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Maharashtra State Board Solutions Class 12 Chemistry

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Maharashtra Board Solutions Class 12 Chemistry

Chemistry is an important subject that Maharashtra State Board class 12 students need to study till they gather all the fundamental knowledge of the subject. Hence, to help students study Chemistry properly here we have provided the links to download Maharashtra State Board Solutions Class 12 Chemistry PDF.

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12th Chemistry 2nd Assignment for July 2021 (with Answers)

12th Chemistry 2nd Assignment for July 2021 (with Answers) :- This is the full answerkey for 12th Chemistry 2nd Assignment for July 2021 Tamilnadu Stateboard Students.Please download using the following links

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Class – 12 Subject:Chemistry (Unit 2and 3)

Unit 2- p block elements –i, part – a, i. choose the correct answer., 1.an aqueous solution of borax is, d) amphoteric.

Answer:- c) Basic

2.Which of the following metals has the largest abundance in the earth’s crust?

A) aluminium, c) magnesium.

Answer:- a) Aluminium

3.In diborane, the number of electrons that accounts for banana bonds is

Answer:- c) Four

4.The element that does not show catenation among the following p-block elements is

D) germanium.

Answer:- c) Lead

5.Carbon atoms in fullerene with formula C 60 have

A) sp 3 hybridised, b) sp hybridised, c) sp 2 hybridised, d) partially sp 2 and partially sp 3 hybridised.

Answer:- c) sp 2 hybridised

6.Oxidation state of carbon in its hydrides

B) – 4.

Answer:- a) +4

7.Which of the following is not sp 2 hybridised

A) graphite, b) grapheme, c) fullerene.

Answer:- d)Dry ice

8.The geometry at which carbon atom in diamond are bonded to each other is

A) tetrahedral, b) hexagonal, c) octahedral, d) none of these.

Answer:- a) Tetrahedral

9.Thermodynamically the most stable form of carbon is

B) graphite.

Answer:- b) Graphite

10 .The compounds that is used in nuclear reactors as protective shields and control rods is

A) metal borides, b) metal oxides, c) metal carbonates, d) metal carbide.

Answer:- a) Metal borides

II.Very Short Answer.

1.what is catenation describe briefly the catenation property of carbon..

Catenation: It is the phenomenon of an atom to form a strong covalent bond with the atoms of itself. Carbon shares the property of catenation to maximum extent because it is small in size and can form pn-pn multiple bonds to itself. The following conditions are necessary for catenation.

The valency of element is greater than or equal to two.
Element should have an ability to bond with itself.
The self bond must be as strong as its bond with other elements.
Kinetic inertness of catenated compound towards other molecules. Carbon possesses all the above properties and forms a wide range of compounds with itself.

2.Give the uses of silicones.

Silicones are used for low temperature lubrication and in vacuum pumps, high temperature oil baths etc.
They are used for making water proofing clothes.
They are used as insulting material in electrical motor and other appliances
They are mixed with paints and enamels to make them resistant towards high temperature, sunlight, dampness and chemicals.

3. AlCl 3 behaves like a lewis acid. Substantiate this statement.

In AlCl 3 , Al an electron-deficient it needs two electrons to complete octet so it acts as lewis acid. AlCl3 usually exists as a dimer to achieve octet by bridged Cl atom electron-deficient compounds are lewis acids.

4.Write a note on metallic nature of p-block elements.

The tendency of an element to form a cation by loosing electrons is known as electro¬positive or metallic character.
This character depends on the ionisation energy.
Generally on descending a group the ionisation energy decreases and hence the metallic character increases.

In p-block, the elements present in lower left part are metals while the elements in the upper right part are non metals. Elements of group 13 have metallic character except the first element boron which is a metalloid, having properties intermediate between the metal and nonmetals. The atomic radius of boron is very small and it has relatively high nuclear charge and these properties are responsible for its nonmetallic character.

In the subsequent groups the non-metallic character increases. In group 14 elements, carbon is a nonmetal while silicon and germanium are metalloids. In group 15, nitrogen and phosphorus are non metals and arsenic & antimony are metalloids. In group 16, oxygen, sulphur and selenium are non metals and tellurium is a metalloid. All the elements of group 17 and 18 are non metals.

5.CO is a reducing agent . Justify with an example.

Both thermodynamic and kinetic factors make carbon monoxide (CO) a better reducing agent. When CO is used to reduce a metal oxide, it gets oxidized to CO 2 Thermodynamically, CO 2 is much more stable than CO. For example, CO + Fe 2 O 3 → 2Fe + 3CO 2

III. Short Answer.

1.how will you identify borate radical.

When boric acid or borate salt is heated with ethyl alcohol in presence of concentrated H 2 SO 4 , an ester triethyl borate is formed. The Vapour of this ester burns with a green edged flame and this reaction is used to identify the presence of borate.

2.Give the uses of Borax.

Borax is used for the identification of coloured metal ions.
In the manufacture optical and borosilicate glass, enamels and glazes for pottery.
It is also used as a flux in metallurgy and also acts as a good preservative.

3.Give the structure of CO and CO 2 .

4.Write a note on Fisher Tropsch Synthesis.

5.Write a short note on anamolous properties of the first element of p-block.

In p-block elements the first member of each group differs from the other elements of the corresponding group. The following factors are responsible for this anomalous behavior.

Small size of the first member.
High ionisation enthalpy and high electronegativity.
Absence of d-orbitals in their valance shell.

The first member of group-13, boron is a metalloid while others are reactive metals. Moreover, boron shows a diagonal relationship with silicon of group -14. The oxides of boron and silicon are similar in their acidic nature.

IV.Write in detail.

1.describe the structure of diborane..

In diborane two BH 2 units are linked by two bridged hydrogens. Therefore, it has eight B-H bonds. However, diborane has only 12 valance electrons and are not sufficient to form normal covalent bonds. The four terminal B-H bonds are normal covalent bonds (two centre – two electron bond or 2c-2e bond). The remaining four electrons have to used for the bridged bonds, i.e. two three centred B-H-B bonds utilise two electrons each.

Hence, these bonds are three centre – two electron bonds. The bridging hydrogen atoms are in a plane as shown in the figure. In dibome, the boron is sp 3 hybridised. Three of the four sp 3 hybridised orbitals contains single electron and the fourth orbital is empty.

Two of the half filled hybridised orbitals of each boron overlap with the two hydrogens to form four terminal 2c-2e bonds, leaving one empty and one half filled hybridised orbitals on each boron. The Three centre – two electron bonds, B-H-B bond formation involves overlapping the half filled hybridised orbital of one boron, the empty hybridised orbital of the other boron and the half filled 1s orbital of hydrogen

2.Write a note on zeolites.

Zeolites are three dimensional crystalline solids containing aluminium, silicon and oxygen in their regular three dimensional framework.
They are hydrated sodium alumino silicates with general formula, Na 2 O. (Al 2 O 3 ). x(SiO 2 )y(H 2 O) (x = 2 to 10; y = 2 to 6)
Zeolites have porous structure in which the monovalent sodium ions and water molecules are loosely held.
The Si and Al atoms are tetrahederally coordinated with each other through shared oxygen atoms.
Zeolites structure looks like a honeycomb consisting of a network of interconnected tunnels and cages.
Zeolite crystal to act as a molecular sieve. They helps to remove permanent hardness of water.

Unit 3- P Block elements –II

1.an element belongs to group 15 and 3 rd period of the periodic table, its electronic configuration would be, a) 1s 2 2s 2 2p 4, b) 1s 2 2s 2 2p 3, c) 1s 2 2s 2 2p 6 3s 2 3p 2, d) 1s 2 2s 2 2p 6 3s 2 3p 3.

Answer:- d) 1s 2 2s 2 2p 6 3s 2 3p 3

2.P 4 O 6 reacts with cold water to give

A) h 3 po 3, b) h 4 p 2 o 7, d) h 3 po 4.

Answer:- a) H 3 PO 3

3.The basicity of ortho phosphorus acid ( H 4 PO 3 ) is

Answer:- b) 2

4.Amon the ollo in hich is he trongest o idizing agen ?

Answer:- b) F 2

5.Most easily liquefiable gas is

Answer:- c) He

6.Which is the strongest acid among the hydrogen halide

Answer:- a) HI

7.When copper is heated with Conc HNO 3 it produces

A) cu(no 3 ) 2 , no and no 2, b) cu(no 3 )2 and n 2 o, c) cu(no 3 ) 2 and no 2, d) cu(no 3 ) 2 and no.

Answer:- c) Cu(NO 3 ) 2 and NO 2

8.The oxidation state of oxygen in OF 2 is………

Answer:- d) 0

9.The radioactive inert gas is ……….

Answer:- d) Rn

10.An example for interhalogen compound is……

D) all the above.

Answer:- d) All the above

1.What is inert pair effect?

In p-block elements, as we go down the group, two electrons present in the valence s-orbital become inert and are not available for bonding (only p-orbital involves chemical bonding). This is called inert pair effect.

2.What are inter halogen compounds? Give examples.

Each halogen combines with other halogens to form a series of compounds called interhalogen compounds. For example, Fluorine reacts readily with oxygen and forms difluorine oxide (F 2 O) and difluorine dioxide (F 2 O 2 ).

3.Mention any two uses of chlorine.

Purification of drinking water
Bleaching of cotton textiles, paper and rayon
It is used in extraction of gold and platinum

4.Give a test for sulphate ion.

5.write down the hybridisation and shape for the compounds of xenon a) xef 4 b) xeo 3, 1.give the reason to support the sulphuric acid (h 2 so 4 ) is a dehydrating agent..

2.Give the uses of helium.

Helium and oxygen mixture is used by divers in place of air oxygen mixture. This prevents the painful dangerous condition called bends.
Helium is used to provide inert atmosphere in electric arc welding of metals
Helium has lowest boiling point hence used in cryogenics (low temperature science).
It is much less denser than air and hence used for filling air balloons

3.Write the structural formula and molecular formula for the following compounds.

A) nitric acid b) phosphoric acid.

4.Why HF is a weak acid, where as the binary acids of all others halogens are strong acids ?

The hydrogen halides are extremely soluble in water due to the ionisation. X + H 2 O → H 3 O + + X – ( X = F, Cl, Br or I )

Solutions of hydrogen halides are therefore acidic and known as hydrohalic acids. Hydrochloric, hydrobromic and hydroiodic acids are almost completely ionised and are therefore strong acids but HF is a weak acid. For HF,

HF + H 2 O ⇌ H 3 O + + F –
HF + F – → HF–2

At high concentration, the equilibrium involves the removal of fluoride ions is important. Since it affects the dissociation of hydrogen fluoride, therefore it is a weak acid.

5.Write the reason for the anamolous behaviour of nitrogen.

1. Due to its small size, high electro negativity, high ionisation enthalpy and absence of d-orbitals.

2. N, has a unique ability to form pπ – pπ multiple bond whereas the heavier members of this group (15) do not form pπ – pπ bond, because their atomic orbitals are so large and diffused that they cannot have effective overlapping.

3. Nitrogen exists a diatomic molecule with triple bond between the two atoms whereas other elements form single bond in the elemental state.

4. N cannot form dπ – pπ bond due to the absence of d – orbitals whereas other elements can.

1.Explain Deacons process.

2.Complete the following reactions. i. Na 2 SiO 3 + 6HF →

Na 2 SiO 3 + 6HF → Na 2 SiF 6 + 3H 2 0

ii. XeO 2 F2 + SiO 2 →

XeO 2 F2 + SiO 2 → 2XeO 3 + SiF 4

iii. XeO 2 F2 + SiO 2 →

XeO 2 F2 + SiO 2 → 3NaH 2 P0 2 + PH 3 ↑

iv. NaCl + MnO 2 + H 2 SO 4 →

NaCl + MnO 2 + H 2 SO 4 → cl 2 + Mncl 2 + 4NaHSO 4 +2H 2 O

v. KClO 3 →

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NCERT Solutions for Class 12 Chemistry in PDF Form

NCERT Solutions for class 12 Chemistry

Download NCERT Solutions for class 12 Chemistry all chapters given in separate PDF. NCERT exercises and intex questions solution are given separately.

12 Chemistry Solutions – All Chapters

Chapter 1: the solid state.

Point Defects – The defects caused by missing or misplaced atoms or ions in the crystal. Schottky, Frenkel and interstitial defects are stoichiometric defects.

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Chapter 2: Solutions

Raoult’s Law – The vapour pressure of a solution is equal to the product of mole fraction of the solvent and its vapour pressure in pure state.

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Chapter 3: Electrochemistry

A galvanic cell is represented by writing the anode (where oxidation occurs) on the left hand side and cathode (where reduction occurs) on the right hand side.

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Chapter 4: Chemical Kinetics

Chemical Kinetics – The branch of chemistry which deals with the study of reaction rates and their mechanism. The rate of change of concentration of any of the reactant or product with time at any particular moment of time is known as Rate of reaction.

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Chapter 5: Surface Chemistry

A colloidal solution in which there is great affinity between the disperse phase and the dispersion medium is called Lyophilic Solutions. A colloidal solution in which there is a little affinity between the disperse phase and the disperse medium is called Lyophobic solutions.

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Chapter 6: General Principles and Processes of Isolation of Elements

Refining is essential for getting pure metal which is depends upon the nature of metal and gangue. Hardness of steel is depend upon the percent of carbon.

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Chapter 7: The p – Block Elements

Each noble gas element, with its saturated shell electronic configuration, marks the completion of the period in which it is placed. They are chemically inert, but can be made to combine by coordination or by excitation of electrons (in case of large atoms).

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Chapter 8: The d – Block Elements

The characteristic properties of transition metals result from the fact that in these metals the differentiation electron enters into penultimate shell d-orbital. Thus the (n-1)d configuration of the metal and its ion plays important part in giving the observation set of properties to the metal or its ion.

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Chapter 9: Coordination Compounds

Formation of coordination compounds results from the Levis acid property of the metal ions with high e/r ratio. Shapes of these compounds decided by the electronic configuration of metal ion and hence the type of hybridisation which the metal ion undergoes during coordination of the ligands.

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Chapter 10: Haloalkanes and Haloarenes

Tri-idomethane (Iodoform) is used as an antiseptic and this nature is due to iodine that is liberates. However because of its very unpleasant smell, it has now been replaced by better antiseptic. It is used in the manufacturing of pharmaceuticals.

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Chapter 11: Alcohols, Phenols and Ethers

Lucas Test – A solution of ZnCl2 prepared in concentrated HCl is called Lucas regent. If this reagent is added to a specimen of alcohol then if white turbidity appears immediately it is tertiary alcohol. This turbidity is due to the formation of alkyl chloride in general.

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Chapter 12: Aldehydes, Ketones and Corboxylic Acids

Acetone does not undergo polymerisation reactions rather two and three molecules of acetone combine together in the presence of dy HCl gas to form mesityl oxide & phorone.

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Chapter 13: Amines

In common system, the cyanides are named as alkyl or aryl cyanide whereas in IUPAC system they are named as alkane or arene nitriles. In common system, they are also named on the basis of the acid produced by their hydrolysis replacing ‘ic acid’ by ‘onitriles’.

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Chapter 14: Biomolecules

Niacin is nicotinic acid and niacin amide nicotinamide. Both contain a pyridine ring system. Niacin is obtain form fish, yeast, beans and peanuts. Milk and fish contain only a low account of this vitamins.

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Chapter 15: Polymers

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Chapter 16: Chemistry in Everyday Life

Artificial Sweeteners are the chemical compounds which give sweetening effect to food and enhance its odour and flavour. A very popular sweetening agent is saccharin.

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3 thoughts on “Class 12 Chemistry question paper 2021 Maharashtra State Board”

Sir can you please provide same paper for maths– 2021

R/sir, Require following questions papers of Maharashtra state board examination of 2021 ,2022 Sub : CHEMISTRY, BIOLOGY AND PHYSICS

yes will upload

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12th Chemistry Public Exam Question Paper 2021

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12th Chemistry Revision Exam Question - Answers

12th chemistry revision exam question - answers, 12 th standard chemistry.

TIME: 3HRS MARKS: 70

CHOOSE THE CORRECT ANSWER: 15 × 1 = 15

1. 𝑆𝑐(𝑍 = 21) is a transition element but 𝑍𝑛(𝑍 = 30) is not because,

a) Both 𝑆𝑐 3+ and 𝑍𝑛 2+ ions are colourless and form white compounds.

b) In case of 𝑆𝑐 , 3𝑑 - orbital are partially filled but in 𝑍𝑛 these are completely filled.

c) Last electron as assumed to be added to 4𝑠 level in case of 𝑍𝑛

d) Both 𝑆𝑐 and 𝑍𝑛 do not exhibit variable oxidation states

Answer: b) In case of 𝑆𝑐 , 3𝑑 - orbital are partially filled but in 𝑍𝑛 these are completely filled.

2. The stability of +1 oxidation state increases in the sequence,

a) 𝐴𝑙 < 𝐺𝑎 < 𝐼𝑛 < 𝑇𝑙 b) 𝑇𝑙 < 𝐼𝑛 < 𝐺𝑎 < 𝐴𝑙

c) 𝐼𝑛 < 𝑇𝑙 < 𝐺𝑎 < 𝐴𝑙 d) 𝐺𝑎 < 𝐼𝑛 < 𝐴𝑙 < 𝑇𝑙

A nswer: a) 𝐴𝑙 < 𝐺𝑎 < 𝐼𝑛 < 𝑇𝑙

3. Solid 𝐶𝑜 2 is an example of,

a) Covalent solid b) Metallic solid c) Molecular solid d) Ionic solid

4. Match items in column-I with the items of column-II and assign the correct code. Column-I Column-II

A) Cyanide process (i) Ultrapure Ge

B) Froth floatation process (ii) Extraction of Al

C) Electrolytic reduction (iii) Dressing of ZnS

D) Zone refining (iv) Extraction of Au code A B C D

a) i ii iii iv

b) iii iv ii i

c) iv iii ii i

d) ii iii i iv

5. 𝛼 − 𝐷(+) glucose and 𝛽 − 𝐷(+) glucose are,

a) Epimers b ) Anomers c) Enantiomers d) Conformational isomers

6. The product formed, when glycol treated with conc 𝐻 2 𝑆𝑜 4 is,

a) 1,4 –dioxane b) 1,2 –epoxy ethane c) Ethene d) Ethanal

7. Which of these is not likely to act as Lewis base?

a) 𝐵𝐹 3 b) 𝑃𝐹 3 c) 𝐶𝑂 d) 𝐹 −

8. The 𝑃 𝐻 of 10 −5 𝑀 𝐾𝑂𝐻 solution will be,

a) 9 b) 5 c) 19 d) 14

9. Which one of the following is the strongest acid?

a) 2-nitrophenol b) 4-chlorophenol c) 4-nitrophenol d) 3-nitrophenol

10. Williamson synthesis of preparing dimethyl ether is a/an,

a) 𝑆 𝑁 1 reaction b) 𝑆 𝑁 2 reaction

c) electrophilic addition d) electrophilic substitution

11. Acid hydrolysis of inulin produces,

a) Fructose b) Glucose c) Maltose d) Sucrose

12. In Froth floatation process, sodium cyanide act as,

a) Frothing agent b) Collector c) Depressing agent d) Flux

13. Which of the following metals has the largest abundance in the earth’s crust?

a) Aluminium b) Calcium c) Magnesium d) Sodium

14. Assertion: 𝐶𝑒 4+ is used as an oxidizing agent in volumetric

analysis Reason: 𝐶𝑒 4+ has the tendency of attaining +3 oxidation state

a) Both assertion and reason are true and reason is the correct explanation of assertion

b) Both assertion and reason are true but reason is not the correct explanation of assertion

c) Assertion is true but reason is false

d) Both assertion and reason are false

15. The number of carbon atoms per unit cell of diamond is,

a) 8 b) 5 c) 1 d) 4

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