1) CH_{3}COOK
2) FeCl_{3}
3) Pb(CH_{3}COO)_{2}
4) Al(CN)_{3}
CH_{3}COOK + H_{2}O > CH_{3}COOH + KOH (basic)
FeCl_{3} – Acidic solution
Al(CN)_{3} – Salt of weak acid and weak base
Pb(CH_{3}COO)_{2} – Salt of weak acid and weak base
CH_{3}COOK is salt of weak acid and strong base.
Hence solution of CH_{3}COOK is basic.
Answer: (1)
Kjeldahl method is not applicable for compounds containing nitrogen in nitro, and azo groups and nitrogen in ring, as N of these compounds does not change to ammonium sulphate under these conditions. Hence only aniline can be used for estimation of nitrogen by Kjeldahl’s method.
Answer: (1)
1) AlCl_{3} and SiCl_{4}
2) PH_{3} and SiCl_{4}
3) BCl_{3} and AlCl_{3}
4) PH_{3} and BCl_{3}
BCl_{3} – electron deficient, incomplete octet
AlCl_{3} – electron deficient, incomplete octet
SiCl_{4} can accept lone pair of electron in dorbital of silicon hence it can act as Lewis acid.
Although the most suitable answer is (3). However, both option (3) and (1) can be considered as correct answers.
Eg. Hydrolysis of SiCl_{4}
Answer: (3)
Answer: (4)
Base 
Acid 
End point 

1. 
Strong 
Strong 
Pinkish red to yellow 
2. 
Weak 
Strong 
Yellow to Pinkish red 
3. 
Strong 
Strong 
Pink to colourless 
4. 
Weak 
Strong 
Colourless to pink 
The pH range of methyl orange is
Weak base is having pH greater than 7. When methyl orange is added to weak base solution, the solution becomes yellow. This solution is titrated by strong acid and at the end point pH will be less than 3.1. Therefore solution becomes pinkish red.
Answer: (2)
1) 3×10^{20}
2) 6×10^{21}
3) 5×10^{19}
4) 5×10^{8}
K_{a1}.K_{a2} = K_{eq}
Answer: (1)
1) 452.46
2) 3260
3) 3267.6
4) 4152.6
C_{6}H_{6} (l) + (15/2)O_{2} (g) > 6CO_{2} (g) + 3H_{2}O (l)
Δn_{g} = 6(15/2)
= 3/2
ΔH = ΔU +Δn_{g}RT
= 3263.9 +(3/2)×8.314×298×10^{3}
= 3263.9 + (3.71)
= 3267.6 kJ mol^{1}
Answer: (3)
1) (NH_{4})_{2}Cr_{2}O_{7}
2) NH_{4}NO_{2}
3) (NH_{4})_{2}SO_{4}
4) Ba(N_{3})_{2}
Among all the given compounds, only (NH_{4})_{2}SO_{4 }do not form dinitrogen on heating, it produces ammonia gas.
Answer: (3)
1) 0.8 hours
2) 3.2 hours
3) 1.6 hours
4) 6.4 hours
B_{2}H_{6} +3O_{2} > B_{2}O_{3} +3H_{2}O
27.66 of B_{2}H_{6} = 1 mole of B_{2}H_{6} which requires three moles of oxygen (O_{2}) for complete burning
6H_{2}O > 6H_{2} +3O_{2} (On electrolysis)
Number of faradays = 12 = amount of charge
12×96500 = i×t
12×96500 = 100×t
t = 12×96500/100 sec
t = 12×96500/(100×3600) hour
t = 3.2 hours
Answer: (2)
1) 6
2) 9
3) 12
4) 3
Structure of I_{3}^{–}
Number of lone pairs in I_{3}^{–} is 9.
Answer: (2)
1) Ca
2) Al
3) Fe
4) Zn
Al_{2}O_{3} is used in column chromatography.
Answer: (2)
1) He_{2}^{+}
2) H_{2}^{–}
3) H_{2}^{2}
4) He_{2}^{2+}
Molecule having zero bond order will not be a viable molecule.
Answer: (3)
1) (b) < (a) < (c) < (d)
2) (b) < (a) < (d) < (c)
3) (d) < (b) < (a) < (c)
4) (a) < (b) < (c) < (d)
Correct order of basicity is: b < a < d < c
Answer: (2)
1) Vacancy defect
2) Frenkel defect
3) Metal deficiency defect
4) Schottky defect
In Frenkel defect, cation is dislocated from its normal lattice site to an interstitial site.
Answer (2)
KCl, PH_{3}, O_{2}, B_{2}H_{6}, H_{2}SO_{4}
1) KCl, H_{2}SO_{4}
2) KCl
3) KCl, B_{2}H_{6}
4) KCl, B_{2}H_{6}, PH_{3}
KCl – Ionic bond between K^{+} and Cl^{–}
PH_{3}– Covalent bond between P and H
O_{2 }– Covalent bond between O atoms
B_{2}H_{6} – Covalent bond between B and H atoms
H_{2}SO_{4}– Covalent bond between S and O and also between O and H.
Compound having no covalent bonds is KCl only.
Answer: (2)
1) +3, +2 and +4
2) +3, 0 and +6
3) +3, 0 and +4
4) +3, +4 and +6
[Cr(H_{2}O)_{6}]Cl_{3} ⇒ x+0×61×3 = 0
x = +3
[Cr(C_{6}H_{6})_{2}] ⇒ x+2×0 = 0x = 0
K_{2}[Cr(CN)_{2}(O)_{2}(O_{2})(NH_{3})] ⇒ 1×2+x1×222×1 = 0
⇒ x6 = 0
So x = +6
Answer: (2)
1) (H_{2}O+O_{2}) and (H_{2}O+OH^{–})
2) H_{2}O and (H_{2}O+O_{2})
3) H_{2}O and (H_{2}O+ OH^{–})
4) (H_{2}O+O_{2}) and H_{2}O
Answer: (2)
1) 1Hexene
2) Hexanoic acid
3) 6iodohexanal
4) nHexane
Answer: (4)
At pH (7.4) major form of histamine is protonated at primary amine.
1) [3Ca(F)_{2}.Ca(OH)_{2}]
2) [3Ca_{3}(PO_{4})_{2}.CaF_{2}]
3) [3{Ca(OH)_{2}}.Ca(F)_{2}]
4) [CaF_{2}]
F^{–} ions make the teeth enamel harder by converting
Answer: (2)
(I) Two isomers are produced if the reactant complex ion is a cisisomer
(II) Two isomers are produced if the reactant complex ion is a transisomer.
(III) Only one isomer is produced if the reactant complex ion is a transisomer.
(IV) Only one isomer is produced if the reactant complex ion is a cisisomer.
The correct statements are:
1) (I) and (III)
2) (III) and (IV)
3) (II) and (IV)
4) (I) and (II)
So option (1) is correct.
Answer: (1)
1) NaBH_{4}
2) Na/liq. NH_{3}
3) Sn – HCl
4) H_{2}Pd/C, BaSO_{4}
So, option (2) is correct.
Answer: (2)
1) C_{2}H_{4}O
2) C_{3}H_{4}O_{2}
3) C_{2}H_{4}O_{3}
4) C_{3}H_{6}O_{3}
Element 
Relative mass 
Relative mole 
Simplest whole number ratio 
C 
6 
6/12 = 0.5 
1 
H 
1 
1/1 = 1 
2 
So, X = 1, Y = 2
Equation for combustion of C_{X}H_{Y}
C_{X}H_{Y} +(X+Y/4)O_{2} > XCO_{2} +(Y/2)H_{2}O
Oxygen atoms required = 2(X+Y/4)
Given 2(X+Y/4) = 2Z
⇒(1+2/4) = Z
⇒ Z = 1.5
Molecule can be written as C_{X}H_{Y}O_{Z}
⇒ C_{1}H_{2}O_{3/2}
⇒ C_{2}H_{4}O_{3}
Answer: (3)
Answer: (2)
CH_{3}O^{–} is a strong base and strong nucleophile, so favourable condition is S_{N}2/E2.
Given alkyl halide is 2^{0} and βC’s are 4^{0} and 2^{0} , so sufficiently hindered, therefore, E2 dominates over S_{N}2.
Also, polarity of CH_{3}OH (solvent) is not as high as H_{2}O so E1 is also dominated by E2.
Answer: (1)
1) B and C
2) C and D
3) A and D
4) A and B
Equilibrium constant K =
ln K = ln (A_{f}/A_{b}) – (ΔH^{0}/R)(1/T)
y = c+mx
comparing with equation of straight line,
slope = ΔH^{0}/R
Since, reaction is exothermic, ΔH^{0 }= ve, therefore,
slope = +ve.
Answer: (4)
Answer: (3)
1) 2 ×10^{9} M
2) 1.1×10^{9} M
3) 1.0×10^{9} M
4) 5 ×10^{9} M
Final concentration of [SO_{4}^{—}] = 50×1/500 = 0.1M
K_{sp} of BaSO_{4},
[Ba^{2+}] [SO_{4}^{2}] = 1×10^{10} [Ba^{2+}] [0.1] = 10^{10}/0.1 = 10^{9} MConcentration of Ba^{2+} in final solution = 10^{9} M
Concentration of Ba^{2+} in original solution.
M_{1}V_{1} = M_{2}V_{2}
M_{1}(50050) = 10^{9}(500)
M_{1} = 1.11×10^{9} M
Answer: (2)
1) 3
2) 1
3) 0
4) 2
Assume the order of reaction with respect to acetaldehyde is x.
Condition 1:
Rate = k[CH_{3}CHO]^{x}
1 = k[363×0.95]^{x}
1 = k[344.85]^{x} ..(i)
Condition 2:
0.5 = k[363×0.67]^{x}
0.5 = k[243.21]^{x} ..(ii)
Divide equation (i) by (ii),
1/0.5 = (344.85/243.21)^{x}
⇒2 = 1.414^{x}
⇒ x = 2
Answer: (4)
1) [Co(H_{2}O)_{5}Cl]Cl_{2}. H_{2}O
2) [Co(H_{2}O)_{4}Cl_{2} ]Cl. 2H_{2}O
3) [Co(H_{2}O)_{3}Cl_{3}]. 3H_{2}O
4) [Co(H_{2}O)_{6 }]Cl_{3}
The solution which shows maximum freezing point must have minimum number of solute particles.
(1) [Co(H_{2}O)_{5}Cl]Cl_{2}. H_{2}O >[Co(H_{2}O)_{5}Cl]^{2+} +2Cl^{–}, i = 3
(2) [Co(H_{2}O)_{4}Cl_{2} ]Cl. 2H_{2}O > [Co(H_{2}O)_{4}Cl_{2}]^{+} +Cl^{–}, i = 2
(3) [Co(H_{2}O)_{3}Cl_{3}]. 3H_{2}O > [Co(H_{2}O)_{3}Cl_{3} ] , i = 1
(4) [Co(H_{2}O)_{6 }]Cl_{3} > [Co(H_{2}O)_{6 }]^{3+} +3Cl^{–} , i = 4
So, solution of 1 molal [Co(H_{2}O)_{3}Cl_{3}]. 3H_{2}O will have minimum number of particles in aqueous state.
Answer: (3)
Video Lessons – January 10 Shift 1 Chemistry
JEE Main 2018 Chemistry Paper With Solutions Shift 1 January 10
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