Bihar Board 12th Chemistry Model Question Paper 5 in English Medium

Bihar Board 12th Chemistry Model Papers

Bihar Board 12th Chemistry Model Question Paper 5 in English Medium

Time : 3 Hours 15 Min
Full Marks: 70

Instructions for the candidates

1. Candidates are required to give answers in their own words as far as practicable.
2. Figures in the right hand margin indicates full marks.
3. While answering the questions, candidate should adhere to the words limit as far as practicable.
4. 15 Minutes of extra time has been allotted for the candidates to read the questions carefully.
5. This question paper is divided into two sections : Section -A and Section-B
6. In Section A, there are 35 objective type questions which are compulsory, each carrying 1 mark. Darken the circle with blue/black ball pen against the correct option on OMR Sheet provided to you. Do not use Whitener/Liquid/Blade/ Nail on OMR Sheet otherwise your result will be treated as invalid.
7. In section-B, there are 18 short answer type questions (each carrying 2 marks), out of which only 10 (ten) questions are to be answered Apart from this there are 06 Long Answer type questions (each carrying 5 marks), out of which 3 questions are to be answered.
8. Use of any electronic device is prohibited.

Objective Type Questions

In the following questions no. from 1 to 35, there is only one correct answer against each question. For each question, mark (darken) the correct answer on the OMR Sheet provided to you. (1 x 35 = 35)

Question 1.
The coordination number of metal crystallising in a hexagonal close packing is ………….
(a) 12
(b) 4
(c) 8
(d) 6
Answer:
(a) 12

Question 2.
A crystalline structure has radius ratio (r, / r) in the range of 0.225 – 0.414. The coordination number and arrangement of anions around the cations are
(a) 3, plane triangular
(b) 6, octahedral
(c) 4, tetrahedral
(d) 8, cubic
Answer:
(c) 4, tetrahedral

Question 3.
A crystal lattice with alternative +ve and -ve ions has radius ratio 0.524. The coordination number of lattice is
(a) 4
(b) 6
(c) 8
(d) 12
Answer:
(b) 6

Question 4.
A solid AB has a rock salt structure. If radius of cation A⁺ is 120 pm, what is the minimum value of radius of B^–anion ?
(a) 120 pm
(b) 240 pm
(c) 290 pm
(d) 360 pm
Answer:
(c) 290 pm

Question 5.
A crystal is formed by two elements X and Y in cubic structure. X atoms are at the comers of a cube while Y atoms are at the face centre. The formula of the compound will be ………..
(a) XY
(b) XY₂
(c) X₂Y₃
(d) XY₃
Answer:
(d) XY₃

Question 6.
What will be the mole fraction of ethanol in a sample of spirit containing 85% ethanol by mass ?
(a) 0.69
(b) 0.82
(c) 85
(d) 0.60
Answer:
(a) 0.69

Question 7.
What is the molarity of a solution containing 10 g of NaOH in 500 mL of solution ?
(a) 0.25 mol L^-1
(b) 0.75 mol L^-1
(c) 5 mol L^-1
(d) 1.25 mol L^-1
Answer:
(c) 5 mol L^-1

Question 8.
In a cell reaction,

If the concentration of Cu²⁺ ions is doubled then Ep_ell will be
(a) doubled
(b) halved
(c) increased by four times
(d) unchanged
Answer:
(d) unchanged

Question 9.
A standard hydrogen electrode thas a zero potential because
(a) hydrogen can be most easily oxidised
(b) hydrogen has only one electron
(c) the electrode potential is assumed to be zero
(d) hydrogen is the lightest element.
Answer:
(c) the electrode potential is assumed to be zero

Question 10.
In a reaction 2X → Y, the concentration of X decreases from 3.0 moles/litre to 2.0 moles/litre in 5 minutes. The rate of reaction is …………
(a) 0.1 mol L^-1 min ¹
(b) 5 mol L^-1 min^-1
(c) 1 mol L^_1 min^-1
(d) 0.5 mol L^-1 min^-1
Answer:
(a) 0.1 mol L^-1 min ¹

Question 11.
For the reaction, 2N₂O₅ → 4NO₂+O rate and rate constant are 1,02 x 10^-4 mol Lr¹ s^-1 and 3.4 x 10^-5 s^-1  The
concentration of N₉O₅ in mol L^-1 will be ………….
(a) 3.4 x 10⁴
(b) 3.0
(c) 2
(d) 3.2 x 10~⁵
Answer:
(b) 3.0

Question 12.
The incorrect statement about physical adsorption is
(a) it lacks specificity
(b) it is generally reversible
(c) porous surfaces are good adsorbent
(d) heat of adsorption is quite high
Answer:
(d) heat of adsorption is quite high

Question 13.
Which is correct in case of vander Waals adsorptions ?
(a) High temperature, low pressure
(b) High temperature, high pressure
(c) Low temperature, low pressure
(d) Low temperature, high pressure
Answer:
(d) Low temperature, high pressure

Question 14.
Which of the following is not an ore of magnesium ?
(a) Carnal lite
(b) Magnesite
(c) Dolomite
(d) Gypsum
Answer:
(d) Gypsum

Question 15.
Which of the following is not an oxide ore ?
(a) Corundum
(b) Zincite
(c) Calamine
(d) Chromite
Answer:
(c) Calamine

Question 16.
Nitrogen is relatively inactive element because
(a) its atoms has a stable electronic configuration
(b) it has low atomic radius
(c) its electronegativity is fairly high
(d) dissociation energy of its molecule is fairly high
Answer:
(d) dissociation energy of its molecule is fairly high

Question 17.
Nitrogen combines with metals to form
(a) nitrites
(b) nitrates
(c) nitrosyl chloride
(d) nitrides
Answer:
(d) nitrides

Question 18.
Reactivity of transition elements decreases almost regularly from Sc to Cu because of
(a) lanthanoid contraction
(b) regular increase in ionisation enthalpy
(c) regular decrease in ionisation enthalpy
(d) increase in number of oxidation states.
Answer:
(b) regular increase in ionisation enthalpy

Question 19.
The first ionisation energies of the elements of the first transition series (Ti to Cu)
(a) increases as the atomic number increases
(b) decreases as the atomic number increases
(c) do not show any change as the addition of electrons takes place in the inner (n – 1) d-orbitals
(d) increases from Ti to Mn and then decreases from Mn to Cu
Answer:
(a) increases as the atomic number increases

Question 20.
The ligand N(CH₂ CH₂ NH₂)₃ is
(a) bidentate
(b) tridentate
(c) tetradentate
(d) pentadentate
Answer:
(c) tetradentate

Question 21.
Which of the following is a tridentate ligand ?
(a) EDTA^4-
(b) (COO)₂^2-
(c) dien
(d)  NO^–₂
Answer:
(c) dien

Question 22.
Which of the following is not correctly matched with its IUPAC name ?         .
(a) CHF₂CBrClf : l-Bromo-l-chloro-1,2, 2 – trifluoro- ethane
(b) (CC1₃)₃ CC1: 2-(Trichloromethyl)-l, 1,1,2,3,3,3 – heptachloropropane
(c) CH₃C(p-ClC₆H₄)₂CH(Br)CH₃ : 2-Bromo-3,3-bis (4- chlorophenyl) butane
(d) o-BrC₆H₄CH(CH₃)CH₂CH₃ : 2-Bromo-l-methyl- propropyl benzene
Answer:
(d) o-BrC₆H₄CH(CH₃)CH₂CH₃ : 2-Bromo-l-methyl- propropyl benzene

Question 23.
Halogen acids react with alchols to form alkyl halides. The reaction follows a nucleophilic substitution mechanism. What will be the major product of the folowing reaction ?

Answer:
(c)

Question 24.
The best method to prepare 3-methylbutan-2-ol from 3- methylbut-l-ene is
(a) addition of water in presence of dil. H₂SO₄ .
(b) addition of HCl followed by reaction with dil. NaOH
(c) hydroboration-oxidation reaction
(d) Reimer-Tiemann reaction
Answer:
(a) addition of water in presence of dil. H₂SO₄ .

Question 25.
An alkene CH₃CH = CH₂ is treated with B₂H₆ in presence of H₂O₂. The final product formed is
(a) CH₃CH₂CHO
(b) CH₃CH(OH)CH₃
(c) CH₃CH₂CH₂OH
(d) (CH₃CH₂CH₂)₃B
Answer:
(c) CH₃CH₂CH₂OH

Question 26.
Benzoyl chloride on reduction with H₂/Pd – BaSO₄ produces
(a) benzoic acid
(b) benzyl alcohol
(c) benzoyl sulphate
(d) benzaldehyde
Answer:
(d) benzaldehyde

Question 27.
The oxidation of toluene to benzaldehyde by chromyl chloride is called
(a) Etard reaction
(b) Riemer-Tiemann reaction
(c) Wurtz reaction
(d) Cannizzaro’s reaction
Answer:
(a) Etard reaction

Question 28.
When excess of ethyl iodide is treated with ammonia, the product is
(a) ethylamine
(b) diethylamine
(c) triethylamine
(d) tetraethylammonium iodide
Answer:
(d) tetraethylammonium iodide

Question 29.
Secondary amines can be prepared by
(a) reduction of nitro compounds
(b) reduction of amides
(c) reduction of isonitriles
(d) reduction of nitriles
Answer:
(c) reduction of isonitriles

Question 30.

(a) 2-Iodoheptane
(b) Heptane-2-ol
(c) 2-lodohexane
(d) Heptanoic acid
Answer:
(d) Heptanoic acid

Question 31.
On oxidation with a mild oxidising agent like Br₂/H₂O, the glucose is oxidised to …………..
(a) saccharic acid
(b) glucaric acid
(c) gluconic acid
(d) valeric acid
Answer:
(c) gluconic acid

Question 32.
Which factor imparts the crystalline nature to a polymer like nylon ?
(a) Strong intermolecular farces the hydrogen bonding between chains.
(b) vander Walls forces between the polymeric chains.
(c) Close packing of the chains due to ionic bonding between the chains.
(d) Three-dimensional network of chains.
Answer:
(a) Strong intermolecular farces the hydrogen bonding between chains.

Question 33.
Arrange the following polymes in an increasing order of intermolecular forces; fibre, plastic, elastomer.
(a) Elastomer < Fibre < Plastic
(b) Elastomer < Plastic < Fibre
(c) Plastic < Elastomer < Fibre
(d) Fibre < Elastomer < Plastic
Answer:
(b) Elastomer < Plastic < Fibre

Question 34.
Antihistamines are not helpful
(a) in curing nasal allegries
(b) in treating rashes caused by itching
(c) in bringing down acute fever
(d) in vasodilation.
Answer:
(c) in bringing down acute fever

Question 35.
The drugs which are given to the patients suffering from anxiety and mental tension are known as
(a) tranquilizers
(b) analgesics
(c) antimicrobials
(d) antibiotics
Answer:
(a) tranquilizers

Short Answer Type Questions

In this Section, there are 18 Short answer type questions (each carrying 2 marks), out of which answer any 10 questions.  2 x 10 = 20

Question 1.
Which methods are usually employed for purifying the following metals ?
(i) Nickel
(ii) Iron. Mention the principles.
Answer:
(i) Silesion process
(ii) Pudding process, cementation process, Bessemer’s rocess, siemens martin open heart process, Linz Donawitz process.

Question 2.
Cl, Br, I or halogens are members of which group in the periodic table ?
Answer:
Group-(ViiA)

Question 3.
Explain, why the valency of inert gases is zero.
Answer:
It is because since the valence shell is saturated with 8 electrons. The hoble gases have so a valence equal to zero as in principle can not make chemical bands with other – atoms.

Question 4.
Explain standard electrode potential.
Answer:
Standard electrode potential : It is the measure of individual potential of a reversible electorde at standard state which is with solutes at an effective concentration of 1 mol dm^-3 and gases at a pressure of 1 atm.

Question 5.
What is carbocation ? Explain.
Answer:
Carbocation : A carbocation is an ion with positively-charged carbon atom. Among the simplest example methenium CH⁺₃ ethanium C₂H⁺₂ . Some carbocations may have two or more positive charges, on the same carbon atom or on different atom such as ethylene dictation (C₂H₄⁺). It is classified in two categories according to the valence of the charged carbon.

Question 6.
What do you mean by acid rain ? Explain.
Answer:
Acid rain : It is a rain or any other form of precipitation that usually acidic, meaning that it possesses elevated levels of hydrogen ions (low pH). It can have harmful effects on plants aquatic animals and infrastructure. Acid rain is caused by emissions of sulphur dioxide and nitrogen oxide, which react with water molecules in the atmosphere to produce acids. In chemical acid rain can cause point to peel, corrosion of steel structures such as bridges and erosion of stone statues.

Question 7.
Explain two important uses of formalin.
Answer:
Two use of formaline are following:

• It is used as preservative
• It is also used as a disinfectation and an antibacterial.

Question 8.
How will you convert Aniline into Benzoic acid ?
Answer:

Question 9.
Any transition series contains only ten elements. Why?
Answer:
There are only 10 members because in period left to right atomic number increases and hence after a certain number when the D subshell is fully filled the electron has to go in S and P sub shell, hence it ends and P blocks

Question 10.
Explain mole fraction.
Answer:
Mole fraction : It is defined as the amount of a constituent n divided by the total amount of all constituents
in a mixture n_total It is also called amount fraction i.e  \(\left(x_{i}=\frac{h_{i}}{n_{\text {total }}}\right)\)

Question 11.
What is salt bridge ? What are its uses ?
Answer:
Salt bridge : In electro chemistry, it is a laboratory device used to connect the oxidation and reduction half-cells of a galvanic cell (voltaic cell, a type of electrochemical cell. Salt bridge usually comes into two types : glass tube and filter paper.
Uses : It is used in voltaic cell. Purpose of salt bridge is only to move electrons from electrotype solution to the other.

Question 12.
(a) Write the IUPAC name of the following compound :
Answer:

(b) Name the product formed when ethyne is oxidised with cold alkaline solution of KMnO₄.
Answer:
(a) 2, 3 -Dimethyl butane
(b) Ethane – 1, 2 – diol
(CH₂OH – CH₂OH)

Question 13.
If 20 g of calcium carbonate is added to a solution containing 20 g of HCl. What substances will be present when the reaction is over and what quantity of each of them will be present there?
Answer:

Thus, HCl will be present when the reaction is over. Rest amount of HCl = 20 – 14.8 = 0.56 g.

Question 14.
Write chemical reaction to obtain the following :
(i) Methane to chloroform
(ii) Chloroform to ethyne.
Answer:

Question 15.
Give reasons for the following:
(i) Ethyne is more acidic than ethane,
(ii) Lower members of aldehyde are more soluble in water.
Answer:
(i) Ethyne contains SP hybridized carbons, while ethane contain SP³ hybridized carbons. Th SP hybrid orbitals have greater S character than SP³ which allows negative charge to be held closer to the nucleus and increasing the acidic character that is why ethyne is more acidic than ethane.

(ii) Because, lower member of aldehyde have capability of forming H-bond with water molecule.

Question 16.
Transition elements form coloured compound. Explain.
Answer:
The value of electrods potential depends upon the heat of sublimation and ionisation energy AH – AH_sib + IE + AH_hyd. Due to.the influence of ligand d-or-bitals of transition metals divided into two unequal energy containing sets, unpaired electrons obsorb sunlight and jumps from one orbit into another orbit and oen colour is reflected.

Question 17.
Why is the bond angle of PH₄ more than PH ₃ ?
Answer:
Because CH₄ has no lone pair of electrons, while PH₃ has so bond angle of CH₄ (1095°) is more than PH₃.

Question 18.
Give I.U.P.A.C. names of the following :

Answer:
(a) Butane-1,4 dioic acid
(b) 2-hydroxy propan-1- oic acid.

Long Answer Type Questions

There are 06 long answer type questions (each carrying 05 marks), out of which answer any there questions.
(5 x 3 = 15)

Question 19.
How does nitric acid react with the following ? Give equation.
(i) Copper
(ii) Iron.
Answer:
(i) Reaction of HNO₃ with copper : (A) With not and concentrated HNO₃, copper reacts to give nitrogen peroxide, copper nitrate and water
Cu + 4HNO₃ → Cu(NO₃ ) + 2NO₂+H₂O.

(ii) With 50% concentrated nitric acid copper reacts to give copper nitrate, nitric oxide and water
3Cu+ HNO₃ → Cu(NO₃)₂ +2NO + 4H₂O

(iii) With 20 – 25% dilute. HNO₃ copper reacts to give copper nitrate, nitrous oxides and water.
4Cu + 10HNO₃ → 4Cu(NO₃)₂ +N₂O + 5H₂O

(iv) On passing vapour of nitric acid on heated copper nitrogen gas and water vapour are obtained
5Cu + 2HNO₃ → 4CuO + N₂+H₂O

(v) Reaction of HNO₃ with Iron : (A) With cold and dilute HNO₃ Iron reacts to give ferrous nitrate and ammonium nitrate
4Fe +10HNO₃ → NH₄NO₃ +4Fe(NO₃)₂ + 3H₂O

(B) Hot & dil nitric acid reacts with iron and gives ferric Nitrate & No gas.
2Fe + SHNO₃ → 2Fe(NO₃)₂ +2NO↑+H₂O
Fe + 6HNO₃ → Fe(NO₃)₃ +3NO₂ +3H₂O

(C) With hot and cone. HNO₃ Iron reacts to give ferric nitrate and nitrogen dioxide gas.
Fe + 6HNO₃ → Fe(NO₃)₃ + 3NO₂ + 3H₂O

Question 20.
What are the main sources of iodine ? How is iodine extracted from sea weeds ?
Answer:
Natural sources of Iodine : Due to its reactivity Iodine is not found in nature in free state its main sources are

• sea weeds
• chile salt puter
• Natural brine.

Extraction of Iodine from sea weeds : Sea weeds lamineria contains iodine, Sea weed is well dried and burnt in deep pite carefully so that iodine donot get destroyed. The obtained ash is called kelp which contains 0.4 to 1.3% Iodine kelp is dissolved in water and solution is partially crystallised when less. Soluble KI and Nal remain in the mother liquor,

cone H₂SO₄ is added when basic sulphides despositeat the bottom, when is filtered and removed Now the filtrate is mixed with MnO₂ and H₂SO₄ and heated iron vessel. Iodine vapourises due to the reaction and is collected Aludel. Iodine is new collected as solid after condensation.

2Nal + MnO₂ +3H₂SO₄ -4 2NaHSO₄ + MnSO₄ + 2H₂O+I₂ ↓

Iodine obtained by this method contains CI₂ and Br₂ as impurifies. It is treted with KI to obtained pure Iodine

Question 21.
(a) Write only the principle for the manufacturing of sulphuric acid by contact process.
(b) Complete the following reactions-
(i) KBr + Cl₂ →…………….. +……….
(ii) I₂ + H₂O + CI₂→………….. +………..
(iii) NaOH + Cl₂ → …………..+…………. +………… (Cold and dilute)
Answer:
(a) Principle-(i) The process involves the oxidation of sulphur dioxide by air in the presence of catalyst V₂O₅.

(ii) Sulphur trioxide is dissolved in 98% sulphuric acid, forms oleum.

(iii) Sulphuric acid of any desire concentration is prepared from oleum with water.
H₂S₂O₇ + H₂O→2H₂SO₄
The oxidation of SO₂ is reversible process, contraction in volume and exochermic.
Hence applying i.e., Chatelier’s principle to obtained greater yield of SO₃.

• Reaction is carried out high pressure.
• At low temperature production of SO₃ should increase. But at lower temperature SO₂ does not oxidise. Hence at optimum temperature 450°C catalyst V₂O₅ is applied.

(b)

• KCI, Br₂
• HIOj, HCl
• NaCl, NaClO₃

Question 22.
(a) State Werner’s coordination theory.
(b) What are ligands ? Classify them with examples.
Answer:
(a)

• In co-ordination compounds metals show two types of linkages (valencies) primary and secondary.
• The primary valencies are normally ionisable and are satisfied by negative ions.
• The secondar valencies are non-ionisable. These are satisfied by neutral molecules or negative ions. The secondary valence is equal to the co-ordination number and is fixed for a neutral.
• The ions/groups bound by the secondary linkages to the metal have characteristic spatial arrangements corresponding to different numbers.
• In the modem formulations, such spatial arrangements are called co-ordination polyhedra. The species with the square brackets are co-ordination entities or complexes and the ions out side the square bracket are called counter ions.
• He further postulated that octahedral, tetrahedral and square planar geometrical shapes are more common in co­ordination compounds of transition metals. Thus [Co(NH₃)₆]^{3 +}, [CoCl(NH₃)₅]²⁺and [CoCl₂(NH₃)₄]⁺ are octahderal entities, while [Ni(CO)₄] and [PtCl₄]^2- are tetrahedral and square planar respectively.

(b) The atoms or molecules or ions which donate pair of electrons to the central metal atom and thus forms co-ordinate bond with the central metal atom. Common ligands are – NH3, CO, CN-, Cl”, Br, I”, H₂O etc. Ligands which can attach themselves (called ligating) through two different atoms of the same molecule are called ambident ligands or group, e.g., NO₂ , SCN^–. They introduce linkage isomerism in the complexes.

Question 23.
Explain why
(i) Aniline dissolves in HCl.
(ii) Amines are stronger base than Ammonia
(iii) Ethylamine is more basic than Aniline
(iv) Cyclohexamine is more basic than Aniline
(v) Phenol is acidic in nature.
Answer:
(i) Aniline is somewhat basic. It is not as basic as amonia, since protonation destroys the interaction between the ring aind the lone pair on the nitrogen atom but even still it is sufficiently basic (PKb ~ 9.4) to dissolve in HCl.

(ii) The inductive effect of attached carbon chain to an amine will raise the energy of the lone pair on the nitrogen. On amines carbon will withdraw less electron density from an atom of interest that hydrogen (in amonia). That explain why alkyl amine are stronger bases than amonia.

(iii) Therefore, less electron is available in the nitrogen atom in C₆H₅NH₂. It can easily dissociate to denote a proton, H⁺ by breaking -N-H bond in the amine group that attach to phenyl group. Hence aniline is a stronger acid than ethylamine.

(iv) In cyclohexamine, the electron donating group is attached to SP³ hybridized cyclohexane ring while in case of aniline it is attached with SP² hybridized carbon containing benzene ring. Due to more S-character in benzene ring, here becomes the de localization of 7t-electrons in the ring and further the NH₂– group donates more electron to the ring so that it reduces the -i-ve charge centers and stablizes the ring and moves it less acidic means basicity increases while in case of cyclohexamine, the lone pair of electron donating NH,- group is not de localised over cyclohexane ring and is available for donation, which makes the compound more basic than anile.

(v) Phenol loses its H⁺ when kept in an acidic medium. Due to which phenote ion is formed. This ion formed is more stable than phenol due to its resonating character. Hence phenote formation is favoured and releases of H⁺ make phenol acidic.

Question 24.
Explain the main points in the extraction of zinc from zinc blende.
Answer:
Explain the main points in the extraction of zinc from zinc blende.
Answer:
Extraction of zinc : Reduction process involves the steps.
(i) Concentration of the ore : The ore is Crushed and made to powder. The crushed ore is concentrated by Froth floation process. The concentrated ore goes to the surface along with the froth. This is removed and dried.

(ii) Roasting: The concentrated ore is roasted in a furnace at high temp. (900°C) in presence of excess of air. In this process zinc blends is converted into ZnO.

There is chance of formation of ZnSO₄ if the reaciton temperature remains low. The ZnSO₄ thus formed is decomposed to give ZnO, SO₂ and O₂

If calamine ore (ZnCO₃) is used then it is converted to ZnO by calcination.
ZnCO₃ → ZnO + CO₂ ↑

(iii) Reduction: ZnO thus formed as a result of roasting or calcination is reduced by carbon. As a result at reduction Zn metal in molten state is formed.

(iv) Purification: Impure Zn is purified by the process of electrolysis, for this impure zinc rod is made anode and a rod of pure zinc is made cathode. These rods are dipped in an electrolytic cell containing ZnSO₄ Zn in deposited at cathode on electrolysis. The following cell reactions occur
Zn Zn⁺⁺ +2e (At anode),Zn⁺⁺ +2e → Zn (At cathode)