NEET PYQs

Hint

(c) In case of hydrogen like atoms, energy depends on the principal quantum number only. Hence, \(2 s\)-orbital will have energy equal to \(2 p\)-orbital.

Hint

(b) For \(n=3\) and \(l=1\), the subshell is \(3 p\) and a particular \(3 p\) orbital can accommodate only 6 electrons.

Hint

(c) \(d_{x^{2}-y^{2}}\) and \(d_{z 2}\) orbitals have electron density along the axes while \(d_{x y}, d_{y z}\) and \(d_{x z}\) orbitals have electron density in between the axes.

Hint

(b) For the two electrons occupying the same orbital values of \(n, l\) and \(m_{l}\) are same but \(m_{s}\) is different, i.e., \(+\frac{1}{2}\) and \(-\frac{1}{2}\).

Hint

(c) \(\operatorname{Ti}(22): 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{2}\)
\(\therefore\) Order of increasing energy is \(3 s, 3 p, 4 s, 3 d\)
.

Hint

(d)  Number of \(d\)-electrons in \(\mathrm{Fe}^{2+}=6\)
Number of \(p\)-electrons in \(\mathrm{Cl}=11\)

Hint

(c) Angular momentum \(=\sqrt{l(l+1)} h\); For \(d\) orbital, \(l=2\)

Angular momentum \(=\sqrt{2(2+1)} h=\sqrt{6} h\)

Hint

(a)  Only one orbital, \(3 p_{z}\) has following set of quantum numbers, \(n=3, l=1\) and \(m_{l}=0\).

Hint

(d) \(E=\frac{h c}{\lambda}\) [Given, \(\lambda=45 \mathrm{~nm}=45 \times 10^{-9} \mathrm{~m}\) ]
On putting the given values in the equation, we get \(E=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{45 \times 10^{-9}}=4.42 \times 10^{-18} \mathrm{~J}\)

Hint

(b)

\(
\begin{array}{cc}
\text { Species } & \text { No. of electrons } \\
\mathrm{Be}^{2+} & 2 \\
\mathrm{H}^{+} & 0 \\
\mathrm{Li}^{+} & 2 \\
\mathrm{Na}^{+} & 10 \\
\mathrm{Mg}^{2+} & 10
\end{array}
\)

Hint

(b)  The orbital associated with \(n=3, l=1\) is \(3 p\). One orbital (with \(m=-1\) ) of \(3 p\)-subshell can accommodate maximum 2 electrons.

Hint

(b)  The electron is more tightly bound in the smallest allowed orbit.

Note: The lowest possible energy state is the one in which the electrons are most tightly bound, and it’s called the ground state. In the ground state there is more force of attraction which means elections in ground state are tightly bounded.

Hint

(a) \(c=f \lambda\)
\(
\lambda=\frac{c}{f}=\frac{3 \times 10^{17}}{6 \times 10^{15}}=50 \mathrm{~nm}
\)

Hint

(c) Half-filled stabled configuration will be more favoured to Gd. Arrange the orbitals according to the increasing energy. The configuration of Gd is \([\mathrm{Xe}] 4 \mathrm{f}^7 5 \mathrm{~d}^1 6 \mathrm{~s}^2\).

Hint

\(
\text { (a) We know that, } E=\frac{h c N_{A}}{\lambda}
\)\(
\begin{aligned}
&=\frac{6.62 \times 10^{-27} \times 3 \times 10^{10} \times 6.02 \times 10^{23}}{\lambda} \\
&=\frac{1.1955 \times 10^{8}}{\lambda}=\frac{1.196 \times 10^{8}}{\lambda} \mathrm{ergs}  \mathrm{~mol}^{-1}
\end{aligned}
\)

Hint

(a)  \(l=3\) and \(n=4\) represent \(4 f\). So, total number of electrons in a subshell \(=2(2 l+1)=2(2 \times 3+1)=14\) electrons. Hence \(f\)-subshell can contain maximum 14 electrons.

Hint

(c) \(\mathrm{Rb}(37): 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{6} 5 s^{1}\)

For \(5 s, n=5, l=0, m=0, s=+1 / 2\) or \(-1 / 2\)

Hint

(a) Orbital angular momentum
\((m)=\sqrt{l(l+1)} \frac{h}{2 \pi}\)
For \(p\)-electrons; \(l=1\)
Thus, \(m=\sqrt{1(1+1)} \frac{h}{2 \pi}=\frac{\sqrt{2} h}{2 \pi}=\frac{h}{\sqrt{2} \pi}\)

Hint

(b)  Total number of atomic orbitals in any energy level is given by \(n^{2}\). It should be 16.

Hint

(b) We know the relation between energy and lambda

\(
E_1=\frac{h c}{\lambda_1} \text { and } E_2=\frac{h c}{\lambda_2}
\)
\(
\begin{array}{l}
\frac{E_1}{E_2}=\frac{h c}{\lambda_1} \times \frac{\lambda_2}{h c}=\frac{\lambda_2}{\lambda_1} \\
\frac{25}{50}=\frac{\lambda_2}{\lambda_1} \text { or } \frac{1}{2}=\frac{\lambda_2}{\lambda_1} \Rightarrow \lambda_1=2 \lambda_2
\end{array}
\)

Hint

(a) Principle quantum number \((n)\) tells us the number of shells in which the last electron enters.
Basically, the principal quantum number tells us about the valence shell.
According to Aufbau’s rule, electrons will fill lower energy atomic levels first and then get filled up in the higher energy levels.
Aufbau’s rule is represented below in the given diagram.

The electrons will start filling up in the order:
\(1s\) then \(2 s\) then \(2 p\) then \(3 s, 3 p, 4 s, 3 d, 4 p, 5 s, 4 d, 5 p, 6 s, 4 f, 5 d, 6 p\) and then \(7 s\) and \(7 p\).
\(1s\) orbital has the least amount of energy.
If the valence electron enters in the sixth orbital then, the electron enters in 6 s orbital followed by \(4 \mathrm{f}\) orbital then \(5 \mathrm{~d}\) orbital and then \(6 \mathrm{p}\) orbital.
Therefore, when \(n=6\), the electron fills in the order \(6 s, 4 f, 5 d\) and \(6 p\).
So, the option \(n s \rightarrow(n-2) f \rightarrow(n-1) d \rightarrow n p\) is correct.

Hint

(c)  We know that
\(\Delta E \propto\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]\), where \(n_{2}>n_{1}\)
\(\therefore n=6\) to \(n=5\) will give least energetic photon.

Hint

(c) According to de-Broglie equation, \(\lambda=\frac{h}{m v}\) Given, \(h=6.6 \times 10^{-34} \mathrm{Js} ; m=0.66 \mathrm{~kg} ; v=100 \mathrm{~m} \mathrm{~s}^{-1}\)
\(
\therefore \quad \lambda=\frac{6.6 \times 10^{-34}}{0.66 \times 100}=1 \times 10^{-35} \mathrm{~m}
\)

Hint

(d) For a given shell, \(l\),
the number of subshells, \(m_{l}=(2 l+1)\)
Since each subshell can accommodate 2 electrons of opposite spin, so maximum number of electrons in a subshell \(=2(2 l+1)=4 l+2\).

Hint

(b)  In an atom, for any value of \(n\), the values of \(l=0\) to \((n-1)\).

For a given value of \(l\), the values of \(m_{l}=-l\) to 0 to \(+l\) and the value of \(s=+1 / 2\) or \(-1 / 2\).
In option (b), \(l=2\) and \(m_{l}=-3\)
This is not possible, as values of \(m_{l}\) which are possible for \(l=2\) are \(-2,-1,0,+1\) and \(+2\) only.

Hint

(c) Uncertainty in momentum
\(
(m \Delta v)=1 \times 10^{-18} \mathrm{~g} \mathrm{~cm} \mathrm{~s}^{-1}
\)
Uncertainty in velocity,
\(
(\Delta v)=\frac{1 \times 10^{-18}}{9 \times 10^{-28}}=1.1 \times 10^{9} \mathrm{~cm} \mathrm{~s}^{-1}
\)

Hint

(b) 
(i) represents an electron in \(3 s\) orbital.
(ii) is not possible as value of \(l\) varies from \(0,1, \ldots(n-1)\).
(iii) represents an electron in \(4 f\) orbital.
(iv) is not possible as value of \(m\) varies from \(-l \ldots+l\)
(v) is not possible as value of \(m\) varies from \(-l \ldots+l\), it can never be greater than \(l\).

Hint

(c)  From Heisenberg uncertainty principle \(\Delta p \cdot \Delta x \geq \frac{h}{4 \pi}\) or \(m \Delta v \times \Delta x \geq \frac{h}{4 \pi}\) or \((m \Delta v)^{2} \geq \frac{h}{4 \pi}\) \((\because \Delta x=\Delta p)\) or \(\Delta v \geq \frac{1}{2 m} \sqrt{\frac{h}{\pi}}\)

Hint

(d) Magnetic quantum number

The orientation of an atomic orbital is governed by magnetic quantum number.

Hint

(b)  \(\Delta x \cdot m \Delta v=h / 4 \pi\)
\(
\begin{array}{r}
0.1 \times 10^{-10} \times 9.11 \times 10^{-31} \times \Delta v=\frac{6.626 \times 10^{-34}}{4 \times 3.143} \\
\therefore \Delta v=\frac{6.626 \times 10^{-34}}{0.1 \times 10^{-10} \times 9.11 \times 10^{-31} \times 4 \times 3.143} \\
=5.79 \times 10^{6} \mathrm{~m} \mathrm{~s}^{-1}
\end{array}
\)

Hint

(b) \(E_{n}=-K\left(\frac{Z}{n}\right)^{2}\)
\(Z=1\) for hydrogen \(; n=2\)
\(
\begin{gathered}
E_{2}=\frac{-K \times 1}{4} \Rightarrow E_{2}=-328 \mathrm{~kJ} \mathrm{~mol}^{-1} ; K=4 \times 328 \\
E_{4}=\frac{-K \times 1}{16} \Rightarrow E_{4}=-4 \times 328 \times \frac{1}{16}=-82 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{gathered}
\)

Hint

(b)

Formula Used: Planck’s equation:
\(\Delta E=h \nu\) where \(\Delta E\) is the change in energy, \(\mathrm{h}\) is thee planck’s constant and \(\nu\) is the frequency.
Complete step by step answer: Ionization energy means it is the energy for the first orbit is given to be
\(2.18 \times 10^{-18}\) Jatom \(^{-1}\). Atomic number of hydrogen atom is 1 i.e. \(Z=1\), then energy for \(n^{\text {th }}\) orbit is given as:
\(
E_n=\frac{-2.18 \times 10^{-18}}{n^2} \dots(1)
\)
For \(n=1\)
\(
E=-2.18 \times 10^{-18}
\)
The change in energy when electrons fall from \(n=4\) to \(n=1\) is given as:
\(
\Delta E=E_4-E_1
\)
Then, using (1):
\(
\begin{array}{l}
\Delta E=\frac{-2.18 \times 10^{-18}}{4^2}-\left(\frac{-2.18 \times 10^{-18}}{1^2}\right) \\
\Rightarrow \Delta E=-2.18 \times 10^{-18}\left(\frac{1}{4^2}-\frac{1}{1^2}\right) \\
\Rightarrow \Delta E=-2.18 \times 10^{-18} \times\left(-\frac{15}{16}\right) \\
\Rightarrow \Delta E=2.0437 \times 10^{-18} \text { Jatom }^{-1}
\end{array}
\)
According to Planck’s equation the change in energy is directly proportional to frequency of electron and Planck’s constant ( \(h\) ) is used as the proportionality constant. So:
\(
\begin{array}{l}
\Delta E=h \nu \\
\Rightarrow \nu=\frac{\Delta E}{h}
\end{array}
\)
Substituting the known values to find the value of frequency, we get:
\(
\begin{array}{l}
\nu=\frac{2.0437 \times 10^{-18}}{6.625 \times 10^{-34}} \\
\Rightarrow \nu=3.084 \times 10^{15} \mathrm{~s}^{-1}
\end{array}
\)

Hint

(c)  Applying \((f)=c / \lambda\),
\(\lambda=\frac{c}{f}=\frac{3 \times 10^{8}}{8 \times 10^{15}}=37.5 \times 10^{-9} \mathrm{~m}=37.5 \mathrm{~nm} \approx 4 \times 10^{1} \mathrm{~nm}
\)

Hint

(a)  Kinetic energy \(=\frac{1}{2} m v^{2}=\left(\frac{\pi e^{2}}{n h}\right)^{2} \times 2 m\)
\(
\left[\because v=\frac{2 \pi e^{2}}{n h}\right]
\)
Total energy \(E_{n}=-\frac{2 \pi^{2} m e^{4}}{n^{2} h^{2}}=-\left(\frac{\pi e^{2}}{n h}\right)^{2} \times 2 m=-K . E\).
\(\therefore\) Kinetic energy \(=-E_{n}\)
Energy of first excited state is \(-3.4 \mathrm{~eV}\)
\(\therefore\) Kinetic energy of same orbit \((n=2)\) will be \(+3.4 \mathrm{~eV}\).

Hint

(a) For \(\pi\) overlap, the lobes of the atomic orbitals are perpendicular to the line joining the nuclei.

Hence, only sidewise overlapping takes place.

Hint

Explanation for the correct option
(a ) 1
– \(\mathrm{n}=3, \mathrm{l}=2, \mathrm{~m}=+2\) denotes a \(3 \mathrm{~d}\) orbital containing the magnetic quantum number +2 .
– For \(\mathrm{n}=3, \mathrm{l}=2\), we know that \(\mathrm{m}=-2,-1,0,+1,+2\)
– No two orbitals have the same magnetic quantum number.
– As a result, the provided quantum number is only achievable for one orbital and two electrons.
– Hence this option is correct.

Explanation for the incorrect options:
(d) 4
– There are no two orbitals with the same quantum number.
– Hence no 4 orbitals can have the same quantum numbers.
– This option is incorrect.
(c) 3
– There are no two orbitals with the same quantum number.
– Hence no 3 orbitals can have the same quantum numbers.
– This option is incorrect.
(b) 2
– There are no two orbitals with the same quantum number.
– Hence no 2 orbitals can have the same quantum numbers.
– This option is incorrect.

Hint

(d)
\(
\begin{aligned}
E=\frac{h c}{\lambda} \Rightarrow \lambda &=\frac{6 \cdot 6 \times 10^{-34} \times 3 \times 10^{8}}{3.03 \times 10^{-19}} \\
&=656 \mathrm{~nm}
\end{aligned}
\)

Hint

(a)  Species having same no. of electrons are called isoelectronics.

The no. of electrons in \(\mathrm{CO}=\mathrm{CN}^{-}=\mathrm{NO}^{+}=\mathrm{C}_{2}^{2-}=14\).
So these are isoelectronics

Hint

(a) \(\Delta x \times \Delta p=\frac{n}{4 \pi}\)
(Heisenberg uncertainty principle)
\(
\Rightarrow \Delta x=\frac{6.62 \times 10^{-34}}{4 \times 3.14 \times 10^{-5}}=5.27 \times 10^{-30} \mathrm{~m}
\)

Hint

(d)  Sommerfield modified Bohr’s theory considering that in addition to circular orbits electrons also move in elliptical orbits

Hint

\(
\begin{aligned}
&\text { (c) } \lambda=\frac{h}{m v}=\frac{6.63 \times 10^{-27} \mathrm{erg} \mathrm{sec}}{1 \mathrm{~g} \times 10^{4} \mathrm{~cm} / \mathrm{s}} \\
&=6.63 \times 10^{-31} \mathrm{~cm}=6.63 \times 10^{-33} \mathrm{~m}
\end{aligned}
\)

Hint

(d) for \(n^{\text {th }}\) orbit of ‘ \(\mathrm{H}^{\prime}\) atom, \(r_{n}=n^{2} \times r_{1}\) \(\Rightarrow\) radius of \(2^{\text {nd }}\) Bohr’s orbit.
\(
r_{2}=4 \times r_{1}=4 \times 0.530=2.120 Å
\)

Hint

(d)  According to uncertainty principle the product of uncertainty in position and uncertainty in momentum is constant for a particle.

i.e., \(\Delta x \times \Delta p=\frac{h}{4 \pi}\)

As, \(\Delta x=1.0 \mathrm{~nm}\) for both electron and helium atom, so \(\Delta p\) is also same for both the particles.
Thus uncertainty in momentum of the helium atom is also \(5.0 \times 10^{-26} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}\).

Hint

(a)  Since both \(\mathrm{CO}\) and \(\mathrm{CN}^{-}\)have 14 electrons, therefore these are isoelectronic (i.e. having same number of electrons).

Hint

(b)  The longest wavelength means the lowest energy. We know that relation for wavelength
\(
\frac{1}{\lambda}=R_{H}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)
\)

here, \(n_1=2, n_2=3\)
\(R_{\mathrm{H}}\left(\right.\) Rydberg constant \(=109677 \mathrm{~cm}^{-1}\)
\(
\frac{1}{\lambda}=109677\left(\frac{1}{(2)^2}-\frac{1}{(3)^2}\right)=15233
\)
or, \(\lambda=\frac{1}{15233}=6.56 \times 10^{-5} \mathrm{~cm}=6.56 \times 10^{-7} \mathrm{~m}=656 \mathrm{~nm}\)

Hint

(d)  Energy of an atom when \(n=1\)
\(
E_{1}=-\frac{1312}{(1)^{2}}=-1312 \mathrm{~kJ} \mathrm{~mol}^{-1}
\)
Similarly energy when \(n=3,\left(E_{3}\right)=-\frac{1312}{(3)^{2}}\) \(=-145.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
The energy absorbed when an electron jumps from \(n=1\) to \(n=3\)
\(E_{3}-E_{1}=-145.7-(-1312)=1166.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
\(
\begin{aligned}
&=\frac{1166.3}{6.023 \times 10^{23}}=193.6 \times 10^{-23} \mathrm{~kJ} \\
&=193.6 \times 10^{-20} \mathrm{~J}\left[1 \mathrm{Joule}=10^{7} \mathrm{ergs}\right] \\
&\Rightarrow 193.6 \times 10^{-13} \mathrm{ergs}=0.1936 \times 10^{-10} \mathrm{ergs}
\end{aligned}
\)

Hint

(c) Mass of an electron \((\mathrm{m})=9.1 \times 10^{-28} \mathrm{~g}\)

Velocity of electron \((v)=3 \times 10^{4} \mathrm{~cm} / \mathrm{s}\)
Accuracy \(=0.001 \%=\frac{0.001}{100}\) and

Planck’s constant \((h)=6.626 \times 10^{-27}\) erg-second
We know that actual velocity of the electron \((\Delta v)=3 \times 10^{4} \times \frac{0.001}{100}=0.3 \mathrm{~cm} / \mathrm{s}\)

Hint

(c)  Due to ground state, state of hydrogen atom \((n)=1\)
Radius of hydrogen atom \((r)=0.53 Å\)
Atomic no. of \(\mathrm{Li}(Z)=3\)
Now, radius of Li \({ }^{2+}\) ion \(=r \times \frac{n^{2}}{7}=0.53 \times \frac{(1)^{2}}{3}=0.17 Å\)

Hint

(b) Energy of electron depends on the value of \((n+l)\). The subshell are \(3 d, 4 d, 4 p\) and \(5 s, 4 d\) has highest energy.

Hint

(a) The number of electrons in \(\mathrm{O}^{2-}, \mathrm{N}^{3-}, \mathrm{~F^{-}}\) and \(\mathrm{Na}^{+}\)is 10 each. but number of electrons in \(\mathrm{Tl}^{+}\)is 80 .

Hint

(b) Atomic No. of \(\mathrm{Ca}=20\)
Electronic configuration of \(\mathrm{Ca}=[\mathrm{Ar}] 4 s^{2}\)

Hint

(c) Energy of an electron in \(n^{\text {th }}\) Bohr orbit of hydrogen atom \(=\frac{-13.6}{n^{2}} \mathrm{eV}\)

Hint

(d) This is a Pauli exclusion principle.

Hint

(d) \(l=3\) means \(f\)-subshell
Maximum no. of electrons in \(f\)-subshell \(=14\)

Hint

(b) As per Aufbau Principle.

The principle states : In the ground state of the atoms, the orbitals are filled in order of their increasing energies.

Hint

(b) Electronic configuration of \(\mathrm{Cu}\)
\(
=1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10} 4 s^1
\)

Hint

(c)  \(n=2, l=1\)
It means \(2 p\)-orbitals
Total no. of electrons that can be accomodated in \(2 p\) orbitals \(=6\)

Hint

(a) Electronic configuration of \(\mathrm{Cu}^{+}=[\mathrm{Ar}] 3 d^{10}\)

Hint

 (d)  It is uncertainty principle and not Bohr’s postulate.

Hint

(c) \(\mathrm{N}^{2+}=1 s^{2} 2 s^{2} 2 p_{x}^{1}\)
No. of unpaired electrons \(=1\)

Hint

(b)  No. of orbitals in a subshell \(=2 l+1\)
No. of electrons \(=2(2 l+1)=4 l+2\)

Hint

 (b) Both \(\mathrm{He}\) and \(\mathrm{Li}^{+}\) contain 2 electrons each.

Hint

(a)  No. of radial nodes in \(3 p\)-orbital \(=n-l-1\)
\(
=3-1-1=1
\)

Hint

(a) Radius of \(n^{\mathrm{th}}\) orbit of H-atom \(=r_{0} n^{2}\) where \(r_{0}=\) radius of the first orbit

Hint

-There are many waves that exist in this universe with different wavelengths and different frequencies. The waves with higher frequency have less wavelength and vice versa.
-Based on this, the spectrum is classified mainly as cosmic, gamma, X-ray, UV, visible light, infrared, microwave and radiowave region. The wavelength of visible light ranges from \(380 \mathrm{~nm}\) to \(780 \mathrm{~nm}\). Only the wavelengths of these ranges can be seen by the human eye.
-Hydrogen atom consists of 1 proton and 1 electron. The energy state of the orbitals of the atom was given by Bohr in his atomic model and the spectral emission is based on Schrodinger’s equation.
-Electrons jump from higher energy level to lower energy level releasing energy in form of spectral emissions. The spectral lines are grouped together to form the spectral series. Hydrogen shows 4 such series named Lyman, Balmer, Brackett and Paschen.
-The energy differences and the wavelengths of the emitted photons is related by Rydberg formula as
\(
\frac{1}{\lambda}=Z^2 R\left(\frac{1}{n_1{ }^2}-\frac{1}{n_2^2}\right)
\)
Where \(\mathrm{Z}\) is the atomic number, \(\mathrm{n}\) is the principal quantum number and the numbers 1 and 2 represent the lower and higher energy levels and \(\mathrm{R}\) is the Rydberg’s constant.
-Different series are obtained by changing the values of \(n\) as 1,2,3 and 4 for hydrogen. The value of \(\mathrm{R}\) for hydrogen is \(1.09677 \times 10^7 \mathrm{~m}^{-1}\).
-Putting the values of \(R\) and \(Z\), we can find the values of the wavelengths for different values of the energy levels and by doing so, it is observed that the range of wavelengths for visible light is obtained in the Balmer series only.
-The range for Lyman series starts from \(121 \mathrm{~nm}\) and decreases further, Balmer series starts from 656nm, Paschen series starts from \(1875 \mathrm{~nm}\) and Brackett series starts from \(4051 \mathrm{~nm}\).
Therefore we see that the Balmer series falls under the visible region in the hydrogen spectrum and the correct option is b.

Hint

Bohr radius, \(a_0=52.9 p m n=2, r_n=n^2 a_0=(2)^2 a_0=4 \times 52.9 p m=211.6 p m\)
The angular momentum of an electron in a given stationary state can be expressed as in equation, \(m v r=n \cdot \frac{h}{2 \pi}=2 \times \frac{h}{2 \pi}=\frac{h}{\pi} \dots(i)\)
de-Broglie equation \(mvr\) \(\pi=h \lambda=\frac{h}{m v} ; {\lambda} m v=h \dots(ii)\)

From equations, (i) and (ii), we get \(\lambda=\pi r\)
Putting the value of \(r, \lambda=211.6 \pi \mathrm{pm}\)

Hint

(d)

Orbital having angular node \((\ell)=3\)
Total node \(=\) Radial node + angular node
\(
\begin{array}{l}
=n-\ell-1+\ell \\
3=n-1 \\
n=4
\end{array}
\)

Subshell ” \(n \ell\) ” \(=4 f\)

Hint

 (a)

The \((n+l)\) values are given below,
\(
\begin{array}{l}
4 d=4+2=6 \\
5 p=5+1=6 \\
5 f=5+3=8 \\
6 p=6+1=7
\end{array}
\)

The correct order of energy would be:
\(
5 f>6 p>5 p>4 d
\)

Hint

(a)
\(
\begin{array}{l}
\lambda=\frac{\mathrm{c}}{\mathrm{v}}\\
\lambda=\frac{3 \times 10^8}{1368 \times 10^3}=219.298 \mathrm{~m}=219.3 \mathrm{~m}
\end{array}
\)

Hint

(b)

\(
\mathrm{n}=5, \ell=2, \mathrm{~m_l}=-2,-1,+1,+2, \mathrm{~m}_{\mathrm{s}}=+\frac{1}{2}
\)

Hint

(a) Statement I is true and Statement II is false (From NCERT)

Hint

(c) Statements B, C & E are correct.
(B) Mass of the electron is \(9.10939 \times 10^{-31} \mathrm{~kg}\)
(C) All the isotopes of given elements show same chemical properties.
(E) Dalton’s atomic theory, regarded the atom as an ultimate particle of matter.

Hint

(d) Number of permissible values of magnetic quantum number for a given value of azimuthal quantum \((l)\)
\(
\begin{aligned}
& \Rightarrow \mathrm{n}_{\mathrm{m}}=2 \ell+1 \\
& \Rightarrow \ell=\frac{\mathrm{n}_{\mathrm{m}}-1}{2}
\end{aligned}
\)

Hint

(a)

\(
\begin{array}{|c|c|c|}
\hline n & l & \text { Subshell notation } \\
\hline 2 & 0 & 2 s \\
\hline 2 & 1 & 2 p \\
\hline 3 & 0 & 3 s \\
\hline 3 & 1 & 3 p \\
\hline 3 & 2 & 3 d \\
\hline
\end{array}
\)

Hint

(a) Energy of one photon \(=\frac{h c}{\lambda}(\lambda=300 \mathrm{~nm})\)
For one mole photons, \(E=\frac{h c}{\lambda} \times N_A\)
\(
\begin{aligned}
& E=\frac{6.626 \times 10^{-34} \times 3 \times 10^8 \times 6.023 \times 10^{23}}{300 \times 10^{-9}} \\
& E=3.99 \times 10^5 \mathrm{~J} \mathrm{~mol}^{-1}
\end{aligned}
\)

Kinetic energy \(=1.68 \times 10^5 \mathrm{~J} \mathrm{~mol}^{-1}\)
\(
\begin{aligned}
& W_0=E-K . E . \\
& =3.99 \times 10^5-1.68 \times 10^5 \\
& =2.31 \times 10^5 \mathrm{~J} \mathrm{~mol}^{-1}
\end{aligned}
\)

Hint

(d) The shapes of \(d_{x y}, d_{y z}\) and \(d\) orbitals are similar to each other; and \(d_{x^2-y^2}\) and \(d_z^2\) are similar to each other.

Hint

(a)

\(
\begin{aligned}
&r_n \propto \frac{n^2}{Z}\\
&\begin{aligned}
& \frac{r_3\left(L i^{2+}\right)}{r_2\left(H e^{+}\right)}=\frac{\left(n_3\right)^2}{Z\left(L i^{2+}\right)} \times \frac{Z\left(H e^{+}\right)}{\left(n_2\right)^2} \\
& \frac{r_3\left(L i^{2+}\right)}{105.8}=\frac{(3)^2}{3} \times \frac{2}{(2)^2} \\
& =105.8 \times \frac{3}{2} \\
& r_3\left(L i^{2+}\right)=158.7 \mathrm{~pm}
\end{aligned}
\end{aligned}
\)

Hint

(b)

(I) \(n=4, l=2, m_l=-2, s=-\frac{1}{2}\); represents \(4 d(n+l=6)\)
(II) \(n=3, l=2, m_l=1, s=+\frac{1}{2}\); represents \(3 d(n+l=5)\)
(III) \(n=4, l=1, m_l=0, s=+\frac{1}{2}\); represents \(4 p(n+l=5)\)
(IV) \(n=3, l=1, m_l=-1, s=+\frac{1}{2}\); represents \(3 p(n+l=4)\)

Order of energy depends on the \((n+l)\), greater is the \((n+l)\) value greater is the energy, if \((n+l)\) is same, then it depends on \(n\); if ‘ \(n\) ‘ is more, energy is more.

Step-1: According to \(( n + l )\)
Energy \(=\) (I) \(>\) (II) \(=\) (III) \(>\) (IV)
Step-2: If \(n \uparrow\), then energy increases
Energy \(=\) (I) \(>\) (III) \(>\) (II) \(>\) (IV)

Hint

(c)

\(
\begin{aligned}
&\text { For Balmer series }\\
&\begin{aligned}
n _1 & =2 \\
n _2 & =3 \\
\bar{v} & =\frac{1}{\lambda}=R_H\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right] \\
\bar{v} & =R_H\left[\frac{1}{4}-\frac{1}{9}\right] \\
\bar{v} & =\frac{5 R_H}{36} cm^{-1}
\end{aligned}
\end{aligned}
\)

Hint

(a)

\(
E_n=-R_H\left(\frac{z^2}{n^2}\right) J
\)
For \(He ^{+}(n=1)\),
\(
\begin{aligned}
& E_n=-x=-R_H\left(\frac{2^2}{1^2}\right)=-4 R_H \\
& \therefore R_H=\frac{x}{4}
\end{aligned}
\)
For \(B e^{3+}(n=2)\),
\(
\begin{aligned}
& E_n=-R_H\left(\frac{Z^2}{n^2}\right) J \\
& =-\frac{x}{4} \times\left(\frac{4 \times 4}{2 \times 2}\right)=-x J
\end{aligned}
\)

Hint

(b) To match List I (Quantum Numbers) with List II (Information provided), we need to understand what each quantum number represents:
1. Principal Quantum Number \((n)\) : This quantum number determines the size and energy level of the orbital. An increase in \(n\) implies a higher energy level and a larger orbital size.
2. Azimuthal Quantum Number ( \(\ell\) ): (not directly listed but related to \(m_l\) because \(m _l\) depends on \(\ell\) ): This quantum number defines the shape of the orbital. Different values of \(\ell\) correspond to different shapes (s, p, d, f, etc.).
3. Magnetic Quantum Number \(\left(m_l\right)\) : This quantum number describes the orientation of the orbital in space. It can take values from \(-\ell\) to \(\ell\), where each value corresponds to a specific orientation of the orbital.
4. Spin Quantum Number \(\left( m _{ s }\right)\) : This quantum number specifies the orientation of the spin of the electron. The two possible values are \(+\frac{1}{2}\) and \(-\frac{1}{2}\), representing the two possible spin orientations (up or down).
Matching the given options:
A. \(m _l\) – should be matched with “Orientation of orbital” (III).
B. \(m _{ s }\) – should be matched with “Orientation of spin of electron” (IV).
C. I – This is likely intended to be \(\ell\), and would be matched with “Shape of orbital” (I).
D. \(n\) – relates to “Size of orbital” (II), as it affects the distance from the nucleus along with the energy level.

Using the understanding above, we can identify the correct matches:
A-III (Orientation of orbital)
B-IV (Orientation of spin of electron)
C-I (Shape of orbital)
D-II (Size of orbital)
Therefore, the correct answer is:

Option B: A-III, B-IV, C-I, D-II