a) (iii) < (i) < (iv) < (ii)
b) (iii) < (iv) < (i) < (ii)
c) (i) < (iii) < (iv) < (ii)
d) (iii) < (iv) < (ii) < (i)
In HCN, CN– acts as a nucleophile, attack first that –CHO group which has a maximum positive charge. The magnitude of the (+ve) charge increases by –M and –I group. So reactivity order will be
So, option (1) is the correct answer.
Answer:(a)
a) Van Arkel Method
b) Ostwald Process
c) Mond Process
d) Bredig’s Arc Method
Bredig’s Arc method
Chapter name surface chemistry
Answer: (d)
a) leaves the vapour increases
b) leaves the solution increases
c) leaves the vapour decreases
d) leaves the solution decreases
Hence, Rate at which water molecules leave the vapour increases.
Answer: (a)
(I) both the complexes can be high spin.
(II) Ni(II) complexes can very rarely be low spin.
(III) with strong field ligands, Mn(II) complexes can be low spin.
(IV) the aqueous solution of Mn(II) ions is yellow in colour.
The correct statements are:
a) (I), (III) and (IV) only
b) (I), (II) and (III) only
c) (II), (III) and (IV) only
d) (I) and (II) only
Mn2+ [Ar]3d5 it can form low spin as well as high spin complex depending upon nature of ligand same of Ni2+ ion with coordination no 4. It can be dsp2 or sp3 i:e low spin or high spin depending on the open nature of the ligand.
Answer: (b)
a) in the stratosphere, it forms a protective shield against UV radiation.
b) in the atmosphere, it is depleted by CFCs.
c) in the stratosphere, CFCs release chlorine-free radicals (Cl) which reacts with O3 to give chlorine dioxide radicals.
d) it is a toxic gas and its reaction with NO gives NO2.
Answer: (c)
‘x’, ‘y’ and ‘z’ in these reactions are respectively.
a) 4, 5 & 5
b) 5, 4 & 5
c) 5, 6 & 5
d) 4, 6 & 5
(CH3CO)2O reacts with –OH group to form acetyl derivative, so as the no. of –OH group no. of eq. of (CH3CO)2O will be used
So, x = 4 y = 6 z = 5
So, option (4) will be the correct answer.
Answer: (d)
a) 2,5-dimethyl-5-carboxy-hex-3-enal
b) 2,5-dimethyl-6-oxo-hex-3-enoic acid
c) 6-formyl-2-methyl-hex-3-enoic acid
d) 2,5-dimethyl-6-carboxy-hex-3-enal
2,5–Dimethyl–6–oxohex–3–enoic acid
Answer: (b)
Reason (R): The equilibrium constant of Cu2+ (aq) + S2– (aq) ? CuS (s) is high because the solubility product is low.
a) (A) is false and (R) is true.
b) Both (A) and (R) are false.
c) Both (A) and (R) are true but (R) is not the explanation for (A).
d) Both (A) and (R) are true but (R) is the explanation for (A).
(A) is (B) true & (R) is the correct explanation of (A)
The answer is 4.
Answer: (d)
d = Density, P = Pressure, T = Temperature
The correct statements are:
a) (i)
b) (iv)
c) (iii)
d) (ii)
d = [P × M] / RT
∴ II’ Graph is incorrect.
Answer: (d)
a) Bunsen burner and measuring cylinder
b) Burette and porcelain tile
c) Clamp and phenolphthalein
d) Pipette and distilled water
Bunsen Burner & measuring cylinder are not Required. As titration is already on the exothermic process
Answer: (a)
Carius method
mass % of ‘Br’ = [0.08 / 0.172] * 100 = (8000 / 172) = 46.51%
option (1) mass % = (80 / 95) * 100 =
(2) mass % = (2 * 80 * 100) / 252
(3) mass % = (1 * 80 * 100) / (80 + 72 + 6 + 14) = [8000 / 172] %
(4) mass % = [(1 * 80 * 100) / 109] %
Option (3) matches with the given mass percentage value
Answer: (c)
a) (NH4)2Cr2O7
b) NaN3
c) NH4NO2
d) Pb(NO3)2
The gas (B) is N2 which is found in the air.
NH3 + H2O → NH4OH (weak base)
(NH4)2Cr2O7 → N2 + Cr2O3 + H2O
NaN3 → N2 + Na
NH4NO2 → N2 + H2O
Pb(NO3)2 → PbO + NO2 + O2
Answer: (d)
Option (3) is the correct answer.
Answer: (3)
a) Ionization enthalpy
b) Electronegativity
c) Atomic radius
d) Electron gain enthalpy
Ionisation energy, electronegativity & electron gain enthalpy increase across a period but atomic radius decreases.
Answer: (c)
The internal energy of ‘Ar’ or any gas, has nothing to do with Quantum nature of atom hence
Answer: (4)
Answer: (3)
a) tetrahedral and –0.6Δt
b) tetrahedral and –1.6Δt + 1P
c) octahedral and –1.6Δ0
d) octahedral and –2.4Δ0 + 2P
Heat of combustion ? 1/ stability
Answer (a)
a) Square planar
b) Tetrahedral
c) Square pyramidal
d) Rectangular planar
Incorrect question Option 1 is more appropriate with respect to the given option (Chemical bonding)
(Options are incorrect)
Answer (a)
In 
attack of OH– is not on chiral carbon, it is adjacent to chiral carbon, so the configuration of chiral carbon remains constant.
Answer: (1)
a) Li
b) Cs
c) Rb
d) Na
‘Cs’ is used in the photoelectric cell as its ionisation energy is lowest.
Answer: (b)
(x / m) = KP1/n
log (x / m) = log (k) + (1 / n) log (p)
y = c + mx
Intercept C = logk = 0.4771
Slope = 1 / n = 2, k = 3
(x / m) = k (P)1/n at P = 4 atm
= 3 (4)2
(x / m) = 3 * 16 = 48 Answer
Answer: (48)
Cu(s) + Sn2+ (aq.) → Cu2+(aq.) + Sn(s)
Cu(s) + Sn+2(aq) ? Cu+2 (aq) + Sn(s)
Eºcell = –0.16 – 0.34
= –0.50
ΔGº = –nF Eºcell
= – 2 × 96500 × (–0.5)
= +96500
ΔG = Gº + RT nQ
= 96500 + (25 / 3) × 298 × 2.303 log (1)
ΔG = 96500 Joules
Answer: (96500 Joules)
(Given: ΔHvap for water at 373 K = 41 kJ/mol, R = 8.314 JK–1 mol–1)
H2O (l) → H2O (g)
ΔEvap = ΔHvap – Δng RT
= 41000 × 5 – 5 × 8.314 × 373
= 189494.39
Answer: (189494.39)
Na4 [Fe+2(CN)5 (NOS)]
Na4 [Fe+4O4] [Fe20 (CO)9]
Answer is 6
Answer: (6)
Total chiral carbon = 5
Answer (5)
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