Oswal Specimen Papers ISC Class 12 Chemistry Solutions (Specimen Paper - 8)


1. (A) (i) chemical, electrical
(ii) increases, lowering
(iii) yellow, iodoform
(iv) 2, 4, 6-trinitrophenol

(B) (i) (a) La(OH)3 is less basic than Lu(OH)3
(ii) (c) Medicine suspensions for children
(iii) (c) ΔG° is negative, K is greater than 1
(iv) (c) Uracil
(v) (d) independent of the initial concentration 

Explanation :    

For a first order reaction, the half-life period (t1/2) is independent of the initial concentration t1/2 = 0693/k 

(vi) (b) Both assertion and reason are true but reason is not the correct explanation for assertion.

Explanation :    

In the complex [Fe(CN)6]3– the iron is in +3 oxidation state having d5 configuration, so even after pairing due to strong field cyanide ligands one electron will remain unpaired and hence it is weakly paramagnetic. While in the complex [Fe(CN)6]4– iron is in +2 oxidation state and having d6 configuration, so after pairing due to strong field cyanide ligand no unpaired electron remains, so it is diamagnetic. Thus, both assertion and reason are true but reason is not the correct explanation of assertion. 

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

Explanation :    

The reaction between (CH3) C—O—CH3, and HI follows SN 1 mechanism. For an SN 1 reaction, the formation of product is decided by the stability of the cabocation formed in the slowest step. As tertbutyl carbonium ion formed after the cleavage of C—O bond in the slowest step is more stable than
methyl carbonium ion therefore, (CH3)3 C—I and CH3OH are formed as main products. Thus, both assertion and reason are true and reason is the correct explanation of assertion.

(C) (i) Primary valency of metal is the oxidation number of metal and the secondary valency of the metal is the coordination number. Therefore primary valency and secondary valency of chromium in the complex are +3 and +6 respectively.

(ii) CN has −1 oxidation state and nickel forms complex in +2 oxidation state (stable oxidation state).
Thus, 2 + 4 × (−1) = x
and x = −2.

(iii) 3 mol of AgCl implies 3 Cl is given in the arrangement hence forth, the equation of the complex will be [Cr(H2O)6]Cl3


2. Given,
Weight of solvent (H2O) (W1) = 1 kg
Weight of solute (C6H12O6) (W2) = 18 g
Molar mass of solute (M2) = 180 g/mol
Kb = 0.52 K kg mol−1
b = 373.15 K
We know that 

$$∴\Delta T_b=\frac{K_b×1000×W_2}{M_2×W_1}\\=\frac{0.52×1000×18}{180×1000}\\=0.052K\\∴ΔT_b=T_b–T°_b=0.052\\0.052=T_b−373.15\\T_b=373.202K\\\text{Hence, boiling point of solution is 373.202 K.}$$

3. (i) 




4. (i) The compound A forms a derivative with hydrazine means it contains a carbonyl group and as the compound A does not give reaction with Tollen’s reagent, it does not have an aldehyde group. So, the compound A is a ketone. Also, the compound A gives the iodoform reaction with iodine and sodium hydroxide. It contains a methyl group next to the carbonyl group. So, the structure of compound A is CH3CO(CH2)3CH3, i.e., Hexan-2-one. The reactions can be explained as follows: 


Brown chromium complex

5. (i) Haloarenes (say chlorobenzene) is a resonance hybrid of the following five structures:

Resonance leads to lowering of energy and hence greater stability. On the other hand, no such resonance is possible in haloalkanes. Halogens directly attached to benzene ring are O, p-directing in electrophilic subsitution reactions. This is due to greater electron density at these positions in resonance.

(ii) CHCl3 is stored in dark coloured bottles to cut off light because CHCl3 is slowly oxidised by air in presence of light to form an extremely poisonous gas, carbonyl chloride, popularly known as phosgene. 

Phosgene or Carbonyl chloride

6. At anode: Fe Fe2+ + 2e-
At cathode: 2H+ + 2e- H2
So, total number of electrons (n) transferred = 2
Given that: E°cell = + 0.44V
Temperature (T) = 298 K
We know, 

$$E_{cell}=E°_{cell}-(\frac{2.303RT}{nF})log\frac{a_{oxi}}{a_{red}}\\E_{cell}=E°_{cell}-(\frac{0.05916V}{n})log\frac{[Fe^{2+}]}{[H^+]^2}\\=0.44-0.02955×(-3)=0.44+0.0885\\=0.53V\\E_{cell}=0.44-(\frac{0.5916}{2})log\frac{0.001}{1}\\\text{Thus, e.m.f. of the cell is 0.53 V.}$$

7. (i) 3d3 (Vanadium) : + 2, + 3, + 4 and + 5
(ii) 3d8 (Nickel) : + 2 and + 3 (in complexes)

8. (i) E° value for Mn3+/Mn2+ is highly positive because Mn2+ has a stable d5 configuration and it is reluctant to lose one electron to achieve the +3 state. Cr2+ has 3d4 configuration and losing another electron to achieve 3d3 configuration is not that difficult, hence the E˚ value is not more positive compared to Mn3+/Mn2+


(ii) (a) Tetrahedral, sp3
The valence shell electronic configuration of ground state Ni atom is 3d8 4s2. All of these 10 electrons are pushed into 3d orbitals and get paired up when strong field CO ligands approach Ni atom. The empty 4s and three 4p orbitals undergo sp3 hybridization and form bonds with CO ligands to give Ni(CO)4. Thus, Ni(CO)4 is diamagnetic in nature. Therefore, according to VBT sp3 hybridization have tetrahedral geometry.
Electronic configuration of Ni atom in ground state. 

electron pairs by CO molecules

Hence, Ni(CO)4 undergoes sp3 hybridisation.

(b) Square planar, dsp2

In [Ni(CN)4]2–, there is Ni2+ ion for which the electronic configuration in the valence shell is 3d8 4s0.
In presence of strong field CN ions, all the electrons are paired up. The empty 4d, 3s and two 4p orbitals undergo dsp2 hybridisation to make bonds with CN ligands in square planar geometry.
Thus, [Ni(CN)4]2– is diamagnetic.
Electronic configuration of Ni atom in ground state. 

4 electron pairs

9. Given :
Weight of benzoic acid (solute) = 5 g
Weight of benzene (solvent) = 35 g
Kf = 4.9 K Kg mol–1
ΔTf = 2.94 K
We know that molecular weight of solute can be calculated as: 

$$M_2=\frac{(K_f×W_2×1000)}{\Delta T_f×W_1}\\\text{Substituting the values,}\\M_2=\frac{(K_f×W_2×1000)}{\Delta T_f×W_1}\\=\frac{4.9×5×1000}{2.94×35}=\frac{24500}{102.9}= 238.09g\\\text{Now let us consider the following equilibrium for the benzoic acid :}\\2C_6H_5COOH\xrightleftharpoons{}(C_6H_5COOH)_2$$

If x represents the degree of association of the solute then it would have (1 – x) mol of benzoic acid left in unassociated form and correspondingly x/2 as associated moles of benzoic acid at equilibrium. Therefore, total number of moles of particles at equilibrium is:
(1 – x) + x/2 = 1– (x/2)
Thus, total number of moles of particles at equilibrium equals to van’t Hoff factor (i)

$$But, i =\frac{\text{normal molar mass}}{\text{abnormal molar mass}}\\\text{So, van’t Hoff factor,}\\i =\frac{122}{238.09}= 0.512\\\text{Hence,}1– (x/2) = 0.512\\(x/2) = 0.512 = 0.488\\x = 0.976\\\text{Hence, the percentage association is} 97.6\% \text{ for benzoic acid.}$$

10. (i) Hydrogen bonds between —CO— and —NH— group of amino acids stabilise the α-helix structure of proteins.
(ii) The Ka value of a-aminoacids are very low because in a-amino acids, the acidic group is —NH3 + instead of —COOH group in a carboxylic acids, similarly the Kb values of a-amino acids are very low because the basic group is —COO instead of —NH2 group in aliphatic amines. 

11. (i) The given reaction sequence gives the following compounds, Acetaldehyde (A), Acetic acid (B), Acetyl chloride (C). But, aniline does not gives Friedel-Crafts reaction because the reagent AlCl3 (the Lewis acid which is used as a catalyst in Friedel Crafts reaction), being electron deficient acts as a Lewis base and attacks on the lone pair of nitrogen present in aniline to form an insoluble complex which precipitates out and the reaction does not proceed.

Anilinium salt


12. (i) Order of reaction=1/2+2=5/2
(ii) For first order reaction,

$$\text{Half life}(t_{½})=\frac{0.693}{k}\\=\frac{0.693}{5.5×10^{-4}}=1.26×10^{13}s$$

13. (i)

$$\underset{\text{Bromomethane}}{CH_3Br}\xrightarrow{KCN}\underset{\text{(A)Ethanenitrile}}{CH_3CN}\xrightarrow{LiAIH_4}\underset{\text{(B)Ethaneamine}}{CH_3CH_2NH_2}\xrightarrow[273K]{HNO_2}\underset{\text{(C)Ethanol}}{C_2H_5OH}+N_2+HCl\\(ii)\underset{\text{Acetic acid}}{CH_3COOH}\xrightarrow{NH_3}\underset{\text{(A)Ethanamide}}{CH_3CONH_2}\xrightarrow[Hoffman's Bromidereaction]{Br+KOH}\underset{\text{(B)Methanamine}}{CH_3NH_2}\xrightarrow[Carbylamine\space reaction]{CHCl_3+NaOH}\underset{\text{(C) Methyl isonitrite (Methyl Carbylamine)}}{CH_3NC}$$

14. (i) Peptide bond (—NHCO—) is the bond formed between the carboxylic group (COOH) of one amino acid with the amino group (NH2) of another amino acid, with elimination of water to form a dipeptide.
(ii) Glycosidic bond (—O—) is the bond formed between the OH group attached to an anomeric carbon atom of a monosaccharide which can easily dehydrate with an —OH group attached to another monosaccharide leading to the formation of a disaccharide. Since the reducing groups of glucose and fructose are involved in glycosidic bond formation, between two monosaccharide units through oxygen atom, it is called as glycosidic linkage.
Sucrose is a biomolecule with glycosidic linkage. 

glycosidic linkage

15. Given: ΔTf = 0.068 K, m = 0.01 m, Kf = 1.86 K kg mol–1
Now we have,

ΔTf = iKfm

$$i=\frac{\Delta T}{k_fm}\\i=\frac{0.068}{1.86×0.01}\\=\frac{0.068}{0.0186}=3.65\\Now, AlCl_3→Al^{3+}+3Cl^–\\\text{Initial modes 1 0 0}\\\text{After time (t) 1 – α α 3α}\\\text{Total number of moles at time (t) = 1 – α + α + 3α = 1 + 3α}\\i =\frac{\text{Total number of particle after dissociation}}{\text{Total number of particles before dissociation}}\\i=\frac{1+3\alpha}{1}\\3.65=1+3α\\3.65–1=3α\\α=\frac{2.65}{3}= 0.8833 or 88.33%\\\text{Thus, the percentage of dissociation is 88.33\%.}$$

Thus, the percentage of dissociation is 88.33%.

16. (i)

hydrogen atoms

(iii) There are four different types of hydrogen atoms present in the substrate, replacing which following four monochlorinated products are formed.

17. (i)





18. The reaction is—
A + 2B → C

(i) It can be seen that when concentration of A is doubled keeping B constant, then the rate increases by a factor of 4 (from 4.2 × 10-2 to 1.68 × 10-1). This indicates that the rate depends on the square of the concentration of the reactant A. Also, when concentration of reactant B is made four times,
keeping the concentration of reactant A constant, the reaction rate also becomes 4 times (2.4 × 10-2 to 6.0 × 10-3). This indicates that the rate of reaction depends on concentration of reactant B to the first power.
Order with respect to A = 2,

(ii) So, the rate equation will be : order with respect to A= 2:
Rate = k [A]2[B]
Overall order of reaction will be 2 + 1 = 3.

(iii) Rate constant can be calculated by putting the values given.
4.2 × 10-2 M min-1 = k (0.2)2(0.3) M

k = (0.042)/(0.012)=3.5 min-1


19. (i) (a) Four different types of hydrogen atoms are present in the substrate, replacing which following products are formed upon monochlorination :

(CH3)2CHCH2CH2Cl, (CH3)2CHCH(Cl)CH3, (CH3)2C(Cl)CH2CH3 , CH3CH(CH2Cl)CH2CH3.

(b) The para-isomer of dihalobenzene, being symmetrical, fits in a better way than meta-isomer in crystal lattice, hence more energy is required to break the lattice and move, so melting point of the paraisomer is eventually more.

(c) –NO2 is electron withdrawing group, so it decreases the electron availability in the ring rendering it less reactive for electrophilic substitution reactions. 

(ii) (a) When ethyl bromide (bromoethane) is boiled with alcoholic solution of potassium cyanide in aqueous ethanol, ethyl cyanide (ethylnitrile) is formed.

Ethyl cyanide

(b) The reaction of toluene with chlorine in the presence of iron catalysts results in the substitution in the benzene ring. The – CH3 group of toluene is o and p-directing therefore, the product formed in the reaction is the mixture of o-chlorotoluene and p-chlorotoluene. 


20. (i) (a) Tetrachloridonickelate (II) ion
(b) sp3 hybridisation
(c) Tetrahedral shape.
(ii) (a) [CoF6]3– is paramagnetic as it has four unpaired electrons.
(b) [Co(en3)]3+ is more stable.
(c) [Co(en3)]3+ is inner orbital complex.
(d) [CoF6]3– is high spin complex. 


(i) We know molar conductivity,

$$(\lambda_m)=\frac{1000×\text{conductivity}(k)}{\text{concentration(c)}}\\\lambda_m=\frac{1000×5.25×10^{-5}}{2.5×10^{-4}}= 210 cm^2 mol^{-1}\\\lambda^0_{HCOOH}=\lambda^0_{H^+}+\lambda^0_{(HCOOH^-)}= (349.5 + 50.5)\\=400Scm^2mol^{-1}\\=\alpha=\frac{\lambda_m}{\lambda_0}=\frac{210}{400}=0.52\\or, a=52.5\%\\\text{Hence, molar conductivity and degree of dissociation is} 210 cm^2 mol^{–1} and 52.5\%\text{respectively.}$$

(ii) (a) Mg(s) + Cu2+(aq) Mg2+(aq) + Cu(s)
rG0 = - nFE°cell = -2 × 96500 × 2.71

= - 523.03 kJ mol-1 

(b) Hydrogen-oxygen fuel cells and solar cells.

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