Oswal Practice Papers CBSE Class 12 Chemistry Solutions (Practice Paper - 7)

Section-A 

1. (b) Allyl chloride

Explanation :    

When chlorine gas is reacted with propene at high temperature (400°C) then substitution takes place in place of addition reaction. Hence, allyl chloride is formed.
$$CH_3CH=CH_2Cl\space\underrightarrow{400°C}\space\underset{Allyl Chloride}{ClCH_2CH}=CH_2+HCl$$

2. (c) + 3

Explanation :    

$$K_2Cr_2O_7+7H_2SO_4+6KI→4K_2SO_4+Cr_2(SO_4)_3+7H_2O+3I_2$$

We have Cr2(SO4)3 in the product in which the oxidation state of Cr is calculated as:
Let, x is the oxidation state of Cr, so,
2x + 3(– 2) = 0
x = + 3 

3. (b) 3.87 BM

Explanation :    

$$Cr→(Z=24)\\Cr^{3+}→(Z=21)\\1s^2,2s^2,2p^6,3s^2,3p^6,3d^3,4s^0\\\text{n=3(3 unpaired electrons)}\\\mu=\sqrt{n(n+2)}=\sqrt{3(3+2)}\\=\sqrt{3(5)}=\sqrt{15}=3.87 BM$$

4. (c)

phenol

Explanation :    

When phenol react with CH3COCl in presence of anhydrous AlCl3 it produce ortho (minor) and para (major) product.

ortho

5. (b) 297 g

Explanation :    

$$\text{Given that,}\\\Delta T_f=10°C\\Now,\Delta T_f=K_f×\text{molality}\\10=1.86×\frac{x×1000}{92×1000}=296.8 g\\\text{Wt. of glycerol taken = 297 g}$$

6. (a) Positive

Explanation :    

$$\text{Gibb’s free energy is given by}\\\Delta_rG=–nFE_{cell}\\\text{For spontaneous reaction, we know}\\\Delta_rG=–ve\\E_{cell}=+ve\space{(... n, F are constant)}$$

7. (c) 1-(q), 2-(p), 3-(r), 4-(t), 5-(s)

Explanation :    

Osmotic pressure is colligative property, dilute solution obeys Raoult’s law, Cottrell’s method uses ebullioscopic constant, temperature is an intensive property and isotonic solutions are solutions having same osmotic pressure.

8. (b) Doubled

Explanation :    

The rate concentration of a reaction does not depend upon concentration of the reactions. Hence, it will remain the same. Even if the equation shows the double concentration level the rate concentration doubles so it is the same throughout.
Following with the equation A + 2B → C
If rate considered as (1) Rate1 = k [A][B]
If rate considered as 2
Then, Rate1 = k [A][2B]
Rate2 = 2 Rate1
Hence, the value of rate constant will be doubled. 

9. (d) Diethyl ether, acetone, ethyl alcohol, water

Explanation :    

The vapour pressure increases with decrease in intermolecular forces and increase in temperature. As when temperature increases the molecular forces are weakend and bonds are broken between molecules. When the forces are weak, the liquid has high volatility and maximum vapour pressure. Diethyl ether has highest vapour pressure while water has lowest vapour pressure.

10. (b) 31.76 g of Cu

Explanation :    

$$\text{At cathode:}Cu^{2+}(aq)+2e^–→Cu(s)\\At \text{anode:}4OH^–(aq)→2H_2O(l)+O_2(g)+4e^{–}\\\text{Faraday’s constant=96500 C/mol}\\\text{To deposite 1 mole of copper, we need 2 × 96500 C.}\\\text{So, 96500 C will deposite 0.5 moles of copper = 0.5×63.5=31.75 g.}\\\text{Hence, 31.76 gm of Cu will be liberated.}$$

11. (d) All of these

Explanation :    

Transition metals have defects in their crystal lattice. The transition metals form interstitial compounds as small atoms like C, H or N are trapped inside the interstitial spaces in the crystal lattice of metals. Hence, transition metals, when they form interstial compounds, the non-metals are accommodated in voids.

12. (c) LiAlH4 ether

Explanation :    

The given reaction is the reduction of primary amides in the presence of the reducing agent LiAlH4 to primary amines.
$$R—CO—NH_2\space\underrightarrow{LiAlH_4}\space RCH_2NH_2$$

Question No. 13 to 16 consist of two statements – Assertion (A) and Reason (R). Answer these questions selecting the appropriate option given below:
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.

13. (a) Both A and R are true and R is the correct explanation of A.

Explanation :    

It is not always convenient to determine the instantaneous rate, as it is measured by determining the slope of the tangent at point ‘t’ in concentration Vs time plot. This generally makes it difficult to determine the rate law and hence, the order of the reaction. Hence, both assertion and reason are true and reason is the correct explanation of assertion.

14. (c) A is true, but R is false.

Explanation :    

Phenol is stronger acid than ethanol as phenoxide ion is stabilized by resonance whereas no such stabilization occurs in ethoxide ion. Sodium ethoxide can be prepared by reaction of ethanol with sodium. Thus, assertion is true but reason is false.

15.(c) A is true, but R is false.

Explanation :    

When NaCl is added to water a depression in freezing point is observed. This is due to lowering of vapour pressure of a solution. Thus, assertion is true but reason is false.

16. (a) Both A and R are true and R is the correct explanation of A.

Explanation :    

Electrical work done in one second is equal to electrical potential multiplied by total charge. If we want to obtain maximum work from a galvanic cell then charge has to be passed reversibly. The reversible work done by a galvanic cell is equal to decrease in its Gibbs energy and therefore, if the emf of cell is E and nF is the amount of charge passed and ΔG is the Gibbs energy of reaction then, 
ΔG = − nFEcell 
Thus, both assertion and reason are correct and reason is the correct explanation of assertion.

Section-B

17. In a molecule of tertiary amine, there are no H-atoms whereas in primary amines, two hydrogen atoms are present. Due to the presence of H-atoms, primary amines undergo extensive intermolecular H-bonding. In a molecule of tertiary amine, there are no H-atoms whereas in primary amines, two hydrogen atoms are present. Due to the presence of H-atoms, primary amines undergo extensive intermolecular H-bonding.

Tertiary amine

As a result, extra energy is required to separate the molecules of primary amines. Hence, primary amines have higher boiling points than tertiary amines.

18. The initial rate of the reaction is
Rate = k[A][B]2
= (2.0 × 10–6 mol–2 L2 s–1) (0.1 mol L–1) (0.2 mol L–1)2
= 8.0 × 10–9 mol L–1 s–1
When [A] is reduced from 0.1 mol L–1 to 0.06 mol L–1, the concentration of [A] reacted
= (0.1 – 0.06) mol L–1 = 0.04 mol L–1
Therefore, concentration of [B] reacted =(1/2)× 0.04
mol L–1 = 0.02 mol L–1.
As [B] used in reaction is 1/2 moles per 1 mole of A.
Then, concentration of [B] available, [B] = (0.2 – 0.02) mol L–1.
= 0.18 mol L–1
After [A] is reduced to 0.06 mol L–1, the rate of the reaction is given by,
Rate = k[A][B]2
= (2.0 × 10–6 mol–2 L2 s–1) (0.06 mol L–1) (0.18 mol L–1)2
= 3.89 × 10–9 mol L–1 s–1

19. (a) 2-Methylpropan-1-ol
(b) 2-Bromo-3-methylbut-2-en-1-ol.

OR

Ethoxybenzene

20. (a) Glucose is obtained commercially by hydrolysis of starch by boiling it with dilute H2SO4 at 393 K under pressure of 2-3 atm. 
$$\underset{Starch or cellulose}{(C_6H_{10}O_5 )_n}+nH_2o\space\xrightarrow[393 K, 2-3 atm]{H^+}\underset{Glucose}{nC_6H_{12}O_6}$$

(b) Aldehyde group is present in the structure of glucose.

21.  (a) 1. Tris(ethane-1, 2-diamine) cobalt(III) sulphate
2. Mercury(I) tetrathiocyanatocobaltate(III)
(b) 1. Trigonal bipyramidal – sp3d
2. Square planar – dsp

Section-C

22. (a)

lone pair

the lone pair of electrons on nitrogen atom is involved in resonance with the carbonyl group as follows.

carbonyl group

Hence, the electron pair of nitrogen is not easily available for donation. But +I – effect of —CH2—CH3 enhances the lone pair availability on 'N' atom in CH3 — CH2— NH2

$$(b)CH_3-CH_2-CI\space\underrightarrow{NaCN}\space-CH_3\underset{\text{\textcircled A}}{-CH_2}-CN\space\underrightarrow{Ni/H_2}CH_3CH_2\underset{\text{\textcircled B}}{-CH_2}-NH_2$$

23. The given reaction is of the first order with respect to A and of zero order with respect to B. Therefore, the rate of the reaction is given by,
Rate = k[A]1[B]0
⇒ Rate = k[A]
From experiment I, we obtain
2.0 ×10–2 mol L–1 min–1 = k(0.1 mol L–1)
⇒ k = 0.2 min–1
From experiment II, we obtain
4.0 ×10–2 mol L–1 min–1 = 0.2 min–1 [A]
⇒ [A] = 0.2 mol L–1
From experiment III, we obtain
Rate = 0.2 min–1 × 0.4 mol L–1
⇒ = 0.08 mol L–1 min–1
From experiment IV, we obtain
2.0 ×10–10 mol L–1 min–1 = 0.2 min–1 [A]
⇒ [A] = 0.1 mol L–1 

24. (a) Cell reaction,
$$Cu(s)+2Ag^+(aq)→2Ag(s)+Cu^{2+}(aq)\\n=2\\\text{Using Nernst equation,}\\E_{cell}=E°_{cell}–\frac{0.0591}{n}log\frac{[Cu^2+]^1}{[Ag^+]^2}(T=298K)\\i.e., E_{cell} = (+0.80V–0.34V)–\frac{0.0591}{2}log\frac{4}{(0.1)^2}\\=+0.46 V–\frac{0.0591}{2}log(400)\\=+0.46 V–0.0295×2.6021\\= 0.46 V–0.0768 V=0.3832V\\Hence,\space E_{cell}=0.3832 V.$$

(b) For 3 moles of Fe3+ to convert to Fe2+, we require 1 mol of electron charge per mol.
3Fe3+ + 3e → 3Fe2+ 

Hence, 3 moles of electrons are required,
i.e., Q = 3 × 1F = 3 ×96500 C.
as, 1F = charge of 1 mole of electron.
Also, Q = It
$$t=\frac{Q}{t}=\frac{3×96500}{2}second\\=144,750 sec\space\text{[1 hour = 3600 second]}\\\text{Hence, time in hours}\frac{144750}{3600}=\text{40.20 hours.}$$

25. (a) As the alkyl groups increase the electron density attached to the carbon bearing the oxygen atom, they destabilise the anion formed after losing proton. Hence, the acidic strength will vary in below order :
2- methylpropan-2-ol < 2-methylpropan-1-ol < n-butanol.
(b) Lucas Test in alcohol is a test used to differentiate between Primary, Secondary and Tertiary alcohols. This test is carried out with the help of Lucas reagent, which is a solution of anhydrous ZnCl2 and conc. HCl.
In case of tertiary alcohols, instantaneous turbidity appears in the solution while in secondary alcohols turbidity appears after 4-5 minutes. Therefore, propanol and 2-methylpropan-2-ol can be distinguished using by Luca’s reagent, (ZnCl2+HCl) solution. 2-methylpropan-2-ol react immediately
at room temperature to give a white precipitate of corresponding alkyl chloride, whereas in case of propanol turbidity appears after heating.

turbidity appears

(c) 2-Methyl propan-2-ol

2-Methyl propan

26. (a) As depression of freezing point (ΔTf) is directly proportional to molality of solution :
ΔTf = Kf m
Now putting the values :
m = ΔTf /Kf m
$$m=\frac{273.0 K-272.4 K}{1.86 K kg mol ^{-1}}=0.323 mol kg{–1}$$

(b) Similarly ΔTb = Kbm
= 0.323 mol kg–1 × 0.512 K kg mol–1 = 0.165 K
As, boiling point of water is 373.15 K, therefore boiling point of solution would be :
373.15 K + 0.165 K = 373.31 K 

27. Compound [A] gives iodoform reaction hence it must have a (CH3CO) group.

iodoform

must be [A]. Hence the reactions are:

28. The galvanic cell in which the given reaction takes place is depicted as:
Zn(s) | Zn2+ (aq) || Ag+ (aq) | Ag(s)
(a) Zn electrode (anode) is negatively charged.
(b) Ions are carriers of current in the cell and in the external circuit, current will flow from silver to zinc. Ions are Zn2+ and Ag+.
(c) The reaction taking place at the anode is given by,
Zn(s) → Zn2+ (aq) + 2e
The reaction taking place at the cathode is given by,
Ag+ (aq) + e → Ag(s) 

Section-D

29.
(a) Zn, Cd, Hg (due to d10 configuration).
(b) Penultionate d-orbital i.e., (n – 1) d-orbital.
(c) Due to incomplete filling of d-orbital, oxidation states vary. 

OR

Mn, + 7.

30.
(a) SN 1 type reaction.
(b) Difference in solvolysis reaction is due to the stability of carbocation formed during the course of reaction.
(c) The carbocation obtained from C—Cl bond cleavage of the compound (A) is a very stable aromatic compound [cyclopropenyl cation having (4n+2)p electrons, where n = 0].
For this reason, the compound (A) undergoes solvolysis (SN₁) at a faster rate. 

Aromatic

OR

The carbocation expected to be obtained on C—Cl bond cleavage of ‘B’ in an unstable anti-aromatic carbocation [cyclopentadienyl cation, having 4np electrons, where n =1]. There, the compound ‘B’ does not undergo solvolysis (SN 1) in ethanol. 

antiaromatic

Section-E

31. (a) Due to the presence of a single carbonyl group glucose forms glucoxime.

Glucoxime

(b) Glucose on reduction with HI in presence of red P at about 373 K produces n-hexane. This reaction confirms the presence of 6-carbon chain in glucose.

glucose

(c) In glucose all the hydroxy (—OH) i.e., 5 groups react with acetic anhydride to form a pentaacetate except the aldehydic group. This product indicates presence of 5, hydroxyl group in normal structure of D-glucose.

D-glucose pentaacetate

(d) The aldehydic group is absent in the a-D-glucose pentaacetate or b-D-glucose pentaacetate. It is involved in the cyclic bond between first and the 5th carbon of the chain. Hence, glucose pentaacetate does not give any reaction with NH2OH.

(e) The open chain form of glucose is responsible for the positive tests of aldehyde given by glucose. But in case of Schiff’s reagent and NaHSO3 which are weak reagents, the reactions are reversible and the equilibrium cannot be shifted to get more and more open chain form, which is smaller in
amount in the mixture of a and β cyclic glucose forms. 

(f) Lactose is a disaccharide composed of β-D-galactose and β-D-glucose. Thus, on hydrolysis, it gives β-D-galactose and β-D-glucose.

(g) Pyranose ring is a six membered heterocyclic ring with an oxygen atom. It is the base structure for both α and β-D-glucose.

32. (a) [Fe(CN)6]4
In the above coordination complex, iron exists in the +2 oxidation state.
Fe2+ : Electronic configuration is 3d6 orbitals of Fe2+ ion. 

orbitals

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 hybridisation is d2sp3.
hybridisations

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

six hybrid

Hence, the geometry of the complex is octahedral and complex is diamagnetic inner orbital complex.

(b) [FeF6]3–
In this complex, the oxidation state of Fe is + 3.
Orbitals of Fe+3 ion: 

complex

There are 6F ions. Thus, it will undergo sp3d2 hybridization as F is a weak field ligand and it does not cause the pairing of the electrons in the 3d orbital. Hence, hybridisation is sp3d2.
sp3d2 hybridized orbitals of Fe3+ are: 

six electron pair

Hence, the geometry of the complex is found to be octahedral, strongly paramagnetic and outer orbital complex.

(c) [Co(C2O4)3]3–
Cobalt exists in the +3 oxidation state in the given complex.
Orbitals of Co3+ ion : 4s03d6 configuration. 

Cobalt

Oxalate is a strong field ligand so, it causes the pairing of the unpaired 3d orbital electrons. As there are 6 ligands, hybridisation has to be d2sp3 hybridisation of Co3+

Oxalate

The 6 electron pairs from 3 oxalate ions (oxalate anion is a bidentate ligand) occupy these d2sp3 orbitals.

Hence, the geometry of the complex is found to be octahedral, inner orbital, diamagnetic.

(d) [CoF6]3–
Cobalt exists in the +3 oxidation state.
Orbitals of Co3+ ion: 

Cobalt exists

Again, fluoride is a weak field ligand. It cannot cause the pairing of the unpaired 3d electrons.
As a result, the Co3+ ion will undergo sp3d2 hybridisation. sp3d2 hybridised orbitals of Co3+ ion are: 

Hence, the geometry of the complex is octahedral and paramagnetic and outer orbital complex.

OR

[Mn(CN)6]3– oxidation state of Mn is (+ 3)
[Co(NH3)6]3+ oxidation state of Co is (+ 3)
[FeCl6]4– oxidation state of Fe is (+ 2)
In [Mn(CN)6]3–, CN is strong field ligand.
In [Co(NH3)6]3+, NH3 is weak field ligand.
In [FeCl6]4–, Cl is a weak field ligand.
Now [Mn(CN)6]3– ; Mn3+(3d4). 

Pairing will take place due to strong field ligand CN

due to strong

(a) Hybridisation d2sp3
(b) Octahedral geometry
(c) Paramagnetic
(d) Inner orbital complex
(e) μ =n(n+2)=√2(2+2)
= √8 = 2.87 BM 

field ligand

(a) Hybridisation d2sp3
(b) Octahedral
(c) Diamagnetic
(d) Inner orbital complex
(e) μ =√0(0+2)=0

cause pairing

(a) Hybridisation d2sp3
(b) Octahedral
(c) Paramagnetic
(d) Outer orbital complex
(e) μ =√n(n+2)=√4(4+2)
=√24= 4.9 BM

33. [A] seems like an organic compound of benzene from formula C7H6Cl2.

benzene

Therefore [A] will be

Nitrobenzoic acid
Benzyl alcohol

OR

(a) (i) Nitro group is an electron withdrawing group. Hence, its presence in para position makes the release of H+ ion from benzoic acid easier than unsubstituted benzoic acid. Hence p-Nitrobenzoic acid is more acidic and have higher Ka value.

(ii) The extent of hydrogen bonding in carboxylic acids is far more than alcohols. The OH bond of –(COOH) group is far more polarized than OH bond of alcohols; because of presence of electron withdrawing carbonyl group. Additional H–bonds are also formed by the negatively charged oxygen atom of carbonyl carbon with a positively charged hydrogen of some adjacent molecule. 

hydrogen bonding

(iii) Acetone can undergo some degree of hydrogen bonding with water molecules by the presence of polar carbonyl

polar carbonyl

group. In case of benzophenone the carbonyl group is sterically hindered by two big phenyl group (C6H5COC6H5), hence the carbonyl oxygen is masked and cannot participate in hydrogen bonding with water. 

(b) (i) Iodoform test is given by compound containing a methyl ketone group i.e., CH3CH3CH2COCH3 will give positive iodoform test.

(ii) The rate of hydrolysis depends on the electron deficiency of the carbonyl carbon of the ester group. Hence presence of electron withdrawing groups increases the rate of reaction whereas electron donating group decreases the rate of hydrolysis. Hence the decreasing order of rate for
the given esters is. 

Iodoform test

CBSE Practice Paper Chemistry Class 12

All Practice Paper for Class 12 Exam 2024

The dot mark field are mandatory, So please fill them in carefully
To download the Sample Paper (PDF File), Please fill & submit the form below.