction can be represented as:Mg(s) + Cu 2+(aq.) Mg 2+ (aq.) + Cu(s)

Q30 (a) How do you prepare:(i) K2MnO 4 from MnO 2?(ii) Na 2Cr 2O7 from Na 2CrO 4?


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(b) Account for the following:(i) Mn 2+ is more stable than Fe 2+ towards oxidation to +3 state.(ii) The enthalpy of atomisation is lowest for Zn in 3d series of the transition elements.(iii) Actinoid elements show wide range of oxidation states.

OR

(i) Name the elements of 3d transition series that show maximum number of oxidationstates. Why does this happen?(ii) Which transition metal of 3d series has positive E (M 2+ /M) value and why?(iii) Out of Cr 3+ and Mn 3+ , which is a stronger oxidising agent and why?(iv) Name a member of the lanthanoid series that is well-known to exhibit +2 oxidationstate.(v) Complete the following equation:

Solution:a)i) K2MnO 4 can be prepared from pyrolusite (MnO 2). The ore is fused with KOH in thepresence of either atmospheric oxygen or an oxidising agent, such as KNO 3 or KClO 4,to give K 2MnO 4.

2 MnO 2 + 4 KOH + O 2 2 K 2MnO 4 +2 H 2Ogreen

ii)Na

2Cr

2O

7can be prepared from Na

2CrO

4 in the following way:

For the preparation of sodium dichromate, the yellow solution of sodium chromate isacidified with sulphuric acid to give a solution from which orange sodium dichromate,Na 2Cr 2O7.2H 2O can be crystallised.Balanced equation for above reactions is as follows:

2 Na 2CrO 4 + 2 H + Na 2Cr 2O7 + 2 Na + + H 2OYellow Orange

b)i) Electronic configuration of Mn 2+ is [Ar] 18 3d 5 .

Electronic configuration of Fe 2+ is [Ar]18 3d 6 .It is known that half-filled and fully-filled orbitals are more stable. Therefore, Mn in +2

state has a stable d 5 configuration. Therefore, Mn 2+ shows resistance to oxidation toMn3+ . Also, Fe 2+ has 3 d 6configuration and by losing one electron, its configurationchanges to a more stable 3 d 5 configuration. Therefore, Fe 2+ gets oxidised to Fe 3+ easily.
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ii) The extent of metallic bonding an element undergoes, decides the enthalpy ofatomisation. The more extensive the metallic bonding of an element, the more will be itsenthalpy of atomisation. In all transition metals (except Zn, electronic configuration:3d 10 4 s 2), there are some unpaired electrons that account for their stronger metallicbonding. Due to the absence of these unpaired electrons, the inter-atomic electronic

bonding is the weakest in Zn and as a result, it has the least enthalpy of atomisation.iii) Actinides exhibit larger oxidation states because of very small energy gap between5f, 6d and 7s sub-shells . The energies are calculated on the basis of (n+l) rule. The(n+l) values of the three orbitals are:

5 f = 5 + 3 = 86 d = 6 + 2 = 87 s = 7 + 0 = 7

Since, all the values are almost same, therefore all orbitals can involved in bonding

resulting in larger oxidation number for actinoids.

OR

1)In 3d-series of transition metals, manganese has an atomic number of 25 that gives theelectronic configuration as [Ar] 3 d 54s 2 ,where we see that the maximum number ofunpaired electrons is found in manganese atom; so, it can show a maximum oxidationstate upto +7.

2)Copper is the transition metal of 3d series that exhibits positive E 0(M2+ /M). The valueof E 0(M2+ /M) for copper is (+0.34). This happens because the E 0(M2+ /M) value of ametal depends on the energy changes involved in the following:1. Sublimation energy: The energy required for converting one mole of an atom fromthe solid state to the gaseous state.

M(s) M(g) sH (Sublimation energy)2. Ionisation energy: The energy required to take out electrons from one mole of atomin the isolated gaseous state.

M(g) M2+ (g) iH (Ionisation energy)3. Hydration energy: The energy released when one mole of ions are hydrated.

M2+ (g) M2+ (aq) hyd H (Hydration energy)Since, copper has a high energy of atomisation and low hydration energy, the E 0(M2+ /M)value for copper is positive.

3) Out of Cr 3+ and Mn 3+ , Mn 3+ is a stronger oxidising agent because it has 4 electrons inits valence shell and when it gains one electron to form Mn 2+ , it results in the half-filled(d 5) configuration that has extra stability.


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4) Europium (Eu) is well-known to exhibit +2 oxidation state due to its half filled f orbitalin +2 oxidation state.

5) MnO 4-

+ 8H+

+ 5e-

Mn2+

+ 4 H 2O

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Class XII Chemistry SET 2 2014