The set of quantum numbers for the 23rd electron of chromium (atomic number \( = 24 \)):

Chromium has an atomic number of 24, meaning it has 24 electrons. Its electronic configuration is:
\( \text{Cr: } (Ar) 3d^5 4s^1 \)

To determine the set of quantum numbers for the 23rd electron:
– The 24th electron goes into the \( 4s \) orbital.
– Therefore, the 23rd electron is in the \( 3d \) subshell.

The quantum numbers for an electron in the \( 3d \) subshell are:
– \( n = 3 \) (principal quantum number)
– \( l = 2 \) (azimuthal quantum number for \( d \) orbitals)
– \( m_l \) can be \(-2, -1, 0, +1, +2\) (magnetic quantum number)
– \( m_s = +\frac{1}{2} \) or \(-\frac{1}{2} \) (spin quantum number)

Therefore, the set of quantum numbers can be:
\( (3, 2, -2, +\frac{1}{2}) \)

So the correct answer is:
\( \boxed{(A) \ 3, 2, -2, +\frac{1}{2}} \)

In two different H-atoms (A and B) electrons are revolving in orbits of radius \( r \) and \( 4r \) respectively. The ratio of the times required for one complete revolution by the electrons in the respective atoms is:

The time period \( T \) of an electron in orbit is proportional to \( r^{3/2} \).

For atom A:
\( r_A = r \)

For atom B:
\( r_B = 4r \)

The ratio of the times required for one complete revolution:
\( \frac{T_A}{T_B} = \frac{r_A^{3/2}}{r_B^{3/2}} = \frac{r^{3/2}}{(4r)^{3/2}} = \frac{r^{3/2}}{4^{3/2} r^{3/2}} = \frac{1}{8} \)

So, the correct answer is:
\( \boxed{(C) \ 1:8} \)

The distance traversed by an electron is numerically equal to its de Broglie wavelength. The velocity of the electron is:

According to the de Broglie hypothesis:
\( \lambda = \frac{h}{mv} \)

If the distance \( s \) traversed by the electron is equal to the de Broglie wavelength \( \lambda \), then:
\( s = \lambda = \frac{h}{mv} \)

We need to express \( v \) in terms of known quantities. Let’s consider the energy relation:
\( KE = \frac{1}{2}mv^2 \)

Rewriting in terms of velocity:
\( v = \sqrt{\frac{2 \cdot KE}{m}} \)

Given that the distance \( s \) is equal to the de Broglie wavelength:
\( \lambda = s = \frac{h}{mv} \)

Therefore, the correct option for velocity is:
\( \boxed{(D) \ \sqrt{\frac{h}{2(K E)}}} \)

The uncertainties in the measurement of position and velocity of a particle are equal. So uncertainty in the measurement of its velocity is:

Using Heisenberg’s uncertainty principle:
\( \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \)

Since the uncertainties in position (\( \Delta x \)) and velocity (\( \Delta v \)) are equal:
\( \Delta x = \Delta v \)

Therefore:
\( \Delta x \cdot m \Delta v \geq \frac{h}{4\pi} \)

Given \( \Delta x = \Delta v \):
\( (\Delta v)^2 \cdot m \geq \frac{h}{4\pi} \)

\( (\Delta v)^2 \geq \frac{h}{4\pi m} \)

\( \Delta v \geq \sqrt{\frac{h}{4\pi m}} \)

So the correct answer is:
\( \boxed{(C) \ \frac{1}{2} \sqrt{\frac{h}{\pi m}}} \)

The ratio of the energies of the electron of H-atom in the ground state and that of the \( \mathrm{Be}^{3+} \) ion in the first excited state is:

The energy of an electron in a hydrogen-like atom is given by:
\( E_n = -13.6 \frac{Z^2}{n^2} \)

For the H-atom in the ground state (\( Z = 1, n = 1 \)):
\( E_H = -13.6 \text{ eV} \)

For the \( \mathrm{Be}^{3+} \) ion in the first excited state (\( Z = 4, n = 2 \)):
\( E_{Be^{3+}} = -13.6 \times \frac{4^2}{2^2} = -13.6 \times \frac{16}{4} = -13.6 \times 4 = -54.4 \text{ eV} \)

The ratio of the energies of the H-atom in the ground state to the \( \mathrm{Be}^{3+} \) ion in the first excited state is:
\( \frac{E_H}{E_{Be^{3+}}} = \frac{-13.6 \text{ eV}}{-54.4 \text{ eV}} = \frac{13.6}{54.4} = \frac{1}{4} \)

So the correct answer is:
\( \boxed{(A) 1:4} \)

The energy of the electron in the ground state of the hydrogen atom is \(-13.6 \text{ eV}\). So, the energy of the electron revolving in the 5th orbit of the H-atom will be:

The energy of an electron in a hydrogen-like atom is given by:
\( E_n = -\frac{13.6 \text{ eV}}{n^2} \)

For the 5th orbit (\(n = 5\)):
\( E_5 = -\frac{13.6 \text{ eV}}{5^2} = -\frac{13.6}{25} \text{ eV} = -0.544 \text{ eV} \)

So the correct answer is:
\( \boxed{(A) \ -0.54 \text{ eV}} \)

The uncertainty in the momentum of an electron is \(1.0 \times 10^{-5} \text{ kg m/s}\). The uncertainty in its position is calculated using Heisenberg’s uncertainty principle:

\( \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \)

Given:
\( \Delta p = 1.0 \times 10^{-5} \text{ kg m/s} \)
\( h = 6.625 \times 10^{-34} \text{ kg m}^2/\text{s} \)

\( \Delta x \geq \frac{h}{4\pi \Delta p} = \frac{6.625 \times 10^{-34}}{4\pi \times 1.0 \times 10^{-5}} \)

\( \Delta x \geq \frac{6.625 \times 10^{-34}}{4 \times 3.14159 \times 10^{-5}} \)

\( \Delta x \geq \frac{6.625 \times 10^{-34}}{12.56636 \times 10^{-5}} \)

\( \Delta x \geq \frac{6.625 \times 10^{-34}}{1.256636 \times 10^{-4}} \)

\( \Delta x \geq 5.27 \times 10^{-30} \text{ m} \)

So the correct answer is:
\( \boxed{(C) \ 5.27 \times 10^{-30} \text{ m}} \)

In \( \mathrm{He}^{+} \) ion the distance of the electron from the nucleus:

The Bohr radius \(a_0\) for hydrogen is 52.9 pm. For a hydrogen-like ion with atomic number \(Z\), the radius of the \(n\)-th orbit is given by:
\( r_n = \frac{n^2 a_0}{Z} \)

For \( \mathrm{He}^{+} \) (\(Z = 2\)), in the ground state (\(n = 1\)):
\( r_1 = \frac{1^2 \times 52.9 \text{ pm}}{2} = \frac{52.9}{2} \text{ pm} = 26.45 \text{ pm} \)

So the correct answer is:
\( \boxed{(C) \ 26.5 \text{ pm}} \)

The ratio between kinetic energy and the total energy of the electron of hydrogen atom according to Bohr’s model:

In Bohr’s model, the total energy \(E\) is given by:
\( E = -\frac{K}{2} \)

Where \(K\) is the kinetic energy. Hence,
\( K = -2E \)

Thus, the ratio of kinetic energy to total energy is:
\( \frac{K}{E} = \frac{-2E}{E} = -2 \)

So the correct answer is:
\( \boxed{(A) \ 2:1} \)

The wavelength of the photons of a radiation of wave number \( 8 \times 10^{15} \text{ s}^{-1} \).

Wave number (\( \nu \)) is typically given in \( \text{m}^{-1} \) but here, it seems like the given value is in \( \text{s}^{-1} \), which is the frequency.

The formula to find the wavelength (\( \lambda \)) from frequency (\( \nu \)) is:
\( \lambda = \frac{c}{\nu} \)

Given:
\( \nu = 8 \times 10^{15} \text{ s}^{-1} \)
\( c = 3.0 \times 10^8 \text{ m/s} \)

Calculate \(\lambda\):
\( \lambda = \frac{3.0 \times 10^8 \text{ m/s}}{8 \times 10^{15} \text{ s}^{-1}} \)
\( \lambda = \frac{3.0}{8} \times 10^{-7} \text{ m} \)
\( \lambda = 0.375 \times 10^{-7} \text{ m} \)
\( \lambda = 3.75 \times 10^{-8} \text{ m} \)
\( \lambda = 37.5 \text{ nm} \)

Given the options, the correct answer is closest to:
\( \boxed{(D) \ 4 \times 10^{1} \text{ nm}} \)

The magnetic moment (\(\mu\)) for an ion is given by the formula:
\( \mu = \sqrt{n(n+2)} \text{ BM} \)
where \(n\) is the number of unpaired electrons.

Let’s determine the number of unpaired electrons for each ion:

\( \mathrm{Mn}^{2+} \): Mn has an atomic number of 25. The electron configuration is \((Ar) 3d^5 4s^2\). For \( \mathrm{Mn}^{2+} \), we lose the 4s electrons and have 3d^5, which has 5 unpaired electrons.
\( \mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \text{ BM} \)

\( \mathrm{Sc}^{3+} \): Sc has an atomic number of 21. The electron configuration is \((Ar) 3d^1 4s^2\). For \( \mathrm{Sc}^{3+} \), we lose all the 3d and 4s electrons, leaving no unpaired electrons.
\( \mu = \sqrt{0(0+2)} = 0 \text{ BM} \)

\( \mathrm{Ti}^{3+} \): Ti has an atomic number of 22. The electron configuration is \((Ar) 3d^2 4s^2\). For \( \mathrm{Ti}^{3+} \), we lose the 4s electrons and one 3d electron, leaving one unpaired electron.
\( \mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \text{ BM} \)

\( \mathrm{Cr}^{2+} \): Cr has an atomic number of 24. The electron configuration is \((Ar) 3d^5 4s^1\). For \( \mathrm{Cr}^{2+} \), we lose the 4s and one 3d electron, leaving 4 unpaired electrons.
\( \mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \text{ BM} \)

So, \( \mathrm{Mn}^{2+} \) has the highest magnetic moment.

The correct answer is:
\( \boxed{(A) \ \mathrm{Mn}^{2+}} \)

The angular momentum of the electron in the first excited state in H-atom is:

The angular momentum \(L\) of an electron in a hydrogen atom is given by:
\( L = n \frac{h}{2\pi} \)

For the first excited state (\(n = 2\)):
\( L = 2 \frac{h}{2\pi} = \frac{h}{\pi} \)

The correct answer is:
\( \boxed{(D) \ \frac{h}{\pi}} \)

If the wavelength of the line having least value in the Lyman series of hydrogen spectrum be \( x \), then the wavelength of the line having highest value in the Balmer series of \( \mathrm{Li}^{2+} \) ion will be:

For the Lyman series in hydrogen, the least wavelength corresponds to the transition from \( n = \infty \) to \( n = 1 \). For the Balmer series, the longest wavelength corresponds to the transition from \( n = 3 \) to \( n = 2 \).

For \( \mathrm{Li}^{2+} \) (\(Z = 3\)), the wavelength is scaled by \( \frac{1}{Z^2} \).

So the ratio of wavelengths in hydrogen to that in \( \mathrm{Li}^{2+} \):
\( \lambda_{Li^{2+}} = \frac{\lambda_{H}}{Z^2} \)

For the Balmer series, using the longest wavelength (which is 4 times the shortest wavelength in the Lyman series):
\( \lambda_{H(Balmer)} = 4 \lambda_{H(Lyman)} = 4x \)

Thus:
\( \lambda_{Li^{2+}} = \frac{4x}{3^2} = \frac{4x}{9} \)

So the correct answer is:
\( \boxed{(D) \ \frac{4x}{9}} \)

The de Broglie wavelength of a cricket ball of mass \( 0.5 \text{ kg} \) moving with a speed of \( 100 \text{ m/s} \) is:

The de Broglie wavelength (\(\lambda\)) is given by:
\( \lambda = \frac{h}{mv} \)

Given:
\( h = 6.625 \times 10^{-34} \text{ J s} \)
\( m = 0.5 \text{ kg} \)
\( v = 100 \text{ m/s} \)

\( \lambda = \frac{6.625 \times 10^{-34}}{0.5 \times 100} \)
\( \lambda = \frac{6.625 \times 10^{-34}}{50} \)
\( \lambda = 1.325 \times 10^{-35} \text{ m} \)

So the correct answer is:
\( \boxed{(B) \ 1.32 \times 10^{-35} \text{ m}} \)

If de Broglie wavelength of an electron revolving in the 4th Bohr orbit be \( \lambda \), then the circumference of this orbit will be:

For an electron in the \( n \)-th Bohr orbit, the circumference is given by:
\( 2\pi r = n \lambda \)

For the 4th Bohr orbit (\(n = 4\)):
\( 2\pi r = 4 \lambda \)

So the correct answer is:
\( \boxed{(D) \ 4 \lambda} \)

When an electron drops from a higher energy state \( n \) to the ground state in an atom, the maximum number of spectral lines that can be obtained is:

The maximum number of spectral lines that can be obtained when an electron transitions from a higher energy state \( n \) to the ground state is given by the formula:

\( \frac{n(n-1)}{2} \)

This formula arises because each possible transition between two energy levels corresponds to a spectral line, and the total number of such transitions is the combination of \( n \) levels taken 2 at a time.

So the correct answer is:
\( \boxed{(D) \frac{n(n-1)}{2}} \)

Wave number corresponding to the spectral line having the highest value is obtained from the Rydberg equation \( \left(\bar{v}=R_{H} Z^{2}\left(\frac{1}{n_{j}^{2}}-\frac{1}{n_{i}^{2}}\right)\right) \) by putting \( n_{i} \) equal to:

The wave number (\(\bar{v}\)) is highest when the difference in energy levels is greatest. For the Rydberg equation:

\( \bar{v} = R_H Z^2 \left( \frac{1}{n_j^2} – \frac{1}{n_i^2} \right) \)

The highest value occurs when \( \frac{1}{n_i^2} \) is minimized, which is when \( n_i \) approaches infinity.

So the correct answer is:
\( \boxed{(D) \ \infty} \)

The wavelength of any spectral line for an electronic transition in the spectrum is inversely proportional to:

The wavelength of a spectral line (\(\lambda\)) is inversely proportional to the difference in energy levels (\(\Delta E\)). According to the energy-wavelength relationship:

\( \Delta E = \frac{hc}{\lambda} \)

So, the correct answer is:
\( \boxed{(C) \ \text{the difference in the energy levels involved in the transition}} \)

Transition of an electron from \( n=3 \) level to \( n=2 \) level results in a spectral line in the:

For hydrogen, transitions to the \( n=2 \) level are part of the Balmer series, which is in the visible region of the electromagnetic spectrum.

So the correct answer is:
\( \boxed{(B) \ \text{visible region}} \)

The value of uncertainty calculated from the equation \( \Delta x \cdot \Delta p = \frac{h}{4\pi} \) is:

The Heisenberg Uncertainty Principle states that the product of the uncertainties in position (\(\Delta x\)) and momentum (\(\Delta p\)) is at least \( \frac{h}{4\pi} \).

So the correct answer is:
\( \boxed{(A) \ \text{lowest value}} \)

In the ground state of \( M^{2+} \) ion, the last electron has the four quantum numbers: \( 3, 2, -2 \) and \( +\frac{1}{2} \). The atomic number of \( M \) is:

The quantum numbers \( n=3, l=2, m_l=-2, m_s=+\frac{1}{2} \) correspond to a 3d orbital. This configuration indicates that the electron was removed from a 3d orbital. To find the atomic number:

– For \( M^{2+} \), the configuration without the two lost electrons would be:
\( (\text{Ar}) 3d^6 \)
– Therefore, \( M \) has 26 electrons (since it lost 2 to become \( M^{2+} \)).
– So, the atomic number is 26.

The correct answer is:
\( \boxed{(D) \ 26} \)

The velocity of an electron with de Broglie wavelength 66 nm is \( \left(h=6.6 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-1}\right)- \

The de Broglie wavelength (\(\lambda\)) is related to the momentum (\(p\)) by:

\( \lambda = \frac{h}{p} \)

And momentum \( p \) is related to mass (\(m\)) and velocity (\(v\)) by:

\( p = mv \)

Combining these, we get:

\( \lambda = \frac{h}{mv} \)
\( v = \frac{h}{m \lambda} \)

For an electron (\( m_e = 9.1 \times 10^{-31} \, \text{kg} \)) and \(\lambda = 66 \times 10^{-9} \, \text{m} \):

\( v = \frac{6.6 \times 10^{-34} \, \text{kg m}^2/\text{s}}{9.1 \times 10^{-31} \, \text{kg} \times 66 \times 10^{-9} \, \text{m}} \)
\( v \approx 1.1 \times 10^7 \, \text{ms}^{-1} \)

So the correct answer is:
\( \boxed{(D) \ 1.1 \times 10^7 \, \text{ms}^{-1}} \)

The line with highest wavelength in the Paschen series of hydrogen spectrum \( \left(R=109670 \, \text{cm}^{-1}\right) \):

The Paschen series corresponds to transitions to the \( n=3 \) level. The highest wavelength (lowest energy transition) in this series is from \( n=4 \) to \( n=3 \).

Using the Rydberg formula:
\( \frac{1}{\lambda} = R \left( \frac{1}{3^2} – \frac{1}{4^2} \right) \)
\( \frac{1}{\lambda} = 109670 \left( \frac{1}{9} – \frac{1}{16} \right) \)
\( \frac{1}{\lambda} = 109670 \left( \frac{16 – 9}{144} \right) \)
\( \frac{1}{\lambda} = 109670 \left( \frac{7}{144} \right) \)
\( \lambda \approx 18760 \, \text{Å} \)

So the correct answer is:
\( \boxed{(C) \ 18760 \, \text{Å}} \)

The wave number of 1st Balmer line in hydrogen spectrum is \( 15200 \, \text{cm}^{-1} \). The wave number of 1st Balmer line in the spectrum of \( \text{Li}^{2+} \) ion will be:

The wave number (\(\bar{v}\)) for \( \text{Li}^{2+} \) ion can be found by multiplying the wave number for hydrogen by \( Z^2 \), where \( Z \) is the atomic number of lithium (3).

\( \bar{v}_{\text{Li}^{2+}} = \bar{v}_{\text{H}} \times Z^2 \)
\( \bar{v}_{\text{Li}^{2+}} = 15200 \, \text{cm}^{-1} \times 3^2 \)
\( \bar{v}_{\text{Li}^{2+}} = 15200 \, \text{cm}^{-1} \times 9 \)
\( \bar{v}_{\text{Li}^{2+}} = 136800 \, \text{cm}^{-1} \)

So the correct answer is:
\( \boxed{(B) \ 136800 \, \text{cm}^{-1}} \)

In the ground state, the radius of H-atom is 0.53 Å. So its radius in the 1st excited state is:

For hydrogen, the radius of the \( n \)-th orbit is given by:

\( r_n = n^2 \times r_1 \)

For the first excited state (\( n=2 \)):

\( r_2 = 2^2 \times 0.53 \, \text{Å} \)
\( r_2 = 4 \times 0.53 \, \text{Å} \)
\( r_2 = 2.12 \, \text{Å} \)

So the correct answer is:
\( \boxed{(D) \ 2.12 \, \text{Å}} \)

\( X \) and \( Y \) are two isotones with mass numbers 70 and 72 respectively. If the atomic number of \( X \) be 34, the atomic number of \( Y \) will be:

Isotones are atoms with the same number of neutrons.

For \( X \) (with mass number 70 and atomic number 34):
\( \text{Number of neutrons in } X = 70 – 34 = 36 \)

For \( Y \) (

with the same number of neutrons):
\( \text{Number of neutrons in } Y = 36 \)
\( \text{Atomic number of } Y = 72 – 36 = 36 \)

So the correct answer is:
\( \boxed{(A) \ 36} \)

In the \( (n+1) \)-th quantum shell, the values of azimuthal quantum number ‘ \( l \) ‘ will be:

For a given principal quantum number \( n \), the azimuthal quantum number \( l \) can take values from 0 to \( n-1 \).

So, for the \( (n+1) \)-th quantum shell:
\( l = 0 \text{ to } n \)

So the correct answer is:
\( \boxed{(A) \ 0 \text{ to } n} \)

Isosters are molecules or ions that have the same number of atoms and the same number of valence electrons.

\( \text{H}_2\text{O} \text{ and } \text{N}_2\text{O} \)

Both have 10 valence electrons and similar structure.

So the correct answer is:
\( \boxed{(A) \ \text{H}_2\text{O}, \text{N}_2\text{O}} \)

The energy difference between levels is given by:

\( \Delta E = 13.6 \left( \frac{1}{n_f^2} – \frac{1}{n_i^2} \right) \text{eV} \)

The maximum energy absorption occurs for the largest difference between initial and final energy levels, which is:

\( n=1 \text{ to } n=2 \)

So the correct answer is:
\( \boxed{(D) \ n=1 \text{ to } n=2} \)

Paramagnetic character is due to the presence of unpaired electrons. The paramagnetic character decreases with a decrease in the number of unpaired electrons.

\( \text{Mn} > \text{Cr} > \text{Zn} \)

Manganese (Mn) has the highest number of unpaired electrons, followed by Chromium (Cr), and Zinc (Zn) has none.

So the correct answer is:
\( \boxed{(A) \ \text{Mn} > \text{Cr} > \text{Zn}} \)

The atomic spectrum of an ion is similar to another ion if both have the same number of electrons. \( \mathrm{He}^{+} \) has only one electron (like a hydrogen atom).

– \( \mathrm{Li}^{2+} \) has 3 protons and 1 electron (similar to \(\mathrm{He}^{+}\)).

So the correct answer is:
\( \boxed{(B) \ \mathrm{Li}^{2+}} \)

The energy of the electron in the ground state of H-atom is -13.6 eV. Its energy in the 4th quantum shell will be:

The energy of an electron in the \( n \)-th quantum shell of a hydrogen atom is given by:

\( E_n = \frac{E_1}{n^2} = \frac{-13.6 \, \text{eV}}{4^2} = \frac{-13.6 \, \text{eV}}{16} = -0.85 \, \text{eV} \)

So the correct answer is:
\( \boxed{(C) \ -0.85 \, \text{eV}} \)

If the radius of 1st Bohr orbit of H-atom is \( a_0 \), the radius of the 3rd Bohr orbit will be:

The radius of the \( n \)-th orbit in the Bohr model is given by:

\( r_n = n^2 \times r_1 \)

For the 3rd Bohr orbit (\( n = 3 \)):

\( r_3 = 3^2 \times a_0 = 9 \times a_0 \)

So the correct answer is:
\( \boxed{(C) \ 9 \times a_0} \)

\( \mathrm{Fe}^{3+} \) and \( \mathrm{Mn}^{2+} \)

\( \mathrm{Fe}^{3+} : (\text{Ar}) \, 3d^5 \)
\( \mathrm{Mn}^{2+} : (\text{Ar}) \, 3d^5 \)

So the correct answer is:
\( \boxed{(B) \ \mathrm{Fe}^{3+}, \mathrm{Mn}^{2+}} \)

For \( \mathrm{Li}^{2+} \) in the \( n = 2 \) state:

\( r_n = \frac{n^2 \times a_0}{Z} = \frac{2^2 \times a_0}{3} = \frac{4a_0}{3} \)

For \( \mathrm{He}^{+} \) in the \( n = 2 \) state:

\( r_n = \frac{2^2 \times a_0}{2} = 2a_0 \)

For \( \mathrm{Be}^{3+} \) in the \( n = 2 \) state:

\( r_n = \frac{2^2 \times a_0}{4} = a_0 \)

So the correct answer is:
\( \boxed{(D) \ \mathrm{Be}^{3+} (n=2)} \)

The electromagnetic spectrum is ordered from longest to shortest wavelength as follows: infrared < visible < ultraviolet < X-rays < gamma rays < cosmic rays.

So the correct answer is:
\( \boxed{(B) \ \text{cosmic ray}} \)

\( E = \frac{hc}{\lambda} \)

\( \lambda = \frac{hc}{E} = \frac{(6.63 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{3.03 \times 10^{-19} \, \text{J}} \)

\( \lambda \approx 656 \, \text{nm} \)

So the correct answer is:
\( \boxed{(A) \ 656 \, \text{nm}} \)

\( \mathrm{Cu}^{2+} : (\text{Ar}) \, 3d^9 \)
\( \mathrm{Ni}^{2+} : (\text{Ar}) \, 3d^8 \)
\( \mathrm{Cr}^{3+} : (\text{Ar}) \, 3d^3 \)
\( \mathrm{Fe}^{3+} : (\text{Ar}) \, 3d^5 \)

The correct sequence is:

\( \boxed{(B) \ \mathrm{Fe}^{3+} > \mathrm{Cr}^{3+} > \mathrm{Ni}^{2+} > \mathrm{Cu}^{2+}} \)

Energy levels are given by:

\( E_n = \frac{E_1}{n^2} \)

Given \( E_2 = -328 \, \text{kJ/mol} \):

\( E_1 = -328 \times 4 = -1312 \, \text{kJ/mol} \)

For \( n=4 \):

\( E_4 = \frac{-1312}{4^2} = \frac{-1312}{16} = -82 \, \text{kJ/mol} \)

So the correct answer is:
\( \boxed{(D) \ -82 \, \text{kJ/mol}} \)

\( d_{z^2} \) has a unique shape with two lobes along the z-axis and a torus around the middle.

So the correct answer is:
\( \boxed{(D) \ d_{z^2}} \)

For the first Balmer line (\( n=3 \) to \( n=2 \)):

\( \bar{\nu} = R \left( \frac{1}{2^2} – \frac{1}{3^2} \right) \)

\( \bar{\nu} = R \left( \frac{1}{4} – \frac{1}{9} \right) \)

\( \bar{\nu} = R \left( \frac{5}{36} \right) \)

So the correct answer is:
\( \boxed{(D) \ \frac{5R}{36} \, \text{cm}^{-1}} \)

An \( f \)-orbital has 3 nodal planes.

So the correct answer is:
\( \boxed{(C) \ 3} \)

The total energy \( E \) in the \( n \)-th orbit is:

\( E_n = -\frac{13.6 \, \text{eV}}{n^2} \)

The kinetic energy \( K \) is half the magnitude of the total energy but positive:

\( K = -\frac{E_n}{2} \)

For \( n=2 \):

\( E_2 = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \)
\( K = -\frac{-3.4 \, \text{eV}}{2} = 1.7 \, \text{eV} \)

In terms of the given options, using the expression for kinetic energy:

\( K = \frac{h^2}{16 \pi^2 m a_0^2} \)

So the correct answer is:
\( \boxed{(B) \ \frac{h^2}{16 \pi^2 m a_0^2}} \)

For \( n=3 \):

– \( l \) can be 0, 1, or 2.
– For each \( l \), \( m_l \) can have \( (2l + 1) \) values.
– Each \( m_l \) can have 2 \( m_s \) values (\( +\frac{1}{2} \) and \( -\frac{1}{2} \)).

The

maximum number of electrons with \( m_s = -\frac{1}{2} \) is half the total number of electrons in \( n=3 \):

\( 2n^2 = 2 \times 3^2 = 18 \)

Half of these have \( m_s = -\frac{1}{2} \):

\( 18/2 = 9 \)

So the correct answer is:
\( \boxed{(C) \ 9} \)

In Sommerfeld’s model, the orbit of an electron is an ellipse.

So the correct answer is:
\( \boxed{(A) \ \text{perfect ellipse}} \)

Protons = 16 (so element is Sulfur, S)
Electrons = 18 (so the ion has 2 extra electrons, \( S^{2-} \))

So the correct answer is:
\( \boxed{(D) \ \mathrm{S}^{2-}} \)

The \( p_x \) orbital is oriented along the x-axis and has zero probability in the \( yz \)-plane.

So the correct answer is:
\( \boxed{(A) \ p_x} \)

\( E = 242 \times 10^3 \, \text{J/mol} \)
\( E_{\text{photon}} = \frac{242 \times 10^3 \, \text{J/mol}}{6.02 \times 10^{23} \, \text{mol}^{-1}} \approx 4.02 \times 10^{-19} \, \text{J} \)

\( \lambda = \frac{hc}{E} = \frac{(6.63 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{4.02 \times 10^{-19} \, \text{J}} \approx 494 \, \text{nm} \)

So the correct answer is:
\( \boxed{(D) \ 494 \, \text{nm}} \)

\( \Delta v = 600 \times 0.005\% = 600 \times 0.00005 = 0.03 \, \text{m/s} \)

\( \Delta x = \frac{h}{4\pi m \Delta v} = \frac{6.63 \times 10^{-34}}{4\pi (9.1 \times 10^{-31})(0.03)} \approx 1.92 \times 10^{-3} \, \text{m} \)

So the correct answer is:
\( \boxed{(A) \ 1.92 \times 10^{-3} \, \text{m}} \)

To find the electronic configuration of each ion, we start with the neutral atom and then remove electrons corresponding to the charge of the ion:

1. \( \mathrm{Co}^{3+} \):
– Neutral Co: \( (\mathrm{Ar}) 3d^7 4s^2 \)
– Remove 3 electrons: \( (\mathrm{Ar}) 3d^6 \)
– Configuration: \( (\mathrm{Ar}) 3d^6 \)

2. \( \mathrm{Ni}^{3+} \):
– Neutral Ni: \( (\mathrm{Ar}) 3d^8 4s^2 \)
– Remove 3 electrons: \( (\mathrm{Ar}) 3d^7 \)
– Configuration: \( (\mathrm{Ar}) 3d^7 \)

3. \( \mathrm{Mn}^{3+} \):
– Neutral Mn: \( (\mathrm{Ar}) 3d^5 4s^2 \)
– Remove 3 electrons: \( (\mathrm{Ar}) 3d^4 \)
– Configuration: \( (\mathrm{Ar}) 3d^4 \)

4. \( \mathrm{Fe}^{3+} \):
– Neutral Fe: \( (\mathrm{Ar}) 3d^6 4s^2 \)
– Remove 3 electrons: \( (\mathrm{Ar}) 3d^5 \)
– Configuration: \( (\mathrm{Ar}) 3d^5 \)

So the correct answer is:
\( \boxed{(A) \ \mathrm{Co}^{3+}} \)