Solution-Class-11-Science-Chemistry-Chapter-4-Structure of Atom-Maharashtra Board

Structure of Atom

Class-11-Science-Chemistry-Chapter -4 Maharashtra State Board

Solution

Question 1.

Choose correct option.

(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

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(B)  The value of Plank’s constant is -

(a)  6.626 × 10-34Js

(b)  6.023 × 10-24Js

(c)  1.667 × 10-28Js

(d)  6.626 × 10-28Js

Answer :

(a)  6.626 × 10-34Js

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(C)  p-orbitals are....... in shape.

(a)  spherical

(b)  dumb bell

(c)  double dumbell

(d)  diagonal

Answer :

(b)  dumb bell

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(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

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(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

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Question 2.

Make the pairs:

‘A’ ‘B’
a. Neutrons i. six electrons
b. p-orbital ii.-1.6 × 10-19 C
c. charge on electron iii. Ultraviolet region
d. Lyman series iv. Chadwick

Answer :

‘A’ ‘B’
a. Neutrons iv. Chadwick
b. p-orbital i. six electrons
c. charge on electron ii.-1.6 × 10-19 C
d. Lyman series iii. Ultraviolet region

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Question 3.

Complete the following information about the isotopes in the chart given below.

(Hint: Refer to Periodic Table if required)

Answer :

Substance Mass Number

(A)

Number of
Protons

(Z)

Neutrons

(A-Z)

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

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Question 4.

Match the following :

Element No. of Neutron
(a) \(_{18}^{40}\text{Ar}\) (i) 7
(b) \(_{6}^{14}\text{C}\) (ii) 21
(c) \(_{19}^{40}\text{K}\) (iii) 8
(d) \(_{7}^{14}\text{N}\) (iv) 22

Answer :

Element No. of Neutron
(a) \(_{18}^{40}\text{Ar}\) (iv) 22
(b) \(_{6}^{14}\text{C}\) (iii) 8
(c) \(_{19}^{40}\text{K}\) (ii) 21
(d) \(_{7}^{14}\text{N}\) (i) 7

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Question 5.

Answer in one sentence.

(A) If an element ‘X’ has mass number 11 and it has 6 neutrons, then write its representation.

Answer :

Atomic number = 11 − 6 = 5

 Element is \(_{5}^{11}\text{B}\)

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(B) Name the element that shows simplest emission spectrum.

Answer :

Since hydrogen atom has one electron it shows the simplest emission spectrum.

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(C) State Heisenberg uncertainty principle.

Answer :

Heisenberg uncertainty principle : This principle states that it is not possible to determine simultaneously the position and momentum of a moving microscopic particle like electron with absolute certainty.

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(D) Give the names of quantum numbers.

Answer :

Quantum numbers are as follows :

  • Principal quantum number denoted by n.
  • Azimuthal quantum number denoted by l.
  • Magnetic quantum number denoted by ml
  • Spin quantum number denoted by ms.

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(E) Identify from the following the isoelectronic species:

Ne, O2, Na+ OR Ar, Cl, K+

Answer :

Species Ne   O2  Na+ Ar    Cl  K+
Number of electrons 10   10   10 18   18   18

∴Isoelectronic species : (i) Ne   O2  Na+  (ii) Ar    Cl  K+

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Question 6.

Answer the following questions.

(A) Differentiate between Isotopes and Isobars.

Answer :

Isotopes Isobars
Isotopes are the atoms of the same element. Isobars are the atoms of different elements.
Isotopes have same atomic number but different mass numbers. Isobars have different atomic numbers but same mass number.
They have same number of electrons. They have different number of electrons.
They have same number of protons but different number of neutrons. They have different number of protons and neutrons.
They have same chemical properties. They have different chemical properties.
For Example \(_{17}^{35}\text{Cl}\) and \(_{17}^{34}\text{Cl}\) For example \(_{18}^{40}\text{Ar}\), \(_{19}^{40}\text{K}\), \(_{20}^{40}\text{Ca}\)

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(B) Define the terms:

(i) Isotones

Answer :

Isotones : The atoms of different elements having same number of neutrons in their nuclei are called isotones.

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(ii) Isoelectronic species.

Answer :

Isoelectronic species : Atoms and ions having the same number of electrons are isoelectronic species.

Example  :

Consider K+ formed by removal of one electron from K atom. Which has 19 electrons (Z = 19). Therefore K+ has 18 electrons.

Species such as Ar, Ca2+ containing 18 electrons are isoelectronic with K+ .

Electronic configuration of all these species with 18 electrons is 1s2, 2s2, 2p6, 3s2, 3p6.

Species  K→ K+ + e Ar, Ca2+
Number of electrons  19   18 18  18

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(iii) Electronic configuration.

Answer :

Electronic configuration of an atom is the distribution of its electrons in orbitals.

The electronic configuration can be written by applying the aufbau principle.

There are two methods of representing electronic configuration:

(a) Orbital notation: nsa npb ndc..........

For example for boron atom 5B 1s2, 2s2, 2p1.

(b) Orbital diagram:

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(C) State and explain Pauli’s exclusion principle.

Answer :

The capacity of an orbital to accommodate electrons is decided by Pauli exclusion  principle.

Statement of Pauli’s principle : No two electrons in an atom can have all the four quantum numbers, (n, l, m and s) same.

OR

Only two electrons may exist in the given orbital having three quantum numbers same but fourth quantum number being different with opposite spins.

This principle describes the capacity of a sub-shell or orbital to accommodate maximum number of electrons.

Consider helium atom, which has two electrons. The four quantum numbers of two electrons in He atom will be,

Hence 1s orbital can accommodate two electrons.

This principle is illustrated with helium atom He (Z = 2). Its electronic configuration is 1s2 as

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(D) State Hund’s rule of maximum multiplicity with suitable example.

Answer :

Statement of Hund’s rule of maximum multiplicty : It states that pairing of electrons in the orbitals belonging to the same subshell does not occur unless each orbital belonging to that subshell has accommodated one electron each.

  • Consider filling of p-subshell which has three degenerated orbitals namely

After filling three electrons, one in each with same spins, the next electrons enter with pairing

It observed that half filled and completely filled set of degenerate orbitals have extra stability.

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(E) Write the drawbacks of Rutherford’s model of an atom.

Answer :

Rutherford’s atomic model has following drawbacks :

  • According to the classical electromagnetic theory, a revolving charged particle like electron should emit radiation and lose energy. Due to this, electron should come closer to the nucleus by following a spiral path and finally fall into the nucleus, giving unstable atom. But in reality atoms are stable.
  • The orbital motion is an accelerated motion accompanied by continuous change in the velocity of electron due to changing direction.
  • Orbital motion of an electron dos not explain the change in energy and hence atomic spectrum.
  • The electron would follow a spiral path and fall into the nucleus.
  • This model does not explain the distribution of electrons around the nucleus and their energies.

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(F) Write postulates of Bohr’s Theory of hydrogen atom.

Answer :

Postulates of Bohr atomic theory

Bohr’s model of hydrogen atom is based on the following postulates.

(i) The electron in the hydrogen atom can move around the nucleus in certain permitted circular orbits arranged concentrically in increasing order of energy.

(ii) The energy of an electron in the orbit does not change with time. Hence orbits are called stationary orbits. On absorption of required energy electron moves from lower orbit to higher orbit.

The transition from higher energy orbit to lower energy orbit is accompanied by emission of energy in the form of electromagnetic radiation.

(iii) The frequency ν of radiation absorbed or emitted on transition between two stationary orbits differing by energy ΔE is given by,

ν = ΔE/h = E2 − E1

where E1 and E2 are the energies of the lower and higher energy states respectively. Above equation represents Bohr’s frequency rule.

(iv) An electron can occupy only those orbits in which its angular momentum is integral (n) multiple of h/2π

Angular momentum = \(\frac{nh}{2π}\)

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(G) Mention demerits of Bohr’s Atomic model.

Answer :

Demerits of Bohr model

  • Bohr’s atomic model failed to account for finer details of the hydrogen atom spectrum observed in sophisticated spectroscope experiments.
  • Bohr model was unable to explain the spectrum of atoms other than hydrogen.
  • Bohr theory could not explain the splitting of spectral lines in the presence of a magnetic field (Zeeman effect) or electric field (Stark effect).
  • Bohr theory failed to explain the ability of atoms to form molecules by chemical bonds.

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(H) State the order of filling atomic orbitals following Aufbau principle.

Answer :

Aufbau is a German word which means building up. This principle explains the sequence of filling up of orbitals with electrons.

Aufbau principle : It states that in the ground state of an atom, the orbitals are filled with electrons in order of the increasing energies.

The orbitals are filled in order of increasing value of (n + l). For example, 4s-orbital

(n + l = 4 + 0 = 4) is filled prior to 3d-orbital (n + l = 3 + 2 = 5).

Among two orbitals having same (n + l) value, that orbital with lower value of n will be filled first. For example, 3d-orbital (n + l = 3 + 2 = 5) is filled prior to 4p-orbital (n + l = 4 + 1 = 5).

The increasing order of energy of different orbitals is as follows :

1s<2s<2p<3s<3p<4sz<3d<4p<5s<4d<5p<6s<4f<5d<6p<7s<5f<6d

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(I) Explain the anomalous behavior of copper and chromium.

Answer :

Anomalous behaviour of copper :

Copper (29Cu) has electronic configuration.

29Cu (Expected) : 1s2, 2s2, 2p6, 3s2, 3p6, 3d9, 4s2

       (Observed) : 1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s1

Explanation :

  • The energy difference between the 3d- and 4s-orbitals is very low.
  • d-orbitals being degenerate, they acquire more stability when they are completely filled.
  • Therefore, there arises a transfer of one electron from 4s-orbital to 3d-orbital in Cu giving completely filled more stable d-orbital.
  • Hence, the configuration of Cu is [Ar] 3d10 4s1 and not [Ar] 3d9 4s2.
  • This shows anomalous behaviour of copper.

Anomalous behaviour of chromium :

Chromium (24Cr) has electronic configuration,

24Cr (Expected) : 1s2, 2s2, 2p6, 3s2, 3p6, 3d4, 4s2

       (Observed) : 1s2, 2s2, 2p6, 3s2, 3p6, 3s2, 4s1

Explanation :

  • The energy difference between 3d- and 4s-orbitals is very low.
  • d-orbitals being degenerate, they acquire more stability when they are half-filled (3d5)
  • Therefore, there arises a transfer of one electron from 4s-orbital to 3d-orbital in Cr giving more stable half-filled orbital. Hence, the configuration of Cr is [Ar] 3d5 4s1 and not [Ar] 3d4 4s2.
  • This shows anomalous behaviour of chromium.

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(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 :

(a) n = 2, l = 1 the orbital is 2p

(b) n = 4, l = 2 the orbital is 4d

(c) n = 3, l = 2, the orbital is 3d

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(K) Write electronic configurations of Fe, Fe2+, Fe3+

Answer :

26Fe  [Ar] 3d6 4s2

Fe2+  [Ar] 3d6

Fe3+  [Ar] 3d5

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(L) Write condensed orbital notation of electronic 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 :

(a) 3Li  [He] 2s1

(b) 6C  [He] 2s2 2p2

(c) 8O  [He] 2s2 2p4

(d) 14Si  [Ne] 3s2 3p2

(8) 17Cl  [Ne] 3s2 3p5

(f) 20Ca  [Ar] 4s2

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(M) Draw shapes of 2s and 2p orbitals.

Answer :

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(N) Explain in brief, the significance of azimuthal quantum number.

Answer :

Azimuthal quantum number (l) : This represents the subsidiary quantum number or the subshell of the orbit to which the electron belongs. It is denoted by l. It is also called as secondary, subsidiary, orbital or angular momentum quantum number. It defines the shape of the orbital and the angular momentum of the electron. The value of l depends on principal quantum number n. It has positive values between 0 to (n — 1) :

l 0 1 2 3
subshell s p d f

This explains the significance of azimuthal quantum number

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(O) If n=3, what are the quantum number l and m?

Answer :

For n = 3, l = 0, 1, 2.

For l = 0, ml = 0

For l = 1, ml = −1, 0, +1

For l = 2, ml = −2, −1, 0, +1

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(P) The electronic configuration of oxygen is written as 1s2 2s2 2px2 2py1 2pz1 and not as 1s2 2s2 2px2 2py2 2pz0, Explain.

Answer :

By Hund's rule,

 1s2 2s2 2px2 2py1

The electronic configuration,

1s2 2s2 2px2 2py2 2pz0  violates Hund's rule

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 (Q) Write note on ‘Principal Quantum number.

Answer :

Principal quantum number (n) : This describes the orbit or shell of an atom to which the electron belongs. It is represented by ‘n’ which has integral values. The energy of an electron depends on the value of n.

Principal quantum number n Shell symbol Allowed number of orbitals n2 Size of shell
1 K 1 Increases

2 L 4
3 M 9
4 N 16

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(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 :

‘M’ shell has,

n = 3

l = 0, 1, 2

ml : When l = 0, ml = 0, ms = \(±\frac{1}{2}\)

ml : When l = 1, ml = −1, 0, +1 ms = \(±\frac{1}{2}\),\(±\frac{1}{2}\),\(±\frac{1}{2}\)

ml : When l = 2, ml = −2, −1, 0, +1, +2 ms = \(±\frac{1}{2}\),\(±\frac{1}{2}\),\(±\frac{1}{2}\),\(±\frac{1}{2}\),\(±\frac{1}{2}\)

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(S) Indicate the number of unpaired electrons in :

(a) Si (Z=14) (b) Cr (Z=24)

Answer :

(a) 14Si  [Ne] 3s2 3px1; 3py1;

Number of unpaired electrons = 2

(b) 24Cr [Ar] 3d5 4s1

Number of unpaired electrons = 6

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(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) Since atom has 29 electrons, it has 29 protons. The atomic number of the element is 29. It is copper.

(b) ∴ Electronic configuration : 29Cu [Ar] 3d10 4s1.

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