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JEE Advanced · Chemistry · 2024

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1
To form a complete monolayer of acetic acid on 1 g of charcoal, 100 mL of 0.5 M acetic acid was used. Some of the acetic acid remained unadsorbed. To neutralize the unadsorbed acetic acid, 40 mL of 1 M NaOH solution was required. If each molecule of acetic acid occupies P×10−23 m2P\times 10^{-23}\ m^2 surface area on charcoal, the value of PP is: [Use given data: Surface area of charcoal =1.5×102 m2g−1= 1.5\times 10^2\ m^2g^{-1}; Avogadro's number (NA)=6.0×1023 mol−1(N_A) = 6.0\times 10^{23}\ mol^{-1}]
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Your answer: —
Correct answer: 2500
2
An aqueous solution of hydrazine (N2H4N_2H_4) is electrochemically oxidized by O2O_2, thereby releasing chemical energy in the form of electrical energy. One of the products generated from the electrochemical reaction is N2(g)N_2(g). Choose the correct statement(s) about the above process:
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A
OH−OH^- ions react with N2H4N_2H_4 at the anode to form N2(g)N_2(g) and water, releasing 4 electrons to the anode.
B
At the cathode, N2H4N_2H_4 breaks to N2(g)N_2(g) and nascent hydrogen released at the electrode reacts with oxygen to form water.
C
At the cathode, molecular oxygen gets converted to OH−OH^-.
D
Oxides of nitrogen are major by-products of the electrochemical process.
3
For a double strand DNA, one strand is given below (written 5′→3′5'\to 3'): 5′-A G T C A C G T A A G T C-3′5'\text{-A G T C A C G T A A G T C-}3' The amount of energy required to split the double strand DNA into two single strands is _____ kcal mol−1mol^{-1}. [Given: Average energy per H-bond for A-T base pair =1.0= 1.0 kcal mol−1mol^{-1}, G-C base pair =1.5= 1.5 kcal mol−1mol^{-1}, and A-U base pair =1.25=1.25 kcal mol−1mol^{-1}. Ignore electrostatic repulsion between the phosphate groups.]
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Your answer: —
Correct answer: 41
4
The option(s) with correct sequence of reagents for the conversion of PP to QQ (structures shown in the figure) is(are):
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reagents
A
i) Lindlar's catalyst, H2H_2; ii) SnCl2/HClSnCl_2/HCl; iii) NaBH4NaBH_4; iv) H3O+H_3O^+
B
i) Lindlar's catalyst, H2H_2; ii) H3O+H_3O^+; iii) SnCl2/HClSnCl_2/HCl; iv) NaBH4NaBH_4
C
i) NaBH4NaBH_4; ii) SnCl2/HClSnCl_2/HCl; iii) H3O+H_3O^+; iv) Lindlar's catalyst, H2H_2
D
i) Lindlar's catalyst, H2H_2; ii) NaBH4NaBH_4; iii) SnCl2/HClSnCl_2/HCl; iv) H3O+H_3O^+
5
An organic compound PP having molecular formula C6H6O3C_6H_6O_3 gives ferric chloride test and does not have intramolecular hydrogen bond. The compound PP reacts with 3 equivalents of NH2OHNH_2OH to produce oxime QQ. Treatment of PP with excess methyl iodide in the presence of KOH produces compound RR as the major product. Reaction of RR with excess iso\textit{iso}-butylmagnesium bromide followed by treatment with H3O+H_3O^+ gives compound SS as the major product. The total number of methyl (−CH3-CH_3) group(s) in compound SS is _____.
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Your answer: —
Correct answer: 12
6
Paragraph\textbf{Paragraph}: An organic compound PP with molecular formula C9H18O2C_9H_{18}O_2 decolorizes bromine water and also shows a positive iodoform test. PP on ozonolysis followed by treatment with H2O2H_2O_2 gives QQ and RR. Compound QQ shows a positive iodoform test, while compound RR does not give a positive iodoform test. QQ and RR on oxidation with pyridinium chlorochromate (PCC) followed by heating give SS and TT, respectively; both SS and TT show a positive iodoform test. Complete copolymerization of 500 moles of QQ and 500 moles of RR gives one mole of a single acyclic copolymer UU. [Given, atomic mass: H=1H=1, C=12C=12, O=16O=16] Q.14\textbf{Q.14}: Sum of number of oxygen atoms in SS and TT is _____.
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Your answer: —
Correct answer: 2
7
Paragraph\textbf{Paragraph}: An organic compound PP with molecular formula C9H18O2C_9H_{18}O_2 decolorizes bromine water and also shows a positive iodoform test. PP on ozonolysis followed by treatment with H2O2H_2O_2 gives QQ and RR. Compound QQ shows a positive iodoform test, while compound RR does not give a positive iodoform test. QQ and RR on oxidation with pyridinium chlorochromate (PCC) followed by heating give SS and TT, respectively; both SS and TT show a positive iodoform test. Complete copolymerization of 500 moles of QQ and 500 moles of RR gives one mole of a single acyclic copolymer UU. [Given, atomic mass: H=1H=1, C=12C=12, O=16O=16] Q.15\textbf{Q.15}: The molecular weight of UU is _____.
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Your answer: —
Correct answer: 93018
8
Among [Co(CN)4]4−[Co(CN)_4]^{4-}, [Co(CO)3(NO)][Co(CO)_3(NO)], XeF4XeF_4, [PCl4]+[PCl_4]^+, [PdCl4]2−[PdCl_4]^{2-}, [ICl4]−[ICl_4]^-, [Cu(CN)4]3−[Cu(CN)_4]^{3-} and P4P_4, the total number of species with tetrahedral geometry is _____.
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Your answer: —
Correct answer: 5
9
Paragraph\textbf{Paragraph}: When potassium iodide is added to an aqueous solution of potassium ferricyanide, a reversible reaction is observed in which a complex PP is formed. In a strong acidic medium, the equilibrium shifts completely towards PP. Addition of zinc chloride to PP in a slightly acidic medium results in a sparingly soluble complex QQ. Q.16\textbf{Q.16}: The number of moles of potassium iodide required to produce two moles of PP is _____.
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Your answer: —
Correct answer: 2
10
Paragraph\textbf{Paragraph}: When potassium iodide is added to an aqueous solution of potassium ferricyanide, a reversible reaction is observed in which a complex PP is formed. In a strong acidic medium, the equilibrium shifts completely towards PP. Addition of zinc chloride to PP in a slightly acidic medium results in a sparingly soluble complex QQ. Q.17\textbf{Q.17}: The number of zinc ions present in the molecular formula of QQ is _____.
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Your answer: —
Correct answer: 3
11
Vessel-1 contains w2w_2 g of a non-volatile solute XX dissolved in w1w_1 g of water. Vessel-2 contains w2w_2 g of another non-volatile solute YY dissolved in w1w_1 g of water. Both vessels are at the same temperature and pressure. The molar mass of XX is 80%80\% that of YY. The van't Hoff factor for XX is 1.21.2 times that of YY for their respective concentrations. The elevation of boiling point for the solution in Vessel-1 is _____ %\% of the solution in Vessel-2.
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Your answer: —
Correct answer: 150
12
In a metal deficient oxide sample, MXY2O4M_XY_2O_4 (MM and YY are metals), MM is present in both +2+2 and +3+3 oxidation states and YY is in +3+3 oxidation state. If the fraction of M2+M^{2+} ions present in MM is 13\dfrac{1}{3}, the value of XX is:
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A
0.25
B
0.33
C
0.67
D
0.75
13
In the following reaction sequence, the major product QQ is: L-Glucose is first treated with (i) HIHI, Δ\Delta (excess HI on prolonged heating), and then (ii) Cr2O3Cr_2O_3 at 775 K775\ K, 1010-20 atm20\ atm, to give the major product PP. Compound PP is then treated with excess Cl2Cl_2 under UV light (hνh\nu) to give the major product QQ. Identify QQ from the structures shown (A)-(D).
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(A) (B) (C) (D)
A
Hexachlorobenzene
B
1,2,4,51,2,4,5-Tetrachlorobenzene
C
1,41,4-Dichlorobenzene
D
1,2,3,4,5,61,2,3,4,5,6-Hexachlorocyclohexane
14
The species formed on fluorination of phosphorus pentachloride in a polar organic solvent are:
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A
[PF4]+[PF6]−[PF_4]^+[PF_6]^- and [PCl4]+[PF6]−[PCl_4]^+[PF_6]^-
B
[PCl4]+[PCl4F2]−[PCl_4]^+[PCl_4F_2]^- and [PCl4]+[PF6]−[PCl_4]^+[PF_6]^-
C
PF3PF_3 and PCl3PCl_3
D
PF5PF_5 and PCl3PCl_3
15
The compound(s) having peroxide linkage is(are):
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A
H2S2O7H_2S_2O_7
B
H2S2O8H_2S_2O_8
C
H2S2O5H_2S_2O_5
D
H2SO5H_2SO_5
16
According to Bohr's model, the highest kinetic energy is associated with the electron in the:
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A
first orbit of H atom
B
first orbit of He+^+
C
second orbit of He+^+
D
second orbit of Li2+^{2+}
17
A sample initially contains only U-238 isotope of uranium. With time, some of the U-238 radioactively decays into Pb-206 while the rest of it remains undisintegrated. When the age of the sample is P×108P\times 10^8 years, the ratio of mass of Pb-206 to that of U-238 in the sample is found to be 7. The value of PP is _____. [Given: Half-life of U-238 is 4.5×1094.5\times 10^9 years; log⁡e2=0.693\log_e 2 = 0.693]
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Your answer: —
Correct answer: 143
18
List-I contains various reaction sequences (starting materials shown in the figure, labelled P, Q, R, S) and List-II contains the possible products (shown in the figure, labelled 1-5). Match each entry in List-I with the appropriate entry in List-II. (P) Cyclohexene →i) O3,Zn; ii) aq. NaOH,Δ; iii) ethylene glycol, PTSA; iv) a) BH3, b) H2O2,NaOH; v) H3O+; vi) NaBH4\xrightarrow{\text{i) }O_3, Zn\text{; ii) aq. }NaOH,\Delta\text{; iii) ethylene glycol, PTSA; iv) a) }BH_3\text{, b) }H_2O_2, NaOH\text{; v) }H_3O^{+}\text{; vi) }NaBH_4} (Q) A methyl-substituted cyclohexene (methyl on the alkene carbon, shown in figure) - same 6-step sequence as (P). (R) 2-methylcyclopent-2-en-1-one →i) ethylene glycol, PTSA; ii) a) Hg(OAc)2,H2O, b) NaBH4; iii) H3O+; iv) NaBH4\xrightarrow{\text{i) ethylene glycol, PTSA; ii) a) }Hg(OAc)_2, H_2O\text{, b) }NaBH_4\text{; iii) }H_3O^{+}\text{; iv) }NaBH_4} (S) 2-methylcyclopent-2-en-1-one →i) ethylene glycol, PTSA; ii) a) BH3, b) H2O2,NaOH; iii) H3O+; iv) NaBH4\xrightarrow{\text{i) ethylene glycol, PTSA; ii) a) }BH_3\text{, b) }H_2O_2, NaOH\text{; iii) }H_3O^{+}\text{; iv) }NaBH_4} (A) P-3, Q-5, R-4, S-1 (B) P-3, Q-2, R-4, S-1 (C) P-3, Q-5, R-1, S-4 (D) P-5, Q-2, R-4, S-1
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P Q R S
A
P-3, Q-5, R-4, S-1
B
P-3, Q-2, R-4, S-1
C
P-3, Q-5, R-1, S-4
D
P-5, Q-2, R-4, S-1
19
List-I contains various reaction sequences (starting materials shown in the figure, labelled P, Q, R, S) and List-II contains different phenolic compounds (shown in the figure, labelled 1-5). Match each entry in List-I with the appropriate entry in List-II. (P) Benzenesulfonic acid →i) molten NaOH,H3O+; ii) conc. HNO3\xrightarrow{\text{i) molten }NaOH, H_3O^{+}\text{; ii) conc. }HNO_3} (Q) Nitrobenzene →i) conc. HNO3/H2SO4; ii) Sn/HCl; iii) NaNO2/HCl,0-5∘C; iv) H2O; v) conc. HNO3/H2SO4\xrightarrow{\text{i) conc. }HNO_3/H_2SO_4\text{; ii) }Sn/HCl\text{; iii) }NaNO_2/HCl, 0\text{-}5^{\circ}C\text{; iv) }H_2O\text{; v) conc. }HNO_3/H_2SO_4} (R) Resorcinol (benzene-1,3-diol) →i) conc. H2SO4; ii) conc. HNO3; iii) H3O+,Δ\xrightarrow{\text{i) conc. }H_2SO_4\text{; ii) conc. }HNO_3\text{; iii) }H_3O^{+}, \Delta} (S) Toluene →i) a) KMnO4/KOH,Δ, b) H3O+; ii) conc. HNO3/H2SO4,Δ; iii) a) SOCl2, b) NH3; iv) Br2,NaOH; v) NaNO2/HCl,0-5∘C; vi) H2O\xrightarrow{\text{i) a) }KMnO_4/KOH,\Delta\text{, b) }H_3O^{+}\text{; ii) conc. }HNO_3/H_2SO_4,\Delta\text{; iii) a) }SOCl_2\text{, b) }NH_3\text{; iv) }Br_2, NaOH\text{; v) }NaNO_2/HCl, 0\text{-}5^{\circ}C\text{; vi) }H_2O} (A) P-2, Q-3, R-4, S-5 (B) P-2, Q-3, R-5, S-1 (C) P-3, Q-5, R-4, S-1 (D) P-3, Q-2, R-5, S-4
Not attempted
P Q R S
A
P-2, Q-3, R-4, S-5
B
P-2, Q-3, R-5, S-1
C
P-3, Q-5, R-4, S-1
D
P-3, Q-2, R-5, S-4
20
In a conductometric titration, a small volume of titrant of higher concentration is added stepwise to a much larger volume of titrate of much lower concentration (so dilution effects can be neglected), and the conductance is measured after each addition. The limiting ionic conductivities Λ0\Lambda^0 (in mS m2 mol−1mS\,m^2\,mol^{-1}) are: Ag+=6.2Ag^{+}=6.2, K+=7.4K^{+}=7.4, Na+=5.0Na^{+}=5.0, H+=35.0H^{+}=35.0, NO3−=7.2NO_3^{-}=7.2, Cl−=7.6Cl^{-}=7.6, SO42−=16.0SO_4^{2-}=16.0, OH−=19.9OH^{-}=19.9, CH3COO−=4.1CH_3COO^{-}=4.1. List-I: (P) Titrate KClKCl, Titrant AgNO3AgNO_3; (Q) Titrate AgNO3AgNO_3, Titrant KClKCl; (R) Titrate NaOHNaOH, Titrant HClHCl; (S) Titrate NaOHNaOH, Titrant CH3COOHCH_3COOH. List-II: conductance-vs-volume-of-titrant graphs (1)-(5), shown in the figure. Match List-I with List-II and choose the correct option. (A) P-4, Q-3, R-2, S-5 (B) P-2, Q-4, R-3, S-1 (C) P-3, Q-4, R-2, S-5 (D) P-4, Q-3, R-2, S-1
Not attempted
(1)ConductanceVolume of titrant (2)ConductanceVolume of titrant (3)ConductanceVolume of titrant (4)ConductanceVolume of titrant (5)ConductanceVolume of titrant
A
P-4, Q-3, R-2, S-5
B
P-2, Q-4, R-3, S-1
C
P-3, Q-4, R-2, S-5
D
P-4, Q-3, R-2, S-1
21
Among V(CO)6V(CO)_6, Cr(CO)5Cr(CO)_5, Cu(CO)3Cu(CO)_3, Mn(CO)5Mn(CO)_5, Fe(CO)5Fe(CO)_5, [Co(CO)3]3−[Co(CO)_3]^{3-}, [Cr(CO)4]4−[Cr(CO)_4]^{4-}, and Ir(CO)3Ir(CO)_3, the total number of species isoelectronic with Ni(CO)4Ni(CO)_4 is ______. [Given atomic number: V=23V=23, Cr=24Cr=24, Mn=25Mn=25, Fe=26Fe=26, Co=27Co=27, Ni=28Ni=28, Cu=29Cu=29, Ir=77Ir=77]
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Your answer: —
Correct answer: 1
22
Consider the reaction 2H2(g)+2NO(g)→N2(g)+2H2O(g)2H_2(g) + 2NO(g) \rightarrow N_2(g) + 2H_2O(g), which follows the mechanism: step 1 (fast equilibrium, rate constants k1k_1 forward / k−1k_{-1} reverse): 2NO(g)⇌N2O2(g)2NO(g) \rightleftharpoons N_2O_2(g); step 2 (slow, rate constant k2k_2): N2O2(g)+H2(g)→N2O(g)+H2O(g)N_2O_2(g) + H_2(g) \rightarrow N_2O(g) + H_2O(g); step 3 (fast, rate constant k3k_3): N2O(g)+H2(g)→N2(g)+H2O(g)N_2O(g) + H_2(g) \rightarrow N_2(g) + H_2O(g). The order of the reaction is ______.
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Your answer: —
Correct answer: 3
23
Consider the following volume-temperature (VV-TT) diagram (shown in the figure) for the expansion of 55 moles of an ideal monoatomic gas: state XX is at T=335 KT=335\text{ K}, V=10 LV=10\text{ L}; state YY is at T=335 KT=335\text{ K}, V=20 LV=20\text{ L}; state ZZ is at T=415 KT=415\text{ K}, V=20 LV=20\text{ L}. Considering only PP-VV work, the total change in enthalpy (in Joule) for the transformation of state in the sequence X→Y→ZX \to Y \to Z is ______. [Use CV,m=12 J K−1mol−1C_{V,m} = 12\text{ J K}^{-1}\text{mol}^{-1}, R=8.3 J K−1mol−1R = 8.3\text{ J K}^{-1}\text{mol}^{-1}]
Not attempted
Your answer: —
Correct answer: 8120
24
The option(s) in which at least three molecules follow the Octet Rule is(are) (A) CO2CO_2, C2H4C_2H_4, NONO and HClHCl (B) NO2NO_2, O3O_3, HClHCl and H2SO4H_2SO_4 (C) BCl3BCl_3, NONO, NO2NO_2 and H2SO4H_2SO_4 (D) CO2CO_2, BCl3BCl_3, O3O_3 and C2H4C_2H_4
Not attempted
A
CO2CO_2, C2H4C_2H_4, NONO and HClHCl
B
NO2NO_2, O3O_3, HClHCl and H2SO4H_2SO_4
C
BCl3BCl_3, NONO, NO2NO_2 and H2SO4H_2SO_4
D
CO2CO_2, BCl3BCl_3, O3O_3 and C2H4C_2H_4
25
Reaction of iso-propylbenzene (cumene) with O2O_2 followed by treatment with H3O+H_3O^{+} forms phenol and a by-product PP. Reaction of PP with 3 equivalents of Cl2Cl_2 gives compound QQ. Treatment of QQ with Ca(OH)2Ca(OH)_2 produces compound RR and calcium salt SS. The correct statement(s) regarding P,Q,RP, Q, R and SS is(are) (A) Reaction of PP with RR in the presence of KOHKOH followed by acidification gives the compound shown in the figure below. (B) Reaction of RR with O2O_2 in the presence of light gives phosgene gas. (C) QQ reacts with aqueous NaOHNaOH to produce Cl3CCH2OHCl_3CCH_2OH and Cl3CCOONaCl_3CCOONa. (D) SS on heating gives PP.
Not attempted
A
PP + RR with KOHKOH then acidification gives the compound shown in the figure (chlorotone).
B
RR with O2O_2/light gives phosgene gas.
C
QQ with aqueous NaOHNaOH gives Cl3CCH2OHCl_3CCH_2OH and Cl3CCOONaCl_3CCOONa.
D
SS on heating gives PP.
26
Complete reaction of acetaldehyde with excess formaldehyde, upon heating with conc. NaOHNaOH solution, gives PP and QQ. Compound PP does not give Tollens' test, whereas QQ on acidification gives a positive Tollens' test. Treatment of PP with excess cyclohexanone in the presence of a catalytic amount of pp-toluenesulfonic acid (PTSAPTSA) gives product RR. Sum of the number of methylene groups (−CH2−-CH_2-) and oxygen atoms in RR is ______.
Not attempted
Your answer: —
Correct answer: 18
27
In the reaction sequence H−C≡C−(CH2)15−CO2Et→i) Hg2+,H3O+; ii) Zn-Hg/HCl; iii) H3O+,ΔPH-C\equiv C-(CH_2)_{15}-CO_2Et \xrightarrow{\text{i) }Hg^{2+}, H_3O^{+}\text{; ii) }Zn\text{-}Hg/HCl\text{; iii) }H_3O^{+}, \Delta} P, the major product PP is formed. Glycerol reacts completely with excess PP in the presence of an acid catalyst to form QQ. Reaction of QQ with excess NaOHNaOH followed by treatment with CaCl2CaCl_2 yields Ca-soap RR, quantitatively. Starting with one mole of QQ, the amount of RR produced in gram is ______. [Given atomic weight: H=1,C=12,N=14,O=16,Na=23,Cl=35,Ca=40H=1, C=12, N=14, O=16, Na=23, Cl=35, Ca=40]
Not attempted
Your answer: —
Correct answer: 909
28
Among the following, the correct statement(s) for electrons in an atom is(are) (A) Uncertainty principle rules out the existence of definite paths for electrons. (B) The energy of an electron in the 2s2s orbital of an atom is lower than the energy of an electron that is infinitely far away from the nucleus. (C) According to Bohr's model, the most negative energy value for an electron is given by n=1n=1, which corresponds to the most stable orbit. (D) According to Bohr's model, the magnitude of velocity of electrons increases with increase in values of nn.
Not attempted
A
Uncertainty principle rules out the existence of definite paths for electrons.
B
The energy of an electron in the 2s2s orbital is lower than that of an electron infinitely far away.
C
The most negative energy value (n=1n=1) corresponds to the most stable orbit (Bohr's model).
D
The magnitude of electron velocity increases with increasing nn (Bohr's model).
29
Among the following options, select the option in which each complex in Set-I shows geometrical isomerism and the two complexes in Set-II are ionization isomers of each other. [en=H2NCH2CH2NH2en = H_2NCH_2CH_2NH_2] (A) Set-I: [Ni(CO)4][Ni(CO)_4] and [PdCl2(PPh3)2][PdCl_2(PPh_3)_2]; Set-II: [Co(NH3)5Cl]SO4[Co(NH_3)_5Cl]SO_4 and [Co(NH3)5(SO4)]Cl[Co(NH_3)_5(SO_4)]Cl (B) Set-I: [Co(en)(NH3)2Cl2][Co(en)(NH_3)_2Cl_2] and [PdCl2(PPh3)2][PdCl_2(PPh_3)_2]; Set-II: [Co(NH3)6][Cr(CN)6][Co(NH_3)_6][Cr(CN)_6] and [Cr(NH3)6][Co(CN)6][Cr(NH_3)_6][Co(CN)_6] (C) Set-I: [Co(NH3)3(NO2)3][Co(NH_3)_3(NO_2)_3] and [Co(en)2Cl2][Co(en)_2Cl_2]; Set-II: [Co(NH3)5Cl]SO4[Co(NH_3)_5Cl]SO_4 and [Co(NH3)5(SO4)]Cl[Co(NH_3)_5(SO_4)]Cl (D) Set-I: [Cr(NH3)5Cl]Cl2[Cr(NH_3)_5Cl]Cl_2 and [Co(en)(NH3)2Cl2][Co(en)(NH_3)_2Cl_2]; Set-II: [Cr(H2O)6]Cl3[Cr(H_2O)_6]Cl_3 and [Cr(H2O)5Cl]Cl2⋅H2O[Cr(H_2O)_5Cl]Cl_2\cdot H_2O
Not attempted
A
Set-I: [Ni(CO)4][Ni(CO)_4], [PdCl2(PPh3)2][PdCl_2(PPh_3)_2]; Set-II: [Co(NH3)5Cl]SO4[Co(NH_3)_5Cl]SO_4, [Co(NH3)5(SO4)]Cl[Co(NH_3)_5(SO_4)]Cl
B
Set-I: [Co(en)(NH3)2Cl2][Co(en)(NH_3)_2Cl_2], [PdCl2(PPh3)2][PdCl_2(PPh_3)_2]; Set-II: [Co(NH3)6][Cr(CN)6][Co(NH_3)_6][Cr(CN)_6], [Cr(NH3)6][Co(CN)6][Cr(NH_3)_6][Co(CN)_6]
C
Set-I: [Co(NH3)3(NO2)3][Co(NH_3)_3(NO_2)_3], [Co(en)2Cl2][Co(en)_2Cl_2]; Set-II: [Co(NH3)5Cl]SO4[Co(NH_3)_5Cl]SO_4, [Co(NH3)5(SO4)]Cl[Co(NH_3)_5(SO_4)]Cl
D
Set-I: [Cr(NH3)5Cl]Cl2[Cr(NH_3)_5Cl]Cl_2, [Co(en)(NH3)2Cl2][Co(en)(NH_3)_2Cl_2]; Set-II: [Cr(H2O)6]Cl3[Cr(H_2O)_6]Cl_3, [Cr(H2O)5Cl]Cl2⋅H2O[Cr(H_2O)_5Cl]Cl_2\cdot H_2O
30
Among the following complexes, the total number of diamagnetic species is ______: [Mn(NH3)6]3+[Mn(NH_3)_6]^{3+}, [MnCl6]3−[MnCl_6]^{3-}, [FeF6]3−[FeF_6]^{3-}, [CoF6]3−[CoF_6]^{3-}, [Fe(NH3)6]3+[Fe(NH_3)_6]^{3+}, and [Co(en)3]3+[Co(en)_3]^{3+}. [Given atomic number: Mn=25Mn=25, Fe=26Fe=26, Co=27Co=27; en=H2NCH2CH2NH2en = H_2NCH_2CH_2NH_2]
Not attempted
Your answer: —
Correct answer: 1
31
Aspartame, an artificial sweetener, is the dipeptide aspartyl phenylalanine methyl ester, built from L-aspartic acid and L-phenylalanine (structures of these two free amino acids, and the four candidate structures A-D for aspartame, are shown in the figure). Which structure correctly represents aspartame?
Not attempted
A B C D
A
Structure A (see figure)
A B C D
B
Structure B (see figure)
A B C D
C
Structure C (see figure)
A B C D
D
Structure D (see figure)
A B C D
32
At room temperature, disproportionation of an aqueous solution of in situ generated nitrous acid (HNO2HNO_2) gives the species
Not attempted
A
H3O+,NO3−H_3O^{+}, NO_3^{-} and NONO
B
H3O+,NO3−H_3O^{+}, NO_3^{-} and NO2NO_2
C
H3O+,NO−H_3O^{+}, NO^{-} and NO2NO_2
D
H3O+,NO3−H_3O^{+}, NO_3^{-} and N2ON_2O
33
Based on the VSEPR model, match the xenon compounds in List-I with the corresponding geometries and number of lone pairs on xenon in List-II, and choose the correct option. List-I: (P) XeF2XeF_2; (Q) XeF4XeF_4; (R) XeO3XeO_3; (S) XeO3F2XeO_3F_2. List-II: (1) Trigonal bipyramidal, 2 lone pairs; (2) Tetrahedral, 1 lone pair; (3) Octahedral, 2 lone pairs; (4) Trigonal bipyramidal, 0 lone pairs; (5) Trigonal bipyramidal, 3 lone pairs. (A) P-5, Q-2, R-3, S-1 (B) P-5, Q-3, R-2, S-4 (C) P-4, Q-3, R-2, S-1 (D) P-4, Q-2, R-5, S-3
Not attempted
A
P-5, Q-2, R-3, S-1
B
P-5, Q-3, R-2, S-4
C
P-4, Q-3, R-2, S-1
D
P-4, Q-2, R-5, S-3
34
A closed vessel contains 10 g10\text{ g} of an ideal gas XX at 300 K300\text{ K}, which exerts 2 atm2\text{ atm} pressure. At the same temperature, 80 g80\text{ g} of another ideal gas YY is added to it and the pressure becomes 6 atm6\text{ atm}. The ratio of root mean square velocities of XX and YY at 300 K300\text{ K} is
Not attempted
A
22:32\sqrt{2} : \sqrt{3}
B
22:12\sqrt{2} : 1
C
1:21 : 2
D
2:12 : 1
Tip\textbf{Tip}: At fixed VV and TT, P∝nP \propto n (from PV=nRTPV=nRT). Combine this with n=w/Mn=w/M to get the molar-mass ratio, then use vrms=3RT/Mv_{rms}=\sqrt{3RT/M}.
Deconstruction\textbf{Deconstruction}: Two ideal gases occupy the same closed vessel (same VV) at the same TT. We know each gas's mass and the pressure before/after mixing, and must find the ratio of rms speeds. Strategy\textbf{Strategy}: Since V,TV,T are common to both gases, Dalton's law lets us read off each gas's partial pressure, and PV=nRTPV=nRT converts that directly into a mole ratio. The given masses then give a molar-mass ratio, which plugs into vrms=3RT/Mv_{rms}=\sqrt{3RT/M}. Execution\textbf{Execution}: For XX alone: PXV=nXRT⇒2V=nXRTP_X V = n_X RT \Rightarrow 2V = n_X RT. After adding YY, total pressure is 6 atm6\text{ atm}, so PY=6−2=4 atmP_Y = 6-2 = 4\text{ atm}: 4V=nYRT4V = n_Y RT. Dividing, nYnX=42=2\dfrac{n_Y}{n_X} = \dfrac{4}{2} = 2. Now nX=10MXn_X = \dfrac{10}{M_X} and nY=80MYn_Y = \dfrac{80}{M_Y}, so 80/MY10/MX=2⇒80 MX10 MY=2⇒8MXMY=2⇒MY=4MX\dfrac{80/M_Y}{10/M_X} = 2 \Rightarrow \dfrac{80\,M_X}{10\,M_Y} = 2 \Rightarrow 8\dfrac{M_X}{M_Y} = 2 \Rightarrow M_Y = 4M_X. Since vrms=3RT/Mv_{rms} = \sqrt{3RT/M}, vrms,Xvrms,Y=MYMX=4=2\dfrac{v_{rms,X}}{v_{rms,Y}} = \sqrt{\dfrac{M_Y}{M_X}} = \sqrt{4} = 2. Verification\textbf{Verification}: Substitute back with MX=mM_X = m, MY=4mM_Y = 4m: nX=10/mn_X = 10/m, nY=80/(4m)=20/mn_Y = 80/(4m) = 20/m, giving nY/nX=2n_Y/n_X = 2, exactly the mole ratio required by the pressure data (4:24:2). Internally consistent. Final Answer\textbf{Final Answer}: Option (D), vrms,X:vrms,Y=2:1v_{rms,X} : v_{rms,Y} = 2:1.