Students can use CBSE Previous Year Question Papers Class 12 Chemistry with Solutions and CBSE Class 12 Chemistry Question Paper 2015 Delhi to familiarize themselves with the exam format and marking scheme.
CBSE Class 12 Chemistry Question Paper 2015 Delhi with Solutions
Time allowed: 3 hours
Maximum Marks: 70
General Instructions
- All questions are compulsory.
- Section A: Questions number 1 to 5 are very short answer questions and carry 1 mark each.
- Section B: Questions number 6 to 12 are short answer questions and carry 2 marks each.
- Section C: Questions number 13 to 24 are also short answer questions and carry 3 marks each.
- Section D: Questions number 25 to 27 are long answer questions and carry 5 marks each.
- There is no overall choice. However, an internal choice has been provided in two questions of one mark, two questions of two marks, four questions of three marks and all the three questions of five marks weightage. You have to attempt only one of the choices in such questions
- Use of log tables, if necessary. Use of calculators is not allowed.
\(\dagger\)Question 1.
What is the basicity of H
3
PO
4
? [1]
Answer:
Basicity is 3.
Question 2.
Write the IUPAC name of the given compound: [1]
Answer:
2, 5-dinitrophenol.
Question 3.
Which would undergo S
N
2 reaction faster in the following pair and Why? [1]
Answer:
CH
3
CH
2
Br reacts faster because it is a primary halide (1° halide).
Question 4.
Out of BaCl
2
and KCl, which one is more effective in causing coagulation of a negatively charged colloidal Sol? Give reason. [1]
Answer:
BaCl
2
is more effective in causing coagulation because it has double +ve charge than K
+
.
\(\dagger\)Question 5.
What is the formula of a compound in which the element Y forms ccp lattice and atoms of X occupy 1/3 rd of tetrahedral voids? [1]
Answer:
Formula is X
2
Y
3
.
Question 6.
What are the transition elements? Write two characteristics of the transition elements. [2]
Answer:
Elements which have partially filled d-orbital in its ground states or any one of its oxidation states are called transition elements.
- They show variable oxidation states.
- They form coloured ions.
- They form complex compounds.
Question 7.
(i) Write down the IUPAC name of the following complex:
[Cr(NH
3
)
2
Cl
2
(en)]Cl (en = ethylenediamine)
(ii) Write the formula for the following complex:
Pentaamminenitrito-o-Cobalt(III). [2]
Answer:
(i) [Cr(NH
3
)
2
Cl
2
(en)]Cl
IUPAC name : Diammine dichlorido ethylenediamine chromium (III) chloride.
(ii) [C0(NH
3
)
5
NO
2
]
2+
Question 8.
Name the reagents used in the following reactions: [2]
Answer:
(i) LiAlH
4
(Lithium Aluminium Hydride)
(ii) KMnO
4
(Alkaline)
Question 9.
What is meant by positive deviations from Raoult’s law? Give an example. What is the sign of ∆
mix
H for positive deviation? [2]
OR
Define azeotropes. What type of azeotrope is for-med by positive deviation from Raoult’s law? Give an example.
Answer:
In positive deviations, the partial vapour pressure of each component A and B of a solution and the total pressure of the solution is higher than the vapour pressure calculated from Raoult’s law.
For example, Water and Ethanol.
In case of positive deviations, ∆
mix
H > 0 (Positive)
Or
Azeotropes : Liquid mixture which distills without change in compositions is called azeotropic mixture.
In positive deviations from Raoult’s law, minimum boiling point azeotropic mixture is formed. For example, 95% ethanol + 5% water.
Question 10.
(a) Following reactions occur at the cathode during the electrolysis of aqueous silver chloride solution:
Ag
+
(aq) + e
–
→ Ag(s); E
0
= +0.80 V
H
+
(aq) + e
–
→ \(\frac { 1 }{ 2 }\) H
2
(g); E
0
= 0.00 V
On the basis of their standard reduction electrode potential (E
0
) values, which reaction is feasible at the cathode and why?
(b) Define limiting molar conductivity. Why the conductivity of an electrolyte solution decreases with decrease in concentration? [2]
Answer:
(a) At the cathode Ag
+
(aq) + e
–
→ Ag(s) reaction is feasible, because Ag
+
ion has higher reduction potential i.e. higher E° value.
(b) Limiting molar conductivity is the molar conductivity of solution at infinite dilution. Conductivity of an electrolyte solution decreases on dilution because number of ions per unit volume decreases.
Question 11.
3.9 g of benzoic acid dissolved in 49 g of benzene shows a depression in freezing point of 1.62 K. Calculate the Van’t Hoff factor and predict the nature of solute (associated or dissociated). (Given: Molar mass of benzoic acid = 122 g mol
-1
, K
f
for benzene = 4.9 K kg mol
-1
) [3]
Answer:
Question 12.
(i) Indicate the principle behind the method used for the refining of zinc.
(ii) What is the role of silica in the extraction of copper?
(iii) Which form of the iron is the purest form of commercial iron? [3]
Answer:
(i) Zinc is refined by distillation method because zinc has low boiling point.
(ii) During roasting copper pyrites are converted into a mixture of FeO and Cu
2
O.
Silica acts as acidic flux in the extraction of copper to remove the iron oxide obtained during the process of roasting.
To remove FeO, silica is added during smelting. FeO then combines with SiO
2
to form ferrous silicate (FeSiO
3
) slag which floats over molten matter and hence can be easily removed.
(iii) Wrought iron.
\(\dagger\)Question 13.
An element with molar mass 27 g mol
-1
forms a cubic unit cell with edge length 4.05 × 10
-8
cm. If its density is 2.7 g cm
-3
, what is the nature of the cubic unit cell? [3]
Answer:
Question 14.
(a) How would you account for the following:
(i) Actinoid contraction is greater than lanthanoid contraction.
(ii) Transition metals form coloured compounds.
(b) Complete the following equation: [3]
2MnO
4
–
+ 6H
+
+ 5NO
2
–
→
Answer:
(a) (i) Actinoid contraction is greater than lanthanoid because 5/ electrons (in actinoids) have a poorer shielding effect than 4f electrons (in lanthanoids).
(ii) Transition metals form coloured compounds because they have unpaired electrons in the d-orbital.
(b) 2Mn\(\mathrm{O}_4^{-}\) + 6H
+
+ 5N\(\mathrm{O}_3^{-}\) → 2Mn
+2
+ 5N\(\mathrm{O}_3^{-}\) + 3H
2
O
Question 15.
(i) Draw the geometrical isomers of a complex [Pt(NH
3
)
2
Cl
2
].
(ii) On the basis of crystal field theory, write the electronic configuration for d
4
ion if ∆
0
< P.
(iii) Write the hybridization and magnetic behaviour of the complex [Ni(CO)
4
].
(At. no. of Ni = 28). [3]
Answer:
(i)
(ii) Electronic configuration for d
4
ion if < P : \(t_{2 g}{ }^3 e_g{ }^1\) (high spin complex is formed)
(iii) Ni(CO)
4
has SP
3
hydbridization. It is diamagnetic in nature.
Question 16.
Calculate emf of the following cell at 25°C: [3]
Fe | Fe
2+
(0.001M) | H
+
(0.01M) | | H
2
(g) (1 bar)
Pt(s) E
0
(Fe
2+
| Fe) = -0.44 V E
0
(H
+
/H
2
) = 0.00 V.
Answer:
Question 17.
Give reasons for the following observations:
(i) Leather gets hardened after tanning.
(ii) Lyophilic sol is more stable than lyophobic sol.
(iii) It is necessary to remove CO when ammonia is prepared by Haber’s process. [3]
Answer:
(i) Due to mutual coagulation of leather by tanning. [Positively charged animal hyde (leather) with negatively charged colloidal particles of tannin].
(ii) Lyophilic sols are more stable because there is strong interaction between dispersed phase and dispersion medium.
(iii) Because CO acts as a poison for catalyst.
Question 18.
Write the names and structures of the monomers of the following polymers:
(i) Nylon-6,6
(ii) PHBV
(iii) Neoprene. [3]
Answer:
(i) Nylon-6, 6 : It has two repeating monomers
(a) Hexa methylene diamin
Question 19.
Predict the products of the following reactions:
Answer:
Question 20.
How do you convert the following:
(i) Phenol to anisole
(ii) Propane 2-ol to 2-methyl propane-2-ol
(iii) Aniline to phenol [3]
OR
(a) Write the mechanism of the following reaction:
(b) Write the equation involved in the acetylation of Salicyclic acid. [3]
Answer:
Or
Question 21.
(i) Which one of the following is a disaccharide:
Starch, Maltose, Fructose, Glucose?
(ii) What is the difference between fibrous protein and globular protein?
(iii) Write the name of the vitamin whose deficiency causes bone deformities in children. [3]
Answer:
(i) Maltose is a disaccharide.
(ii)
| Globular Proteins | Fibrous Proteins |
| 1. Globular proteins have almost spheroidal shape due to folding of the polypeptide chain. | 1. Polypeptide chains of fibrous proteins consist of thread like molecules which tend to lie side by side to form fibres. |
| 2. Globular proteins are soluble in water. | 2. Fibrous proteins are insoluble in water. |
| 3. Globular proteins are sensitive to small changes of temperature and pH. Therefore they undergo denaturation on heating or on treatment with acids/bases. | 3. Fibrous proteins are stable to moderate changes of temperature and pH. |
|
4. They possess biological activity that’s why they act as enzymes.
Example : Maltase, invertase etc., hormones (insulin) antibodies, transport agents (haemoglobin), etc. |
4. They do not have any biological activity but serve as chief structural material of animal tissues.
Example : Keratin in skin, hair, nails and wool, collagen in tendons, fibroin in silk etc. |
(iii) Vitamin D
Question 22.
Give reasons:
(a) n-Butyl bromide has a higher boiling point than t-butyl bromide.
(b) A racemic mixture is optically inactive.
(c) The presence of the nitro group (-NO
2
) at o/p positions increases the reactivity of haloarenes towards nucleophilic substitution reactions. [3]
Answer:
(a) n-Butyl bromide has higher boiling point than t-butyl bromide because it has larger surface area hence have more Van der Waals’ forces.
(b) Rotation due to one enantiomer is cancelled by another enantiomer.
(c) The presence of nitro group (— NO
2
) at ortho and para positions withdraws the electron density from benzene ring and thus facilitating the attack of nucleophile.
Question 23.
Mr Roy, the principal of one reputed school organized a seminar in which he invited parents and principals to discuss the serious issue of diabetes and depression in students. They all resolved this issue by strictly banning junk food in school and to introduce healthy snacks and drinks like soup, lassi, milk etc. in school canteens. They also decided to make compulsory half an hour of physical activities for the students in the morning assembly daily. After six months, Mr Roy conducted the health survey in most of the schools and discovered a tremendous improvement in the health of students.
After reading the above passage, answer the following:
(i) What are the values (at least two) displayed by Mr Roy?
(ii) As a student, how can you spread awareness about this issue?
(iii) What are tranquillizers? Give an example.
(iv) Why is the use of aspartame limited to cold foods and drinks? [4]
Answer:
*(i) As per latest CBSE curriculum, Value Based Questior will not be asked in the examination.
(ii) As a student I can spread awareness through social media, debates and discussions in school, through school plays and other such means.
As a student I wifi also advise my friends and people of society to take up physical activities daily for good health and set an example myself by exercising regularly.
(iii) Tranquilizers are the chemical substances used for the treatment of anxiety, stress or mild and severe mental disorders, etc. For example, Equanil.
(iv) Aspartame decomposes at the cooking temperature therefore its use is limited to cold foods and drinks.
Question 24.
(a) Account for the following:
(i) Acidic character increases from HF to HI.
(ii) There is a large difference between melting and boiling points of oxygen and sulphur.
(iii) Nitrogen does not form pentahalide.
(b) Draw the structures of the following:
(i) ClF
3
(ii) XeF
4
[5]
OR
(i) Which allotrope of phosphorus is more reactive and why?
(ii) How the supersonic jet aeroplanes are responsible for the depletion of ozone layers?
(iii) F
2
has a lower bond dissociation enthalpy than Cl
2
. Why?
(iv) Which noble gas is used in filling balloons for meteorological observations?
(v) Complete the equation:
XeF
2
+ PF
5
→
Answer:
(a) (i) Acidic character increases from HF to HI because bond dissociation enthalpy decreases from HF to HL
(ii) Oxygen exists as diatomic O
2
molecule while sulphur exists as polyatomic S
8
molecule which has very high molecular mass therefore sulphur has much high melting and boiling points.
(iii) Nitrogen does not contain ‘d’ orbitals.
(b) (i) ClF
3
(ii) XeF
4
Or
(i) White phosphorus is more reactive because it is less stable due to angular strain;
(ii) The oxides of nitrogen released by the exhausts of supersonic jetplanes are causing the depletion of ozone layer.
(iii) F
2
has lower dissociation enthalpy as due to small size of F there is strong repulsive forces between F—F electrons. Bond dissocition enthalpy of F
2
is less than that of 2 because of large electron-electron repulsion among the lone pairs in small size F
2
molecule.
(iv) Helium is filled in the balloons for meteorological observations.
(v) XeF
2
+ PF
5
→ [XeF]
+
[PF
6
]
–
Question 25.
An aromatic compound ‘A’ of molecular formula C
7
H
7
ON undergoes a series of reactions as shown below. Write the structures of A, B, C, D and E in the following reactions.
OR
(a) Write the structures of main products when aniline reacts with the following reagents:
(i) Br
2
water
(ii) HCl
(iii) (CH
3
CO)
2
O/pyridine.
(b) Arrange the following in the increasing order of their boiling point:
C
2
H
5
NH
2
, C
2
H
5
OH, (CH
3
)
3
N
(c) Give a simple chemical test to distinguish between the following pair of compounds:
(CH
3
)
2
NH and (CH
3
)
3
N.
Answer:
Or
(a)
(b) Increasing order of boiling point :
(CH
3
)
3
N < C
2
H
5
NH
2
< C
2
H
5
OH
(c) By Hinsberg test, secondary amines or (CH
3
)
2
NH shows precipitate formation which is insoluble in KOH. Tertiary amines or (CH
3
)
3
N do not react with Hinsberg’s reagent (benzene sulphonyl chloride).
Question 26.
For the hydrolysis of methyl acetate in aqueous solution, the following results were obtained:
(i) Show that it follows pseudo-first-order reaction, as the concentration of water remains constant.
(ii) Calculate the average rate of reaction between the time interval 30 to 60 seconds. (Given log 2 = 0.3010, log 4 = 0.6021) [5]
OR
(a) For a reaction A + B → P, the rate is given by
Rate = k [A] [B]
2
(i) How is the rate of reaction affected if the concentration of B is doubled?
(ii) What is the overall order of reaction if A is present in large excess?
(b) A first-order reaction takes 30 minutes for 50% completion. Calculate the time required for 90% completion of this reaction. (log 2 = 0.3010)
Answer:
(i)
As k is constant in both the readings, hence it is a pseudo first order reaction.
(ii) Rate = -∆[R]/∆t, Average rate between 30 to 60 seconds
= \(\) = \(\)
Or
(a) For the reaction A + B → P rate is given by Rate = k[A]
1
[B]
2
(i) r
1
= k[A]
1
[B]
2
r
2
= k[A]
1
[2B]
2
= 4k[A]
1
[B]
2
r
1
= 4r
2
, rate will increase four times of actual rate.
(ii) When A is present in large amount, order w.r.t. A is zero.
Hence overall order = 0 + 2 = 2, second order reaction.
(b)
