1. The potential energy of an electron can be denoted by:
a Ze² / 8πε₀r
b -Ze² / 8πε₀r
c -Ze² / 4πε₀r
d None of them
✅Correct Answer: (b)
The potential energy (PE) of an electron in a hydrogen-like atom is negative, because the electron is bound to the nucleus. From Coulomb’s law and Bohr model:
PE = − Ze²/8πε₀r
Positive sign would indicate repulsion, so option (a) and (c) are incorrect.
2. If the volume occupied by oxygen gas (O₂) at STP is 44.8dm³, the number molecules of O₂ in the vessels are:
a 3.01 x 10²³
b 12.04 x 10²³
c 6.02 x 10²³
d 24.08 x 10²³
✅Correct Answer: (b)
Number of moles of O₂ = 44.8 dm³/22.4 dm³/mol = 2 mol
Number of molecules = 2 mol × 6.022×10²³ molecules/mol = 12.04×10²³ molecules
3. The number of carbon atoms in half mole of sugar (C₁₂H₂₂O₁₁) are approximately:
a 6 x 10²³
b 36 x 10²³
c 60 x 10²³
d 72 x 10²³
✅Correct Answer: (b)
No. of atoms = n x Nᴀ x atomicity = 0.5 x 6.02 x 10²³ x 12 = 36.12 x 10²³
4. Significant figures in 0.0080900 are
a Four
b Three
c Seven
d Five
✅Correct Answer: (d)
The number 0.0080900 has five significant figures.
Here's why:
Leading zeros (before 8) are not significant.
8, 9 are non-zero digits, so they are significant.
The trailing zeros after 9 (in the decimal part) are significant because they appear after the decimal point.
So, the significant figures are 8, 9, 0, 0, 0, which gives a total of five significant figures.
5. This molecule has zero dipole moment
a CCl₄
b CO₂
c Both a and b
d SO₂
✅Correct Answer: (c)
🔎 CCl₄ and CO₂ are highly symmetrical molecules, so their bond dipoles cancel out, giving zero dipole moment. SO₂ is bent, so it has a net dipole.
6. The number of bonds in ethene (C₂H₄) is:
a 5 sigma, 1 pi
b 5 sigma, 2 pi
c 3 sigma, two pi
d 7 sigma, 1 pi
✅Correct Answer: (a)
🔎 Ethene has a double bond between carbons (1 sigma + 1 pi) and four C–H sigma bonds, giving 5 sigma and 1 pi bonds.
7. The number of orbitals in 3rd energy level is
a 4
b 16
c 9
d 32
✅Correct Answer: (c)
🔎 The number of orbitals in any shell is 2n². For n = 3, that gives 9 (3²) orbitals.
8. This molecule has minimum bond angle
a CS₂
b NH₃
c H₂S
d BF₃
✅Correct Answer: (c)
🔎 H₂S is bent with a bond angle (92°) much smaller than NH₃ (107°) or BF₃ (120°), and CS₂ is linear (180°). Hence, H₂S has the minimum bond angle.
9. According to Hund’s rule the electronic configuration of carbon (C = 6) is:
a 1s² ,2s² 2pₓ¹ 2pᵧ²
b 1s² ,2s² 2pₓ¹ 2pᵧ¹
c 1s² ,2s² 2pₓ¹ 2pᵧ¹ 2pz¹
d 1s² ,2s² 2pₓ²
✅Correct Answer: (b)
🔎 Hund’s Rule states that electrons fill degenerate orbitals singly first. Thus, carbon’s ground-state configuration is 1s² 2s² 2pₓ¹ 2pᵧy¹.
10. Under similar conditions CH₄ gas diffuses faster than SO₂ gas:
a 1.5 times
b 16 times
c 4 times
d 2 times
✅Correct Answer: (d)
CH₄ diffuses faster than SO₂ because its molar mass is lower.
Using Graham's law:
Molar mass of CH₄ = 16
Molar mass of SO₂ = 64
The ratio of diffusion rates is the square root of the ratio of the molar masses:
Rate of CH₄ = √64÷16 = √4 = 2
So, CH₄ diffuses 2 times faster than SO₂.
11. It is an example of crystalline solid :
a Glass
b Rubber
c Plastic
d table salt
✅Correct Answer: (d)
🔎 Table salt (sodium chloride) forms a crystalline solid because its atoms or ions are arranged in a highly ordered, repeating structure. Crystalline solids have a regular and repeating arrangement of particles, which is characteristic of substances like salt, diamonds, and quartz.
The other options—glass, rubber, and plastic—are amorphous solids, meaning their particles are arranged more randomly.
12. One mole of H₂O contains this number of hydrogen atoms
a 3.01 x 10²³
b 6.02 x 10²³
c 1.204 x 10²³
d 1.204 x 10²⁴
✅Correct Answer: (d)
Since each water molecule has 2 hydrogen atoms, one mole of H₂O contains twice Avogadro’s number of hydrogen atoms (2Nᴀ) → 1.204 x 10²⁴
13. The number of molecules in this pair are same
a 10g H₂ & 10 g CH₄
b 10g H₂ & 50 g CH₄
c 10g H₂ & 16 g CH₄
d 10g H₂ & 80 g CH₄
✅Correct Answer: (d)
🔎 10 g H₂ corresponds to 5 moles, and 80 g CH₄ also corresponds to 5 moles. Since moles × Avogadro’s number gives molecules, both samples contain the same number of molecules.
14. The rate of diffusion of CO₂ is equal to
a CH₄
b CO
c C₃H₈
d SO₂
✅Correct Answer: (c)
🔎 According to Graham’s Law, diffusion rate ∝ √1/M. Since CO₂ and C₃H₈ both have molar mass 44 g/mol, their diffusion rates are equal.
15. The (n+l) value for 5d orbital is
a 7
b 4
c 5
d 6
✅Correct Answer: (a)
🔎 For the 5d orbital, n = 5 and l = 2. Thus, n+l = 7.
16. By emitting 𝛂-particles, ₉₂U²³⁸ converts into
a ₈₄Po²¹⁰
b ₉₀Th²³⁴
c ₈₉Ac²²⁷
d ₈₂Pb²⁰⁸
✅Correct Answer: (b)
Emission of an α-particle reduces the mass number by 4 and atomic number by 2. Thus, Uranium-238 becomes Thorium-234.
17. e/m value is minimum for positive rays when discharge tube contains
a Helium
b Nitrogen
c Oxygen
d Hydrogen
✅Correct Answer: (c)
The e/m ratio represents the charge-to-mass ratio of a particle, which is crucial in determining the behavior of ions in a magnetic field, such as in the case of positive rays or ions in a discharge tube.
For positive rays (positive ions), the e/m value will be minimum when the ions have the largest mass for a given charge. Since the charge is the same for all ions of the same charge state, the heavier the ion, the smaller the e/m ratio.
So, the e/m value depends inversely on mass → heavier ions have smaller e/m.
Molar masses:
Hydrogen (H⁺): 1 g/mol → highest e/m
Helium (He⁺): 4 g/mol
Nitrogen (N⁺): 14 g/mol
Oxygen (O⁺): 16 g/mol → lowest e/m
👉 Therefore, the minimum e/m value corresponds to the heaviest ion, i.e. Oxygen.
🔎 Since e/m ∝ 1/mass, the heaviest ion (O⁺, 16 g/mol) gives the minimum e/m value among the options. .
18. The bond energy of this molecule is maximum
a O₂
b N₂
c CH₄
d Cl₂
✅Correct Answer: (b)
🔎 Nitrogen molecule (N₂) has a triple bond, making it the strongest and giving it the highest bond energy among the options.
Approximate bond energies:
👉 N₂: ~945 kJ/mol (triple bond, very strong) → maximum bond energy
O₂: ~498 kJ/mol (double bond)
CH₄ (C–H bond): ~412 kJ/mol (single bonds)
Cl₂: ~242 kJ/mol (weak single bond)
19. This hydrogen halides hast the highest percentage of ionic character
a HCl
b HBr
c HF
d HI
✅Correct Answer: (c)
🔎 The ionic character increases with greater electronegativity difference. Since fluorine is the most electronegative element, HF has the maximum ionic character among hydrogen halides.Among hydrogen halides, HF has the highest percentage of ionic character (~43%), followed by HCl (~17%), HBr (~12%), and HI (~11%). This trend is due to the decreasing electronegativity difference between hydrogen and the halogen as we move down the group
20. This colour has the shortest wavelength
a Blue
b Violet
c Red
d Orange
✅Correct Answer: (b)
Visible light ranges from about 400 nm (violet) to 700 nm (red). Wavelength order (shortest → longest): Violet < Blue < Green < Yellow < Orange < Red Therefore, violet light has the shortest wavelength. 🔎 In the visible spectrum, violet light (~400 nm) has the shortest wavelength, while red (~700 nm) has the longest.
21. In an exothermic reaction, increase of temperature favours
a Forward reaction
b To remain in equilibrium
c Irreversible reaction
d Reverse reaction
✅Correct Answer: (d)
🔎 For exothermic reactions, increasing temperature adds heat to the products, so equilibrium shifts backward to consume the excess heat → favouring the reverse reaction.
22. This law is applied in collection of gases over water:
a Graham's Law
b Avogadro's law
c Dalton's Law
d Charles’s law
✅Correct Answer: (c)
🔎 In collection of gases over water, the measured pressure includes both the gas and water vapour. Dalton’s Law explains this by adding partial pressures.
23. Volume-volume relationship is based on:
a Dalton's Law
b Avogadro's Law
c Boyle's Law
d Gay-Lussac's Law
✅Correct Answer: (d)
🔎 The volume–volume relationship in gaseous reactions is explained by Gay-Lussac’s Law of Combining Volumes which states that When gases react together, they do so in simple ratios by volume, provided conditions of temperature and pressure are constant, not Dalton’s, Avogadro’s, or Boyle’s Law.
24. In James Chadwick's experiment alpha particles were bombarded on:
a Boron
b Beryllium
c Bromine
d Carbon
✅Correct Answer: (b)
🔎 Chadwick bombarded beryllium with α-particles, which emitted neutral radiation. This radiation was shown to consist of neutrons, leading to their discovery.
25. 7.5 g of a gas occupies 5.6 liters at STP. The gas is
a CO
b NO
c CO₂
d N₂O
✅Correct Answer: (d)
🔎 At STP, 7.5 g gas occupying 5.6 L corresponds to a molar mass of 30 g/mol, which matches nitric oxide (NO).
26. In the hydrogen atom spectrum, the series of line obtained when electron jumps from higher orbits to the first orbit, is called:
a Balmer Series
b Lyman Series
c Brackett Series
d Paschen Series
✅Correct Answer: (b)
🔎 The Lyman Series corresponds to transitions ending at the first orbit (n = 1) in hydrogen, producing spectral lines in the ultraviolet region.
In the hydrogen atom spectrum, different series are named based on the final orbit (n) to which the electron falls:
Lyman Series: electron falls to n = 1 (first orbit) → lies in UV region.
Balmer Series: electron falls to n = 2 → visible region.
Paschen Series: electron falls to n = 3 → infrared region
Brackett Series: electron falls to n = 4 → infrared region.
27. The range of bond energy of hydrogen bond is:
a 10-20 kJ/mol
b 40 - 50 kJ/mol
c 20 - 40 kJ/mol
d 50 - 60 kJ/mol
✅Correct Answer: (c) 20 – 40 kJ/mol
🔎 Hydrogen bonds are stronger than van der Waals interactions (< 10 kJ/mol) but weaker than covalent bonds (~200–400 kJ/mol), with bond energies typically in the 20–40 kJ/mol range (intermediate range).
28. The octet rule is not applied on this molecule:
a N₂
b H₂
c O₂
d F₂
✅Correct Answer: (b)
🔎 Hydrogen is an exception to the octet rule because it requires only 2 electrons (duplet) for stability, unlike N₂, O₂, and F₂ which obey the octet rule.
29. This molecule has linear structure:
a CH₄
b C₂H₂
c BF₃
d NH₃
✅Correct Answer: (b)
🔎 Acetylene (C₂H₂) has sp-hybridized carbons, giving a linear structure (180° bond angle). The others are tetrahedral (CH₄), trigonal planar (BF₃), or pyramidal (NH₃).
30. In this reaction Kₚ > Kc:
a H₂ + I₂ ⇌ 2HI
b 2NO₂ ⇌ N₂O₄
c PCl₅ ⇌PCl₃+ Cl₂
d 2SO₂+O₂ ⇌2SO₃
✅Correct Answer: (c) PCl₅ ⇌ PCl₃ + Cl₂
🔎 Relationship between Kₚ and Kc: Kₚ = Kc(RT)▵ⁿ
where Δn = moles of gaseous products- moles of gaseous reactants.
If Δn > 0, then Kₚ > Kc.
If Δn = 0, then Kₚ = Kc.
If Δn < 0, then Kₚ < Kc.
🔎 Since Δn = +1 in this reaction, Kₚ = Kc(RT) ⁺¹, meaning Kₚ>Kc.
31. The reaction CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O is called:
a Neutralization
b Esterification
c Hydrolysis
d Electrolysis
✅Correct Answer: (b)
🔎 The combination of acetic acid and ethanol produces ethyl acetate (an ester) and water, which is the classic esterification reaction.
32. According to principle of Le-Chatelier the yield of the products cannot be controlled with the help of:
a Concentration
b Catalyst
c Pressure
d Temperature
✅Correct Answer: (b)
🔎 Catalysts accelerate both forward and reverse reactions equally, so they do not alter equilibrium composition. Therefore, product yield cannot be controlled by a catalyst according to Le Chatelier’s principle.
33. This compound does not contain hydrogen bond:
a H₂O
b HF
c CH₄
d NH₃
✅Correct Answer: (c) CH₄
Hydrogen bonding occurs when H is covalently bonded to highly electronegative atoms (N, O, F) and interacts with lone pairs on nearby molecules.
Check each option:
CH₄ (Methane): H is bonded to C (not highly electronegative, no lone pairs) → ❌ no hydrogen bonding.
H₂O (Water): O is highly electronegative, has lone pairs → ✅ strong hydrogen bonding.
HF (Hydrogen fluoride): F is highly electronegative, has lone pairs → ✅ hydrogen bonding.
NH₃ (Ammonia): N is electronegative, has lone pairs → ✅ hydrogen bonding.
🔎 Methane does not form hydrogen bonds because carbon is not sufficiently electronegative and lacks lone pairs, unlike H₂O, HF, and NH₃.
34. Motor Oil grading is based on:
a Surface Tension
b Viscosity
c Vapour Pressure
d Boiling Point
✅Correct Answer: (b) Viscosity
Motor oil is graded by the SAE (Society of Automotive Engineers) system.
The grading depends on how easily the oil flows at different temperatures.
This property is called viscosity (resistance to flow).
Surface tension, vapour pressure, and boiling point are not used for grading motor oils.
🔎 Motor oil grades (like SAE 10W-30, SAE 20W-50) are based on viscosity, which determines how the oil performs at low and high temperatures.
35. Two solid compounds having the same crystal structure are called
a Allotropes
b Isotopes
c Polymorphous
d Isomorphous
✅Correct Answer: (d) Isomorphous
Allotropes: Different forms of the same element (e.g., diamond and graphite for carbon).
Isotopes: Atoms of the same element with different numbers of neutrons.
Polymorphous: A single compound existing in more than one crystal form.
Isomorphous: Different compounds that crystallize in the same crystal structure.
👉 The question asks: Two solid compounds having the same crystal structure → that is isomorphous.
🔎 Compounds are called isomorphous when they have similar crystal structures, even though their chemical compositions differ
36. This particle having mass 1836 times that of electron is:
a Hyperon
b Proton
c Meson
d Neutron
✅Correct Answer: (b) Proton
The electron mass is taken as the reference.
The proton has a mass about 1836 times that of an electron.
The neutron is slightly heavier than the proton (~1839 times electron mass).
Hyperons and mesons are heavier subatomic particles but not typically described with this ratio.
🔎 The proton’s mass is approximately 1836 times the electron’s mass, making it the correct choice
37. This one is not same in Na⁺ (Z = 11), Mg²⁺ (Z = 12) and Al³⁺ (Z 13):
a number of shells
b number of electrons
c electronic configuration
d number of protons
✅ Correct Answer:(d) number of protons
📘Na⁺ (Z = 11):
Protons = 11
Electrons = 10 (lost 1)
Electronic configuration = 1s² 2s² 2p⁶ → 2 shells
📘Mg²⁺ (Z = 12):
Protons = 12
Electrons = 10 (lost 2)
Electronic configuration = 1s² 2s² 2p⁶ → 2 shells
📘Al³⁺ (Z = 13):
Protons = 13
Electrons = 10 (lost 3)
Electronic configuration = 1s² 2s² 2p⁶ → 2 shells
Comparison
Number of electrons: All have 10 → same.
Electronic configuration: All are 1s² 2s² 2p⁶ → same.
Number of shells: All have 2 shells → same.
Number of protons: Different (Na⁺ = 11, Mg²⁺ = 12, Al³⁺ = 13).
🔎 Na⁺, Mg²⁺, and Al³⁺ are isoelectronic ions (same electrons, shells, and configuration), but they differ in proton number.
38. The (n + 1) value for 4f orbital is:
a 4
b 5
c 6
d 7
✅Correct Answer: (d)
🔎 For the 4f orbital, n=4 and l=3. Therefore, n+l = 7.
39. This molecule has two lone pair and two bond pair electrons:
a NH₃
b CCl₄
c BeCl₂
d H₂O
✅ Correct Answer:(d) H₂O
📘NH₃ (Ammonia):
Central atom N has 5 valence electrons.
Forms 3 bonds with H → 3 bond pairs + 1 lone pair. → Not correct.
📘CCl₄ (Carbon tetrachloride):
Central atom C has 4 bonds with Cl → 4 bond pairs, no lone pairs. → Not correct
📘BeCl₂ (Beryllium chloride):
Central atom Be has 2 bonds with Cl → 2 bond pairs, no lone pairs. → Not correct.
📘H₂O (Water):
Central atom O has 6 valence electrons
Forms 2 bonds with H → 2 bond pairs
Remaining 4 electrons → 2 lone pairs.
So total = 2 bond pairs + 2 lone pairs ✅
🔎 Water has two bond pairs (O–H bonds) and two lone pairs on oxygen, giving a bent molecular geometry.
40. Principal and azimuthal quantum number values for 3d orbital are:
a n = 2, l =1
b n = 2, l =3
c n =3, l = 3
d n = 3, l = 2
✅ Correct Answer:(d) n = 3, l = 2
The notation 3d orbital means:
Principal quantum number n = 3.
For d-orbital, the azimuthal quantum number l = 2.
So for 3d orbital: n = 3, l = 2
🔎 The principal quantum number is taken directly from the orbital label (3), and the azimuthal quantum number for a d-orbital is always 2.
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