Text Book Back Questions and Answers - Chapter 2 Quantum Mechanical Model of Atom 11th Chemistry Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Quantum Mechanical Model of Atom
Textual Evaluation Solved
I. Choose the correct answer
Question 1.
Electronic configuration of species $\mathrm{M}^{2+}$ is $1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^6$ and its atomic weight is 56 . The number of neutrons in the nucleus of species $M$ is
(a) 26
(b) 22
(c) 30
(d) 24
Answer:
(c) 30
Solution:
$
\begin{aligned}
& \mathrm{M}^{2+}: 1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^6 \\
& M: 1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^8
\end{aligned}
$
Atomic number $=26$
Mass number $=56$
No. of neutrons $=56-26=30$.
Question 2.

he energy of light of wavelength $45 \mathrm{~nm}$ is
(a) $6.67 \times 10^{15} \mathrm{~J}$
(b) $6.67 \times 10^{11} \mathrm{~J}$
(c) $4.42 \times 10^{18} \mathrm{~J}$
(d) $4.42 \times 10^{-15} \mathrm{~J}$
Answer:
(c) $4.42 . \times 10^{18} \mathrm{~J}$
Solution:
$
\begin{aligned}
& \mathrm{E}=\mathrm{hv}=\mathrm{hc} / \lambda \\
& \frac{6.626 \times 10^{-34} \mathrm{~J} \times 3 \times 10^8 \mathrm{~ms}^{-1}}{45 \times 10^{-9} \mathrm{~m}}=4.42 \times 10^{18} \mathrm{~J} .
\end{aligned}
$
Question 3.
The energies $E_1$ and $E_2$ of two radiations are $25 \mathrm{eV}$ and $50 \mathrm{eV}$ respectively. The relation between their wavelengths i.e. $\lambda_1$ and $\lambda_2$ will be
(a) $\frac{\lambda_1}{\lambda_2}=1$
(b) $\lambda_1=2 \lambda_2$
(c) $\lambda_1=\sqrt{25 \times 50} \lambda_2$
(d) $2 \lambda_1=\lambda_2$
Answer:
(b) $\lambda_1=2 \lambda_2$

Solution:
$
\begin{aligned}
& \frac{E l}{E 2}=\frac{25 e V}{50 V}=\frac{1}{2} \\
& \frac{\mathrm{hc}}{\lambda_1} \times \frac{\lambda_2}{\mathrm{hc}}=\frac{1}{2} \\
& 2 \lambda_2=\lambda_1 .
\end{aligned}
$
Question 4.
Splitting of spectral lines in an electric field is called
(a) Zeeman effect
(b) Shielding effect
(c) Compton effect
(d) Stark effect
Answer:
(d) Stark effect
Solution:
Splitting of spectral lines in magnetic field is called Zeeman effect and splitting of spectral lines in electric field, is called Stark effect.
Question 5.
Based on equation $\mathrm{E}=-2.178 \times 10^8 \mathrm{~J}\left(\frac{z^2}{n^2}\right)$ certain conclusions are written. Which of them is not correct ?
(a) Equation can be used to calculate the change in energy when the electron changes orbit
(b) For $n=1$, the electron has a more negative energy than it does for $n=6$ which means that the electron is more loosely bound in the smallest allowed orbit
(c) The negative sign in equation simply means that the energy of electron bound to the nucleus is lower than it would be if the electrons were at the infinite distance from the nucleus.
(d) Larger the value of $n$, the larger is the orbit radius.
Answer:
(b) For $n=1$, the electron has a more negative energy than it does for $n=6$ which means that the electron is more loosely bound in the smallest allowed orbit
Solution:
Correct statement:
For $n=6$, the electron has more negative energy than it does for $n=6$ which means that the electron is strongly bound in the smallest allowed orbit.
Question 6.
According to the Bohr Theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon?
(a) $n=6$ to $\mathrm{n}=1$
(b) $n=5$ to $n=4$
(c) $\mathrm{n}=5$ to $\mathrm{n}=3$
(d) $n=6$ to $n=5$
Answer:
(d) $n=6$ to $n=5$
Solution:
$
\begin{aligned}
& \mathrm{n}=6 \text { to } \mathrm{n}=5 \\
& \mathrm{E}_6=-13.6 / 6^2 ; E_5=-13.6 / 5^2
\end{aligned}
$

$
\begin{aligned}
& \mathrm{E}_6-\mathrm{E}_5=\left(-13.6 / 6^2\right)-\left(-13.6 / 5^2\right) \\
& =0.166 \mathrm{eV} \text { atom }^{-1}
\end{aligned}
$
$
\begin{aligned}
& E_5-E_4=\left(-13.6 / 5^2\right)-\left(-13.6 / 4^2\right) \\
& =0.306 \mathrm{eV} \text { atom }^{-1}
\end{aligned}
$
Question 7.
Assertion : The spectrum of $\mathrm{He}^{+}$is expected to be similar to that of hydrogen
Reason : $\mathrm{He}^{+}$is also one electron system,
(a) If both assertion and reason are true and reason is the correct explanation of assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
(c) If assertion is true but reason is false
(d) If both assertion and reason are false
Answer:
(a) If both assertion and reason are true and reason is the correct explanation of assertion.
Question 8.
Which of the following pairs of d-orbitals will have electron density along the axes ? (NEET Phase - II)
(a) $\mathrm{d}_{\mathrm{z}}, \mathrm{d}_{\mathrm{xz}}$
(b) $\mathrm{d}_{\mathrm{xz}}, \mathrm{d}_{\mathrm{yz}}$
(c) $\mathrm{d}_{z^2}, d_{x^2-y^2}$
(d) $\mathrm{d}_{\mathrm{xy}}, d_{x^2-y^2}$
Answer:
(c) $\mathrm{d}_{\mathrm{z}}{ }^2, d_{x^2-y^2}$

Question 7.
Assertion : The spectrum of $\mathrm{He}^{+}$is expected to be similar to that of hydrogen
Reason : $\mathrm{He}^{+}$is also one electron system,
(a) If both assertion and reason are true and reason is the correct explanation of assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
(c) If assertion is true but reason is false
(d) If both assertion and reason are false
Answer:
(a) If both assertion and reason are true and reason is the correct explanation of assertion.
Question 8.
Which of the following pairs of d-orbitals will have electron density along the axes ? (NEET Phase - II)
(a) $\mathrm{d}_{\mathrm{z}}, \mathrm{d}_{\mathrm{xz}}$
(b) $\mathrm{d}_{\mathrm{xz}}, \mathrm{d}_{\mathrm{yz}}$
(c) $\mathrm{d}_{\mathrm{z}}, d_{x^2-y^2}$
(d) $\mathrm{d}_{\mathrm{xy}}, d_{x^2-y^2}$
Answer:
(c) $\mathrm{d}_{\mathrm{z}}, d_{x^2-y^2}$
Question 9.
Two electrons occupying the same orbital are distinguished by
(a) azimuthal quantum number
(b) spin quantum number
(c) magnetic quantum number
(d) orbital quantum number
Answer:
(b) spin quantum number
Solution:
Spin quantum number For the first electron $\mathrm{ms}=+\frac{1}{2}$
For the second electron $\mathrm{ms}=\frac{1}{2}$.
Question 10.
The electronic configuration of Eu (atomic no. 63) Gd (atomic no. 64) and $\mathrm{Tb}$ (atomic no. 65) are (NEET Phase II)
(a) $[\mathrm{Xe}] 4 \mathrm{f}^7 5 \mathrm{~d}^1 6 \mathrm{~s}^2,[\mathrm{Xe}] 4 \mathrm{f}^7 5 \mathrm{~d}^1 6 \mathrm{~s}^2$ and $[\mathrm{Xe}] 4 \mathrm{f}^8 5 \mathrm{~d}^1 6 \mathrm{~s}^2$
(b) $\left.[\mathrm{Xe}] 4 \mathrm{f}^7, 6 \mathrm{~s}^2, \mathrm{Xe}\right] 4 \mathrm{f}^7 5 \mathrm{~d}^1 6 \mathrm{~s}^2$ and $[\mathrm{Xe}] 4 \mathrm{f}^9 6 \mathrm{~s}^2$
(c) $[\mathrm{Xe}] 4 \mathrm{f}^7, 6 \mathrm{~s}^2,[\mathrm{Xe}] 4 \mathrm{f}^8 6 \mathrm{~s}^2$ and $[\mathrm{Xe}] 4 \mathrm{f}^8 5 \mathrm{~d}^1 6 \mathrm{~s}^2$
(d) $[\mathrm{Xe}] 4 \mathrm{f}^8 5 \mathrm{~d}^1 6 \mathrm{~s}^2[\mathrm{Xe}] 4 \mathrm{f}^7 5 \mathrm{~d}^1 6 \mathrm{~s}^2$ and $[\mathrm{Xe}] 4 \mathrm{f}^9 6 \mathrm{~s}^2$
Answer:
(b) $[\mathrm{Xe}] 4 \mathrm{f}^7, 6 \mathrm{~s}^2,[\mathrm{Xe}] 4 \mathrm{f}^7 5 \mathrm{~d}^1 6 \mathrm{~s}^2$ and $[\mathrm{Xe}] 4 \mathrm{f}^9 6 \mathrm{~s}^2$
Solution:
$\mathrm{Eu}:[\mathrm{Xe}] 4 \mathrm{f}^7, 5 \mathrm{~d}^0, 6 \mathrm{~s}^2$
$\mathrm{Gd}:[\mathrm{Xe}] 4 \mathrm{f}^7, 5 \mathrm{~d}^1, 6 \mathrm{~s}^2$
$\mathrm{Tb}:[\mathrm{Xe}] 4 \mathrm{f}^9, 5 \mathrm{~d}^0, 6 \mathrm{~s}^2$

Question 11.
The maximum number of electrons in a sub shell is given by the expression
(a) $2 n^2$
(b) $21+1$
(c) $41+2$
(d) none of these
Answer:
(c) $41+2$
Solution:
$
2(21+1)=41+2
$
Question 12.
For d-electron, the orbital angular momentum is
(a) $\frac{\sqrt{2} h}{2 \pi}$
(b) $\frac{\sqrt{2 h}}{2 \pi}$
(c) $\sqrt{2 \times 4}$
(d) $\frac{\sqrt{6} h}{2 \pi}$
Answer:
(d) $\frac{\sqrt{6} h}{2 \pi}$
Solution:
Orbital angular momentum
$=\sqrt{(1(1+1)}) \mathrm{h} / 2 \pi$
For $\mathrm{d}$ orbital $=\sqrt{(2 \times 3)} \mathrm{h} / 2 \pi=\sqrt{6} \mathrm{~h} / 2 \pi$.
Question 13.
What is the maximum numbers of electrons that can be associated with the following set of quantum numbers ? $\mathrm{n}=3,1=1$ and $\mathrm{m}=-1$
(a) 4
(b) 6
(c) 2
(d) 10
Answer:
(c) 2
Solution:
$\mathrm{n}=3 ; 1=1 ; \mathrm{m}=-1$ either $3 \mathrm{p}_{\mathrm{x}}$ or $3 \mathrm{p}_{\mathrm{y}}$
Question 14.
Assertion: Number of radial and angular nodes for $3 \mathrm{p}$ orbital are 1,1 respectively. Reason: Number of radial and angular nodes depends only on principal quantum number.
(a) both assertion and reason are true and reason is the correct explanation of assertion.
(b) both assertion and reason are true but reason is not the correct explanation of assertion.
(c) assertion is true but reason is false
(d) both assertion and reason are false
Answer:
(c) assertion is true but reason is false

Solution:
No. of radial node $=n-1-1$
No. of angular node $=1$ for $3 \mathrm{p}$ orbital
No. of angular node $=1=1$
No. of radial node $=n-1-1=3-1-1=1$.
Question 15.
The total number of orbitals associated with the principal quantum number $\mathrm{n}=3$ is
(a) 9
(b) 8
(c) 5
(d) 7
Answer:
(a) 9
Solution:
$\mathrm{n}=3 ; 1=0 ; \mathrm{m}_1=0$ - one $\mathrm{s}$ orbital $\mathrm{n}=3 ; 1=1 ; \mathrm{m}_1=-1,0,1-$ three $\mathrm{p}$ orbitals $\mathrm{n}=3 ; 1=2 ; \mathrm{m}_1=-2,-1,0,1$,
2 - five $\mathrm{d}$ orbitals, overall nine orbitals are possible.
Question 16.
If $n=6$, the correct sequence for filling of electrons will be,
(a) $\mathrm{ns} \rightarrow(\mathrm{n}-2) \mathrm{f} \rightarrow(\mathrm{n}-1) \mathrm{d} \rightarrow \mathrm{np}$
(b) $\mathrm{ns} \rightarrow(\mathrm{n}-1) \mathrm{d} \rightarrow(\mathrm{n}-2) \mathrm{f} \rightarrow \mathrm{np}$
(c) $\mathrm{ns} \rightarrow(\mathrm{n}-2) \mathrm{f} \rightarrow \mathrm{np} \rightarrow(\mathrm{n}-1) \mathrm{d}$
(d) none of these are correct
Answer:
(a) $\mathrm{ns} \rightarrow(\mathrm{n}-2) \mathrm{f} \rightarrow(\mathrm{n}-1) \mathrm{d} \rightarrow \mathrm{np}$
Solution:
$\mathrm{n}=6$ According Aufbau principle,
$6 \mathrm{~s} \rightarrow 4 \mathrm{f} \rightarrow 5 \mathrm{~d} \rightarrow 6 \mathrm{p}$
$\mathrm{ns} \rightarrow(\mathrm{n}-1) \mathrm{f} \rightarrow(\mathrm{n}-2) \mathrm{d} \rightarrow \mathrm{np}$.
Question 17.
Consider the following sets of quantum numbers:

Which of the following sets of quantum number is not possible?
(a) (i), (ii), (iii) and (iv)
(b) (ii), (iv) and (v)
(c) (z) and (iii)
(d) (ii), (iii) and (iv)
Answer:
(b) (ii), (iv) and (v)
Solution:
(ii) 1 can have the values from 0 to $\mathrm{n}-1 \mathrm{n}=2$; possible 1 values are 0,1 hence $1=2$ is not possible.
(iv) for $1=0 ; m=-1$ not possible
(v) for $\mathrm{n}=31=4$ and $\mathrm{m}=3$ not possible.

Question 18.
How many electrons in an atom with atomic number 105 can have $(n+1)=8$ ?
(a) 30
(6) 17
(c) 15
(d) unpredictable
Answer:
(b) 17
Solution:
$\mathrm{n}+1=8$
Electronic configuration of atom with atomic number 105 is $[\operatorname{Rn}] 5 \mathrm{f}^{14} 6 \mathrm{~d}^3 7 \mathrm{~s}^2$

Question 19.
Electron density in the yz plane of $3 \mathrm{~d}_{\mathrm{x}}{ }^2-\mathrm{y}^2$ orbital is
(a) zero
(b) 0.50
(c) 0.75
(d) 0.90
Answer:
(a) zero

Question 20.
If uncertainty in position and momentum are equal, then minimum uncertainty in velocity is
(a) $\frac{1}{m} \sqrt{\frac{h}{\pi}}$
(b) $\sqrt{\frac{h}{\pi}}$
(c) $\frac{1}{2 m} \sqrt{\frac{h}{\pi}}$
(d) $\frac{\mathrm{h}}{4 \pi}$
Answer:
(c) $\frac{1}{2 m} \sqrt{\frac{h}{\pi}}$
Solution:
$
\begin{aligned}
\Delta \mathrm{x} \cdot \Delta \mathrm{p} & \geq \frac{\mathrm{h}}{4 \pi} \\
\Delta \mathrm{p} \cdot \Delta \mathrm{p} & \geq \frac{\mathrm{h}}{4 \pi} \\
\Delta \mathrm{p}^2 & \geq \frac{\mathrm{h}}{4 \pi}
\end{aligned}
$

$
\begin{gathered}
\mathrm{m}^2(\Delta \mathrm{v})^2 \geq \frac{\mathrm{h}}{4 \pi} \\
(\Delta \mathrm{v}) \geq \sqrt{\frac{\mathrm{h}}{4 \pi \mathrm{m}^2}} \\
\Delta \mathrm{v} \geq \frac{1}{2 \mathrm{~m}} \sqrt{\frac{\mathrm{h}}{\pi}}
\end{gathered}
$
Question 21.
A macroscopic particle of mass $100 \mathrm{~g}$ and moving at a velocity of $100 \mathrm{~cm} \mathrm{~s}^{-1} \mathrm{~d}$ will have a de Broglie wavelength of
(a) $6.6 \times 10^{-29} \mathrm{~cm}$
(b) $6.6 \times 10^{-30} \mathrm{~cm}$
(c) $6.6 \times 10^{-31} \mathrm{~cm}$
(d) $6.6 \times 10^{-32} \mathrm{~cm}$
Answer:
(c) $6.6 \times 10^{-31} \mathrm{~cm}$
Solution:
$
\begin{aligned}
& \mathrm{m}=100 \mathrm{~g}=100 \times 10^{-3} \mathrm{~kg} \\
& \mathrm{v}=100 \mathrm{~cm} \mathrm{~s}^{-1}=100 \times 10^{-2} \mathrm{~m} \mathrm{~s}^{-1} \\
& \lambda=\frac{h}{m v}=\frac{6.626 \times 10^{-34} \mathrm{Js}^{-1}}{100 \times 10^{-3} \mathrm{~kg} \times 100 \times 10^{-2} \mathrm{~ms}^{-1}} \\
& =6.626 \times 10^{-31} \mathrm{~ms}^{-1} \\
& =6.626 \times 10^{-31} \mathrm{~cm} \mathrm{~s}^{-1} \\
&
\end{aligned}
$
Question 22.
The ratio of de Broglie wavelengths of a deuterium atom to that of an a - particle, when the velocity of the former is five times greater than that of later, is ...........

(a) 4
(b) 0.2
(c) 2.5
(d) 0.4
Answer:
(d) 0.4
Question 23.
The energy of an electron in the 3rd orbit of hydrogen atom is -E. The energy of an electron in the first orbit will be
(a) $-3 \mathrm{E}$
(b) $-\mathrm{E} / 3$
(c) $-\mathrm{E} / 9$
(d) $-9 \mathrm{E}$
Answer:
(c) $-\mathrm{E} / 9$
Solution:
$
\begin{aligned}
& \mathrm{E}_{\mathrm{n}}=\frac{-13.6}{n^2} \mathrm{eV} \text { atom }^{-1} \\
& \mathrm{E}_1=\frac{-13.6}{1^2} 13.6=\frac{-13.6}{9} \\
& \text { Given that, }
\end{aligned}
$

Given that,
$
\begin{aligned}
& \mathrm{E}_3=-\mathrm{E} \\
& \frac{-13.6}{9}=-\mathrm{E} \\
& 13.6=-9 \mathrm{E}=\mathrm{E}_1=-9 \mathrm{E} \\
& \mathrm{E}_1=-9 \mathrm{E}
\end{aligned}
$
Question 24 .
Time independent Schnodinger wave equation is
(a) $\widehat{\mathrm{H}} \psi=\mathrm{E} \psi$
(b) $\nabla^2 \psi+\frac{8 \pi^2 m}{h^2}(\mathrm{E}+\mathrm{V}) \psi=0$
(c) $\frac{\partial^2 \psi}{\partial \mathrm{x}^2}+\frac{\partial^2 \psi}{\partial \mathrm{y}^2}+\frac{\partial^2 \psi}{\partial \mathrm{z}^2}+\frac{2 \mathrm{~m}}{\mathrm{~h}^2}(\mathrm{E}-\mathrm{V}) \psi=0$
(d) all of these
Answer:
(a) $\widehat{\mathrm{H}} \psi=\mathrm{E} \psi$.
Question 25 .
Which of the following does not represent the mathematical expression for the Heisenberg uncertainty principle?
(a) $\Delta \mathrm{E} . \Delta \mathrm{p} \geq \mathrm{h} / 4 \pi$
(b) $\Delta \mathrm{E} \cdot \Delta \mathrm{v} \geq \mathrm{h} / 4 \pi \mathrm{m}$
(c) $\Delta \mathrm{E} . \Delta \mathrm{t} \geq \mathrm{h} / 4 \pi$
(d) $\Delta \mathrm{E} \cdot \Delta \mathrm{x} \geq \mathrm{h} / 4 \pi$
Answer:
(d) $\Delta \mathrm{E} . \Delta \mathrm{x} \geq \mathrm{h} / 4 \pi$.
II. Write brief answer to the following questions

Question 26.
Which quantum number reveal information about the shape, energy, orientation and size of orbitals?
Answer:
Magnetic quantum number reveal information about the shape, energy, orientation and size of orbitals.
Question 27.
How many orbitals are possible for $n=4$ ?
Answer:
If $n=4$, the possible number of orbitals are calculated as follows -
$\mathrm{n}=4$, main shell $=\mathrm{N}$
If $\mathrm{n}=4,1$ values are $0,1,2,3$
If $1=0,4$ s orbital $=1$ orbital
If $1=1, \mathrm{~m}=-1,0,+1=3$ orbitals
If $1=2, \mathrm{~m}=-2,-1,0,+1,+2=5$ orbitals
If $1=3, \mathrm{~m}=-3,-2,-1,0,+1,+2,+3=7$ orbitals
$\therefore$ Total number of orbitals $=16$ orbitals

Question 28.
How many radial nodes for $2 \mathrm{~s}, 4 \mathrm{p}, 5 \mathrm{~d}$ and $4 \mathrm{f}$ orbitals exhibit? How many angular nodes?
Answer:
Formula for total number of nodes $=n-1$
1. For 2 s orbital: Number of radial nodes $=1$.
2. For $4 \mathrm{p}$ orbital: Number of radial nodes $=n-1-1 .=4-1-1=2$
Number of angular nodes $=1$
$\therefore$ Number of angular nodes $=1$
So, $4 \mathrm{p}$ orbital has 2 radial nodes and 1 angular node.
3. For $5 \mathrm{~d}$ orbital:
Total number of nodes $=n-1=5-1=4$ nodes
Number of radial nodes $=n-1-1=5-2-1=2$ radial nodes.
Number of angular nodes $=1=2$
$\therefore 5 \mathrm{~d}$ orbital have 2 radial nodes and 2 angular nodes.
4. For $4 \mathrm{f}$ orbital:
Total number of nodes $=n-1=4-1=3$ nodes
Number of radial nodes $=n-7-1=4-3-1=0$ node.
Number of angular nodes $=1=3$ nodes
$\therefore 4$ orbital have 0 radial node and 3 angular nodes.
Question 29.
The stabilization of a half filled d-orbital is more pronounced than that of the p-orbital why?
Answer:
The exactly half filled orbitals have greater stability. The reason for their stability are -
1. symmetry
2. exchange energy.

(1) Symmetry:
The half filled orbitals are more symmetrical than partially filled orbitals and this symmetry leads to greater stability.
(2) Exchange energy:
The electrons with same spin in the different orbitals of the same sub shell can exchange their position. Each such exchange release energy and this is known as exchange energy. Greater the number of exchanges, greater the exchange energy and hence greater the stability. In d-orbital, 10 exchanges are possible but in p-orbital 6 exchanges are possible. So, $d$ - orbital with 5 unpaired electrons ( 10 exchanges)n i.e. half filled is more stable than $p$-orbital with 3 unpaired electrons ( 6 exchanges).
Question 30.
Consider the following electronic arrangements for the $\mathrm{d} 5$ configuration.

(1) Which of these represents the ground state
(2) Which configuration has the maximum exchange energy.

Answer:

(1)

(2)

Question 31.
State and explain Pauli's exclusion principle.
Answer:
Pauli's exclusion principle states that "No two electrons in an atom can have the same set of values of all four quantum numbers".
Illustration: $\mathrm{H}(\mathrm{Z}=1) 1 \mathrm{~s}^1$.
One electron is present in hydrogen atom, the four quantum numbers are $\mathrm{n}=1,1=0, \mathrm{~m}=0$ and $\mathrm{s}=+\frac{1}{2}$.
For helium $\mathrm{Z}=2 . \mathrm{He}: 1 \mathrm{~s}^2$. In this one electron has the quantum number same as that of hydrogen, $\mathrm{n}=1,1=$ $0, \mathrm{~m}=0$ and $\mathrm{s}=+1 / 2$ For other electron, fourth quantum number is different, i.e. $\mathrm{n}=1,1=0, \mathrm{~m}=0$ and $\mathrm{s}=$ $-1 / 2$
Question 32.
Define orbital? What are the $n$ and 1 values for $3 \mathrm{p}_{\mathrm{x}}$ and $4 \mathrm{~d}_{\mathrm{x}}{ }^2 \mathrm{y}^2$ electron?
Answer:
(i) Orbital is a three dimensional space which the probability of finding the electron is maximum.
(ii) For $3 \mathrm{p}_{\mathrm{x}}$ electron $\mathrm{n}$ value $=3$
$l$ value $=1$
(iii) For $4 \mathrm{~d}_{\mathrm{x}^2-\mathrm{y}^2}$ electron $\mathrm{n}$ value $=4$
$l$ value $=2$
Question 33.
Explain briefly the time independent Schrodinger wave equation?
Answer:
The time independent Schrodinger equation can be expressed as
$\widehat{\mathrm{H}} \Psi=\mathrm{E} \Psi$
Where $\widehat{\mathrm{H}}$ is called Hamiltonian operator.
$\Psi$ is the wave function.

$E$ is the energy of the system.
$
\widehat{H}=\left[\frac{-\mathrm{h}^2}{8 \pi^2 \mathrm{~m}}\left(\frac{\partial^2}{\partial \mathrm{x}^2}+\frac{\partial^2}{\partial \mathrm{y}^2}+\frac{\partial^2}{\partial \mathrm{z}^2}\right)+\mathrm{V}\right]
$
Since $\Psi$ is a function of position coordinates of the particle and is denoted by $\Psi(x, y, z)$ $\therefore$ Equation (1) can be written as,
$
\left[\frac{-\mathrm{h}^2}{8 \pi^2 \mathrm{~m}}\left(\frac{\partial^2 \psi}{\partial \mathrm{x}^2}+\frac{\partial^2 \psi}{\partial \mathrm{y}^2}+\frac{\partial^2 \psi}{\partial \mathrm{z}^2}\right)+\mathrm{V} \psi\right]=\mathrm{E} \psi
$
Multiply the equation (3) by $\widehat{\mathrm{H}}$ and rearranging
$
\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
$
The above equation (4) Schrodinger wave equation does not contain time as a variable and is referred to as time independent Schrodinger wave equation.
Question 34.
Calculate the uncertainty in position of an electron, if $\Delta \mathrm{v}=0.1 \%$ and $\mathrm{n}=2.2 \times 10^6 \mathrm{~ms}^{-1}$.
Answer:
Mass of an electron $=\mathrm{m}=9.1 \times 10^{-31} \mathrm{~kg}$.
$\Delta \mathrm{v}=$ Uncertainty in velocity $=\frac{0.1}{100} \times 2.2 \times 10^3 \mathrm{~ms}^{-1}$.
$\Delta \mathrm{v}=0.22 \times 10^4=2.2 \times 10^3 \mathrm{~ms}^{-1}$
$\Delta \mathrm{x} \cdot \Delta \mathrm{v} \cdot \mathrm{m}=\frac{h}{4 \pi}$
$\Delta \mathrm{x}=\frac{h}{\Delta v, m x 4 \pi}$
$=\frac{6.626 \times 10^{-34}}{2.2 \times 10^3 \times 9.1 \times 10^{-31} \times 4 \times 3.14}$
$=\frac{6.626 \times 10^{-34} \times 10^{-3} \times 10^{31}}{2.2 \times 9.1 \times 4 \times 3.14}$

$
\begin{aligned}
& =\frac{6.626 \times 10^{-6}}{251.45} \\
& =0.02635 \times 10^{-6} \\
& \Delta x=2.635 \times 10^{-8} \\
& \text { Uncertainty in position }=2.635 \times 10^{-8}
\end{aligned}
$
Question 35 .
Determine the values of all the four quantum numbers of the 8th electron in $\mathrm{O}$-atom and $15^{\text {th }}$ electron in $\mathrm{Cl}$ atom and the last electron in chromium.
Answer:
(1) $O(Z=8) 1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}_{\mathrm{x}}^2 2 \mathrm{p}_{\mathrm{y}}{ }^1 2 \mathrm{p}_{\mathrm{z}}^1$
Four quantum numbers for $2 \mathrm{p}_{\mathrm{x}}{ }^1$ electron in oxygen atom:
$\mathrm{n}=$ principal quantum number $=2$
$1=$ azimuthal quantum number $=1$
$\mathrm{m}=$ magnetic quantum number $=+1$
$s=$ spin quantum number $=+\frac{1}{2}$

(2) $\mathrm{Cl}(\mathrm{Z}=17) 1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}_{\mathrm{x}}^2 3 \mathrm{p}_{\mathrm{y}}^2 3 \mathrm{p}_{\mathrm{z}}{ }^1$
Four quantum numbers for $15^{\text {th }}$ electron in chlorine atom: $\mathrm{n}=3,1=1, \mathrm{~m}=0, \mathrm{~s}=+1 / 2$
(3) $\mathrm{Cr}(\mathrm{Z}=24) 1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^2 3 \mathrm{~s}^2 3 \mathrm{p}^2 3 \mathrm{~d}^2 4 \mathrm{~s}^1$ $\mathrm{n}=3,1=2, \mathrm{~m}=+2, \mathrm{~s}=+1 / 2$
Question 36.
The quantum mechanical treatment of the hydrogen atom gives the energy value: $\mathrm{E}_{\mathrm{n}}=\frac{-13.6}{n^2} \mathrm{eV}$ atom $^{-1}$
1. use this expression to find $\Delta E$ between $n=3$ and $n=4$
2. Calculate the wavelength corresponding to the above transition.
Answer:
(1) When $\mathrm{n}=3$
$E_3=\frac{-13.6}{3^2}=\frac{-13.6}{9}=-1.511 \mathrm{eV}$ atom $^{-1}$
When $\mathrm{n}=4 \mathrm{E}_4=\frac{-13.6}{4^2}=-0.85 \mathrm{eV}$ atom $^{-1}$
$\Delta \mathrm{E}=\mathrm{E}_4-\mathrm{E}_3=-0.85-(-1.511)=+0.661 \mathrm{eV}$ atom
$\Delta \mathrm{E}=\mathrm{E}_3-\mathrm{E}_4$
$=-1.511-(-0.85)$
$=-0.661 \mathrm{eV}$ atom $^{-1}$
(2) Wave length $=\lambda$
$\Delta \mathrm{E}=\frac{h c}{\lambda}$
$\lambda=\frac{h c}{\Delta E}$
$\mathrm{h}=$ Planck's constant $=6.626 \times 10^{-34} \mathrm{Js}^{-1}$
$\mathrm{c}=3 \times 10^8 \mathrm{~m} / \mathrm{s}$
$\lambda=\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{0.661}$

$
\begin{aligned}
& =10.02 \times 10^{-34} \times 3 \times 10^8 \\
& =30 \times 10^{-26} \\
& \lambda=3 \times 10^{-25} \mathrm{~m} .
\end{aligned}
$
Question 37
How fast must a 54 g tennis ball travel in order to have a de Broglie wavelength that is equal to that of a photon of green light $5400 \AA$ ?
Answer:
$
\begin{aligned}
& \mathrm{m}=\text { mass of tennis ball }=54 \mathrm{~g}=5.4 \times 10^{-2} \mathrm{~kg} . \\
& \lambda=\text { de Broglie wavelength }=5400 \AA .=5400 \times 10^{-10} \mathrm{~m} . \\
& \mathrm{V}=\text { velocity of the ball }=? \\
& \lambda=\frac{h}{m V} \\
& \mathrm{~V}=\frac{h}{\lambda \cdot m}
\end{aligned}
$

$
\begin{aligned}
& =\frac{6.626 \times 10^{-34}}{5400 \times 10^{-10} \times 5.4 \times 10^{-2}} \\
& =\frac{6.626 \times 10^{-34} \times 10^{10} \times 10^2}{5400 \times 5.4} \\
& =\frac{6.626 \times 10^{-24}}{54 \times 5.4} \\
& =\frac{6.626 \times 10^{-24}}{29.6} \\
& =0.2238 \times 10^{-24} \\
& =2.238 \times 10^{-25} \mathrm{~m} .
\end{aligned}
$
Question 38.
For each of the following, give the sub level designation, the allowable $\mathrm{m}$ values and the number of orbitals.
$
\begin{aligned}
& \text { 1. } \mathrm{n}=4,1=2 \text {, } \\
& \text { 2. } \mathrm{n}=5,1=3 \\
& \text { 3. } \mathrm{n}=7,1=0
\end{aligned}
$
Answer:
1. $\mathrm{n}=4,1=2$
If $1=2$, ' $m$ ' values are $-2,-1,0,+1,+2$
So, 5 orbitals such as $\mathrm{d}_{\mathrm{xy}}, \mathrm{d}_{\mathrm{yz}}, \mathrm{d}_{\mathrm{xz}}, d_{x^2-y^2}$ and $\mathrm{d} z$
2. $\mathrm{n}=5,1=3$
If $1=3$, ' $m$ ' values are $-3,-2,-1,0,+1,+2,+3$
So, 7 orbitals such as $f z, f x z, f y z, f x y z, f z(x 2 y 2)^{\prime} \wedge x(x 2-3 y 2)^{\prime} \wedge y(3 \times 2 ? ? y$
3. $\mathrm{n}=7,1=0$
If $1=0$, , $m$ ' values are 0 . Only one value.
So, 1 orbital such as 7 s orbital.

Question 39.
Give the electronic configuration of $\mathrm{Mn}^{2+}$ and $\mathrm{Cr}^{3+}$
Answer:
1. $\mathrm{Mn}(Z=25)$
$\mathrm{Mn} \rightarrow \mathrm{Mn}^{2+}+2 \mathrm{e}^{-}$
$\mathrm{Mn}^{2+}$ electronic configuration is $1 \mathrm{~s} 1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^5$
2. $\mathrm{Cr}(\mathrm{Z}=24)$
$\mathrm{Cr} \rightarrow \mathrm{Cr}^{3+}+3 \mathrm{e}^{-}$
$\mathrm{Cr}^{3+}$ electronic configuration is $\mathrm{Is}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^3$
Question 40 .
Describe the Aufbau principle.
Answer:
In the ground state of the atoms, the orbitals are filled in the order of their increasing energies. That is the electrons first occupy the lowest energy orbital available to them. Once the lower energy orbitals are completely filled, then the electrons enter the next higher energy orbitals.
The order of filling of various orbitals as per Aufbau principle is -
$1 \mathrm{~s}<2 \mathrm{~s}<2 \mathrm{p}<3 \mathrm{~s}<3 \mathrm{p}<4 \mathrm{~s}<3 \mathrm{~d}<4 \mathrm{p}<5 \mathrm{~s}<4 \mathrm{~d} \ldots \ldots \ldots$.
For e.g., $\mathrm{K}(\mathrm{Z}=19)$
The electronic configuration is $1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 4 \mathrm{~s}^1$.
After filling 4 s orbital only we have to fill up $3 \mathrm{~d}$ orbital.
Question 41.
A $n$ atom of an element contains 35 electrons and 45 neutrons. Deduce
1. the number of protons
2. the electronic configuration for the element
3. All the four quantum numbers for the last electron
Answer:
An element $X$ contains 35 electrons, 45 neutrons
1. The number of protons must be equal to the number of electrons. So the number of protons $=35$.
2. Number of electrons $=35$. So the electronic configuration is $1 s^2 2 s^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 4 \mathrm{~s}^2 3 \mathrm{~d}^{10} 4 \mathrm{p}^5$.
3. The last electron i.e. $5^{\text {th }}$ electron in $4 p$ orbital has the following quantum numbers. $n=4,1=1, \mathrm{~m}$ $=+1, \mathrm{~s}=+\frac{1}{2}$
Question 42.
Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wave length associated with the electron revolving around the nucleus.
Answer:
In order for the electron wave to exist in phase, the circumference of the orbit should be an integral multiple of the wavelength of the electron wave. Otherwise, the electron wave is out of phase.
$\mathrm{mvr}=\mathrm{nh} / 2 \pi, 2 \pi \mathrm{r}=\mathrm{n} \lambda$
where $\mathrm{mvr}=$ angular momentum
where $2 \pi \mathrm{r}=$ circumference of the orbit

Question 43.
Calculate the energy required for the process.
$
\mathrm{He}_{(\mathrm{g})}^{+} \rightarrow \mathrm{He}_{(\mathrm{g})}^{2+}+\mathrm{e}^{-}
$
The ionization energy for the $\mathrm{H}$ atom in its ground state is $-13.6 \mathrm{eV}$ atom $^{-1}$.
Answer:
The ionization energy for the $\mathrm{H}$ atom in its ground state $=-13.6 \mathrm{eV}$ atom $^{-1}$.
Ionization energy $=\frac{13.6 z^2}{n^2} \mathrm{eV}$
$\mathrm{Z}=$ atomic number
$\mathrm{n}=$ principal quantum number or shell number
For $\mathrm{He}, \mathrm{n}=1, \mathrm{z}=2$
$
\mathrm{IE}=\frac{-13.6 \times 2^2}{1^2} \mathrm{eV}
$
Question 44.
An ion with mass number 37 possesses unit negative charge. If the ion contains $11.1 \%$ more neutrons than electrons. Find the symbol of the ion.
Answer:
Let the number of electrons in an ion $=\mathrm{x}$
number of neutrons $=\mathrm{n}=\mathrm{x}+\frac{11.1}{100} \mathrm{eV}=1.111 \mathrm{x}$
(As the number of neutrons are $11.1 \%$ more than the number of electrons)
In the neutral of atom, number of electron.

$\mathrm{e}^{-}=\mathrm{x}-1$ (as the ion carries -1 charge)
Similarly number of protons $=\mathrm{P}=\mathrm{x}-1$ Number of protons + number of neutrons $=$ mass number $=37$ $(\mathrm{x}-1)+1.111 \mathrm{x}=37$
$2.111 \mathrm{x}=37+1$
$2.111 \mathrm{x}=38$
$\mathrm{x}=\frac{38}{2.111}=18.009=18$
$\therefore$ Number of protons $=$ atomic number $-1=18-1=17$
$\therefore$ The symbol of the ion $={ }_{17}^{37} \mathrm{Cl}$.
Question 45.
The $\mathrm{Li}^{2+}$ ion is a hydrogen like ion that can be described by the Bohr model. Calculate the Bohr radius of the third orbit and calculate the energy of an electron in $4^{\text {th }}$ orbit.
Answer:
$\mathrm{Li}^{2+}$ hydrogen like ion.
Bohr radius of the third orbit $=\mathrm{r}_3=$ ?
$
\mathrm{r}_3=\frac{(0.529) n^2}{Z} \mathrm{~A}
$
Where $\mathrm{n}=$ shell number, $\mathrm{Z}=$ atomic number.
$
\begin{aligned}
& \mathrm{r}_3=\frac{(0.529) 3^2}{3} \mathrm{~A}[\therefore \text { for lithium } \mathrm{Z}=3, \mathrm{n}=3] \\
& =\frac{0.529 x 9}{3} \\
& \mathrm{r}_3=1.587 \AA \\
& \mathrm{E}_{\mathrm{n}}=\frac{(-13.6) Z^2}{n^2} \mathrm{eV} \text { atom }{ }^{-1} \\
& \mathrm{E}_4=\text { Energy of the fourth orbit }=? \\
& \mathrm{E}_4=\frac{(-13.6) \times 3^2}{4^2}=\frac{-13.6 \times 9}{16}=-7.65 \mathrm{eV} \text { atom }^{-1} \\
& \mathrm{E}_4=-7.65 \mathrm{eV} \text { atom }^{-1}
\end{aligned}
$
Question 46.
Protons can be accelerated in particle accelerators. Calculate the wavelength (in $\AA$ )of such accelerated proton moving at $2.85 \times 108 \mathrm{~ms}^{-1}$ (the mass of proton is $1.673 \times 10^{-27} \mathrm{Kg}$ ).

Answer:
$
\begin{aligned}
& \mathrm{m}=\text { mass of the proton }=1.673 \times 10^{-27} \mathrm{Kg} \\
& \mathrm{v}=\text { velocity of the proton }=2.85 \times 10^8 \mathrm{~ms}^{-1} \\
& \lambda=\frac{h}{m v} \\
& \mathrm{~h}=\text { Planck's constant }=6.626 \times 10^{34} \mathrm{Kg} \mathrm{m}^2 \mathrm{~s}^{-1}
\end{aligned}
$
$
\begin{aligned}
& =\frac{6.626 \times 10^{-34} \mathrm{Kg} \mathrm{m}^{-2} 8^{-1}}{1.673 \times 10^{-27} \mathrm{Kg} \times 2.85 \times 10^8 \mathrm{~mm}^{-1}} \\
& =\frac{6.626 \times 10^{-34} \times 10^{27} \times 10^{-8}}{1.673 \times 2.85} \mathrm{~m} \\
& =\frac{6.626 \times 10^{-15}}{4.768} \\
& =1.389 \times 10^{-15} \mathrm{~m}
\end{aligned}
$
Wavelength of proton $=\lambda=1.389 \times 10^{-15} \mathrm{~m}$.

Question 47.

$\text { What is the de Broglie wavelength (in } \mathrm{cm} \text { ) of a } 160 \mathrm{~g} \text { cricket ball travelling at } 140 \mathrm{Km} \mathrm{hr}^{-1} \text {. }$
Answer:
$
\begin{aligned}
& \mathrm{m}=\text { mass of the cricket ball }=160 \mathrm{~g}=0.16 \mathrm{~kg} \text {. } \\
& \mathrm{v}=\text { velocity of the cricket ball }=140 \mathrm{Km} \mathrm{h}^{-1} \\
& =\frac{140 x 5}{18}=38.88 \mathrm{~ms}^{-1} \\
& \text { de Broglie equation }=\lambda=\frac{h}{m v} \\
& \mathrm{~h}=\text { Planck's constant }=6.626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1} \\
& =\frac{6.626 \times 10^{-34} \mathrm{Kg} \mathrm{m}^2 \mathrm{~s}^{-1}}{0.16 \mathrm{Kg} \times 38.88 \mathrm{mns}^{-1}} \\
& \lambda=\frac{6.626 \times 10^{-34}}{6.2208} \\
& \lambda=1.065 \times 10^{-34} \mathrm{~m} \\
&
\end{aligned}
$$
Wave length in $\mathrm{cm}=1.065 \times 10^{-34} \times 100$
$$
=1.065 \times 10^{-32} \mathrm{~cm}
$$
Question 48.
Suppose that the uncertainty in determining the position of an electron in an orbit is $0.6 \mathrm{~A}$. What is the uncertainty in its momentum?.
Answer:
$\Delta \mathrm{x}=$ uncertainty in position of an electron $=0.6 \AA=0.6 \times 10^{-10} \mathrm{~m}$.
$\Delta \mathrm{p}=$ uncertainty in momentum $=$ ?
Heisenberg's uncertainty principle states that,
$$
\begin{aligned}
& \Delta \mathrm{x} . \Delta \mathrm{p} \geq \frac{h}{4 \pi} \\
& \Delta \mathrm{p}=\frac{h}{4 \pi \cdot \Delta x} \\
& \mathrm{~h}=\text { Planck's constant }=6.626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1} \\
& \therefore \Delta \mathrm{p}=\frac{6.626 \times 10^{-34} \mathrm{Kg} \mathrm{m}^2 \mathrm{~s}^{-1}}{4 \times 3.14 \times 0.6 \times 10^{-10} \mathrm{~m}}
\end{aligned}
$

$
=\frac{6.626 \times 10^{-34} \times 10^{10}}{7.536}
$
Uncertainty in momentum $=0.8792 \times 10^{-24} \mathrm{~kg} \mathrm{~ms}^{-1}$ (or) $=8.792 \times 10^{-25} \mathrm{~kg} \mathrm{~ms}^{-1}$
Question 49 .
Show that if the measurement of the uncertainty in the location of the particle is equal to its de Broglie wavelength, the minimum uncertainty in its velocity is equal to its velocity $/ 4 \pi$
Answer:
If, uncertainty in position $=\Delta \mathrm{x}=\lambda$, the value of uncertainty in velocity $=\frac{v}{4 \pi}$
Heisenberg's principle states that
$\Delta \mathrm{x} . \Delta \mathrm{v} \cdot \mathrm{m}=\frac{h}{4 \pi}$
de Broglie equation states that
$\lambda=\frac{h}{m v}$
$\therefore \mathrm{h}=\lambda \cdot \mathrm{m} \cdot \mathrm{v}$
$\Delta \mathrm{x}=\frac{h}{\Delta v .4 \pi}$
Substituting the value of $h$ in equation (4)
$
\Delta \mathrm{x}=\frac{\lambda x m \cdot v}{\Delta v \cdot 4 \pi \cdot m}
$

$
\begin{aligned}
\text { if } \Delta \mathrm{x} & =\lambda \\
\Delta \mathrm{v} & =\frac{\not \mathrm{X} \cdot \mathrm{m} \cdot \mathrm{v}}{\not \mathrm{X} \cdot 4 \pi \cdot \mathrm{m}}=\frac{v}{4 \pi}
\end{aligned}
$
Question 50.
What is the de Broglie wave length of an electron, which is accelerated from the rest, through a potential difference of $100 \mathrm{~V}$ ?
Answer:
Potential difference $=\mathrm{V}=100 \mathrm{~V}$
Potential energy $=\mathrm{eV}=1.609 \times 10^{-19} \mathrm{c} \times 100 \mathrm{~V}$
$\frac{v}{4 \pi} \mathrm{m} \mathrm{v}{ }^2=1.609 \times 10^{-19} \times 100$
$\frac{v}{4 \pi} \mathrm{m} \mathrm{v}{ }^2=1.609 \times 10^{-19} \mathrm{~J}$
$\mathrm{v}^2=\frac{2 \times 1,609 \times 10^{-17}}{m}$
$\mathrm{m}=$ mass of electron $=9.1 \times 10^{-31} \mathrm{Kg}$
$\therefore \mathrm{v}^2=\frac{2 \times 1.609 \times 10^{-17}}{9.1 \times 10^{-31}}$
$\therefore \mathrm{v}=\sqrt{\frac{2 \times 1.609 \times 10^{-17}}{9.1 \times 10^{-31}}}$
$=\sqrt{\frac{2 \times 1.609 \times 10^{-17} \times 10^{31}}{9.1}}$
$v=\sqrt{\frac{3.218 \times 10^{14}}{9.1}}$
$\mathrm{v}=5.93 \times 10^6 \mathrm{~m} / \mathrm{s}$
$\lambda=\frac{h}{m v}$ where $\mathrm{h}=6.62 \times 10^{-34} \mathrm{JS}$
$=\frac{6.62 \times 10^{-34}}{9.1 \times 10^{-31} \times 5.93 \times 10^6}$

$
\begin{aligned}
& =1.2 \times 10^{-10} \mathrm{~m} \\
& \mathrm{~A}=1.2 \AA
\end{aligned}
$
Question 51 .
Identify the missing quantum numbers and the sub energy level

Answer:

In-Text Questions -Evaluate Yourself
Question 1.
Calculate the de Broglie wavelength of an electron that has been accelerated from rest through a potential difference of $1 \mathrm{keV}$.
Answer:
$
\lambda=\frac{h}{m v}
$
Potential difference of an electron $=\mathrm{V}=1 \mathrm{keV}$.
Potential energy $=\frac{1}{2} \mathrm{mv}^2=\mathrm{eV}$
$\mathrm{e}=$ charge of an electron $=1.609 \times 10^{-19} \mathrm{c}$
$1 \mathrm{k} \mathrm{V}=1000 \mathrm{~V}$
$\therefore$ Potential energy $=1.609 \times 10^{-19} \times 1000=1.609 \times 10^{-19}$
$\frac{1}{2} \mathrm{mv}^2=1.609 \times 10^{-16} \mathrm{~V}$
$\mathrm{m}=9.1 \times 10^{-31} \mathrm{~kg}$
$\lambda=\frac{h}{m v}$
$\mathrm{v}^2=\frac{2 \times 1.609 \times 10^{-16}}{9.1 \times 10^{-31}}$
$v=\sqrt{\frac{2 \times 1.609 \times 10^{-16}}{9.1 \times 10^{-31}}}$
$=5.9 \times 10^7 \mathrm{~ms}^{-1}$
$\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 5.93 \times 10^7}$
$=1.2 \times 10^{-11} \mathrm{~m}$
$\lambda=1.2 \times 10^{-11} \mathrm{~m}$.

Question 2.
Calculate the uncertainty in the position of an electron, if the uncertainty in its velocity is $5.7 \times 10 \mathrm{~s} \mathrm{~ms}^{-1}$.
Answer:
Uncertainty in velocity $=\mathrm{Av}=5.7 \times 10^5 \mathrm{~ms}^{-1}$
Mass of an electron $=\mathrm{m}=9.1 \times 10^{-31} \mathrm{~kg}$.
Uncertainty in position $=\Delta \mathrm{x}=$ ?
$
\begin{aligned}
& \Delta \mathrm{x} \cdot \mathrm{m} \cdot \Delta \mathrm{v}=\frac{h}{4 \pi} \\
& \Delta \mathrm{x}=\frac{\mathrm{h}}{\mathrm{m} . \Delta \mathrm{v} \times 4 \pi}=\frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 5.7 \times 10^5 \times 4 \pi} \\
& \Delta \mathrm{x}=\frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 5.7 \times 10^5 \times 4 \times 3.14} \\
& =1 \times 10^{-10} \mathrm{~m} .
\end{aligned}
$
Uncertainty in position $=1 \times 10^{-10} \mathrm{~m}$
Question 3.
How many orbitals are possible in the 4th energy level? $(n=4)$
Answer:
$
\mathrm{n}=4
$
Number of orbitals in $4^{\text {th }}$ energy level $=$ ?
When $\mathrm{n}=4,1=0,1,2,3$
If $1=0$ orbital $=4 \mathrm{~s}=1$
If $1=1$ orbital $=4 \mathrm{p}_{\mathrm{x}}, 4 \mathrm{p}_{\mathrm{y}}, 4 \mathrm{p}_z=2$
If $1=2$ orbital $=4 \mathrm{~d}_{\mathrm{xy}}, 4 \mathrm{~d}_{\mathrm{yz}}, 4 \mathrm{~d}_{\mathrm{zx}}, 4 \mathrm{~d}_{\mathrm{x}} 2_{\mathrm{y}}, 2,4 \mathrm{~d}_{\mathrm{z}^2}=5$
If $1=3$ orbital $=-3,-2,-1,0,+1,+2,+3=7$
Number of orbitals in 4th energy level $=16$.

Question 4.
Calculate the total number of angular nodes and radial nodes present in $3 \mathrm{~d}$ and $4 \mathrm{f}$ orbitals.
Answer:
Number of angular nodes in $3 \mathrm{~d}$ orbital $=$ ?
Number of radial nodes in $3 \mathrm{~d}$ orbital $=$ ?
Number of angular nodes $=1$
Number of radial nodes $=\mathrm{n}-1-1$
1. For $3 \mathrm{~d}$ orbital:
Number of angular nodes $=2$ because $1=2$
Number of radial nodes $=3-2-1=0$
Total number of nodes in $3 \mathrm{~d}$ orbital $=2$
2. For $4 \mathrm{f}$ orbital:
Number of angular nodes $=3$ because $1=3$
Number of radial nodes $=n-1-1=4-3-1=0$
Total number of nodes in $4 \mathrm{f}$ orbital $=3$.
Question 5.
Energy of an electron in hydrogen atom in ground state is $-13.6 \mathrm{eV}$. What is the energy of the electron in the second excited state?
Answer:

Energy of an electron in ground state $=-13.6 \mathrm{eV}$.
$\therefore$ Energy of an electron in the second excited state $=\mathrm{E}_2$.
$
\begin{aligned}
& n=2 \\
& E_2=\frac{-13.6 \mathrm{eV}}{\mathrm{n}^2}=\frac{-13.6}{2^2}=\frac{-13.6}{4}=-3.4 \mathrm{eV}
\end{aligned}
$
Question 6.
How many unpaired electrons are present in the ground state of $\mathrm{Fe}^{3+}(z=26), \mathrm{Mn} 2+(z=25)$ and argon $(z=18) ?$
Answer:
$1 s^2 2 s^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^6 4 \mathrm{~s}^2$ for $\mathrm{Fe}$ atom.
$1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^6 3 \mathrm{~d}^5$ for $\mathrm{Fe}^{3+}$ ion.
So, it contain 5 unpaired electrons.
$\mathrm{Mn}(\mathrm{Z}=25)$. Electronic configuration is $1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^6 3 \mathrm{~d}^5$
$\mathrm{Mn} \rightarrow \mathrm{Mn}^{2+}+2 \mathrm{e}^{-}$
Number of unpaired electrons in $\mathrm{Mn}^{2+}=5$
$\operatorname{Ar}(Z=18)$. Electronic configuration is $1 s^2 2 s^2 2 p^6 3 s^2 3 p^6$.
All orbitals are completely filled. So, no unpaired electrons in it.
Question 7.
Explain the meaning of the symbol $4 f^2$. Write all the four quantum numbers for these electrons.
Answer:
$4 f^2$ : It means that the element has 2 electrons in outermost $4 \mathrm{f}$ shell.

$\mathrm{n}=$ principal quantum number $=4$
$1=$ azimuthal quantum number $=3$
$\mathrm{m}=$ magnetic quantum number $=-3,-2$
$s=$ spin quantum number $=+\frac{1}{2}-\frac{1}{2}$.
Question 8.
Which has the stable electronic configuration? $\mathrm{Ni}^{2+}$ or $\mathrm{Fe}^{3+}$
Answer:
$\mathrm{Ni}(\mathrm{Z}=28) \cdot 1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 4 \mathrm{~s}^2 3 \mathrm{~d}^8$
$\mathrm{Ni}^{2+}$ electronic configuration $=\mathrm{Is}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^8$
$\mathrm{Fe}(\mathrm{Z}=26) \cdot 1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 4 \mathrm{~s}^2 3 \mathrm{~d}^6$
$\mathrm{Fe}^{3+} \mathrm{Is}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^5$
If $\mathrm{d}$ orbital is half filled, according to Aufbau principle, it is more stable. So $\mathrm{Fe}^{3+}$ is more stable than $\mathrm{Ni}^{2+}$.