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Chemistry
Single correct answer type:
1. With what velocity should an πΌ-particle travel towards the nucleus of a copper atom to
arrive at a distance of 10β13π from the nucleus of the copper atom?
(πΎ = 9 Γ 109ππ2 πΆ2β Mass of πΌ β particle = 6.64 Γ 10β27ππ)
(A) 6.34 Γ 106 ππ β1 (B) 6.34 Γ 105 ππ β1
(C) 5.34 Γ 106 ππ β1 (D) 5.34 Γ 105 ππ β1
Solution: (A)
The potential energy of the πΌ-particle at a distance of 10β13π from the nucleus of the
copper atom is
π =πΎπ1 π2π
πΎ = 9 Γ 109 π.π2 πΆ2β
π1 = 29 Γ 16 Γ 10β19 Coulomb (For copper atom)
π2 = 2 Γ 16 Γ 10β19 Coulomb (For πΌ-particles)
π = 10β13 π
πΈ =9 Γ 109 Γ 29 Γ 16 Γ 10β19 Γ 2 Γ 16 Γ 10β19
10β13
= 1.336 Γ 10β13 π.π
The πΌ-particle must have such kinetic energy.
β΄ πΎ. πΈ = πΈ
or 1
2ππ£2 = 1.336 Γ 10β13
or π£2 =2Γ1.336Γ10β13
6.64Γ10β27
or π£ = 6.34 Γ 106ππ β1
2. What is the ratio of the velocities of πΆπ»4 and π2 molecules so that they are associated
with de-Broglie waves of equal wavelengths?
(A) 1 βΆ 2 (B) 2 βΆ 1 (C) 3 βΆ 2 (D) 1 βΆ 3
Solution: (B)
According to de-Broglie equation, π =β
ππ£
For methane (πΆπ»4)
ππΆπ»4 =β
ππΆβ4 Γ π£πΆπ»4
For oxygen (π2),
ππ2 =β
ππΆ2 Γ π£π2
Since, the wavelength of πΆπ»4 and π2 is equal hence,
β
ππΆπ»4 Γ π£πΆπ»4=
β
ππ2 Γ π£π2
β π£πΆπ»4π£π2
=ππ2ππΆπ»4
=32
16=2
1= 2 βΆ 1
3. What is the value of the spin quantum number of the last electron for π9-
configuration?
(A) 0 (B) β1
2 (C)
1
2 (D) 1
Solution: (B)
The value of the spin quantum number (s) can be +1
2 or β
1
2, because an electron can
rotate along its axis either in clockwise or in anticlockwise direction. But one quantum
number depicts are electron and thus its values will be β1
2 for π9-configuration.
4. Nitrogen forms stable π2 molecule, but phosphorus is converted to π4 from π2
because
(A) ππ β ππ boding is strong in phosphorus
(B) ππ β ππ bonding is weak in phosphorus
(C) Double bond is present in phosphorus
(D) Single P β P bond is weaker than N β N bond
Solution: (B)
ππ β ππ bonding is weak in phosphorus ππ β ππ bonding in nitrogen is strong. Hence, it
can form triple bond with another N. Single N β N is weaker than P β P bond due to high
interionic repulsion of non-bonding electrons. Hence, π β‘ π is table and π2 is not stable
molecule.
5. An amorphous solid (π) burns in air to form a gas (π) which turns lime water milky.
This gas decolarises aqueous solution of acidified πΎπππ4. gas (π) reacts with oxygen
to give another gas (π) which is responsible for acid rain. π, π and π are
(A) π π ππΆ πΆπ πΆπ2
(B) π π ππ ππ2 ππ3
(C) π π ππ π2π3 π2π5
(D) π π ππ ππ3 π»2ππ4
Solution: (B)
6. A black power when heated with conc. π»πΆπ gives a greenish yellow gas. The gas acts
as an oxidizing and a bleaching agent. When it is passed are slaked lime, a white power
is formed which is a ready source of gas. The black powder and white power
respectively are
(A) πΎπΆππ3 and πππΆππ3 (B) πππ2 and πΆππΆππΆπ2
(C) πππ2and πΎπΆππ3 (D) πππΆπ4 and πΆππΆπ2
Solution: (B)
πΆπ2 + π»2π β 2π»πΆπ + [π]
Coloured matter + [π] β Colourless matter
7. The compound that is both paramagnetic and coloured is
(A) πΎ2πΆπ2π7 (B) (ππ»4)2 [πππΆπ6]
(C) π0ππ4 (D) πΎ3[πΆπ’(πΆπ)4]
Solution: (C)
πΎ2πΆπ2π7 contains πΆπ6+(3π) ion which is diamagnetic and coloured due to charge
transfer. (ππ»4)2, [πππΆπ6] contains ππ4+ (3π) which is diamagnetic and colourless. π0ππ4
contains π4+(3π1) which is paramagnetic and coloured.
πΎ3[πΆπ’(πΆπ)4] contains πΆπ2+(3π10) which is diamagnetic and colourless.
8. A solution of πΎπππ4 is reduced to a various products depending upon its pH. At pH >
7, it is reduced to a colourless solution (π΄), at pH = 7 it forms a brown precipitate (π΅)
and at pH > 7, it gives a green solution (πΆ). (π΄), (π΅) and (πΆ) are
(A) π΄ π΅ πΆ
ππ2+ πππ2 πππ42β (B)
π΄ π΅ πΆπππ2 ππ
2+ πππ42β
(C) π΄ π΅ πΆ
ππ2+ πππ42β πππ2
(D) π΄ π΅ πΆ
πππ42β ππ2+ πππ2
Solution: (A)
At ππ» < 7, in acidic medium
At ππ» = 7, in neutral medium
At ππ» > 7, in alkaline medium
9. Although zirconium belongs to 4d and hafnium belongs to 5d transition series even
they show similar physical and chemical properties because both
(A) Belong to d-block
(B) Have same number of electrons
(C) Belongs to the same group of the periodic table
(D) Have similar atomic radius
Solution: (D)
This is because the atomic radii of 4d and 5d transition elements are nearly same. This
similarity in size (π§π = 160,π»π = 159 ππ). Due to tanthanoid contraction, because of
this contraction the radii of Hf becomes nearly to that of Zr and as the properties of an
element depends on its atomic radii they show similar properties.
10. Which of the following good reducing agent?
(A) ππ2+ (B) πΆπ4+ (C) ππ4+ (D) πΏπ3+
Solution: (A)
ππ2
ππ2 β [ππ] 4π14 5π06π 0
πΆπ4+ β [ππ] 4π0 5π0 6π 0
ππ4+ β [ππ] 4π7 5π0 6π 0
πΏπ3+ β 5π1 6π 2
Only ππ2+ can lose πβ and act as reducing agent
ππ3+ β [ππ] 4π13 5π0 6π 0
11. Which of the following will not give tests for free transition metal ion in solution?
(A) πΎ2[ππ(πΆπ)4] (B) πΉπππ4 . πΎ2ππ4. 24π»2π
(C) Both of the above (D) None of the above
Solution: (A)
9n solution, πΎ2[ππ(πΆπ)4] dissociates as
πΎ2[ππ(πΆπ)4] β πΎ+ + [ππ(πΆπ)4]
2β
12. Which of the following is an organometallic compound?
(A) Lithium methoxide (B) Lithium acetate
(C) Lithium dimethylamide (D) Methyl lithium
Solution: (D)
Those compounds in which metal atom is directly bonded to carbon atom are called
organometallic compounds.
13. Among the following metal carbonyls, the πΆ β π bond order is lowest in
(A) [ππ(πΆπ)6]+ (B) [πΉπ(πΆπ)5]
(C) [πΆπ(πΆπ)6] (D) [π(πΆπ)6]β
Solution: (D)
[π(πΆπ)6]β, the anionic carbonyl complex can delocalize more electron density to anti
bonding π = orbital (ππ β ππ back bonding) of CO and thus lower the bond order.
14. The major product obtained in the reaction
is
(A)
(B)
(C)
(D)
Solution: (C)
Ease of substitution of hydrogen is 3π > 2π > 1π .
15. Which of the following compounds can exhibit both geometrical isomerism and
enantiomerism?
Solution: (B)
16. The unit cell length of sodium chloride crystal is 564 pm. Its density would be
(A) 1.082 π ππβ3 (B) 2.165 π ππβ3
(C) 3.247 π ππβ3 (D) 4.330 π ππβ3
Solution: (B)
In sodium chloride unit cell, there are four ππ+ and four πΆπβ ions.
Hence,
density =mass of atoms in a unit cell
volume of a unit cell=π2π3ππ
Each unit cell of πππΆπ has
4ππβ and 4 πΆπβ 2 ππππ β π = 4
=4(23 + 35.5)π πππβ1
(6.023 Γ 1023πππβ1) Γ (564 Γ 10β12 ππ)3
= 2.165 π ππβ3
17. Zinc oxide loses oxygen on heating according to the reaction
πππ Heatβ ππ2+ +
1
2π2 + 2οΏ½Μ οΏ½
It becomes yellow on heating because
(A) ππ2+ ions and electrons move too interstitial site and F β centres are created
(B) Oxygen and electrons move at the crystal and ions become yellow
(C) ππ2+ again combine with oxygen to give yellow oxide
(D) ππ2+ are replaced by oxygen
Solution: (A)
ππ2+ ions and electrons move to interstitial sites and F-centres are created which impart
yellow colour to πππ.
18. In which of the following reactions will Ξπ be equal to Ξπ»?
(A) π»2(π) +1
2π2(π) β π»2π(π)
(B) π»2(π) + π2(π) β 2π»π(π)
(C) π2π4(π) β 2ππ2(π)
(D) 2ππ2(π) + π2(π) β 2ππ3(π)
Solution: (B)
π»2(π) + π2(π) β 2π»π(π)
Ξπ = 2 β (1 + 1) = 0
Ξπ» = Ξπ + Ξππ π
Ξπ» = Ξπ + (0) π π
β΄ Ξπ» = Ξπ
19. Which of the following reactions depict the oxidizing behavior of π»2ππ4?
(A) 2ππΆπ5 + π»2ππ4 β 2πππΆπ3 + 2π»πΆπ + ππ2πΆπ2
(B) 2ππππ» + π»2ππ4 β ππ2ππ4 + 2π»2π
(C) πππΆπ + π»2ππ4 β πππ»ππ4 + π»πΆπ
(D) 2π»π + π»2ππ4 β π2 + ππ2 + 2π»2π
Solution: (D)
π2 is liberated by loss of electron and gain by ππ42β in π»2ππ4.
ππ42β β ππ2
20. When 96.5C of electricity is passed through a solution of silver nitrate (at wt. of Ag =
107.87 β 108), the amount of silver deposted is
(A) 5.8 ππ (B) 10.8 ππ (C) 15.8 ππ (D) 20.8 ππ
Solution: (B)
ππ΄π =πΈπ΄π + π
96500
=108 Γ 9.65
96500
1.08 Γ 10β2π = 10.8 ππ
21. The standard reduction potentials at 298 K for the following half reactions are given
against each
ππ2+(ππ) + 2οΏ½Μ οΏ½ β π(π ) β 0.762 π
πΆπ3+(ππ) + 2οΏ½Μ οΏ½ β πΆπ(π ) β 0.740 π
2π»+(ππ) + 2οΏ½Μ οΏ½ β π»2(π )0.000 π
πΉπ3+(ππ) + 2οΏ½Μ οΏ½ β πΉπ2+(ππ)0.770 π
Which is the strongest reducing agent?
(A) ππ(π ) (B) πΆπ(π ) (C) π»π(π) (D) πΉπ2+(ππ)
Solution: (A)
More the negative the value of reduction potential, stronger is the reducing property,
i.e., power to accept electrons.
22. For the reaction, 2ππ2 + πΉ2 β 2ππ2πΉ, following mechanism has been provided.
ππ2 + πΉ2Slowβ ππ2πΉ + πΉ
ππ2 + πΉ Fastβ ππ2πΉ
Thus, rate expression of the above reaction can be written as
(A) π = π[ππ2]2 [πΉ2] (B) π = π [ππ2] [πΉ2]
(C) π = π [ππ2] (D) π = π [πΉ2]
Solution: (B)
Slowest step is the rate determining step.
π = π[ππ2] [πΉ2]
23. For a reaction, πΌβ + ππΆπβ β πΌπβ + πΆπβ in an aqueous medium, the rate of reaction is
given by:
π[πΌπβ]
ππ‘= π
[πΌβ] [ππΆπβ]
[ππ»β]
The overall order of reaction is
(A) β1 (B) 0 (C) 1 (D) 2
Solution: (C)
ππ‘[πβ]
ππ‘=π[πβ]1 [ππΆπβ]1
[ππ»β]β1
Order of reaction = 1 + 1 β 1 = 1
24. Give the IUPAC name of π β πΆππΆπ»2πΆ6π»4πΆπ»2πΆ(πΆπ»3)3
(A) 1 β (3 β chloro β 3 β methylphenyl) - 2, 2 β diethyl propane
(B) 2 β (3 β chloromethyl propyl) β 2, 2 β dimethyl propane
(C) 1 β (3 β chloromethyl phenyl) β 3, 3 β dimethyl propane
(D) 1 β chloromethyl β 3 β (2, 2 β dimethyl propyl) benzene
Solution: (D)
25. The reaction of toluene with chlorine in presence of ferric chloride gives
predominatly
(A) Benzoyl chloride (B) m β chlorotoluene
(C) Benzyl chloride (D) o β and p β chlorotoluene
Solution: (D)
Since, βπΆπ»3 group is o and p-directing group, so o-and p-chlorotoluene are obtained.
26. Identify (π·) in the following reaction series:
Solution: (C)
27. Phenol is functional compound because
(A) It is acidic and contain β OH
(B) It reacts with Na to give phenoxide
(C) It reacts with both Na and Zn to give phenoxide and benzene respectively
(D) Both (It is acidic and contain β OH) and (It reacts with both Na and Zn to give
phenoxide and benzene respectively)
Solution: (B)
28. Which of the following has lowest boiling point?
(A) p β nitrophenol (B) m β nitrophenol
(C) o β nitrophenol (D) Phenol
Solution: (C)
o-nitrophenol has intramolecular H-bonding.
29. An organic compound X is oxidized by using acidified πΎ2πΆπ2π7. The product
obtained reacts with phenyl hydrazine, but does not answer silver mirror test. The
possible structure of X is
(A) πΆπ»3πΆππΆπ»3 (B) (πΆπ»3)2πΆπ»ππ» (C) πΆπ»3πΆπ»π (D)
πΆπ»3πΆπ»2ππ»
Solution: (B)
The oxidation product of X reacts with phenyl hydrazine, thus it contains > πΆ = 0 group.
The same product does not (give) silver mirror test. Thus, it is a keton, because only
aldehydes give silver mirror test.
Thus, the compound X must be 2π alcohol, as only secondary alcohols give ketones on
oxidation and hence, X is (πΆπ»3)2πΆπ»ππ».
Propanone β Silver mirror not formed.
30. One mole of an organic compound A with the formula πΆ3π»6π reacts completely with
two moles of HI to form X and Y. When Y is boiled with aqueous alkali, it form Z. Z
answers the iodoform test. The compound A is
(A) propan β 2 β ol (B) propan β 1 β ol
(C) ethoxyethane (D) methoxyethane
Solution: (D)
If A reacts with Hl, it could be ether.
31. A compound (π΄) πΆ4π»8 πΆπ2 on alkaline hydrolysis to give compound (π΅) πΆ4π»8 π which
gives an oxime and positive Tollenβs reagent test. What is the structure of (π΄)?
(A) πΆπ»3πΆπ»2πΆπ»2πΆπ»πΆπ2 (B) πΆπ»3πΆπΆπ2πΆπ»2πΆπ»3
(C) πΆπ»3πΆπ»(πΆπ)πΆπ»(πΆπ)πΆπ»3 (D) πΆπ»2πΆππΆπ»2πΆπ»2πΆπ»2πΆπ
Solution: (A)
Positive Tollenβs test β B is an aldehyde. Compared A as hydrolysis gives aldehyde β
A is a germinal dihalide with halogen atoms as terminal C-atom.
32. The increasing order of the rate of HCN addition to compound π΄ β π· is
π΄. π»πΆπ»π π΅. πΆπ»3πΆππΆπ»3
πΆ. πβπΆππΆπ»3 π·. πβπΆππβ
(A) π΄ < π΅ < πΆ < π· (B) π· < π΅ < πΆ < π΄
(C) π· < πΆ < π΅ < π΄ (D) πΆ < π· < π΅ < π΄
Solution: (C)
It is nucleophilic addition whose reactivity depends upon the elecgtrophilic character of
carbonyl carbon and steric hindrance only. Hence, the ease of the reaction would be
π»πΆπ»π > πΆπ»3πΆππΆπ»3 >(π΄) (π΅)
πβπΆππΆπ»3 > πβπΆππβ
(πΆ) (π·)
33. Which of the following is highly acidic in nature?
(A)
(B)
(C)
(D)
Solution: (D)
Methoxy benzoic acid is more acidic because of methoxy (πΆπ»3π β) group.
34. What are the organic products formed in the following reaction?
(A) πΆ6π»5 β πΆπππ» and πΆπ»4
(B) πΆ6π»5 β πΆπ»2 β ππ» and CH4
(C) πΆ6π»5 β πΆπ»3 and CH3 β ππ»
(D) πΆ6π»5 β πΆπ»2 β ππ» and CH3 β ππ»
Solution: (D)
35. In a set of reactions m β bromobenzoic acid gave a product D. Identify the product
D.
(A)
(B)
(C)
(D)
Solution: (D)
36. If β²π΄β² is πΆ2π»5ππ»2. β²π΅β² is (πΆ2π»5)ππ», β²πΆβ² is (πΆ2π»5)3π, then order of solubility in water is
(A) π΄ > π΅ > πΆ (B) π΅ < π΄ < πΆ (C) πΆ < π΅ < π΄ (D)
πΆ > π΅ < π΄
Solution: (C)
Primary amines has βππ»2 due to which H-bonding easily possible. Hence, the solubility
order amines 3π < 2π < 1π .
37. Aniline is resonance hybrid of __ structures.
(A) 1 (B) 3 (C) 5 (D) 7
Solution: (C)
38. What is the conjugate base of ποΏ½Μ οΏ½?
(A) π2 (B) π»2π (C) πβ (D) πβ2
Solution: (D)
ππ»β β π2β + π»+
39. Glucose when treated with conc. π»ππ3 gives
(A) Acetic acid (B) Saccharic acid (C) Gluconic acid (D)
Sorbitol
Solution: (B)
40. Which of the following is a peptide linkage?
(A) βπΆπ β ππ»
(C) βπΆπ β ππ»2
(D) βπΆπ β π β ππ»4
Solution: (A)