Maharashtra Board Class 11 Chemistry Chapter 4 Structure of Atom Solutions

11th Chemistry Chapter 4 Exercise Structure Of Atom Solutions Maharashtra Board

Structure Of Atom Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Chemistry Chapter 4 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 4 Exercise Solutions

1. Choose Correct Option.

Question A. The energy difference between the shells goes on - when moved away from the nucleus.
(a) Increasing
(b) decreasing
(c) equalizing
(d) static
Answer: (b) decreasing
In simple words: As you move further away from the nucleus, the energy levels of electron shells get closer together, meaning the difference in energy between successive shells decreases.

🎯 Exam Tip: Understand the relationship between electron shell distance from the nucleus and energy level differences, a key concept in atomic structure.

Question B. The value of Plank's constant is
(a) \( 6.626 \times 10^{-34} \) Js
(b) \( 6.023 \times 10^{-24} \) Js
(c) \( 1.667 \times 10^{-28} \) Js
(d) \( 6.626 \times 10^{-28} \) Js
Answer: (a) \( 6.626 \times 10^{-34} \) Js
In simple words: Planck's constant (h) is a fundamental physical constant representing the quantum of action in quantum mechanics, crucial for calculating the energy of a photon.

🎯 Exam Tip: Memorize fundamental constants like Planck's constant, as they are often required in calculations and theoretical questions.

Question C. p-orbitals are - in shape.
(a) spherical
(b) dumb bell
(c) double dumb bell
(d) diagonal
Answer: (b) dumb bell
In simple words: P-orbitals have a distinct dumbbell shape, with two lobes on opposite sides of the nucleus, which differentiates them from the spherical s-orbitals.

🎯 Exam Tip: Knowing the shapes of different orbitals (s, p, d, f) is essential for understanding electron distribution and chemical bonding.

Question D. "No two electrons in the same atoms can have identical set of four quantum numbers". This statement is known as
(a) Pauli's exclusion principle
(b) Hund's rule
(c) Aufbau rule
(d) Heisenberg uncertainty principle
Answer: (a) Pauli's exclusion principle
In simple words: Pauli's exclusion principle states that each electron in an atom occupies a unique quantum state, meaning no two electrons can have the exact same set of four quantum numbers.

🎯 Exam Tip: Differentiate clearly between Pauli's exclusion principle, Hund's rule, and the Aufbau principle; they are distinct rules governing electron configuration.

Question E. Principal Quantum number describes
(a) shape of orbital
(b) size of the orbital
(c) spin of electron
(d) orientation of in the orbital electron cloud
Answer: (b) size of the orbital
In simple words: The principal quantum number (n) primarily determines the energy level and the average distance of an electron from the nucleus, thus indicating the size of the orbital.

🎯 Exam Tip: Understand what each of the four quantum numbers (principal, azimuthal, magnetic, spin) describes about an electron's state and orbital properties.

2. Make The Pairs:

A		B
a.	Neutrons	i.	six electrons
b.	p-orbital	ii.	\( -1.6 \times 10^{-19} \) C
c.	charge on electron	iii.	Ultraviolet region
d.	Lyman series	iv.	Chadwick

Answer: a - iv,
b - i,
c - ii,
d - iii
In simple words: This matching exercise connects key atomic concepts: Neutrons were discovered by Chadwick, p-orbitals can hold six electrons, the charge on an electron is \( -1.6 \times 10^{-19} \) C, and the Lyman series in the hydrogen spectrum is found in the ultraviolet region.

🎯 Exam Tip: Remember the discoverers of subatomic particles, characteristics of orbitals, fundamental physical constants, and the regions associated with different spectral series.

3. Complete The Following Information About The Isotopes In The Chart Given Below:

Substance	Mass Number	Number of
		Protons	Neutrons	Electrons
Carbon-14
Lead-208
Chlorine-35
Uranium-238
Oxygen-18
Radium-223

(Hint: Refer to Periodic Table if required)
Answer:

Substance	Mass number	Number of
		Protons	Neutrons	Electrons
Carbon-14	14	6	8	6
Lead-208	208	82	126	82
Chlorine-35	35	17	18	17
Uranium-238	238	92	146	92
Oxygen-18	18	8	10	8
Radium-223	223	88	135	88

In simple words: To complete this table, identify the atomic number (number of protons) from the periodic table, which is also the number of electrons in a neutral atom. Then, subtract the number of protons from the mass number to find the number of neutrons.

🎯 Exam Tip: Remember that for a neutral atom, the number of protons equals the number of electrons. The number of neutrons is calculated as Mass Number - Atomic Number (protons).

4. Match The Following :

Element		No. of Neutron
a. \( ^{40}_{18}Ar \)	i.	7
b. \( ^{14}_6C \)	ii.	21
c. \( ^{40}_{19}K \)	iii.	8
d. \( ^{14}_7N \)	iv.	22

Answer: a - iv,
b - iii,
c - ii,
d - i
In simple words: To match elements with their number of neutrons, subtract the atomic number (bottom number) from the mass number (top number) for each element.

🎯 Exam Tip: Practice calculating the number of neutrons quickly from isotopic notation (atomic number and mass number).

5. Answer In One Sentence :

Question A. If an element 'X' has mass number 11 and it has 6 neutrons, then write its representation.
Answer: The representation of the given element is \( ^{11}_5X \).
In simple words: The atomic number (number of protons) is found by subtracting the number of neutrons from the mass number, which then allows for the proper isotopic notation.

🎯 Exam Tip: Remember that isotopic notation is \( ^A_ZX \), where A is the mass number and Z is the atomic number (number of protons).

Question B. Name the element that shows simplest emission spectrum.
Answer: The element that shows simplest emission spectrum is hydrogen.
In simple words: Hydrogen has only one electron, making its electron transitions and resulting emission spectrum the simplest to analyze.

🎯 Exam Tip: Hydrogen's simple spectrum was crucial for the development of quantum mechanics and Bohr's model.

Question C. State Heisenberg uncertainty principle.
Answer: Heisenberg uncertainty principle states that "It is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron".
In simple words: This principle means you cannot precisely know both where an electron is and where it's going at the same time. The more accurately you know one, the less accurately you know the other.

🎯 Exam Tip: The Heisenberg uncertainty principle is a cornerstone of quantum mechanics, highlighting the probabilistic nature of electron behavior.

Question D. Give the names of quantum numbers.
Answer: The four quantum numbers are: principal quantum number (n), azimuthal or subsidiary quantum number (l), magnetic quantum number (\( m_l \)) and electron spin quantum number (\( m_s \)).
In simple words: Quantum numbers are a set of four values that describe the unique state of an electron in an atom, covering its energy, shape, spatial orientation, and spin.

🎯 Exam Tip: Be able to name all four quantum numbers and briefly explain what each one signifies.

Question E. Identify from the following the isoelectronic species:
Ne, \( O^{2-} \), \( Na^+ \) OR Ar, \( Cl^{2-} \), \( K^+ \)
Answer: Atoms and ions having the same number of electrons are isoelectronic.

Species	No. of electrons
Ne	10
\( O^{2-} \)	8 + 2 = 10
\( Na^+ \)	11 - 1 = 10
Ar	18
\( Cl^{2-} \)	17 + 2 = 19
\( K^+ \)	19 - 1 = 18

Hence, Ne, \( O^{2-} \), \( Na^+ \) are isoelectronic species.
In simple words: Isoelectronic species are atoms or ions that possess the same total number of electrons, leading to similar electron configurations.

🎯 Exam Tip: To identify isoelectronic species, calculate the total number of electrons for each atom or ion (atomic number for neutral atoms, adjust for charge for ions).

6. Answer The Following Questions.

Question A. Differentiate between Isotopes and Isobars.
Answer:

No.	Isotopes	Isobars
i.	Isotopes are atoms of same element.	Isobars are atoms of different elements.
ii.	They have same atomic number but different atomic mass number.	They have same atomic mass number but different atomic numbers.
iii.	They have same number of protons but different number of neutrons.	They have different number of protons and neutrons.
iv.	They have same number of electrons.	They have different number of electrons.
v.	They occupy same position in the modern periodic table.	They occupy different positions in the modern periodic table.
vi.	They have similar chemical properties.	They have different chemical properties.
e.g.	\( ^{12}_6C \) and \( ^{14}_6C \)	\( ^{14}_6C \) and \( ^{14}_7N \)

In simple words: Isotopes are variations of the same element with the same number of protons but different numbers of neutrons, while isobars are atoms of different elements that share the same mass number but have different atomic numbers.

🎯 Exam Tip: Focus on the definitions: isotopes have the same Z, different A; isobars have the same A, different Z. This distinction is crucial.

Question B. Define the terms:
(i) Isotones
(ii) Isoelectronic species
(iii) Electronic configuration
Answer:
(i) Isotones: Isotones are defined as the atoms of different elements having same number of neutrons in their nuclei. e.g. \( ^{11}_5B \) and \( ^{12}_6C \) having 6 neutrons each are isotones.

(ii) Isoelectronic species:
Isoelectronic species are defined as atoms and ions having the same number of electrons.
e. g. Ar, \( Ca^{2+} \) and \( K^+ \) containing 18 electrons each.

(iii) Electronic configuration:
Electronic configuration of an atom is defined as the distribution of its electrons in orbitals.
In simple words: Isotones are atoms with the same neutron count but different protons; isoelectronic species have the same number of electrons; and electronic configuration describes how electrons are arranged in an atom's orbitals.

🎯 Exam Tip: Understand the precise definitions and provide clear examples for each term to demonstrate full comprehension.

Question C. State and explain Pauli's exclusion principle.
Answer:
Pauli's exclusion principle:
(i) Statement: "No two electrons in an atom can have the same set of four quantum numbers". OR "Only two electrons can occupy the same orbital and they must have opposite spins. "
(ii) The capacity of an orbital to accommodate electrons is decided by Pauli's exclusion principle.
(iii) According to this principle, for an electron belonging to the same orbital, the spin quantum number must be different since the other three quantum numbers are the same.
(iv) The spin quantum number can have two values: \( + \frac{1}{2} \) and \( - \frac{1}{2} \).
(v) Example, consider helium (He) atom with electronic configuration \( 1s^2 \).
For the two electrons in 1s orbital, the four quantum numbers are as follows:

Electron number	Quantum number				Set of values of quantum numbers
	n	l	m	s
\( 1^{st} \) Electron	1	0	0	\( + \frac{1}{2} \)	\( (1, 0, 0, + \frac{1}{2}) \)
\( 2^{nd} \) Electron	1	0	0	\( - \frac{1}{2} \)	\( (1, 0, 0, - \frac{1}{2}) \)

Thus, in an atom, any two electrons can have the same three quantum numbers, but the fourth quantum number must be different.
(vi) This leads to the conclusion that an orbital can accommodate maximum of two electrons and if it has two electrons, they must have opposite spin.
In simple words: Pauli's exclusion principle states that no two electrons in an atom can have the same exact set of quantum numbers, meaning if two electrons are in the same orbital, they must have opposite spins. This limits an orbital to a maximum of two electrons.

🎯 Exam Tip: Clearly state the principle and use an example like Helium to illustrate how the four quantum numbers differ for electrons within the same orbital.

Question D. State Hund's rule of maximum multiplicity with suitable example.
Answer:
Hund's rule of maximum multiplicity:
(i) Statement: "Pairing of electrons in the orbitals belonging to the same subshell does not occur unless each orbital belonging to that subshell has got one electron each."
(ii) Example, according to Hund's rule, each of the three-degenerate p-orbitals must get one electron of parallel spin before any one of them receives the second electron of opposite spin. Therefore, the configuration of four electrons occupying p-orbitals is represented as
\( \uparrow\downarrow\uparrow\uparrow \)
and not as
\( \uparrow\downarrow\uparrow\downarrow \)
(iii) As a result of Hund's rule, the atom with fully filled and half-filled set of degenerate orbitals has extra stability.
In simple words: Hund's rule states that electrons will individually occupy each orbital within a subshell with parallel spins before any orbital is doubly occupied. This maximizes the total spin and contributes to atomic stability.

🎯 Exam Tip: Provide a clear statement of Hund's rule and use orbital diagrams (e.g., for p-orbitals) to illustrate the correct and incorrect ways of filling electrons.

Question E. Write the drawbacks of Rutherford's model of an atom.
Answer:
Drawbacks of Rutherford's model of an atom:
(i) Rutherford's model of an atom resembles the solar system with the nucleus playing the role of the massive sun and the electrons are lighter planets. Thus, according to this model, electrons having negative charge revolve in various orbits around the nucleus. However, the electrons revolving about the nucleus in fixed orbits pose a problem. Such orbital motion is an accelerated motion accompanied by a continuous change in the velocity of electron as noticed from the continuously changing direction. According to Maxwell's theory of electromagnetic radiation, accelerated charged particles would emit electromagnetic radiation. Hence, an electron revolving around the nucleus should continuously emit radiation and lose equivalent energy. As a result, the orbit would shrink continuously and the electron would come closer to the nucleus by following a spiral path. It would ultimately fall into the nucleus. Thus, Rutherford's model has an intrinsic instability of atom. However, real atoms are stable.
(ii) Rutherford's model of an atom does not describe the distribution of electrons around the nucleus and their energies.
In simple words: Rutherford's model predicted that electrons orbiting the nucleus would continuously lose energy and spiral into the nucleus, making atoms unstable, which contradicts observed atomic stability. It also failed to explain the specific energy levels of electrons.

🎯 Exam Tip: The main drawback to highlight is the instability of the atom and its failure to explain atomic spectra or electron energy distribution.

Question F. Write postulates of Bohr's Theory of hydrogen atom.
Answer:
Postulates of Bohr's theory of hydrogen atom:
(i) The electron in the hydrogen atom can move around the nucleus in one of the many possible circular paths of fixed radius and energy. These paths are called orbits, stationary states or allowed energy states. These orbits are arranged concentrically around the nucleus in an increasing order of energy.
(ii) The energy of an electron in the orbit does not change with time. However, the electron will move from a lower stationary state to a higher stationary state if and when the required amount of energy is absorbed by the electron. Energy is emitted when electron moves from a higher stationary state to a lower stationary state. The energy change does not take place in a continuous manner.
(iii) The frequency of radiation absorbed or emitted when transition occurs between two stationary states that differ in energy by \( \Delta E \) is given by the following expression:
\( \nu = \frac{\Delta E}{h} = \frac{E_2-E_1}{h} \)
Where \( E_1 \) and \( E_2 \) are the energies of the lower and higher allowed energy states respectively. This expression is commonly known as Bohr's frequency rule.
(iv) The angular momentum of an electron in a given stationary state can be expressed as \( mvr = n \times \frac{h}{2\pi} \)
where, n = 1, 2, 3
Thus, an electron can move only in those orbits for which its angular momentum is integral multiple of \( h/2\pi \).
Thus, only certain fixed orbits are allowed.
In simple words: Bohr's theory proposed that electrons exist in specific, quantized orbits with fixed energies, do not radiate energy while in these orbits, and can only jump between orbits by absorbing or emitting discrete amounts of energy.

🎯 Exam Tip: Key points include stationary states, quantized energy levels, non-continuous energy changes, and the quantization of angular momentum (\( mvr = n h/2\pi \)).

Question G. Mention demerits of Bohr's Atomic model.
Answer:
Demerits of Bohr's atomic model:
- Bohr's atomic model (theory) failed to account for finer details of the atomic spectrum of hydrogen as observed in sophisticated spectroscopic experiments.
- Bohr's atomic model (theory) was unable to explain the spectrum of atoms other than hydrogen.
- Bohr's atomic model (theory) could not explain the splitting of spectral line in the presence of a magnetic field (Zeeman effect) or electric field (Stark effect).
- Bohr's atomic model (theory) failed to explain the ability of atoms to form molecules by chemical bonds.
In simple words: Bohr's model couldn't explain complex spectra of multi-electron atoms, the splitting of spectral lines in magnetic/electric fields (Zeeman/Stark effect), or how atoms form chemical bonds.

🎯 Exam Tip: Focus on the limitations related to multi-electron atoms, spectral phenomena, and chemical bonding, which led to the development of quantum mechanics.

Question H. State the order of filling atomic orbitals following Aufbau principle.
Answer:
Aufbau principle:
(i) Aufbau principle gives the sequence in which various orbitals are filled with electrons.
(ii) In the ground state of an atom, the orbitals are filled with electrons based on increasing order of energies of orbitals, Pauli's exclusion principle and Hund's rule of maximum multiplicity.
(iii) Increasing order of energies of orbitals:
- Orbitals are filled in order of increasing value of (n + l)
- In cases where the two orbitals have same value of (n + l), the orbital with lower value of n is filled first.
(iv) The increasing order of energy of different orbitals in a multi-electron atom is:
\( 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s \)
and so on.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र ऑर्बिटल्स में इलेक्ट्रॉनों के भरने के क्रम (ऑफ़बाऊ सिद्धांत) को दर्शाता है। इसमें विभिन्न ऊर्जा स्तरों (जैसे 1s, 2s, 2p, 3s, 3p, 3d, आदि) को तिरछे तीरों द्वारा जोड़ा गया है, जो इलेक्ट्रॉनों के बढ़ते ऊर्जा क्रम में ऑर्बिटल्स में प्रवेश करने की दिशा दिखाते हैं।
In simple words: The Aufbau principle dictates that electrons fill atomic orbitals in order of increasing energy, generally from lower (n+l) values to higher ones, and using 'n' as a tie-breaker, to achieve the most stable electron configuration.

🎯 Exam Tip: Memorize the Aufbau sequence for orbital filling and understand the (n+l) rule for determining orbital energy order.

Question I. Explain the anomalous behavior of copper and chromium.
Answer:
(i) Copper:
- Copper (Cu) has atomic number 29.
- Its expected electronic configuration is \( 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^9 \).
- The 3d orbital is neither half-filled nor fully filled. Hence, it has less stability.
- Due to interelectronic repulsion forces, one 4s electron enters into 3d orbital. This makes 3d orbital completely filled and 4s orbital half-filled which gives extra stability and the electronic configuration of Cu becomes, \( 1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^{10} \).
(ii) Chromium:
- Chromium (Cr) has atomic number 24.
- Its expected electronic configuration is \( 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^4 \).
- The 3d orbital is less stable as it is not half-filled.
- Due to inter electronic repulsion forces, one 4s electron enters into 3d orbital. This makes 4s and 3d orbitals half-filled which gives extra stability and the electronic configuration of Cr becomes, \( 1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5 \).
In simple words: Copper and chromium show anomalous electron configurations where an electron from the 4s orbital shifts to the 3d orbital to achieve a more stable half-filled (d5) or fully-filled (d10) d-subshell.

🎯 Exam Tip: Always remember that half-filled and fully-filled subshells (especially d and f) confer extra stability, leading to exceptions in electron configurations for elements like Cr and Cu.

Question J. Write orbital notations for electrons in orbitals with the following quantum numbers.
(a) n = 2, l = 1
(b) n = 4, l = 2
(c) n = 3, l = 2
Answer:
(i) 2p
(ii) 4d
(iii) 3d
In simple words: The principal quantum number (n) indicates the main energy level, and the azimuthal quantum number (l) indicates the subshell type (l=0 for s, l=1 for p, l=2 for d, l=3 for f), allowing for the determination of orbital notation.

🎯 Exam Tip: Associate the 'l' values (0, 1, 2, 3) directly with the subshell letters (s, p, d, f) for quick notation conversions.

Question K. Write electronic configurations of Fe, \( Fe^{2+} \), \( Fe^{3+} \)
Answer:

Species	Orbital notation
Fe	\( 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6 \) OR [Ar] \( 4s^2 3d^6 \)
\( Fe^{2+} \)	\( 1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 \) OR [Ar] \( 3d^6 \)
\( Fe^{3+} \)	\( 1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 \) OR [Ar] \( 3d^5 \)

In simple words: For transition metals like Iron, electrons are removed first from the outermost s-orbital before being removed from the d-orbital when forming positive ions.

🎯 Exam Tip: When writing configurations for transition metal ions, always remove electrons from the highest 'n' value orbital first (e.g., 4s before 3d for Fe).

Question L. Write condensed orbital notation of electonic configuration of the following elements:
(a) Lithium (Z = 3)
(b) Carbon (Z=6)
(c) Oxygen (Z = 8)
(d) Silicon (Z = 14)
(e) Chlorine (Z = 17)
(f) Calcium (Z = 20)
Answer:

No.	Element	Condensed orbital notation
i.	Lithium (Z = 3)	[He] \( 2s^1 \)
ii.	Carbon (Z = 6)	[He] \( 2s^2 2p^2 \)
iii.	Oxygen (Z = 8)	[He] \( 2s^2 2p^4 \)
iv.	Silicon (Z = 14)	[Ne] \( 3s^2 3p^2 \)
v.	Chlorine (Z = 17)	[Ne] \( 3s^2 3p^5 \)
vi.	Calcium (Z = 20)	[Ar] \( 4s^2 \)

In simple words: Condensed orbital notation uses the symbol of the preceding noble gas in brackets to represent the core electrons, followed by the configuration of the valence electrons.

🎯 Exam Tip: Be proficient in using noble gas configurations for condensed notation, as it simplifies writing longer electronic configurations.

Question M. Draw shapes of 2s and 2p orbitals.
Answer:
2s orbital:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र 1s और 2s ऑर्बिटल्स के आकार को दर्शाता है। दोनों गोलाकार (spherical) होते हैं, लेकिन 2s ऑर्बिटल 1s ऑर्बिटल से बड़ा होता है और इसमें एक नोडल सतह (nodal surface) होती है जहाँ इलेक्ट्रॉन घनत्व शून्य होता है।

2p orbital:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र 2p ऑर्बिटल्स (2px, 2py, 2pz) के आकार को दर्शाता है। सभी p-ऑर्बिटल्स डंबल के आकार के होते हैं और एक दूसरे के लंबवत (perpendicular) अक्षों (x, y, z) पर संरेखित होते हैं।
In simple words: 2s orbitals are spherical, larger than 1s, with an internal node, while 2p orbitals are dumbbell-shaped, oriented along the x, y, and z axes, with a nodal plane through the nucleus.

🎯 Exam Tip: Be prepared to sketch and label the basic shapes of s and p orbitals, indicating their orientation and relative sizes.

Question N. Explain in brief, the significance of azimuthal quantum number.
Answer:
Azimuthal quantum number (l):
- Azimuthal quantum number is also known as subsidiary quantum number and is represented by letter l.
- It represents the subshell to which the electron belongs. It also defines the shape of the orbital that is occupied by the electron.
- Its value depends upon the value of principal quantum number 'n'. It can have only positive values between 0 and (n - 1).
- Atomic orbitals with the same value of 'n' but different values of 'l' constitute a subshell belonging to the shell for the given 'n' The azimuthal quantum number gives the number of subshells in a principal shell. The subshells have l to be 0, 1, 2,3 - which are represented by symbols s, p, d, f, - respectively.

Principal shell	Value of n	Permissible value of l	Possible subshell	Number of subshells in shell
K	1	0	s	1
L	2	0, 1	s, p	2
M	3	0, 1, 2	s, p, d	3
N	4	0, 1, 2, 3	s, p, d, f	4

In simple words: The azimuthal quantum number (l) determines the shape of an orbital (s, p, d, f) within a given principal shell and its value ranges from 0 to (n-1).

🎯 Exam Tip: Clearly link the 'l' values to specific orbital shapes (s=spherical, p=dumbbell, d=complex) and their dependency on 'n'.

Question O. If n = 3, what are the quantum number l and \( m_l \)?
Answer:
: For a given n, l = 0 to (n - 1) and for given l, \( m_l = -l....., 0...... +l \)
Therefore, the possible values of l and \( m_l \) for n = 3 are:

Value of n	Value of l	Values of \( m_l \)
3	0	\( m_l = 0 \)
	1	\( m_l = -1 \)
		\( m_l = 0 \)
		\( m_l = +1 \)
	2	\( m_l = -2 \)
		\( m_l = -1 \)
		\( m_l = 0 \)
		\( m_l = +1 \)
		\( m_l = +2 \)

In simple words: For a principal quantum number n=3, the azimuthal quantum number (l) can be 0, 1, or 2, corresponding to s, p, and d subshells respectively. For each 'l' value, the magnetic quantum number (\( m_l \)) can range from -l to +l.

🎯 Exam Tip: Systematically list the possible 'l' values (0 to n-1) and for each 'l', list the possible '\( m_l \)' values (-l to +l) to ensure completeness.

Question P. The electronic configuration of oxygen is written as \( 1s^2 2s^2 2p_x^2 2p_y^1 2p_z^1 \) and not as \( 1s^2 2s^2 2p_x^2 2p_y^2 2p_z^0 \). Explain.
Answer:
- According to Hund's rule of maximum multiplicity "Pairing of electrons in the orbitals belonging to the same subshell does not occur unless each orbital belonging to that subshell has got one electron each.”
- Oxygen has 8 electrons. The first two electrons will pair up in the 1s orbital, the next two electrons will pair up in the 2s orbital and this leaves 4 electrons, which must be placed in the 2p orbitals.
- Each of the three degenerate p-orbitals must get one electron of parallel spin before any one of them receives the second electron of opposite spin. Therefore, two p orbitals have one electron each and one p-orbital will have two electrons. Thus, the electronic configuration of oxygen is written as \( 1s^2 2s^2 2p_x^2 2p_y^1 2p_z^1 \) and not as \( 1s^2 2s^2 2p_x^2 2p_y^2 2p_z^0 \).
In simple words: The correct configuration for oxygen's 2p subshell follows Hund's rule, meaning electrons first singly occupy each degenerate 2px, 2py, 2pz orbital with parallel spins before any pairing occurs, leading to \( 2p_x^2 2p_y^1 2p_z^1 \) rather than filling one orbital completely and leaving another empty.

🎯 Exam Tip: Always apply Hund's rule for degenerate orbitals: maximize unpaired electrons with parallel spins before pairing electrons.

 

Question Q. Write note on 'Principal Quantum number.

Answer:Principal quantum number (n):
(i) Principal quantum number indicates the principal shell or main energy level to which the electron belongs.
(ii) It is denoted by 'n' and is a positive integer with values 1, 2, 3, 4, 5, 6, ....
(iii) A set of atomic orbitals with given value of 'n' constitutes a single shell. These shells are also represented by the letters K, L, M, N, etc.
(iv) With increase of 'n', the number of allowed orbitals in that shell increases and is given by \(n^2\).
(v) The allowed orbitals in the first four shells are given below:

Principal quantum number (n)	Shell symbol	Allowed number of orbitals (\(n^2\))	Size of shell
1	K	1	increases
2	L	4
3	M	9
4	N	16

(vi) As the value of 'n' increases, the distance of the shell from the nucleus increases and the size of the shell increases. Its energy also goes on increasing.
In simple words: The principal quantum number (n) describes the main energy level or shell of an electron, indicating its distance from the nucleus and overall energy. Higher 'n' values mean larger, higher-energy shells farther from the nucleus, and the number of orbitals in a shell is \(n^2\).

🎯 Exam Tip: Understanding the significance of each quantum number is crucial for describing electron states and orbital properties in exams.

 

Question R. Using concept of quantum numbers, calculate the maximum numbers of electrons present in the 'M' shell. Give their distribution in shells, subshells and orbitals.

Answer:
(i) Each main shell contains a maximum of \(2n^2\) electrons. For 'M' shell, n = 3.
Therefore, the maximum numbers of electrons present in the 'M' shell = \(2 \times (3)^2\) = 18.
(ii) The distribution of these electrons in shells, subshells and orbitals can be given as follows:

Value of n	Values of l	Values of \(m_l\)	Values of \(m_s\)
3	0	0	±½
	1	-1	±½
		0	±½
		+1	±½
	2	-2	±½
		-1	±½
		0	±½
		+1	±½
		+2	±½

Note: Orbital distribution in the first four shells:

Symbol of Shell	Value of Principal quantum number (n)	Number of subshells	Value of Azimuthal Quantum number (l)	Symbol of subshell	Total Number of orbitals in the subshell = \(2l+1\)	Values of the magnetic quantum number \(m_l\) for the subshell
K	n=1	1	l=0	1s	\(2 \times 0 + 1 = 1\)	\(m_l\) = 0
L	n=2	2	l=0	2s	\(2 \times 0 + 1 = 1\)	\(m_l\) = 0
			l=1	2p	\(2 \times 1 + 1 = 3\)	\(m_l\) = -1 \(m_l\) = 0 \(m_l\) = +1
M	n=3	3	l=0	3s	\(2 \times 0 + 1 = 1\)	\(m_l\) = 0
			l=1	3p	\(2 \times 1 + 1 = 3\)	\(m_l\) = -1 \(m_l\) = 0 \(m_l\) = +1
			l=2	3d	\(2 \times 2 + 1 = 5\)	\(m_l\) = -2 \(m_l\) = -1 \(m_l\) = 0 \(m_l\) = +1 \(m_l\) = +2
N	n=4	4	l=0	4s	\(2 \times 0 + 1 = 1\)	\(m_l\) = 0
			l=1	4p	\(2 \times 1 + 1 = 3\)	\(m_l\) = -1 \(m_l\) = 0 \(m_l\) = +1
			l=2	4d	\(2 \times 2 + 1 = 5\)	\(m_l\) = -2 \(m_l\) = -1 \(m_l\) = 0 \(m_l\) = +1 \(m_l\) = +2
			l=3	4f	\(2 \times 3 + 1 = 7\)	\(m_l\) = -3 \(m_l\) = -2 \(m_l\) = -1 \(m_l\) = 0 \(m_l\) = +1 \(m_l\) = +2 \(m_l\) = +3

In simple words: For the M-shell (n=3), there are a maximum of 18 electrons. These electrons are distributed among the 3s (l=0), 3p (l=1), and 3d (l=2) subshells, occupying one s-orbital, three p-orbitals, and five d-orbitals, respectively, with each orbital holding a maximum of two electrons with opposite spins.

🎯 Exam Tip: Remember the \(2n^2\) rule for maximum electrons in a shell and how quantum numbers (n, l, \(m_l\)) dictate orbital and subshell distribution.

 

Question S. Indicate the number of unpaired electrons in :
a. Si (Z = 14)
b. Cr (Z = 24)

Answer:
(i) Si (Z = 14): \(1s^2 2s^2 2p^6 3s^2 3p^2\)
Orbital diagram:
ℹ️ चित्र व्याख्या (Diagram Explanation): सिलिकॉन (Si) के 1s, 2s, 2p, 3s और 3p ऑर्बिटल्स में इलेक्ट्रॉनों का वितरण दिखा रहा है। 3p ऑर्बिटल में दो अयुग्मित इलेक्ट्रॉन ऊपर की दिशा में हैं।
Number of unpaired electrons = 2
(ii) Cr (Z = 24): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5\)
Orbital diagram:
ℹ️ चित्र व्याख्या (Diagram Explanation): क्रोमियम (Cr) के 1s, 2s, 2p, 3s, 3p, 4s और 3d ऑर्बिटल्स में इलेक्ट्रॉनों का वितरण दिखा रहा है। 4s ऑर्बिटल में एक और 3d ऑर्बिटल में पांच अयुग्मित इलेक्ट्रॉन ऊपर की दिशा में हैं।
Number of unpaired electrons = 6
In simple words: Silicon (Z=14) has an electronic configuration of \(1s^2 2s^2 2p^6 3s^2 3p^2\), leading to two unpaired electrons in its 3p orbitals. Chromium (Z=24) has an anomalous configuration of \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5\) (due to stability of half-filled subshells), resulting in one unpaired electron in 4s and five unpaired electrons in 3d, totaling six unpaired electrons.

🎯 Exam Tip: For elements like Cr and Cu, remember their anomalous electronic configurations due to the stability gained from half-filled or fully-filled d-orbitals. This is a common exam question.

 

Question T. An atom of an element contains 29 electrons and 35 neutrons. Deduce-
a. the number of protons
b. the electronic configuration of that element.

Answer:
a. In an atom, number of protons is equal to number of electrons. The given atom contains 29 electrons.
\( \implies \) Number of protons = 29
b. The electronic configuration of an atom of an element containing 29 electrons is:
\(1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^{10}\)
[Note: Given element is copper (Cu) with Z = 29]
In simple words: Since a neutral atom has an equal number of protons and electrons, an element with 29 electrons also has 29 protons. The electronic configuration for this element (Copper, Z=29) is \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^{10}\), showing the anomalous configuration for extra stability.

🎯 Exam Tip: Remember that for a neutral atom, atomic number (Z) equals the number of protons and electrons. Anomalous configurations for transition metals like Copper and Chromium are important exceptions to the Aufbau principle.

11th Chemistry Digest Chapter 4 Structure Of Atom Intext Questions And Answers

Can you recall? (Textbook Page No. 35)

 

Question i. What is the smallest unit of matter?

Answer: The smallest unit of matter is atom.
In simple words: An atom is the most basic and smallest particle of an element that still retains the chemical identity of that element.

🎯 Exam Tip: This is a fundamental concept in chemistry; ensure you know the basic definition of an atom.

 

Question ii. What is the difference between molecules of an element and those of a compound?

Answer: The molecules of an element are made of atoms of same element while the molecules of a compound are made of atoms of different elements.
In simple words: Molecules of an element contain only one type of atom (e.g., \(O_2\)), whereas molecules of a compound contain two or more different types of atoms chemically bonded together (e.g., \(H_2O\)).

🎯 Exam Tip: Distinguishing between elements and compounds at the molecular level is a key concept for understanding chemical composition.

 

Question iii. Does an atom have any internal structure or is it indivisible?

Answer: Yes, an atom has internal structure. Different subatomic particles such as protons, electrons and neutrons constitute an atom. So, it is divisible.
In simple words: Atoms are not indivisible; they have an internal structure composed of subatomic particles like protons, neutrons, and electrons.

🎯 Exam Tip: This question tests your knowledge of the modern atomic theory, moving beyond Dalton's original indivisible atom concept.

 

Question iv. Which particle was identified by J. J. Thomson in the cathode ray tube experiment?

Answer: Electron was identified by J.J. Thomson in the cathode ray tube experiment.
In simple words: J.J. Thomson's cathode ray tube experiment led to the discovery of the electron, a negatively charged subatomic particle.

🎯 Exam Tip: Know the key experiments and scientists associated with the discovery of subatomic particles (Thomson-electron, Rutherford-nucleus, Chadwick-neutron).

 

Question v. Which part of an atom was discovered by Ernest Rutherford from the experiment of scattering of \( \alpha \)-particles by gold foil?

Answer: Nucleus of an atom was discovered by Ernest Rutherford from the experiment of scattering of \( \alpha \)-particles by gold foil.
In simple words: Ernest Rutherford's gold foil experiment, involving the scattering of alpha particles, revealed the presence of a small, dense, positively charged center within the atom, which he named the nucleus.

🎯 Exam Tip: Be able to describe Rutherford's experiment, its observations, and the conclusions drawn about the atomic nucleus.

Just Think (Textbook Page No. 41)

 

Question 1. What does the negative sign of electron energy convey?

Answer: Negative sign for the energy of an electron in any orbit in a hydrogen atom indicates that the energy of the electron in the atom is lower than the energy of a free electron at rest. A free electron at rest is an electron that is infinitely far away from the nucleus and is assigned the energy value of zero. As the electron gets close to the nucleus, value of 'n' decreases and \(E_n\) becomes large in absolute value and more negative. The negative sign corresponds to attractive forces between electron and nucleus.
In simple words: The negative sign for electron energy signifies that the electron is bound to the nucleus by attractive forces and its energy is lower than that of a free electron (which is assigned zero energy). The more negative the energy, the more stable and strongly bound the electron is to the nucleus.

🎯 Exam Tip: The concept of negative energy indicating a bound state is fundamental in atomic structure and quantum mechanics. A free electron is considered to have zero energy.

Internet my friend (Textbook Page No. 44)

 

Question 1. Collect information about structure of atom.

Answer: Students can use links given below as reference and collect information about structure of atom on their own.
https://www.livescience.com/65427-fundamental-elementary-particles.html
http://www.chemistryexplained.com/Ar-Bo/Atomic-Structure.html
https://www.thoughtco.com/basic-model-of-the-atom-603799
In simple words: To understand the atom's structure, students should research subatomic particles, their properties, and how they are arranged to form atoms, using reliable online resources.

🎯 Exam Tip: While direct information gathering from the internet isn't an exam question, knowing the basic components (protons, neutrons, electrons) and their arrangement in an atom is essential.

11th Std Chemistry Questions And Answers

• Some Basic Concepts of Chemistry Class 11 Chemistry Questions And Answers
• Introduction to Analytical Chemistry Class 11 Chemistry Questions And Answers
• Basic Analytical Techniques Class 11 Chemistry Questions And Answers
• Structure of Atom Class 11 Chemistry Questions And Answers
• Chemical Bonding Class 11 Chemistry Questions And Answers
• Redox Reactions Class 11 Chemistry Questions And Answers
• Modern Periodic Table Class 11 Chemistry Questions And Answers
• Elements of Group 1 and 2 Class 11 Chemistry Questions And Answers