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Q: 1. What is the basic postulate of Bohr's atomic model?

Ans. Bohr's atomic model is based on the following postulates: Electrons revolve around the nucleus in a specific circular path called an orbit. Electrons can only exist in specific energy levels or orbits. Electrons emit or absorb energy when they change from one allowed orbit to another. The energy of a photon emitted or absorbed is equal to the difference between the energies of the two energy levels involved.

Q: 2. How do you calculate the number of valence electrons in an atom?

Ans. Valence electrons are the electrons present in the outermost shell of an atom. To calculate the number of valence electrons in an atom, you need to look at the group number of an element in the periodic table. The number of valence electrons is equal to the group number for elements in groups 1, 2, and 13-18. For transition metals, the number of valence electrons can vary.

Q: 3. What is the significance of the de Broglie wavelength?

Ans. The de Broglie wavelength is significant because it relates the momentum of a particle to its wavelength. It is useful for understanding the behavior of particles that have both wave-like and particle-like properties, such as electrons. The de Broglie wavelength is given by the equation λ = h/p, where h is Planck's constant and p is the momentum of the particle.

Q: 4. What is the difference between an atomic orbital and an electron cloud?

Ans. An atomic orbital is a region of space around the nucleus where there is a high probability of finding an electron. It is described by a set of quantum numbers that specify its energy, shape, and orientation. An electron cloud, on the other hand, is a visual representation of the probability distribution of the electron density in an atom. It shows the regions of space where electrons are most likely to be found.

Q: 5. How does the Aufbau principle explain the electron configuration of atoms?

Ans. The Aufbau principle states that electrons fill atomic orbitals in order of increasing energy, starting from the lowest energy level. This principle explains the electron configuration of atoms by predicting the order in which the orbitals are filled. For example, the electron configuration of carbon (1s2 2s2 2p2) can be explained by the fact that the 1s orbital is filled first, followed by the 2s and then the 2p orbitals.

Previous Year Questions - Structure of Atom, Class 11 (IIT - JEE Advanced)

Q 1. With what velocity should an α-particle travel towards the nucleus of a Cu atom so as to arrive at a distance 10^-13 m. [JEE 1997]

Ans. 6.3×10⁸ m/s

Q 2. A compound of Vanadium has magnetic moment of 1.73 BM work out electronic configuration of Vanadium Ion in the compound. [JEE 1997]

Ans. ₂₃V⁺²: [Ar] 3d¹

Q 3. Calculate the energy required to excite one litre of hydrogen gas at 1 atm and 298 K to the first excited state of atomic hydrogen. The energy for the dissociation of H - H is 436 kJ mol^-1.

Ans. 98.2 kJ

Q4. The Schrodinger wave equation for hydrogen atom is

Previous Year Questions - Structure of Atom, Class 11 (IIT - JEE Advanced)
Where a₀ is Bohr's radius. If the radial node is 2s be at r₀, then find r₀ in term of a₀. [JEE 2004]

Ans. 2a₀

Q 5. A ball of mass 100 g is moving with 100 m/s. Find its wavelength [JEE 2004]

Ans. 6.6 × 10^-35 s^-1

Q 6. Find the velocity (m/s) of electron in first Bohr's orbit of radius a₀. Also find the de Broglie's wavelength (in m). Find the orbital angular momentum of 2p orbital of hydrogen atom in units of Previous Year Questions - Structure of Atom, Class 11 (IIT - JEE Advanced) [JEE 2004]

Ans. 3.34 × 10^-10 m

Q 7. The maximum number of electrons that can have principal quantum number, n = 3, and spin quantum number, m_s = – 1/2, is: [JEE 2011]

Ans. 0009

Q 8. The work function (φ) of some metals is listed below. The number of metals which will show photoelectric effect when light of 300 nm wavelength falls on the metal is [JEE 2011]

Previous Year Questions - Structure of Atom, Class 11 (IIT - JEE Advanced)

Ans. 0004

Q 9. The work functions of Silver and Sodium are 4.6 and 2.3 eV, respectively. The ratio of the slope of the stopping potential versus frequency plot for Silver to that of Sodium is [JEE 2013]

Ans. 1

Q 10. The atomic masses of He and Ne are 4 and 20 a.m.u. respectively. The value of the de Broglie wavelength of He gas at -73 ºC is "M" times that of the de Broglie wavelength of Ne at 727 ºC. M is [JEE 2013]

Ans. 0005

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Previous Year Questions - Organic Chemistry, Class 11 (IIT - JEE Advanced)

Q 1. Out of anhydrous AlCl3 and hydrous AlCl3 which is more soluble in diethyleither ? Explain with reason. [JEE-2003]

Sol. Anhydroush AlCl₃ is more soluble in diethyl ether

Previous Year Questions - Structure of Atom, Class 11 (IIT - JEE Advanced)

Q 2. Match K_a values with suitable acid: [JEE-2003]

K_a
(i) 3.3 × 10^-5
(ii) 4.2 × 10^-5
(iii) 6.3 × 10^-5
(iv) 6.4 × 10^-5
(v) 30.6 × 10^-5

Previous Year Questions - Structure of Atom, Class 11 (IIT - JEE Advanced)

Ans. (i) d

(ii) b

(iii) a

(iv) c

(v) e

Q 3. Give resonating structures of following compound:

Previous Year Questions - Structure of Atom, Class 11 (IIT - JEE Advanced) [JEE-2003]

Ans.

Previous Year Questions - Structure of Atom, Class 11 (IIT - JEE Advanced)

Q 4. Which of the following is more acidic and why ?
Previous Year Questions - Structure of Atom, Class 11 (IIT - JEE Advanced) [JEE-2004]

Ans.

Previous Year Questions - Structure of Atom, Class 11 (IIT - JEE Advanced)

Q 5. Predict whether the following molecules are iso structural or not. Justify your answer. [JEE-2005]
(i) NMe(3

(ii) N(SiMe₃)₃

Ans.

Previous Year Questions - Structure of Atom, Class 11 (IIT - JEE Advanced)

Given componds are not isostrucutral
Delocalised of l.p. of nitrogen in vacant d-orbital of silicon makes compound planar.

Q 6. The total number of basic group in the following form of lysine is [JEE-2010] Previous Year Questions - Structure of Atom, Class 11 (IIT - JEE Advanced)

Ans. 2

Q 7. The total number of contrubting structure showing hyperconjugation (involving C-H bonds) for the following carbocation is [JEE-2011]
Previous Year Questions - Structure of Atom, Class 11 (IIT - JEE Advanced)

Ans. 6

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Previous Year Questions - States of Matter - Gases and Liquids, Class 11 (IIT - JEE Advanced)

Q. 1 The density of the vapour of a substance at 1 atm pressure and 500 K is 0.36 Kg m^-3. The vapour effuses through a small hole at a rate of 1.33 times faster than oxygen under the same condition.
[JEE 2002]

(a) Determine
(i) mol. wt.;
(ii) molar volume;
(iii) compression factor (z) of the vapour and
(iv) which forces among the gas molecules are dominating, the attractive or the repulsive

(b) If the vapour behaves ideally at 1000K , determine the average translational K.E. of a molecule.

Ans.

(a)

(i) 18.1 g/mol ,

(ii) 50.25 L mol^–1 ,

(iii) 1.224 ,

(iv) repulsive,

(b) 2.07 × 10^–20 J

Q 2. The average velocity of gas molecules is 400 m/sec. Calculate its (rms) velocity at the same temperature. [JEE 2003]

Ans. 434.17 m/sec

Q 3. C_V value of He is always 3R/2 but C_V value of H₂ is 3R/2 at low temperature and 5R/2 at moderate temperature and more than 5R/2 at higher temperature explain in two to three lines. [JEE 2003]

Ans. Since H₂ is diatomic and He is monoatomic degree of freedom for mono is 3 and only translational
but for diatomic, vibrational and rotational are also to be considered

Q 4. For a real gas obeying van der Waal's equation a graph is plotted between PVm (y-axis) and P(x-axis) where V_m is molar volume. Find y-intercept of the graph. [JEE 2004]

Ans. RT

Q 5. The atomic masses of He and Ne are 4 and 20 a.m.u, respectively. The value of the de Broglie wavelength of He gas at -73°C is “M” times that of the de Broglie wavelength of Ne 727°C. M is:
[JEE(ADVANCED)-2013, 4/120]

Ans. 5

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Previous Year Questions - Thermochemistry, Class 11 (IIT - JEE Advanced)

Q 1. Estimate the average S - F bond energy in SF6. The Previous Year Questions - Structure of Atom, Class 11 (IIT - JEE Advanced) values of SF₆(g), S(g), and F(g) are -1100, 275 and 80 kJ/mol respectively. [JEE 1999]

Ans. 309.6 kJ/mol

Q 2. Diborane is potential rocket fuel which undergoes combustion according to the reaction,
B₂H₆(g) + 3O₂(g) → B₂O₃(s) + 3H₂O (g)
From the following data, calculate the enthalpy change for the combustion of diborane :

2B(s) + Previous Year Questions - Structure of Atom, Class 11 (IIT - JEE Advanced) O₂(g) → B₂O₃(s) ;                  ΔH = –1273 kJ
H₂(g) + O₂(g) → H₂O(l) ;                 ΔH = –286 kJ
H₂O (l) → H₂O(g) ;                                 ΔH = 44 kJ
2B(s) + 3H₂(g) → B2H₆(g) ;                    ΔH = 36 kJ                      [JEE 2000]

Ans. -2035 kJ mol^-1

Q 3. In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.43 K due to the combustion process. Given that the heat capacity of the calorimeter is 2.5 kJ K^-1, the numerical value for the enthalpy of combustion of the gas in KJ mol^–1 is [JEE 2009]

Ans. 9

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Previous Year Questions - Isomerism, Class 11 (IIT - JEE Advanced)

Q 1

(i) Previous Year Questions - Structure of Atom, Class 11 (IIT - JEE Advanced)

whre µ_i is the dipole moment of a stable conformer of the molcule, Z - CH₂ - CH₂- Z and x_i is the mole fraction of the stable conformer.

Given : µ_obs = 1.0 D and x (Anti) = 0.82

Draw all the stable conformers of Z - CH₂ - CH₂- Z and calculate the value of µ_(Gauche).

(ii) Draw the stable conformer of Y - CHD - CHD - Y (meso form), when Y = CH₃ (rotation about C₂ - C₃) and Y = OH (rotation about C₁ - C₂) in Newmann projection. [JEE 2005]

Ans. Previous Year Questions - Structure of Atom, Class 11 (IIT - JEE Advanced)

(ii) Anti form when Y = CH₃ & Gauche when Y = -OH

Q 2. The total number of cyclic structural as well as stereo isomers possible for compound with the molecular formular C₅H₁₀ is [JEE 2009]

Ans. 7

Q 3. The total number of cyclic isomers possible for a hydrocarbon with the molecular formula C₄H₆ is [JEE 2010]

Ans. 5

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Previous Year Questions - Chemical Equilibrium, Class 11 (IIT - JEE Advanced)

Q 1. The degree of dissociation is 0.4 at 400K & 1.0 atm for the gasoeus reaction PCl₅ PCl₃+ Cl₂(g). Assuming ideal behaviour of all gases. Calculate the density of equilibrium mixture at 400K & 1.0 atm pressure. [JEE 1999]

Ans. 4.54 g dm^–3

Q 2. When 3.06 g of solid NH₄HS is introduced into a two litre evacuated flask at 27°C, 30% of the solid decomposes into gaseous ammonia and hydrogen sulphide. [JEE 2000]

(i) Calculate K_C & K_P for the reaction at 27°C.

(ii) What would happen to the equilibrium when more solid NH₄HS is introduced into the flask ?

Ans. (i) Kc = 8.1 × 10^–5 mol² L² ; Kp = 4.91 × 10^–2 atm²

(ii) No effect ;

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Previous Year Questions - Ionic Equilibrium, Class 11 (IIT - JEE Advanced)

Q 1. What will be the resultant pH when 200 ml aqueous solution of HCl (pH = 2.0) is mixed with 300 ml of an aqueous solution of NaOH (pH=12.0)? [JEE 1998]

Ans. pH=11.3010

Q 2. The solubility of Pb(OH)₂ in water is 6.7 × 10^-6 M. Calculate the solubility of Pb(OH)₂ in a buffer solution of pH= 8. [JEE '1999]

Ans. s = 1.203×10^-3 M

Q 3. 500 ml of 0.2 M aqueous solution of acetic acid is mixed with 500 mL of 0.2 M HCl at 25°C

(a) Calculate the degree of dissociation of acetic acid in the resulting solution and pH of the solution.

(b) If 6 g of NaOH is added to the above solution, determine final pH. Assume there is no change in volume on mixing. K_a of acetic acid is 1.75 × 10^-5 M. [JEE 2002]

Ans. (a) 0.0175%

(b) 4.757

Q 4. Will the pH of water be same at 4°C and 25°C ? Explain. [JEE 2003]

Ans. No it will be > 7

Q 5. 0.1 M of HA is titrated with 0.1 M NaOH, calculate the pH at end point. Given Ka(HA) = 5 × 10^-6 and a << 1. [JEE 2003]

Ans. pH = 9

Q 6. The dissociation constant of a substituted benzoic acid at 25ºC is 1.0 × 10^-4. The pH of a 0.01 M solution of its sodium salt is : [JEE 2009]

Ans. pH = 8

Q 7. The total number of diprotic acids among the following is [IIT-2010]

H₃PO₄ H₂SO₄ H₃PO₃ H₂CO₃H₂S₂O₇

H₃BO₃H₃PO₂H₂CrO₄H₂SO₃

Ans. 6

Q 8. In 1 L saturated solution AgCl [K_sp(AgCl) = 1.6 × 10^-10], 0.1 mol of CuCl [K_sp(CuCl) = 1.0 × 10^-6] is added. The resultant concentration of Ag⁺ in the solution is 1.6 × 10^-x. The value of "x" is

[JEE 2011]

Ans. 7

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Previous Year Questions - Stoichiometry, Class 11 (IIT - JEE Advanced)

Q 1. How many moles of e^_ weigh one Kg

Ans .

Q 2. Calculate the molarity of pure water using its density to be 1000 kg m^-3. [JEE' 2003]

Ans. 55.5 mol L^-1

Q 3. One gm of charcoal absorbs 100 ml 0.5 M CH₃COOH to form a monolayer, and there by the molarity of CH₃COOH reduces to 0.49. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal = 3.01 × 10² m²/gm.

Ans. 5 × 10^-19 m²

Q 4. Calculate the amount of Calcium oxide required when it reacts with 852 gm of P₄O₁₀. [JEE 2005]

Ans. 1008 gm

Q 5. 20% surface sites have adsorbed N₂. On heating N₂ gas evolved from sites and were collected at 0.001 atm and 298 K in a container or volume is 2.46 cm³. Density of surface sites is 6.023 × 10¹⁴/cm² and surface area is 1000 cm², find out the no. of surface sites occupied per molecule of N₂.

Ans. 0002

Q 6. Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gave a solution of density 1.15 g/mL. The molarity of the solution is: [JEE 2011]

Ans. 2.05 M

Q 7. Reaction of Br₂ with Na₂CO₃ in aqueous solution gives sodium bromide and sodium bromate with evolution of CO₂ gas. The number of sodium bromide molecules involved in the balanced chemical equation is: [JEE 2011]

Ans. 0005

Q 8. A decapeptide (Mol. wt. 796) on complete hydrolysis gives glycine (Mol. Wt. 75), alanine and phenylalanine. Glycine contributes 47.0% to the total weight tto the hydrolysed products. The number of glycine units present in the decapeptideis : [JEE 2011]

Ans. 0006

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FAQs on Previous Year Questions - Structure of Atom, Class 11 (IIT - JEE Advanced)

1. What is the basic postulate of Bohr's atomic model?

Ans. Bohr's atomic model is based on the following postulates: Electrons revolve around the nucleus in a specific circular path called an orbit. Electrons can only exist in specific energy levels or orbits. Electrons emit or absorb energy when they change from one allowed orbit to another. The energy of a photon emitted or absorbed is equal to the difference between the energies of the two energy levels involved.

2. How do you calculate the number of valence electrons in an atom?

Ans. Valence electrons are the electrons present in the outermost shell of an atom. To calculate the number of valence electrons in an atom, you need to look at the group number of an element in the periodic table. The number of valence electrons is equal to the group number for elements in groups 1, 2, and 13-18. For transition metals, the number of valence electrons can vary.

3. What is the significance of the de Broglie wavelength?

Ans. The de Broglie wavelength is significant because it relates the momentum of a particle to its wavelength. It is useful for understanding the behavior of particles that have both wave-like and particle-like properties, such as electrons. The de Broglie wavelength is given by the equation λ = h/p, where h is Planck's constant and p is the momentum of the particle.

4. What is the difference between an atomic orbital and an electron cloud?

Ans. An atomic orbital is a region of space around the nucleus where there is a high probability of finding an electron. It is described by a set of quantum numbers that specify its energy, shape, and orientation. An electron cloud, on the other hand, is a visual representation of the probability distribution of the electron density in an atom. It shows the regions of space where electrons are most likely to be found.

5. How does the Aufbau principle explain the electron configuration of atoms?

Ans. The Aufbau principle states that electrons fill atomic orbitals in order of increasing energy, starting from the lowest energy level. This principle explains the electron configuration of atoms by predicting the order in which the orbitals are filled. For example, the electron configuration of carbon (1s2 2s2 2p2) can be explained by the fact that the 1s orbital is filled first, followed by the 2s and then the 2p orbitals.

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