Additional Questions - Chapter 2 Quantum Mechanical Model of Atom 11th Chemistry Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Additional Questions Solved
I. Choose the correct answer

Question 1.
Which of the following experiment proves the presence of an electron in an atom?
(a) Rutherford's $\alpha$-ray scattering experiment
(b) Davisson and Germer experiment
(c) J.J. Thomson cathode ray experiment
(d) G.R Thomson gold foil experiment
Answer:
(c) J.J. Thomson cathode ray experiment.
Question 2.
Consider the following statements regarding Rutherford's $\alpha$-ray scattering experiment.
i. Most of the $\alpha$-particles were deflected through a small angle.
ii. Some of $\alpha$-particles passed through the foil.
iii. Very few $\alpha$-particles were reflected back by $180^{\circ}$.
Which of the above statements is/are not correct.
(a) $i$ and ii
(b) ii and iii
(c) $i$ and iii
(d) $\mathrm{i}$ ii and iii
Answer:
(a) $i$ and ii.
Question 3.
Considering Bohr's model which of the following statements is correct?
(a) The energies of electrons are continuously reduced in the form of radiation.
(b) The electron is revolving around the nucleus in a dynamic orbital.
(c) Electrons can revolve only in those orbits in which the angular momentum (mvr) of the electron must be equal to an integral multiple of $\mathrm{h} / 2 \pi$.
(d) In an atom, electrons are embedded like seeds in watermelon.
Answer:
(c) Electrons can revolve only in those orbits in which the angular momentum (mvr) of the electron must be equal to an integral multiple of $\mathrm{h} / 2 \pi$.

Question 4.
The energy of an electron of hydrogen atom in 2nd main shell is equal to
(a) $-13.6 \mathrm{eV}$ atom $^{-1}$
(b) $-6.8 \mathrm{eV}$ atom $^{-1}$
(c) $-0.34 \mathrm{eV}$ atom $^{-1}$
(d) $-3.4 \mathrm{eV}^{\mathrm{atom}} \mathrm{m}^{-1}$
Answer:
(d) $-3.4 \mathrm{eV}$ atom $^{-1}$
Hints:
Energy of an electron in $2 \mathrm{nd}$ main shell $=\frac{(-13.6) Z^2}{n^2} ; \mathrm{Z}=1, \mathrm{n}=2$
$
\mathrm{E}=\frac{-13.6}{2^2}=\frac{-13.6}{4}=-3.4 \text { atom }^{-1} \text {. }
$
Question 5.
The energy of an electron of $\mathrm{Li}^{2+}$ in the $3^{\text {rd }}$ main shell is
(a) $-1.51 \mathrm{eV}^{\mathrm{atom}} \mathrm{m}^{-1}$
(b) $-6.8 \mathrm{eV}$ atom $^{-1}$

(c) $+1.51 \mathrm{eV}$ atom $^{-1}$
(d) $-3.4 \mathrm{eV}$ atom $\mathrm{m}^{-1}$
Answer:

(a) $-1.51 \mathrm{eV}^{\mathrm{atom}} \mathrm{m}^{-1}$
Hints:
$
\begin{aligned}
& \mathrm{E}=\frac{(-13.6) Z^2}{n^2} \mathrm{eV} \text { atom }^{-1} \\
& \mathrm{Li}^{2+}=\mathrm{H} \text { atom. So } \mathrm{Z}=1, \mathrm{n}=3 \text {. } \\
& \mathrm{E}=\frac{(-13.6) 1^2}{3^2}=\frac{-13.6}{9}=-1.51 \mathrm{eV} \text { atom }^{-1} \\
&
\end{aligned}
$
Question 6.
The energy of an electron of hydrogen atom in main shell in terms of $U$ mold is
(a) $-1312.8 \mathrm{k} \mathrm{J} \mathrm{mol} \mathrm{m}^{-1}$
(b) $-82.05 \mathrm{k} \mathrm{J} \mathrm{mol}^{-1}$
(c) $-328.2 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(d) $-656.4 \mathrm{k} \mathrm{J} \mathrm{mol-1}$
Answer:
(b) $-82.05 \mathrm{k} \mathrm{J} \mathrm{mol}^{-1}$
Hints:
$
\begin{aligned}
& \mathrm{E}=\frac{(-13.6) Z^2}{n^2} \mathrm{~kJ} \mathrm{~mol}^{-1}, \mathrm{Z}=1, \mathrm{n}=4 \\
& \therefore \mathrm{E}=\frac{-1312.8}{16}=-82.50 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$
Question 7.
The Bohr's radius of $\mathrm{Li} 20 \mathrm{f} 21 \mathrm{~d}$ orbit is
(a) $0.529 \AA$
(b) $0.0753 \AA$
(c) $0.7053 \AA$
(d) $0.0529 \AA$
Answer:
(c) $0.7053 \AA$
Hints:

$
\begin{aligned}
& \mathrm{r}_{\mathrm{n}}=\frac{(0.529) Z^2}{n^2} \AA, \mathrm{n}=2, \mathrm{Z}=3\left(\text { for } \mathrm{Li}^{2+}\right) \\
& \mathrm{r}=\frac{(0.529) 3^2}{2^2}=\frac{0.529 x 4}{3}=0.7053 \AA .
\end{aligned}
$
Question 8.
The formula used to calculate the Boh's radius is
(a) $r_n=\frac{(-13.6) Z^2}{n^2} \mathrm{eV}$ atom $^{-1}$
(b) $r_{\mathrm{n}}=\frac{(0.529) Z^2}{n^2} \mathrm{~A}$
(c) $\mathrm{r}_{\mathrm{n}}=\frac{(-1312.8) Z^2}{n^2} \mathrm{~kJ} \mathrm{~mol}^{-1}$
(d) $\mathrm{r}_{\mathrm{n}}=\frac{(+1312.8) Z^2}{n^2} \mathrm{~kJ} \mathrm{~mol}{ }^{-1}$
Answer:
(b) $\mathrm{r}_{\mathrm{n}}=\frac{(0.529) Z^2}{n^2} \mathrm{~A}$

Question 9 .
Who proposed the dual nature of light to all forms of matter?
(a) John Dalton
(b) Neils Bohr
(c) Albert Einstein
(d) J.J. Thomson
Answer:
(c) Albert Einstein
Question 10.
dc Brogue equation is
(a) $\mathrm{E}=\mathrm{h} \gamma$
(b) $\mathrm{E}=\mathrm{mc}^2$
(c) $\gamma=\frac{\mathrm{E}_2-\mathrm{E}_1}{\mathrm{~h}}$
(d) $\lambda=\frac{h}{m v}$
Answer:
(d) $\lambda=\frac{h}{m v}$.
Question 11.
The crystal used in Davison and Germer experiment is
(a) nickel
(b) zinc suiphide
(c) gold foil
(d) $\mathrm{NaCl}$
Answer:
(a) nickel.
Question 12.
Which one of the following is the time independent Schrodinger wave equation?
(a) $\Delta \mathrm{x} . \Delta \mathrm{p} \geq \frac{\mathrm{h}}{4 \pi}$
(b) $\frac{\partial^2 \psi}{\partial \mathrm{x}^2}+\frac{\partial^2 \psi}{\partial \mathrm{y}^2}+\frac{\partial^2 \psi}{\partial \mathrm{z}^2}+\frac{8 \pi^2 \mathrm{~m}}{\mathrm{~h}^2}=0$

(c) $\frac{\partial^2 \psi}{\partial \mathrm{x}^2}+\frac{\partial^2 \psi}{\partial \mathrm{y}^2}+\frac{\partial^2 \psi}{\partial \mathrm{z}^2}+\frac{8 \pi^2 \mathrm{~m}}{\mathrm{~h}^2}(\mathrm{E}-\mathrm{V}) \psi=0$
(d) $\widehat{\mathrm{H}} \psi-\mathrm{E} \psi=-\frac{8 \pi^2 \mathrm{~m}}{\mathrm{~h}^2}$
Answer:
(c) $\frac{\partial^2 \psi}{\partial \mathrm{x}^2}+\frac{\partial^2 \psi}{\partial \mathrm{y}^2}+\frac{\partial^2 \psi}{\partial \mathrm{z}^2}+\frac{8 \pi^2 \mathrm{~m}}{\mathrm{~h}^2}(\mathrm{E}-\mathrm{V}) \psi=0$
Question 13.
Match the list-I and list-II correctly using the code given below the list.
List - I
A. Principal quantum number
B. Azimuthal quantum number
C. Magnetic quantum number
D. Spin quantum number
List-II
1. represents the directional orientation of orbital
2. represents the spin of the electron

3. represents the main shell
4 renresents the sub shell

Answer:
(a) $\mathrm{A}-3 \mathrm{~B}-4 \mathrm{C}-1 \mathrm{D}-2$
Question 14.
The maximum number of electrons that can be accommodated in $\mathrm{N}$ shell is
(a) 8
(b) 18
(c) 32
(d) 36
Answer:
(c) 32
Hints:
Number of electrons in the main shell $=2 n^2 n=4$, for $N$ shell.
$\therefore$ Maximum number of electrons in $\mathrm{N}$ shell $=2(4)^2=32$.
Question 15 .
The maximum number of electrons that can be accommodated in forbital is
(a) 10
(b) 14
(c) 16
(d) 6
Answer:
(6) 14
Hints:
forbital $-1=3$.
Maximum number of electrons in sub shell $=2(21+1)$
$\therefore$ For ' $\mathrm{f}$ ' orbital, the maximum number of electrons $=2(2 \times 3+1)=14$.
Question 16.
When $1=0$, the number of electrons that can be accommodated in the sub shell is
(a) 0

(b) 2
(c) 6
(d) 8
Answer:
(b) 2
Hints:
If $1=0$, number of electrons $=(21+1)$ $=2(2 \times 0+1)=2$.
Question 17.
Which one of the quantum number is used to calculate the angular momentum of an atom?

(a) $\mathrm{n}$
(b) $m$
(c) 1
(d) $\mathrm{s}$
Answer:
(c) 1
Question 18.
What is the formula used to calculate the angular momentum?
(a) $\sqrt{l^{(l+1)}} \frac{h}{2 \pi}$
(b) $\frac{\mathrm{mvr}}{2 \pi}$
(c) $\frac{m v r}{2}$
(d) $\mathrm{m} . \Delta \mathrm{v}$
Answer:
(a) $\sqrt{l^{(l+1)}} \frac{h}{2 \pi}$.
Question 19.
Which of the following provides the experimental justification of magnetic quantum number?
(a) Zeeman effect
(b) Stark effect
(c) Uncertainty principle
(d) Quantum condition
Answer:
(a) Zeeman effect.
Question 20.
What are the values of $\mathrm{n}, 1, \mathrm{~m}$ and $\mathrm{s}$ for $3 \mathrm{p}_{\mathrm{x}}$ electron?
(a) $3,2,1,0$
(b) $3,1,-1,+1 / 2$
(c) $3,2,+1,-1 / 2$
(d) $3,0,0,+1 / 2]$
Answer:

(b) $3,1,-1,+1 / 2$
Hint:
$3 \mathrm{p}_{\mathrm{x}}$ electron ; $\mathrm{n}=3$ (main shell)
for $p_x$ orbitaI, $1=1, m=-1, s=\frac{1}{2}$.
Question 21.
Identify the quantum number for $4 d_{x^2-y^2}$ electron.
(a) $4,2,-2,+1 / 2$
(b) $4,0,0,+1 / 2$
(c) $4,3,2,+1 / 2$
(d) $4,3,2,-1 / 2$
Answer:
(a) $4,2,-2,+1 / 2$
Question 22.
How many orbitals are possible in $3^{\text {rd }}$ energy level?
(a) 16
(6) 9

(c) 3
(d) 27
Answer:
(b) 9
Hints:
$3^{\text {rd }}$ energy level Number of orbitals $=$ ?
$\mathrm{n}=3$ main shell $=\mathrm{m}$
$1=0,1,2 \mathrm{~m}=0,-1,0,+1$
Total $=9$ orbitals.
Question 23.
The region where the probability density function of electron reduces to zero is called
(a) orbit
(b) orbital
(c) nodal surface
(d) sub shell
Answer:
(c) nodal surface.
Question 24.
Consider the following statements.
(i) The region where the probability density of electron is zero, called nodal surface.
(ii) The probability of finding the electron is independent of the direction of the nucleus.
(iii) The number of radial nodes is equal to $n+1+1$ Which of the above statements is/are correct?
(a) (i) and (iii)
(b) (i) and (ii)
(c) (iii) only
(d) (ii) and (iii)
Answer:
(b) (i) and (ii).
Question 25.
Match the list-I and list-II correctly using the code given below the list.
List-I

A. s-orbital
B. $\mathrm{p}-$ orbital
C. $d$-orbital
D. $f$ - orbital
List-II
1. complex three-dimensional shape
2. symmetrical sphere
3. dumb-bell shape
4. clover leaf shape

Answer:
(c) $\mathbf{A}-2 \mathrm{~B}-3 \mathrm{C}-4 \mathrm{D}-1$
Question 26.
Which one of the following is the correct increasing order of effective nuclear charge felt by an electron?
(a) $s>p>d>f$
(b) $s<\mathrm{p}<\mathrm{d}<\mathrm{f}$
(c) $\mathrm{s}>\mathrm{p}>\mathrm{f}>\mathrm{d}$
(d) $\mathrm{f}<$ p $<$ d $<$ s
Answer:
(a) $\mathrm{s}>\mathrm{p}>\mathrm{d}>\mathrm{f}$.
Question 27.
The value of $\mathrm{n}, 1, \mathrm{~m}$ and $\mathrm{s}$ of 8th electron in an oxygen atom are respectively
(a) $1,0,0,+1 / 2$
(b) $2,1,+1,-1 / 2$
(c) $2,1,-1,-1 / 2$
(d) $2,1,0,+1 / 2$
Answer:
(a) $2,1,+1,-1 / 2$.

Question 28.
The number of impaired electrons in carbon atom in the gaseous state is
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(c) 2

Question 29.
Which of the following is not used in writing electronic configuration of an atom?
(a) Aufbau principle
(b) Hund's rule
(c) Pauli's exclusion principle
(d) Heisenberg's uncertainty principle
Answer:
(d) Heisenberg's uncertainty principle.
Question 30 .
Which of the following is the expected configuration of $\mathrm{Cr}(\mathrm{Z}=24)$ ?
(a) $1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^4 4 \mathrm{~s}^2$
(b) $1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^5 4 \mathrm{~s}^1$
(c) $1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^6$
(d) $1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^5 4 \mathrm{~s}^3$
Answer:
(a) $1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^4 4 s^2$
Question 31.
Which of the following is the actual configuration of $\mathrm{Cr}(\mathrm{Z}=24)$ ?

(a) $1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^4 4 \mathrm{~s}^2$
(b) $1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^5 4 \mathrm{~s}^1$
(c) $1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^6$
(d) $1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^5 4 \mathrm{~s}^3$
Answer:
(b) $1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^5 4 s^1$
Question 32.
Assertion (A) : Cr with electronic configuration [Ar] $3 \mathrm{~d}^5 4 \mathrm{~s}^1$ is more stable than [Ar] $3 \mathrm{~d}^4 4 \mathrm{~s}^1$ Reason(R): Half filled orbitals have been found to have extra stability than partially filled orbitals.
(a) $A$ and $R$ are correct and $R$ is the correct explanation of $A$.
(b) $A$ and $R$ are correct but $R$ is not the correct the explanation of $A$.
(c) $A$ is correct but $R$ is wrong.
(d) $A$ is wrong but $R$ is correct.
Answer:
(a) $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$.
Question 33.
Assertion (A): Copper $(Z=29)$ with electronic configuration $[\mathrm{Ar}] 4 s^1 3 \mathrm{~d}^{10}$ is more stable than $[\mathrm{Ar}] 4 s^1$ $3 \mathrm{~d}^{10}$
Reason(R): Copper with [Ar] $4 s^2 3 d^9$ is more stable due to symmetrical distribution and exchange energies of $d$ electrons.
(a) $A$ and $R$ are correct and $R$ is the correct explanation of $A$.
(b) $A$ and $R$ are correct but $R$ is not the correct explanation of $A$.
(c) $A$ is correct but $R$ is wrong.
(d) $A$ is wrong but $R$ is correct.
Answer:
(a) $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$.

Question 34.
In a sodium atom (atomic number $=11$ and mass number $=23$ ) and the number of neutrons is
(a) equal to the number of protons
(b) less than the number of protons
(c) greater than the number of protons
(d) none of these
Answer:
(c) greater than the number of protons.
Question 35.
The idea of stationary orbits was first given by
(a) Rutherford
(b) J.J. Thomson
(c) Nails Bohr
(d) Max Planck
Answer:
(c) Niels Bohr.
Question 36.
de Broglie equation is
(a) $\lambda=\frac{h}{m v}$

(b) $\lambda=\frac{h v}{m}$
(c) $\lambda=\frac{m v}{h}$
(d) $\lambda=\mathrm{hmv}$
Answer:
(a) $\lambda=\frac{h}{m v}$.
Question 37.
The orbital with $\mathrm{n}=3$ and $1=2$ is
(a) $3 \mathrm{~s}$
(b) $3 \mathrm{p}$
(c) $3 \mathrm{~d}$
(d) $3 \mathrm{~J}$
Answer:
(c) $3 \mathrm{~d}$
Question 38.
The outermost electronic configuration of manganese (at. no. $=25$ ) is
(a) $3 d^5 4 s^2$
(b) $3 \mathrm{~d}^6 4 \mathrm{~s}^1$
(c) $3 \mathrm{~d}^7 4 s^0$
(d) $3 \mathrm{~d}^6 4 \mathrm{~s}^2$
Answer:
(a) $3 d^5 4 s^2$
Question 39.
The maximum number of electrons in a sub-shell is given by the equation
(a) $n^2$
(b) $2 \mathrm{n}^2$
(c) 21-1
(d) $21+1$
Answer:
(d) $21+1$

Question 40 .
Which of the following statements is correct for an electron that has the quantum numbers $\mathrm{n}=4$ and $\mathrm{m}=$ $-2$.
(a) The electron may be in $2 \mathrm{p}$ orbital
(b) The electron may be in 4 d orbital
(c) The electron is in the second main shell
(d) The electron must have spin quantum number as $+\frac{1}{2}$.
Answer:
(b) The electron may be in 4 d orbital.
2 - Marks Questions
Question 1.
Write a note about J.J. Thomson's atomic model.
Answer:
- J.J. Thomson's cathode ray experiment revealed that atoms consist of negatively charged particles called electrons.
- He proposed that atom is a positively charged sphere in which the electrons are embedded like the seeds in the watermelon.

Question 2.
Explain about theory of electromagnetic radiation.
Answer:
- The theory of electromagnetic radiation states that a moving charged particle should continuously loose its energy in the form of radiation.
- O Therefore, the moving electron in an atom should continuously loose its energy and finally collide with nucleus resulting in the collapse of the atom.
Question 3.
Explain how matter has dual character?
Answer:
- Albert Einstein proposed that light has dual nature, i.e. like photons behave both like a particle and as a wave.
- Louis de Broglie extended this concept and proposed that all forms of matter showed dual character.
- He combined the following two equations of energy of which one represents wave character (hv) and the other represents the particle nature $\left(\mathrm{mc}^2\right)$.
Question 4.
Explain about the significance of de Broglie equation.
Answer:
- $\mathrm{X}=\frac{h}{m v}$. This equation implies that a moving particle can be considered as a wave and a wave can exhibit the properties of a particle.
- For a particle with high linear momentum (mv) the wavelength will be so small and cannot be observed.
- For a microscopic particle such as an electron, the mass is of the order of $10^{-31} \mathrm{~kg}$, hence the wavelength is much larger than the size of atom and it becomes significant.
- For the electron, the de Broglie wavelength is significant and measurable while for the iron ball it is too small to measure, hence it becomes insignificant.

Question 5 .
How many electrons can be accommodated in the main shell $1, \mathrm{~m}$ and $\mathrm{n}$ ?
Answer:

Question 6.
How many electrons that can be accommodated in the sub shell $s, \mathrm{p}, \mathrm{d}, \mathrm{f}$ ?
Answer:

Question 7.
What are quantum numbers?
Answer:
- The electron in an atom can be characterized by a set of four quantum numbers, namely principal quantum number (n), azimuthal quantum number (1), magnetic quantum number (m) and spin quantum number $(\mathrm{s})$
- When Schrodinger equation is solved for a wave function $T$, the solution contains the first three quantum numbers $n, 1$
- and $\mathrm{m}$.
- The fourth quantum number arises due to the spinning of the electron about its own axis.
Question 8.
How many orbitals are possible in the 3rd energy level?
Answer:
$\mathrm{n}=3$, main shell is $\mathrm{m}$.
Total number of orbitals in 3rd energy level $=?$

Total number of orbitals $=9$.
Question 9.
What are $\Psi$ and $\Psi^2$ ?
Answer:
- $\Psi$ itself has no physical meaning but it represents an atomic orbital.
- $\Psi^2$ is related to the probability of finding the electrons within a given volume of space.
Question 10.
What is meant by nodal surface?
Answer:
- The. region where there is probability density function reduces to zero is called nodal surface or a radial node.
- For ns orbital, (n-1) nodes are found in it.

Question 11.
Mention the shape of $s, p, d$ orbitals.
Answer:
- Shape of $s$ - orbital-sphere
- Shape of $\mathrm{p}$ - orbital - dumb bell
- Shape of $\mathrm{d}$-orbital - clover leaf
Question 12.
Calculate the total number of angular nodes and radial nodes present in $4 \mathrm{p}$ and $4 \mathrm{~d}$ orbitals.
Answer:
1. For $4 \mathrm{p}$ orbital:
Number of angular nodes $=1$
For $4 \mathrm{p}$ orbital $7=1$
Number of angular nodes $=1$
Number of radial nodes $=n-1-1=4-1-1=2$
Total number of nodes $=n-1=4-1=3$
1 angular node and 2 radial nodes.
2. For 4 d orbital:
Number of angular nodes $=1$
For 4 d orbital $1=2$
Number of angular nodes $=2$
Number of radial nodes $=n-1-1=4-2-1=1$
Total number of nodes $=n-1=4-1=3$
1 radial nodes and 2 angular node.
Question 13.
Write the equation to calculate the energy of nth orbit.
Answer:
$
\mathrm{E}_{\mathrm{n}}=\frac{(-1312.8) Z^2}{n^2} \mathrm{~kJ} \mathrm{~mol}^{-1}
$
Where $\mathrm{Z}=$ atomic number, $\mathrm{n}=$ principal quantum number.
Question 14.
what are degenerate orbitals?
Answer:
- Three different orientations in space that are possible for a p-orbital. All the three p-orbitals, namely $\mathrm{p}_{\mathrm{x}}, \mathrm{p}_{\mathrm{y}}$ and $\mathrm{p}_{\mathrm{z}}$ have same energies and are called degenerate orbitals.
- In the presence of magnetic or electric field, the degeneracy is lost.
Question 15.
Energy of an electron in hydrogen atom in ground state is $-13.6 \mathrm{eV}$. What is the energy of the electron in the third excited state?

Answer:
$
\begin{aligned}
& \mathrm{E}_1=-13.6 \mathrm{eV} \\
& \mathrm{E}_3=\frac{-13.6}{n^2} \text { Where } \mathrm{n}=3 \\
& \mathrm{E}_3=\frac{-13.6}{9}=1.511 \mathrm{eV}
\end{aligned}
$
Energy of the electron in the third excited state $=1.511 \mathrm{eV}$.
Question 16.
The energies of the same orbital decreases with an increase in the atomic number. Justify this statement. Answer:
The energy of the $2 \mathrm{~s}$ orbital of hydrogen atom is greater than that of $2 \mathrm{~s}$ orbital of lithium and that of lithium is greater than that of sodium and so on because $\mathrm{H}(\mathrm{Z}=1), \operatorname{Li}(Z=3)$ and $\mathrm{Na}(Z=11)$. When atomic number increases, the energies of the same orbital decreases. $E_{2 s}(\mathrm{H})>\mathrm{E}_{2 s}(\mathrm{Li})>\mathrm{E}_{2 s}(\mathrm{Na})>\mathrm{E}_{2 s}(\mathrm{~K})$
Question 17
State Hund's rule of maximum multiplicity.
Answer:
It states that electron pairing in the degenerate orbitals does not take place until all the available orbitals contain one electron each.
Question 18.
How many unpaired electrons are present in the ground state of -
1. $\mathrm{Cr}^{3+}(\mathrm{Z}=24)$
2. $\mathrm{Ne}(Z=10)$
Answer:
1. $\mathrm{Cr}^{3+}(\mathrm{Z}=24) 1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^5 4 \mathrm{~s}^1$
$\mathrm{Cr}^{3+}-1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^4$
It contains 4 unpaired electrons.
2. Ne $(Z=10) 1 s^2 2 s^2 2 \mathrm{p}^6$. No unpaired electrons in it.

Question 19.
What is meant by electronic configuration? Write the electronic configuration of $N(Z=7)$.
Answer:
The distribution of electrons into various orbitals of an atom is called its electronic configuration.

Question 20.
Which is the actual configuration of $\mathrm{Cr}(\mathrm{Z}=24)$ Why?
Answer:
$\mathrm{Cr}(\mathrm{Z}=24) 1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6$
The reason for this is, $\mathrm{Cr}_r$ with $3 \mathrm{~d}^5$ configuration is half filled and it will be more stable. Chromium has [Ar] $3 \mathrm{~d}^5 4 \mathrm{~s}^1$ and not [Ar] $3 \mathrm{~d}^4 4 \mathrm{~s}^2$ due to the symmetrical distribution and exchange energies of $d$ electrons.
Question 21.
What is the actual configuration of copper $(\mathrm{Z}=29)$ ? Explain about its stability.
Answer:
Copper $(Z=29)$
Expected configuration : $1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^9 4 \mathrm{~s}^2$
Actual configuration : $1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^{10} 4 \mathrm{~s}^1$
The reason is that fully filled orbitals have been found to have extra stability. Copper has the electronic configuration [Ar] $3 \mathrm{~d}^{10} 4 \mathrm{~s}^1$ and not $[\mathrm{Ar}] 3 \mathrm{~d}^9 4 \mathrm{~s}^2$ due the symmetrical distribution and exchange energies of $d$ electrons. Symmetry leads to stability. The full filled configuration have symmetrical distribution of electrons and hence they are more stable than unsymmetrical configuration.

3 - Mark Questions
Question 1.
What are the conclusions of Rutherford's $\alpha$ - rays scattering experiment?
Answer:
- Rutherford bombarded a thin gold foil with a stream of fast moving $\alpha$-particles.
- It was observed that most of the a-particles passed through the foil.
- Some of them were deflected through a small angle.
- Very few $\alpha$ - particles were reflected back by $180^{\circ}$.
- Based on these observations, he proposed that in an atom, there is a tiny positively charged nucleus and the electrons are moving around the nucleus with high speed.
Question 2.
What are the limitations of Bohr's atom model?
Answer:
- The Bohr's atom model is applicable only to species having one electron such as hydrogen, $\mathrm{Li}^{2+}$ etc and not applicable to multi - electron atoms.
- It was unable to explain the splitting of spectral lines in the presence of magnetic field (Zeeman effect) or an electric field (Stark effect).
- Bohr's theory was unable to explain why the electron is restricted to revolve around the nucleus in a fixed orbit in which the angular momentum of the electron is equal to $\mathrm{nh} / 2 \pi$.
Question 3.
Illustrate the significance of de Broglie equation with an iron ball and an electron.
1. $6.626 \mathrm{~kg}$ iron ball moving with $10 \mathrm{~ms}^{-1}$.
2. An electron moving at $72.73 \mathrm{~ms}^{-1}$.
Answer:
$
\begin{aligned}
& \text { 1. } \lambda_{\text {iron ball }}=\frac{h}{m v} \\
& =\frac{6.626 \times 10^{-34} \mathrm{Kg} \mathrm{m}^2 \mathrm{~s}^{-1}}{6.626 \mathrm{Kg} \times 10 \mathrm{~ms}^{-1}} \\
& =1 \times 10^{-35} \mathrm{~m}
\end{aligned}
$
2. $\lambda_{\text {iron ball }}=\frac{h}{m v}$
$
\begin{aligned}
& =\frac{6.626 \times 10^{-34} \mathrm{Kg} \mathrm{m}^2 \mathrm{~s}^{-1}}{9.11 \times 10^{-31} \mathrm{Kg} \times 72.73 \mathrm{~ms}^{-1}} \\
& =\frac{6.626}{662.6} \times 10^{-3} \mathrm{~m}=1 \times 10^5 \mathrm{~m}
\end{aligned}
$
For an electron, the de Broglie wavelength is significant and measurable while for iron ball it is too small to measure, hence it becomes insignificant.

Question 4.
Explain Davisson and Germer experiment.
Answer:
- The wave nature of electron was experimentally confirmed by Davisson and Germer.
- They allowed the accelerated beam of electrons to fall on a nickel crystal and recorded the diffraction pattern.
- The resultant diffraction pattern is similar to the $\mathrm{X}$ - ray diffraction pattern.
- The finding of wave nature of electron leads to the development of various experimental ' techniques such as electron microscope, low energy electron diffraction etc.
Question 5 .
Bohr radius of 1 st orbit of hydrogen atom is $0.529 \AA$. Assuming that the position of an electron in this orbit is determined with the accuracy of $0.5 \%$ of the radius, calculate the uncertainty in the velocity of the electron in hydrogen atom.
Answer:
Uncertainty in position $=\Delta \mathrm{x}$
$=\frac{0.5}{100} \times 0.529 \AA$
$=\frac{0.5}{100} \times 10^{-10} \times 0.529 \mathrm{~m}$
$\Delta \mathrm{x}=2.645 \times 10^{-13} \mathrm{~m}$
From Heisenberg's uncertainty principle,
$\Delta \mathrm{x} . \Delta \mathrm{p} \geq \frac{h}{4 \pi}$
$\Delta \mathrm{x} \cdot \mathrm{m} \cdot \Delta \mathrm{p} \geq \frac{h}{4 \pi}$
$\Delta \mathrm{v} \geq \frac{h}{\Delta x, m, 4 \pi}$
$\Delta \mathrm{v}=\frac{6.626 \times 10^{-34} \mathrm{Kg} \mathrm{m}^2 \mathrm{~s}^{-1}}{2.645 \times 10^{-13} \mathrm{~m} \times 9.11 \times 10^{-31} \mathrm{Kg} \times 4 \times 3.14}$
$\Delta \mathrm{v}=2.189 \times 10^8 \mathrm{~m}$
Question 6 .
Write a note about principal quantum number.
Answer:

- The principal quantum number represents the energy level in which electron revolves around the nucleus and is denoted by the symbol ' $n$ '.
- The ' $n$ ' can have the values $1,2,3, \ldots n=1$ represents $K$ shell; $n=2$ represents $L$ shell and $n=3,4,5$ represent the $\mathrm{M}, \mathrm{N}, \mathrm{O}$ shells, respectively.
- The maximum number of electrons that can be accommodated in a given shell is $2 \mathrm{n}^2$.
- ' $n$ ' gives the energy of the electron,
$\mathrm{E}_{\mathrm{n}}=\frac{(-1312.8) Z^2}{n^2} \mathrm{KJ} \mathrm{mol}^{-1}$ and the distance of the electron from the nucleus is given by $\mathrm{r}_{\mathrm{n}}=\frac{(-0.529) n^2}{Z} \mathrm{~A}$.
Question 7.
Explain about azimuthal quantum number.
Answer:
- It is represented by the letter $7^{\prime}$ and can take integral values from zero to $\mathrm{n}-1$, where $\mathrm{n}$ is the principal quantum number.
- Each 1 value represents a subshell (orbital). $1=0,1,2,3$ and 4 represents the $\mathrm{s}, \mathrm{p}, \mathrm{d}, \mathrm{f}$ and g orbitals respectively.
- The maximum number of electrons that can be accommodated in a given subshell (orbital) is 2 (21 + 1).
It is used to calculate the orbital angular momentum using the expression Angular momentum $=$ $\sqrt{l(l+1)} \frac{h}{2 \pi}$
Question 8.
Draw the shapes of $1 \mathrm{~s}, 2 \mathrm{~s}$ and $3 \mathrm{~s}$ orbitals
Answer:

Question 9.
Explain how effective nuclear charge is related with stability of the orbital.

Answer:
- In a multi-electron atom, in addition to the electrostatic attractive force between the electron and nucleus, there exists a repulsive force among the electrons.
- These two forces are operating in the opposite direction. This results in the decrease in the nuclear force of attraction on electron.
- The net charge experienced by the electron is called effective nuclear charge.
- The effective nuclear charge depends on the shape of the orbitals and it decreases with increase in azimuthal quantum number 1 .
- The order of the effective nuclear charge felt by a electron in an orbital within the given shell is $\mathrm{s}>\mathrm{p}$ $>\mathrm{d}>\mathrm{f}$.
- Greater the effective nuclear charge, greater is the stability of the orbital. Hence, within a given energy level, the energy of the orbitals are in the following order $s Question 10.
Calculate the wavelength of an electron moving with a velocity of $2.05 \times 10^7 \mathrm{~ms}^{-1}$.
Answer:
According to de Broglie's equation, $\lambda=\frac{h}{m v}$
Mass of electron $(\mathrm{m})=9.1 \times 10^{-31} \mathrm{~kg}$
Velocity of electron $(v)=2.05 \times 10^7 \mathrm{~ms}^{-1}$
Planck's constant $(\mathrm{h})=6.626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}$
$
\lambda=\frac{\left(6.626 \times 10^{-34} \mathrm{Kg} \mathrm{m}^2 \mathrm{~s}^{-1}\right)}{\left(9.1 \times 10^{-31} \mathrm{Kg}\right) \times\left(2.05 \times 10^7 \mathrm{~ms}^{-1}\right)}=355 \times 10^{-4} \mathrm{~m}
$
Question 11.
The mass of an electron is $9.1 \times 10^{-31} \mathrm{~kg}$. If its kinetic energy is $3.0 \times 10^{-25} \mathrm{~J}$, calculate its wavelength.

Answer:
Step I.
Calculation of the velocity of electron

Kinetic energy $=1 / 2 \mathrm{mv}^2=3.0 \times 10^{-25} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-2}$
$
\begin{aligned}
& =\frac{2 \times \mathrm{K} . \mathrm{E} .}{\mathrm{m}}=\frac{2 \times\left(3.0 \times 10^{-25} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-2}\right)}{\left(9.1 \times 10^{-31} \mathrm{~kg}\right)}=65.9 \times 10^4 \mathrm{~m}^2 \mathrm{~s}^{-2} \\
& v^2 \\
& v=\left(65.9 \times 10^4 \mathrm{~m} \mathrm{~s}^{-2}\right)=8.12 \times 10^2 \mathrm{~ms}^{-1}
\end{aligned}
$
Step II.
Calculation of wavelength of the electron
According to de Broglie's equation,
$
\begin{aligned}
& \lambda=\frac{h}{m v}=\frac{\left(6.626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}\right)}{\left(9.1 \times 10^{-31} \mathrm{~kg}\right) \times\left(8.12 \times 10^2 \mathrm{~ms}^{-1}\right)} \\
& =0.08967 \times 10^{-5} \mathrm{~m}=8967 \times 10^{-10} \mathrm{~m}=8967 \AA\left(\therefore 1 \AA=10^{-10} \mathrm{~m}\right) .
\end{aligned}
$
Question 12.
From the following sets of quantum numbers, state which are possible. Explain why the others are not possible.
$
\begin{aligned}
& \text { 1. } \mathrm{n}=0,1=0, \mathrm{~m}_1=0, \mathrm{~m}_{\mathrm{s}}=+\frac{1}{2} \\
& \text { 2. } \mathrm{n}=1,1=0, \mathrm{~m}_1=0, \mathrm{~m}_{\mathrm{s}}=-\frac{1}{2} \\
& \text { 3. } \mathrm{n}=1,1=1, \mathrm{~m}_1=0, \mathrm{~m}_{\mathrm{s}}=+\frac{1}{2} \\
& \text { 4. } \mathrm{n}=1,1=0, \mathrm{~m}_1=+1, \mathrm{~m}_{\mathrm{s}}=+\frac{1}{2} \\
& \text { 5. } \mathrm{n}=3,1=3, \mathrm{~m}_1=-3, \mathrm{~m}_{\mathrm{s}}=+\frac{1}{2} \\
& \text { 6. } \mathrm{n}=3,1=1, \mathrm{~m}_1=0, \mathrm{~m}_{\mathrm{s}}=+\frac{1}{2}
\end{aligned}
$
Answer:
1. The set of quantum numbers is not possible because the minimum value of $n$ can be 1 and not zero.
2. The set of quantum numbers is possible.
3. The set of quantum numbers is not possible because, for $\mathrm{n}=1,1$ cannot be equal to 1 . It can have 0 value.
4. The set of quantum numbers is not possible because for $1=0, \mathrm{~m}_1$; cannot be +1 . It must be zero.
5. The set of quantum numbers is not possible because, for $n=3,1=3$.
6. The set of quantum numbers is possible.
Question 13.
How many electrons in an atom may have the following quantum numbers?
(a) $\mathrm{n}=4 ; \mathrm{m}_{\mathrm{s}}=-1 / 2$

(b) $\mathrm{n}=3,1=0$.
Answer:
(a) For $\mathrm{n}=4$
1 Total number of electrons $=2 n^2=2 \times 16=32$
Half out of these will have $\mathrm{ms}=-\frac{1}{2}$
Total electrons with $\mathrm{m}_{\mathrm{s}}(-1 / 2)=16$.
(b) For $n=3$
$1=0 ; m_1=0, m_s=+1 / 2-1 / 2\left(\right.$ two $\left.e^{-}\right)$.
Question 14 .
Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.
Answer:
According to Bohr's theory,
$\operatorname{mvr}=\frac{n h}{2 \pi}(\mathrm{n}=1,2,3, \ldots \ldots$ so on $)$
or $2 \pi \mathrm{r}=\frac{n h}{m v}$ or $\mathrm{mv}=\frac{n h}{2 \pi r}$
According to de Brogue equation,
$\lambda=\frac{h}{m v}$ or $m v=\frac{h}{\lambda} \ldots \ldots \ldots$ (ii)
Comparing (i) and (ii),
$\frac{n h}{2 \pi r}=\frac{h}{\lambda}$ or $2 \pi \mathrm{r}=\mathrm{n} \lambda$
Thus, the circumference ( $2 \pi \mathrm{r})$ of the Bohr orbit for hydrogen atom is an into the de Broglie wave length.
Question 15.
An ion with mass number 56 contains 3 units of positive charge and $30.4 \%$ more neutrons than electrons.
Assign symbol to the ion.
Answer:
Let the no. of electrons in the ion $=\mathrm{x}$
the no. of the protons $=x+3$ (as the ion has three units positive charge)
and the no. of neutrons $=\mathrm{x}+\frac{30.4 x}{100}=\mathrm{x}+0.304 \mathrm{x}$
Now, mass number of ion $=$ Number of protons + Number of neutrons
$=(\mathrm{x}+3)+(\mathrm{x}+0.304 \mathrm{x})$
$\therefore 56=(\mathrm{x}+3)+(\mathrm{x}+0.304 \mathrm{x})$ or $2.304 \mathrm{x}=56-3=53$
$\mathrm{x}=\frac{53}{2.304}=23$
Atomic number of the ion (or element) $=23+3=26$
The element with atomic number 26 is iron $(\mathrm{Fe})$ and the corresponding ion is $\mathrm{Fe}^{3+}$.
Question 16.
The uncertainty in the position of a moving bullet of mass $10 \mathrm{~g}$ is $10 \mathrm{~s} \mathrm{~m}$. Calculate the uncertainty in its velocity?
Answer:
According to uncertainty principle,
$\Delta \mathrm{x} . \mathrm{m} \Delta \mathrm{v}=\frac{h}{4 \pi}$ or $\Delta \mathrm{v}=\frac{h}{4 \pi m \Delta x}$
$\mathrm{h}=6.626 \times 1 \mathrm{o}^{-34} \mathrm{~kg} \mathrm{~m}_2 \mathrm{~s}^{-1} ; \mathrm{m}=10 \mathrm{~g}=10^{-2} \mathrm{~kg}$

$
\Delta \mathrm{x}=10^{-5} \mathrm{~m} ; \Delta \mathrm{v}=\frac{\left(6.626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}\right)}{4 \times 3.143 \times\left(10^{-2} \mathrm{~kg}\right) \times\left(10^{-5} \mathrm{~m}\right)}=5.27 \times 10^{-28} \mathrm{mv}
$
$
\begin{aligned}
& =1.6 \times 10^{-15} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-15} \\
& \mathrm{Or} \\
& \frac{1}{2} \mathrm{mv}^2=1.6 \times 10^{-15} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-2} \\
& \qquad\left(\frac{2 \times 1.6 \times 10^{-15} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-2}}{9.1 \times 10^{-31} \mathrm{~kg}}\right)^{1 / 2}=5.93 \times 10^7 \mathrm{~m}^{-1}
\end{aligned}
$
Question 17.
The uncertainty in the position and velocity of a particle are $10^{-10} \mathrm{~m}$ and $5.27 \times 10^{-24} \mathrm{~ms}^{-1}$ respectively. Calculate the mass of the particle.
Answer:
According to uncertainty principle.
$
\Delta \mathrm{x} \cdot \mathrm{m} \Delta \mathrm{v}=\frac{h}{4 \pi}
$
or
$
\begin{aligned}
& \mathrm{m}=\frac{h}{4 \pi \Delta x \cdot \Delta v} ; \\
& \mathrm{h}=6.626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1} \\
& \Delta \mathrm{x}=10^{-10} \mathrm{~m} ; \Delta \mathrm{x}=5.27 \times 10^{-24} \mathrm{~ms}^{-1} \\
& \quad=\frac{\left(6.626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}\right)}{4 \times 3.143 \times\left(10^{-10} \mathrm{~m}\right) \times\left(5.27 \times 10^{-24} \mathrm{~ms}^{-1}\right)}=0.1 \mathrm{~kg} .
\end{aligned}
$
Question 18.
With what velocity must an electron travel so that its momentum Is equal to that of a photon of wave length $=5200 \mathrm{~A}$ ?
Answer:
According to de Brogue equation, $\lambda=\frac{h}{m v}$
$
\text { Momentum of electron, } \mathrm{mv}=\frac{h}{\lambda}=\frac{\left(6.626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}\right)}{\left(5200 \times 10^{-10} \mathrm{~m}\right)}
$

$=1.274 \times 10^{-27} \mathrm{~kg} \mathrm{~ms}^{-1}$
The momentum of electron can also be calculated as $=\mathrm{mv}=\left(9.1 \times 10^{-31} \mathrm{~kg}\right) \mathrm{x} \mathrm{v}$
Comparing (i) and (ii)
$
\begin{aligned}
& \left(9.1 \times 10^{-31} \mathrm{~kg}\right) \mathrm{v}=\left(1.274 \times 10^{-27} \mathrm{~kg} \mathrm{~ms}^{-1}\right) \\
& =\frac{\left(1.274 \times 10^{-27} \mathrm{~kg} \mathrm{~ms}^{-1}\right)}{\mathrm{v} \quad\left(9.1 \times 10^{-31} \mathrm{~kg}\right)}=1.4 \times 10^3 \mathrm{~ms}^{-1}
\end{aligned}
$
Question 19.
Using Aufbau principle, write the ground state electronic configuration of following atoms.
1. Boron $(Z=5)$
2. Neon $(Z=10)$
3. Aluminium $(Z=13)$
4. Chlorine $(Z=17)$
5. Calcium $(Z=20)$
6. Rubidium $(Z=37)$
Answer:
1. Boron $(Z=5) ; 1 s^2 2 s^2 2 p^1$
2. Neon $(Z=10) ; 1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6$
3. Aluminium $(Z=13) ; 1 s^2 2 s^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^1$
4. Chlorine $(Z=17) ; 1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^5$
5. Calcium $(Z=20) ; 1 s^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 4 \mathrm{~s}^2$
6. Rubidium $(Z=37) ; 1 s^2 2 s^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^{10} 4 \mathrm{~s}^2 4 \mathrm{p}^6 5 \mathrm{~s}^1$

Question 20.
Calculate the de Broglie wavelength of an electron moving with $1 \%$ of the speed of light?
Answer:
According to de Brogue equation, $\mathrm{A}=\frac{h}{m v}$
Mass of electron $=9.1 \times 10^{-31} \mathrm{~kg}$; Planck's constant $6.626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}$
Velocity of electron $=1 \%$ of speed of light $=3.0 \times 10^8 \times 0.01=310^6 \mathrm{~ms}^{-1}$
Wavelength of electron $(\lambda)=\frac{h}{m v}=\frac{\left(6.626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}\right)}{\left(9.1 \times 10^{-31} \mathrm{~kg}\right) \times\left(3 \times 10^6 \mathrm{~ms}^{-1}\right)}$ $=2.43 \times 10^{-10} \mathrm{~m}$
Question 21.
What is the wavelength for the electron accelerated by $1.0 \mathrm{X}$ i0 volts?
Answer:
Step I.
Calculation of the velocity of electron
Energy (kinetic energy) of electron $=1.0 \times 10^4$ volts.
$=1.0 \times 10^4 \times 1.6 \times 10^{-19} \mathrm{~J}=1.6 \times 10^{-15} \mathrm{~J}$.
Step II.
Calculation of wavelength of the electron
According to de Broglie equation,
$
\begin{aligned}
& \lambda=\frac{h}{m v} ; \lambda \frac{\left(6.626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}\right)}{\left(9.1 \times 10^{-31} \mathrm{~kg}\right) \times\left(5.93 \times 10^7 \mathrm{~ms}^{-1}\right)} \\
& =1.22 \times 10^{-11} \mathrm{~m} .
\end{aligned}
$
Question 22.
In a hydrogen atom, the energy of an electron in first Bohr's orbit is $13.12 \times 10^5 \mathrm{~J} \mathrm{~mol}^{-1}$. What is the energy required for its excitation to Bohr's second orbit?
Answer:
$
E_n=-\frac{2 \pi^2 m_e^4}{n^2 h^2}
$
When $\mathrm{n}=1, \mathrm{E}_1=-\frac{2 \pi^2 \mathrm{~m}_{\mathrm{e}}^4}{(1)^2 \mathrm{~h}^2}=-13.12 \times 10^5 \mathrm{~J} \mathrm{~mol}^{-1}$
When $n=2, \mathrm{E}_2=-\frac{2 \pi^2 \mathrm{~m}_{\mathrm{e}}^4}{(2)^2 \mathrm{~h}^2}=-\frac{13 \cdot 12 \times 10^5}{4} \mathrm{~J} \mathrm{~mol}^{-1}$ $=3.28 \times 10^5 \mathrm{~J} \mathrm{~mol}^{-1}$.

The energy required for the excitation is:

\Delta \mathrm{E}=\mathrm{E}_2-\mathrm{E}_1=\left(-3.28 \times 10^5\right)-\left(-13.12 \times 10^5\right)=9.84 \times 10^5 \mathrm{~J} $\mathrm{~mol}^{-1}$
Question 23.
Dual behavior of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is $1.6 \times 10^6 \mathrm{~ms}^{-1}$, calculate de Broglie wavelength associated with this electron.
Answer:
$
\begin{aligned}
& \lambda=\frac{h}{m v} ; \lambda=\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)}{\left(9.1 \times 10^{-31} \mathrm{~kg}\right) \times\left(1.6 \times 10^6 \mathrm{~ms}^{-1}\right)} \\
& =0.455 \times 10^{-34}+25 \mathrm{~m}=0.455 \mathrm{~nm}=455 \mathrm{pm} .
\end{aligned}
$
Question 24.
An element with mass number 81 contains $31.7 \%$ more neutrons as compared to protons. Assign the symbol to the element.
Answer:
An element can be identified by its atomic number only. Let us find the atomic number.
Let the number of protons $=\mathrm{x}$
$\therefore$ Number of neutrons $=\mathrm{x}+\frac{x \times 31.7}{100}=(\mathrm{x}+0.317 \mathrm{x})$
Now, Mass no. of element $=$ No. of protons + No. of neutrons $81=\mathrm{x}+\mathrm{x}+0.317 \mathrm{x}=2.317 \mathrm{x}$
Or
$
\mathrm{x}=\frac{81}{2.317}=35
$
$\therefore$ No. of protons $=35$, No. of neutrons $=81-35=46$
Atomic number of element $(Z)=$ Number of protons $=35$
The element with atomic number (Z) 35 is bromine ${ }_{35}^{81} \mathrm{Br}$.
Question 25.
The electron energy in hydrogen atom is given by $\mathrm{E}_{\mathrm{n}}=\left(-2.18 \times 10^{-18}\right) / \mathrm{n}^2 \mathrm{~J}$. Calculate the energy required to remove an electron completely from the $n=2$ orbit. What is the longest wavelength of light in $\mathrm{cm}$ that can be used to cause this transition?
Answer:
Step I.
Calculation of energy required
The energy required is the difference in the energy when the electron jumps from orbit with $\mathrm{n}=\infty$ to orbit with $\mathrm{n}=2$.
The energy required $(\Delta \mathrm{E})=\mathrm{E}_{\infty}-\mathrm{E}_2$
$
=0-\left(-\frac{2.18 \times 10^{-18}}{4} \mathrm{~J}\right)=5.45 \times 10^{-19} \mathrm{~J} \text {. }
$

Step II.
Calculation of the longest wavelength of light in $\mathrm{cm}$ used to cause the transition
$
\begin{aligned}
& \Delta \mathrm{E}=\mathrm{hv}=\mathrm{hc} / \lambda \\
& \lambda=\frac{h c}{\Delta E}=\frac{\left(6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right) \times\left(3 \times 10^8 \mathrm{~ms}^{-1}\right)}{\left(5.45 \times 10^{-19} \mathrm{~J}\right)} \mathrm{n}=3.644 \times 10^{-7} \\
& \mathrm{~m}=3.644 \times 10^{-7} \times 10^2=3.645 \times 10^{-5} \mathrm{~cm} .
\end{aligned}
$

5-Mark Questions
Question 1.
Describe about Bohr atom model.
Answer:
Assumptions of Bohr atom model.
1. The energies of electrons are quarantined
2. The electron is revolving around the nucleus in a certain fixed circular path called the stationary orbit.
3. Electron can revolve only in those orbits in which the angular momentum (mvr) of the electron must be equal to an integral multiple of $\mathrm{h} / 2 \pi$ $\mathrm{mvr}=\frac{n h}{2 \pi}$ where $\mathrm{n}=1,2,3, \ldots$ etc.
4. As long as an electron revolves in a fixed stationary orbit, it doesn't lose its energy. But if an electron jumps from a higher energy state $\left(E_2\right)$ to a lower energy state $\left(E_1\right)$, the excess energy is emitted as radiation. The frequency of the emitted radiation is $E_2-E_1=h v$.
$
\therefore \mathrm{v}=\frac{\mathrm{E}_2-\mathrm{E}_1}{\mathrm{~h}}
$
Conversely, when suitable energy is supplied to an electron, it will jump from lower energy orbit to a higher energy orbit.
5. Bohr's postulates are applied to a hydrogen like atom $\left(\mathrm{H}, \mathrm{He}^{+}\right.$and $\mathrm{Li}^{2+}$ etc..) the radius of the nth orbit and the energy of the electron revolving in the th orbit were derived.
$
\begin{aligned}
& \mathrm{r}_{\mathrm{n}}=\frac{(0.529) n^2}{Z} \mathrm{~A}(0.529) \mathrm{n}^2 \\
& \mathrm{E}_{\mathrm{n}}=\frac{(-13.6) Z^2}{n} \mathrm{eV} \mathrm{atom}^{-1} \\
& \mathrm{E}_{\mathrm{n}}=\frac{(1312.8) Z^2}{n} \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$
Question 2.
Derive de Brogue equation and give its significance.
Answer:
1. Louis de Brogue extended the concept of dual nature of light to all forms of matter. To quantify this relation, he derived an equation for the wavelength of a matter-wave.
2. He combined the following two equations of the energy of which one represents wave character (hu) and the other represents the particle nature $\left(\mathrm{mc}^2\right)$.
Planck's quantum hypothesis:
$
\mathrm{E}=\mathrm{h} \mathrm{v}
$
Einsteins mass-energy relationship:
$
\mathrm{E}=\mathrm{mc}^2 \ldots \ldots \ldots(2)
$
From (1) and (2)
$
\begin{aligned}
& \mathrm{hv}=\mathrm{mc}^2 \\
& \mathrm{hc} / \lambda=\mathrm{mc}^2 \\
& \therefore \lambda=\frac{h}{m c} \ldots \ldots \ldots \text { (3) }
\end{aligned}
$
The equation (3) represents the wavelength of photons whose momentum is given by mc. (Photons have zero rest mass).
3. For a particle of matter with mass $m$ and moving with a velocity $y$, the equation (3) can be written as $\lambda=$ $\frac{h}{m c}$
4. This is valid only when the particle travels at speed much less than the speed of Light.
5. This equation implies that a moving particle can be considered as a wave and a wave can exhibit the properties of a particle (i.e momentum).
6. Significance of de Brogue equation:
For a particle with high linear momentum, the wavelength will be too small and cannot be observed. For a microscopic particle such as an electron, the mass is $9.1 \times 10^{-31} \mathrm{~kg}$. Hence the wavelength is much larger than the size of atom and it becomes significant.
Question 3.
What are the main features of the quantum mechanical model of an atom.
Answer:
1. The energy of electrons in an atom is quarantined.
2. The existence of quarantined electronic energy levels is a direct result of the wave-like properties of electrons. The solutions of the Schrodinger wave equation gives the allowed energy levels (orbits).
3. According to Heisenberg's uncertainty principle, the exact position and momentum of an electron cannot be determined with absolute accuracy. As a consequence, quantum mechanics introduced the concept of orbital. Orbital is a three-dimensional space in which the probability of finding the electron is maximum.
4. The solution of the Schrodinger wave equation for the allowed energies of an atom gives the wave function $\Psi$, which represents an atomic orbital. The wave nature of the electron present in an orbital can be well defined by the wave function $\Psi$.
5. The wave function $\Psi$ itself has no physical meaning. However, the probability of finding the electron in a small volume $\mathrm{dx}, \mathrm{dy}, \mathrm{dz}$ around a point $(\mathrm{x}, \mathrm{y}, \mathrm{z})$ is proportional to $|\Psi(x, y, z)|^2 \mathrm{dx} d y d z|\Psi(x, y, z)|^2$ is known as probability density and is always positive.

Question 4.
Explain about -
(1) Magnetic quantum number
(2) Spin quantum number
Answer:
(1) Magnetic quantum number:
- It is denoted by the letter $m_1$. It takes integral values ranging from -1 to +1 through 0 . i.e. if $1=1 ; \mathrm{m}=-1,0$ and +1 .
- The Zeeman Effect (the splitting of spectral lines in a magnetic field) provides the experimental justification for this quantum number.
- The magnitude of the angular momentum is determined by the quantum number 1 while its direction is given by magnetic quantum number.
(2) Spin quantum number:
- The spin quantum number represents the spin of the electron and is denoted by the letter ' $\mathrm{m}_{\mathrm{s}}$.'
- The electron in an atom revolves not only around the nucleus but also spins. It is usual to write this as electron spins about its own axis either in a clockwise direction or in anti-clockwise direction.
- Corresponding to the clockwise and anti-clockwise spinning of the electron, maximum two values are possible for this quantum number.
- The values of " $\mathrm{m}_{\mathrm{s}}$ " is equal to $-\frac{1}{2}$ and $+\frac{1}{2}$.
Question 5 .
Explain about the shape of orbitals.
Answer:
Orbital: The solution to Schrodinger equation gives the permitted energy values called eigen values and the wave functions corresponding to the eigen values are called atomic orbitals.
Shape of orbital:
s- orbital:
For Is orbital, $1=0, m=0, f(\theta)=1 \sqrt{ } 2$ and $g(\varphi)=1 / \sqrt{ } 2 \pi$. Therefore, the angular distribution function is equal to $1 / \sqrt{ } 2 \pi$. i.e. it is independent of the angle $\theta$ and $\varphi$. Hence, the probability of finding the electron is independent of the direction from the nucleus. So, the shape of the s orbital is spherical.

$\mathrm{p}$ - orbital:
For $\mathrm{p}$ orbitals $1=1$ and the corresponding $\mathrm{m}$ values are $-1,0$ and +1 . The three different $\mathrm{m}$ values indicates that there are three different orientations possible for $\mathrm{p}$ orbitals. These orbitals are designated as $\mathrm{p}_{\mathrm{x}}, \mathrm{p}_{\mathrm{y}}$ and $p_z$. The shape of $p$ orbitals are dumb bell shape.

$\mathrm{d}$ - orbital:
For ' $\mathrm{d}$ ' orbital $1=2$ and the corresponding $m$ values are $-2,-1,0,+1,+2$. The shape of the orbital looks like a clover leaf. The five $\mathrm{m}$ values give rise to five $\mathrm{d}$ orbitals namely $\mathrm{d}_{\mathrm{xy}}, \mathrm{d}_{\mathrm{yz}}, \mathrm{d}_{\mathrm{zx}}, \mathrm{d}_{\mathrm{x}}^2-\mathrm{y}^2$ and $\mathrm{d}_{\mathrm{z}}^2$ The $3 \mathrm{~d}$ orbitals contain two nodal planes.

$f$ - orbital
For $f$ orbital, $1=3$ and the $m$ values are $-3,-2,-1,0,+1,+2,+3$ corresponding to seven $f$ orbitals,
$f_{z^3}, f_{x z^2}, f_{y z^2}, f_{x y z}, f_{z\left(x^2-y^2\right)}, f_{x\left(x^2-3 y^2\right)}, f_{y\left(3 x^2-y^2\right)}$. They contain 3 nodal planes.

Question 6.
What is exchange energy? How it is related with stability of atoms? Explain with suitable examples.
Answer:
1. If two or more electrons with the same spin are present in degenerate orbitals, there is a possibility for exchanging their positions. During exchange process, the energy is released and the released energy is called exchange energy.
2. If more number of exchanges are possible, more exchange energy is released. More number of exchanges are possible only in the case of half filled and fully filled configurations.
3. For example, in chromium, the electronic configuration is [ $\mathrm{Ar}] 3 \mathrm{~d}^5 4 \mathrm{~s}^1$. The $3 \mathrm{~d}$ orbital is half filled and there are ten possible exchanges.

4. On the other hand only six exchanges are possible for [Ar] $3 \mathrm{~d}^4 4 \mathrm{~s}^2$ configuration.
5. Hence, exchange energy for the half filled configuration is more This increases the stability of half filled $3 \mathrm{~d}$ orbitals.