Chemistry NCERT Exemplar Solutions Class 11th Chapter Two Structure of Atom

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Ans: The electronic configuration of nickel is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸ 4s² , thus it loses electrons from its 4s orbital as its energy is higher compared to the other orbital.

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Ans: In a multielectron atomic system the energy of an electron depends not only on its principal quantum number (shell), but also on its azimuthal quantum number (subshell). Electrons having the same shells and same subshells have the same energy and they are known as degenerate orbitals.

Thus 3d_xy ,3dz² , 3d_yz and 4d_xy, 4d_yz, 4dz² are degenerate.

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Ans: The number of radial nodes is given by n-l-1, where n is principal quantum number, l is azimuthal quantum number. The number of angular nodes is given by n-l, where n is principal quantum number, l is azimuthal quantum number. Here n =3 and l =1 Thus, angular nodes = 3-1 = 2 and radial node = 3-1-1 = 1.

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Ans: Based upon the above information, arrange the following orbitals in the increasing order of energy.

(a) 1s, 2s, 3s, 2p

(b) 4s, 3s, 3p, 4d

(c) 5p, 4d, 5d, 4f, 6s

(d) 5f, 6d, 7s, 7p

     Thus, the increasing order of energy is 3s<3p<4s<4d

Thus, the increasing order of energy is 4d<5p<6s<4f<5d

Thus, the increasing order of energy is 7s<5f<6d<7p

(II) Based upon the above information, solve the questions given below :

(a) Which of the following orbitals has the lowest energy?

4d, 4f, 5s, 5p

Ans -

           The lowest energy orbital is 5s

(b) Which of the following orbitals has the highest energy?

5p, 5d, 5f, 6s, 6p

Ans-

The highest energy orbital is 5f

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Ans: The charged particles get deflected by the electric field. As among the given option only neutron is the particle that is neutral thus it does not get deflected by the electric field.

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Ans: The mass number (A) is defined as the sum of the number of protons and neutrons present in the nucleus and the atomic number is defined as the number of protons or electrons present in an atom. Thus A=13, the number of neutrons is 7 so the number of protons is 6. Thus, the atomic number is 6.

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Ans:  

We know that energy is inversely proportional to the wavelength

Thus, the increasing order of energy is B

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Ans: As per the Hund's rule the half filled and fully filled orbital leads to the extra stability due to the symmetry thus fully filled 3d and half filled 4s is preferred.

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Ans: We have V = 109677 [1/ni ² - 1/nf ² ] cm^-1

 Given, n_i = 2, n_f=4

V =109677 [1/4 - 1/16] = 20564.44 cm^-1

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Ans: As per de Broglie, every object in motion has a wave character i.e. every object/matter have dual nature both particle and wave nature. But the wavelengths associated with ordinary objects are so short (because of their large masses) that their wave properties cannot be detected.

Given, m (mass) = 100g, v=100km/hr

 We have

λ = h/mv= 238.5 × 10-³⁴ m

As the wavelength associated with the cricket ball is so small that it can’t be detected.

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Ans: The line spectrum associated with any element possesses lines corresponding to specific wavelengths and these lines are obtained as a result of electronic transitions between the energy levels. Hence, the electrons in these levels have fixed energy i.e., quantized values.

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Ans: We have

λ = h/mv

Thus, the equation signifies that in order to have the same wavelength the electron should have higher velocity as the mass of the proton is higher than that of the electron

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Ans: The wavelength is defined as the distance between two consecutive crests or troughs of a wave, and it is denoted by  l .

l = 4*2.16 pm = 8.64 pm

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Ans: We know that

λ= c/υ Given,

υ = 4.620 × 10¹⁴ Hz

Thus, λ= c/υ = (3.0 × 10⁸ m/s)/ (4.620×10¹⁴ Hz) = 649.4nm

This frequency falls under visible range

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Ans:

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Ans: Given, m (mass) = 10g, v (speed) = 90 m/s and accuracy = 4%

Uncertainty in speed = $\frac{90\times 4}{100}=$  3.6 ms^-1

Uncertainty in position = h/4  πmΔv = 6.626 × 10^-34/4 × 3.14 × 10 × 3.6

=1.36 × 10^-33m

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Ans: The uncertainty principle is significantly only for the microscopic particles and not for the macroscopic particles can be concluded by considering the following example. Let us consider a particle or an object of mass 1 milligram i.e. 10^-6 kg Then its uncertainty can be calculated as,

? x ? v = 6.626 10^-34 / 4x 3.14 $\times$ 10⁶

= 10^-28 m^-2 s^-1

Thus, the value obtained is negligible and insignificant for the uncertainty principle to be applied to this particle.

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Ans: The energy of electrons is determined by the value of n in the hydrogen atom and by n + l in the multielectron atom. Thus for a given principal quantum number the electrons of different orbitals would have different energy.

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Ans: The drawbacks of Bohr’s model were

(i) It was unable to explain the spectra for multi-electron systems

(ii) It could not explain the molecule formation through chemical bonds. The two important developments that contributed significantly towards the change of concept of movement of an electron in an orbit was replaced by, the concept of probability of finding an electron in an orbital were

(i) Dual nature of matter

(ii) Uncertainty Principle. Quantum mechanical model of the atom is the name of the new model.

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Ans: The points of the Bohr's model that can be consider are as follows: -

(i) Electrons revolve around the nucleus in a fixed orbit with fixed energy

(ii) The energy is absorbed or released when the electron moves from one energy level to another. The energy for the n^th stationary state is given by

En = -2 p ²me⁴ /n²h²

Where

 m = mass of the electron

e = charge of the electron

 h= Planck's constant If an electron jumps from n_i to n_f then we have

D E= E_f - E_i = 2p²me⁴ /h² [ (1/n_i ² ) - (1/n_f² )]

v = DE /hc = 109677 [ (1/n_i ² ) - (1/n_f² )]

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Ans: We have

v= 109677 [1/n_i ² - 1/n_f ² ]

Given, n_i = 3 and n_f =2

D E= hcv = 109677 [1/n_i ² - 1/n_f ² ]

DE = - 3.052 × 10^-19J

n =DE /h= 4.606 × 10¹⁶Hz

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Ans:

(i) Hund's Rule             (c) Pairing of electrons in the orbitals belonging to the same subshell does not take place until each

                                     orbital is singly occupied.

(ii) Aufbau Principle    (e) In the ground state of atoms, orbitals are filled in the order of their increasing energies.

(iii) Pauli Exclusion      (a) No two electrons in an atom can have the same set of four quantum numbers.

Principle

(iv) Heisenberg's         (d) It is impossible to determine the exact position and exact momentum of a subatomic particle

       Uncertainty       simultaneously

       Principle

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Ans:

(i) X-rays  (d) ν = 10¹⁸Hz

(ii) UV  (c) ν = 10¹⁶Hz

(iii) Long radio waves (a) ν = 10 -10⁴ Hz

(iv)Microwave  (b) ν = 10¹⁰ Hz

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Ans:

(i) Photon  (d) Exhibits both momentum and wavelength

(ii) Electron  (d) Exhibits both momentum and wavelength

(iii) ψ 2 (b) Probability density

                                          (c) Always positive value

(iv)  Principal quantum  (a) Value is 4 for N shell 

        number n  (c) Always positive value

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Ans:

(i) Cr                  (d)  [Ar] 3d⁵4s²

(ii) Fe²⁺             (c) [Ar]3d⁶4s⁰

(iii)Ni²⁺               (a) [Ar]3d⁸4s⁰

(iv)Cu  (b) [Ar]3d¹⁰4s¹

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Ans: option  (iii)

According to Heisenberg's uncertainty principle it is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron. Thus, it implies that determining the trajectory of

an electron is impossible as it requires exact position and velocity which is not possible as per the uncertainty principle.

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Ans:  Option (iii) 1:3

The chlorine is a mixture of two isotopes with atomic masses 37U and 35U and its atomic mass is 35.5U. The atomic mass depicts that % composition of Cl-35 is much higher than that of Cl-37 and they are present in the ratio 1:3

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ANS – Option (iv)

The probability density of electrons for 2s orbital decreases uniformly as distance from the nucleus increases. Electrons at the 1s orbital decreases as we move far from the nucleus, however in case of 2s the probability decreases initially then it increases with the distance and thereafter at a certain point it starts decreasing with the distance.

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Ans: Option (iv)

Characteristics of cathode rays depend upon the nature of gas present in the cathode ray tube. As per the result obtained from the cathode ray discharge tube experiment the characteristics of cathode rays (electrons) does not depend upon the material of electrodes and the nature of the gas present in the cathode ray tube. Which concludes that electrons are the basic constituent of all the atoms

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Ans:  Option (ii)

The mass of an electron is equal to the mass of a neutron.

The neutron is heavier than the electron as

Mass of the neutron = 1.67 x 10^-27 kg

Mass of the electron = 9.11 x 10^-31kg

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Ans:  Option (i)

Overall neutrality of atom.

According to the J.J Thomson model the positive charge is uniformly distributed and the electrons are embedded into it in such a manner as to give the most stable electrostatic arrangement just as watermelon of positive charge with plums or seeds (electrons) embedded into it. Thus this model is able to explain the overall neutrality of the atom

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Ans:  Option (iv)

sum of the number of protons and neutrons is same but the number of protons is different

Isobars are the atoms which have same mass number (sum of number of neutrons and protons) but different atomic number (i.e. different proton number)

For example, ₁₄⁶C and ₁₄⁷N.

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Ans:  option (iv)

The number of radial nodes is given by n-l-1

where n is principal quantum number, l is azimuthal quantum number

For 3p orbital, n=3 and l=1

Thus, the number of radial nodes

= n-l-1 

= 3-1-1

 = 1

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Ans: option (iii) 2

The number of angular nodes is given by n-l

where n is principal quantum number, l is azimuthal quantum number

For 4d orbital, n=4 and l=2

Thus, the number of angular nodes

 = n-l

 = 4-2

 = 2

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Ans: (ii) Heisenberg's uncertainty principle.

According to Heisenberg's uncertainty principle it is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron. Thus it implies that determining the trajectory of an electron is impossible as it requires exact position and velocity which is not possible as per the uncertainty principle.

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Ans:  option (iii) 9

The total number of orbitals is given by n2, where n is the principal quantum number or the principal shell.

Thus for the third shell number of orbitals is 32 i.e. 9

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Ans: Option (i)

 “l” which is known as Azimuthal quantum number or orbital angular momentum or subsidiary quantum number depicts the three dimensional shape of the orbital.

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Ans: option  (ii)

Fe³⁺, Mn²⁺ Fe³⁺ electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s² Mn²⁺electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s

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Ans: option (iv) The two electrons present in the 2s orbital have spin quantum numbers MS but of opposite sign.

According to Pauli’s Exclusion Principle no two electrons should possess the same set of four quantum numbers. Thus the two electrons present in the 2s orbital have spin quantum numbers MS but of opposite sign.

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Ans:  option (ii) Alpha particle (He²⁺)

As the wavelength is inversely proportional to the mass of the particles thus the alpha particles would possess the shortest wavelength among the above given option

$\mathit{\lambda }=\frac{\mathit{h}}{\mathit{m}\mathit{v}}$

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Ans: option  (iii) & (iv)

The isotopes are defined as atoms with identical atomic numbers but different mass numbers are known as isotopes.

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Ans:  option (i) & (iv)

In a multielectron atomic system the energy of an electron depends not only on its principal quantum number (shell), but also on its azimuthal quantum number (subshell). Electrons having the same shells and same subshells have the same energy and they are known as degenerate orbitals.

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Ans: option (ii) & (iii)

For the given value of n (principal quantum number) the value of l (Azimuthal quantum number) varies from 0 to n-1. However for the given value of l the m_l (magnetic quantum number) varies from -1 to +1.

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Ans:  option (i) & (iii)

Isoelectronic species are those species that possess the same number of electrons. Here both Na⁺, Mg²⁺ possess 10 electrons and both Na⁺, O^2– possess 10 electrons.

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Ans:  option (i)& (iv)

1. Principal quantum number (n): It represents the size and energy of orbitals.
2. Azimuthal quantum number (l): It represents the subshell and shape of orbitals.
3. Magnetic quantum number (ml): It represents the orientation of the orbitals.
4. Spin quantum number (ms): It represents the direction of spin of electrons in the orbitals.

Quantum numbers describe the unique position and energy of an electron in an atom. These are a set of four numbers and are important for predicting chemical behavior and understanding the distribution of electrons in orbitals.

The hydrogen atom can be accurately explained by Bohr's model. However, it does not account for shielding effects, electron-electron interactions, and the wave nature of electrons for multi-electron atoms.

According to the de Broglie equation, the particles like electrons have wave-like properties. The quantum mechanical model foundation is laid by this concept and it also explains the stability of electron orbits using wave behavior.

No. of equivalents of H₂SO₄ = 100 × 0.1 × 2 = 20

No. of equivalents of NaOH = 50 × 0.1 = 5

No. equivalents of H₂SO₄ left = 20 – 5 = 15

150 × x = 15

$x=\frac{1}{10}=0.1N=1\times 10^{-1}N$

For real gas under high pressure,

$Z=1+\frac{Pb}{RT}\Rightarrow b=\frac{RT}{P}=\frac{0.083\times 298}{99}$        

= 0.25 × 10^-2 L mol^-1

Let x gm is burnt

Moles = x/280

Heat released by x/280 mol = 2.5 × 0.45 kJ

Heat released by 1 mol =  $\frac{2.5\times 0.45\times 280}{x}=kJ$

$\begin{array}{l}\Delta H=\Delta U+\Delta ngRT\\ \Delta H=\Delta U\end{array}$            

$9=\frac{2.5\times 280\times 0.45}{x}$            

x = 35 gm

$\frac{P^0-P_S}{P^0}=\frac{n_{solute}}{n_{solvent}}$

$\frac{P^0-\frac{P^0}{2}}{P^0}=\frac{n_{solute}}{n_{solvent}}$

$n_{solute}=\frac{n_{solvent}}{2}=\frac{100}{18\times 2}=2.78mol$

$K=\frac{0.693}{t_{1/2}}=\frac{0.693}{70\times 60}=\frac{6930}{7\times 6}\times 10^{-6}$

= 165 × 10^-6 s^-1

Bauxite = AlOx (OH)_3-2x (where O < x < 1)

Siderite = FeCO₃

Cuprite = Cu₂O

Calamine = ZnCO₃

Haemetite = Fe₂O₃

Kaolinite = Al₂ (OH)₄Si₂O₅

Malachite = CuCO₃ Cu (OH)₂

Magneite = Fe₃O₄

Sphalerite = ZnS

Limonite = Fe₂O₃.3H₂O

  $2KMn{O}_4+3H_2{O}_2\underset{}{\overset{Basicmedium}{\to }}2Mn{O}_2+3O_2+2H_{2}O+2KOH$

SO₃ -> Sp² Planar

BF₃ -> Sp² Planar

$N{O}_3^{-}$ -> Sp² Planar

SF₄ -> Sp³d non-planar

H₂O₂ -> Sp³ Non-planar

PCl₃ -> Sp³ Non –planar

[Al (OH)₄]^{- ->} Sp³ Non-planar

XeF₄ -> Sp³d² planar

XeO₃ -> Sp³ Non-planar

$P{H}_4^{-}$ -> Sp³ Non-planar

Fehling's solution is a complex of   $C{u}^{2+},C{u}^{2+}=3d^9$

No. of unpaired e^- = 1

MM =  $\sqrt[]{1\left(1+2\right)}=\sqrt[]{3}=1.73BM$

No. of moles =  $\frac{5}{92}\left(toluene\right)$

Moles of benzaldehyde = $\frac{5}{92}\times \frac{92}{100}=5\times 10^{-2}$  

Mass of benzaldehyde = 5 × 10^-2 ×106 = 5.3 gm

= 530 × 10^-2 gm

Kindly consider the following Image

Sol. E? cell = E? (Sn²? |Sn) - E? (Cu²? |Cu)
= -0.16 - 0.34 = -0.50V
ΔG? = -nFE? cell
= -2 × 96500 × (-0.5) = 96500 J
= 96.5 kJ = 96500 J

The oxidation states of iron in these compounds will be
A = +2
B = +4
C = 0
The sum of oxidation states will be = 6.

189000 to 190000
Sol. ΔH = ΔU + Δn? RT
41000 × 5 = ΔU + 5 × 8.314 × 373
205000 = ΔU + 15505.61
ΔU = 189494.39 J = 189494 J

Sol. (x/m) = k (P)¹/?
log (x/m) = logk + 1/n logP
Slope = 1/n = 2 So n = 1/2
Intercept ⇒ logk = 0.477 So k = Antilog (0.477) = 3
So (x/m) = k (P)¹/? = 3² = 48