Structure of Atom

To compare orbital energies, we analyze hydrogen (single-electron system) and lithium (multi-electron system) separately.

(i) Hydrogen atom (single-electron system):
In hydrogen-like species, the energy of an orbital depends only on the principal quantum number $ n $, not on $ l $. Therefore, all orbitals having the same $ n $ possess equal energy.

Hence, for hydrogen:

$ E_{2s}(H) = E_{2p}(H) $

So, Option B is correct.

(ii) Lithium atom (multi-electron system):
In multi-electron atoms, orbital energies depend on the $ (n + l) $ rule. The orbital with lower $ (n + l) $ value has lower energy. If two orbitals have the same $ n + l $, the one with lower $ n $ is lower in energy.

For lithium:

• For $ 2s $: $ n = 2, \, l = 0 \Rightarrow n + l = 2 $
• For $ 2p $: $ n = 2, \, l = 1 \Rightarrow n + l = 3 $

Since $ (n + l)_{2s} < (n + l)_{2p} $, we get:

$ E_{2s}(Li) < E_{2p}(Li) $

This is further supported by the fact that the $ 2s $ orbital penetrates closer to the nucleus and experiences higher effective nuclear charge, making it more stable (lower energy).

Thus, Option A is correct.

(iii) Comparison between hydrogen and lithium:
Energy of an orbital is given by:

$ E \propto -\frac{Z_{\text{eff}}^2}{n^2} $

In lithium, even after shielding by two inner $ 1s $ electrons, the effective nuclear charge experienced by the $ 2s $ electron is greater than that in hydrogen. As a result, the $ 2s $ electron in lithium is more strongly attracted and has lower (more negative) energy.

Therefore:

$ E_{2s}(Li) < E_{2s}(H) $

So, Option D is also correct.

(iv) Checking Option C:
Since $ E_{2p}(H) = E_{2s}(H) $ and $ E_{2s}(Li) $ is lower than $ E_{2s}(H) $, we have:

$ E_{2p}(H) > E_{2s}(Li) $

Thus, Option C is incorrect.

Final Answer: A, B and D

According to Bohr's model of the hydrogen atom, the kinetic energy of an electron in a particular orbit is given by the expression:

$ K.E. = \frac{1}{2}mv^2 $

Where $m$ is the mass of the electron and $v$ is its velocity. However, a more useful expression involving only constants is:

$ K.E. = \frac{13.6 \, \text{eV} \, Z^2}{n^2} $

Where $Z$ is the atomic number and $n$ is the principal quantum number of the orbit. This formula shows that the kinetic energy of the electron is directly proportional to the square of the atomic number $Z$ and inversely proportional to the square of the orbit number $n$.

Let's analyze each option keeping the above formula in mind:

Option A: First orbit of hydrogen ($\mathrm{H}$, $Z = 1$, $n = 1$)

$ K.E. = \frac{13.6 \, \text{eV} \, (1)^2}{(1)^2} = 13.6 \, \text{eV} $

Option B: First orbit of $ \mathrm{He}^{+} $ ($Z = 2$, $n = 1$)

$ K.E. = \frac{13.6 \, \text{eV} \, (2)^2}{(1)^2} = 54.4 \, \text{eV} $

Option C: Second orbit of $ \mathrm{He}^{+} $ ($Z = 2$, $n = 2$)

$ K.E. = \frac{13.6 \, \text{eV} \, (2)^2}{(2)^2} = 13.6 \, \text{eV} $

Option D: Second orbit of $ \mathrm{Li}^{2+} $ ($Z = 3$, $n = 2$)

$ K.E. = \frac{13.6 \, \text{eV} \, (3)^2}{(2)^2} = 30.6 \, \text{eV} $

From the above calculations, it is evident that the highest kinetic energy is associated with the electron in the first orbit of $ \mathrm{He}^{+} $ (Option B), which is $54.4 \, \text{eV}$.

Let's analyze each statement one by one:

Option A: Uncertainty principle rules out the existence of definite paths for electrons.

This statement is correct. The Heisenberg Uncertainty Principle states that it is impossible to simultaneously know both the exact momentum and the exact position of a particle. Therefore, the concept of a definite path for electrons is not applicable according to quantum mechanics.

Option B: The energy of an electron in $2s$ orbital of an atom is lower than the energy of an electron that is infinitely far away from the nucleus.

This statement is also correct. The energy of an electron in any bound state (such as the $2s$ orbital) is always lower (more negative) than that of an electron at an infinite distance from the nucleus (which would have zero energy). This is because an electron infinitely far away from the nucleus is considered to be free from nuclear attraction and thus its potential energy is zero.

Option C: According to Bohr's model, the most negative energy value for an electron is given by $n=1$, which corresponds to the most stable orbit.

This statement is correct as well. In Bohr’s model of the atom, the energy levels of electrons are given by $E_n = -\frac{13.6 eV}{n^2}$. The energy is most negative when $n = 1$, indicating the most stable orbit and thus the lowest energy state (the ground state).

Option D: According to Bohr's model, the magnitude of velocity of electrons increases with increase in values of $n$.

This statement is incorrect. According to Bohr’s model, the speed of an electron in a circular orbit around the nucleus decreases with an increase in the principal quantum number $n$. The velocity $v_n$ is given by the formula: $v_n = \frac{2.18 \times 10^6 m/s}{n}$, where $n$ is the principal quantum number. Hence, as $n$ increases, $v_n$ decreases.

Therefore, the correct statements are:

Option A, Option B, and Option C.

Option (A) is correct.

Option (B) is incorrect.

$\pi^*$ orbital has zero node is the plane which is perpendicular to the molecular axis and goes through the center of molecule.

So, option (C) is incorrect.

$\pi^*$ Orbital has one node in the $x y-$ plane containing the molecular axis.

Option (D) is correct.

(III) Kinetic energy of the electron in nth orbit,

$K.E. = + 13.6 \times {{{Z^2}} \over {{n^2}}}$ or $K.E. \propto {1 \over {{n^2}}}$ or $K.E. \propto {n^{ - 2}}$

From List-II, correct match is (III, P)

(IV) Potential energy of the electron in the nth orbit,

$P.E. = - 2 \times 13.6 \times {{{Z^2}} \over {{n^2}}} \Rightarrow P.E. \propto {1 \over {{n^2}}} \Rightarrow P.E. \propto {n^{ - 2}}$

From List-II, correct match is (IV, P).

Hence, correct matching from List-I and List-II on the basis of given options is (III, P).

(I) Radius of the nth orbit, r = $0.529 \times {{{n^2}} \over Z}$

Here, $r \propto {n^2}$

From List-II, correct match is (I, T)

(II) Angular momentum of the electron, $mvr = {{nh} \over {2\pi }}$ or $mvr \propto n$

From List - II, correct match (II, S)

Hence, correct matching from List-I and List-II on the basis of given options is (I, T).

Given, ground state energy of hydrogen atom = $-$13.6 eV

Energy of He⁺ = $-$3.4 eV, Z = 2

Energy of He⁺, E = $ - {{13.6 \times {Z^2}} \over {{n^2}}}eV$

$ - 3.4eV = {{ - 13.6 \times {{(2)}^2}} \over {{n^2}}}$

$ \Rightarrow $ n = $\sqrt {{{ - 13.6 \times 4} \over {34}}} $ $ \Rightarrow $ n = 4

Given, azimuthal quantum number (l) = 2 (d $-$ subshell)

Magnetic quantum number (m) = 0

$ \therefore $ Angular nodes (l) = 2

Radial node = n $-$ l $-$ 1 = 4 $-$ 2 $-$ 1 = 1

nl = 4d state

Hence, options (a), (c) are correct.

- Option (A) : Incorrect. The total number of radial nodes is given by $n-l-1$. For $1 s$ orbital, the radial node is zero (i.e. $1-0-1=0$ ).

- Option (B) : Incorrect. $d_{z^2}$ orbital has two lobes that point in opposite directions along $z$-axis plus a doughnut-shaped ring of electron density around the centre that lies in the $x y$-plane. Therefore, it does not have any nodal plane.

- Option (C) : Correct. The total number of radial nodes is given by $n-l-1$.

For $2 s$ : Radial nodes $=2-0-1=1$

The orbital 2s of hydrogen like species can be diagrammatically represented as :


There is a spherical region of zero electron density between $1 s$ and the $2 s$ orbital of hydrogen like species. This spherical region is called the node. Only one radial node exist between $1 s$ and $2 s$.

The plot of wave function for $2 s$ verses radial distance from the nucleus is represented as



Point $P$ represents the region of zero electron density at a distance $r$ from the nucleus. This represents spherical node.

- Option (D) : Incorrect. The probability density at nucleus is for $2p$ orbital is zero.

The ion $\mathrm{He}^{+}$having 1 electron (in $1 s$ orbital) is hydrogen like species; hence, its $1 s$ orbital will have:

(i) Wave function same as that of hydrogen, i.e.,

$ \psi_{1,0,0}=\left(\frac{Z}{a_0}\right)^{3 / 2} e^{-Z r / a_0} $

(ii) The probability density for the electron in the $1 s$ orbital is represented as:


For $1 s$ orbital probability density is maximum at the nucleus.

(iii) Probability density for $2 s$ orbitals is given by

$ =\frac{1}{2 a_0^{3 / 2}}\left(1-\frac{r}{2 a_0}\right) e^{-r / 2 a_0} $

Probability density $\propto \frac{1}{a_0^3}$

(iv) For electron in the $1 s$ orbital, the wave function of electron is not the function of angular wave function i.e., $\theta$ and $\phi$. Hence, 1 s orbital cannot have the wave function

$ \psi_{n, l, m_l} \propto\left(\frac{Z}{a_0}\right)^{5 / 2} r e^{\left(-Z r / 2 a_0\right) \cos \theta} $

(i) Wave function for 1s orbital of hydrogen atom with $n, l$ and $m_l$ as quantum numbers is given

$ \psi_{n, l, m_l}=\frac{1}{\sqrt{\pi}}\left(\frac{Z}{a_0}\right)^{3 / 2} e^{-Z r / a_0} $

Where, $\mathrm{Z}$ is atomic number of the element, $r$ is the variable distance of electron from the nucleus, $a_0$ is the Bohr's radius

For $1 \mathrm{~s}$ orbital, $n=1, l=0, m_l=0$ and wave function is represented as :

$ \begin{aligned} & \psi_{1,0,0}=\frac{1}{\sqrt{\pi}}\left(\frac{Z}{a_0}\right)^{3 / 2} e^{-Z r / a_0} \\\\ \Rightarrow \psi_{1,0,0} & \propto\left(\frac{Z}{a_0}\right)^{3 / 2} e^{-Z r / a_0} \end{aligned} $

(ii) Energy of excitation of electron from $n_1=2$ to $n_2=4$ is given as :

$ \begin{aligned} & \mathrm{E}_{2 \rightarrow 4} \propto-\mathrm{R}\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right) \\\\ & \mathrm{E}_{2 \rightarrow 4} \propto-\mathrm{R}\left(\frac{1}{4^2}-\frac{1}{2^2}\right) \\\\ & \mathrm{E}_{2 \rightarrow 4} \propto-\mathrm{R}\left(\frac{-3}{16}\right)=\frac{3 R}{16} \end{aligned} $

(iii) Energy of excitation of electron from $n_1=2$ to $n_2=4$ is given as:

$ \begin{aligned} \mathrm{E}_{2 \rightarrow 6} & \propto-\mathrm{R}\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right) \\\\ \mathrm{E}_{2 \rightarrow 6} & \propto-R\left(\frac{1}{6^2}-\frac{1}{2^2}\right) \\\\ & =-\left(\frac{1}{36}-\frac{1}{4}\right) R \\\\ E_{2 \rightarrow 6} & \propto \frac{+8 R}{36} \end{aligned} $

The ratio of the energy of transition in the two cases is given by :

$ \begin{aligned} & \frac{E_{2 \rightarrow 4}}{E_{2 \rightarrow 6}}=\frac{3 R}{16} \times \frac{6}{8 R}=\frac{27}{32} \\\\ & E_{2 \rightarrow 4}=\frac{27}{32} \times E_{2 \rightarrow 6} \end{aligned} $

the energy of transition from 2 to 4 excited state is $\frac{27}{32}$ times the energy of transition from 2 to 6 excited state.

The probability of finding an electron of hydrogen atom in a spherical shell of infinitesimal thickness, $d r$, at a distance $r$ from the nucleus, with volume $d V=4 \pi r^2 d r$ is

$ P=\psi^2 \cdot d V=\psi^2 \cdot 4 \pi r^2 d r=R^2(r) 4 \pi r^2 d r $

Here, $R^2(r)$ is the radial density function.

For $1 s$ subshell, $n=1, l=0$ and $n-l-1=0$. Thus, the number of radial and angular nodes $=0$. For 1s orbital, the probability of finding an electron will be maximum at near to the the nucleus, as 1s orbital is the nearest to the nucleus as it is the lowest orbital in terms of energy. Therefore, the plot of radial probability $P=R^2(r) 4 \pi r^2$ versus $r$ is as follows :

For Bohr orbit, angular momentum is

$mvr_n = \frac{nh}{2\pi}$

Velocity, $v = \frac{nh}{2\pi mr_n} \quad \text{... (i)}$

Kinetic energy, $K.E. = \frac{1}{2}mv^2 \quad \text{... (ii)}$

By putting the value of $v$ from (i) into (ii),

$K.E. = \frac{1}{2}m \times \frac{n^2h^2}{4\pi^2m^2r_n^2} = \frac{n^2h^2}{8\pi^2mr_n^2}$

For second Bohr orbit, $n = 2$

$r_n = a_0 \times n^2 \quad (a_0 = \text{Bohr radius})$

$r_n = 4a_0$

$K.E. = \frac{(2)^2h^2}{8\pi^2m(4a_0)^2}$

Thus, $K.E. = \frac{h^2}{32\pi^2ma_0^2}$

For hydrogen-like species, the energy levels depend on the principal quantum number $n$ and are inversely proportional to the square of $n$, and they can be characterized by their nuclear charge $Z$. The energy for a hydrogen-like ion can be expressed as:

$E_n = -\dfrac{Z^2R_H}{n^2}$

where $R_H$ is the Rydberg constant for hydrogen, $Z$ is the atomic number (for Li²⁺, $Z=3$), and $n$ is the principal quantum number associated with the energy level.

Given that S₂ has energy equal to the ground state energy of the hydrogen atom ($E_n$ for hydrogen when $n=1$ and $Z=1$ is $E_1 = -R_H$), and given that S₂ has one radial node. The radial node information tells us about the principal quantum number $n$, as the number of radial nodes is given by $n-l-1$, where $l$ is the azimuthal quantum number.

The ground state of hydrogen corresponds to $n=1$. For the given species Li²⁺ to have the same energy as the ground state of the hydrogen atom but for state S₂, we can use the energy relation. Since the energy is specified to be the same as hydrogen's ground state, let's set up the equality according to the equation given and solve for $n$ specific to the condition (Li²⁺ is considered here, but the condition is about energy equivalence).

Given that S₂ has one radial node, it cannot be the ground state energy level for Li²⁺ (which would directly correspond to $n=1$), it implies a different $n$. For one radial node, the condition $n-l-1 = 1$ must be met.

The options presented are:

• 1s ($n=1$, $l=0$) - No radial nodes
• 2s ($n=2$, $l=0$) - One radial node
• 2p ($n=2$, $l=1$) - No radial nodes, due to the different $l$
• 3s ($n=3$, $l=0$) - Two radial nodes

S₁ is described as having one radial node. Based on the rule for the number of radial nodes ($n-l-1$), the only states that fits this condition directly from the options provided are $2s$ (since for a $2s$ orbital, $n=2$, $l=0$, yielding $2-0-1=1$ radial node).

Hence, the correct option for S₁ is:

Option B: 2s

For a hydrogen-like ion, the energy levels can be given by the formula:

$E_n = -\frac{Z^2}{n^2} E_0$

where $E_n$ is the energy of the nth level, $Z$ is the atomic number (for Li$^{2+}$, $Z=3$), $n$ is the principal quantum number, and $E_0$ is the ground state energy of the hydrogen atom ($-13.6 \, eV$).

Given that state $S_2$ has energy equal to the ground state energy of the hydrogen atom and one radial node, we identify that $S_2$ corresponds to $n=2$ for a hydrogen atom. This is because for hydrogen-like species, the number of radial nodes is given by $n-1$, where $n$ is the principal quantum number. The ground state ($n=1$) has 0 nodes, the first excited state ($n=2$) has 1 radial node, etc.

Since the energy of $S_2$ is equal to the ground state energy of the hydrogen atom, we can directly compare the energies. For hydrogen ($Z=1$), the ground state energy ($n=1$) is:

$E = -E_0$

For the Li$^{2+}$ ion in state $S_2$ (which we established is equivalent to $n=2$ in terms of energy for hydrogen), we use the formula $E_n = -\frac{Z^2}{n^2} E_0$. Since $Z=3$ for Li$^{2+}$, and given that $S_2$ has the energy equivalent to the ground state of hydrogen ($E_0$), we solve for the energy ratio rather than the specific energy of $S_2$.

We then look at $S_1$, which we know must be the ground state for Li$^{2+}$ since it is the state before $S_2$ and has one radial node (indicating $n=2$ for $S_1$).

Thus, for $S_1$, which actually corresponds to $n=2$ for Li$^{2+}$, the energy in units of the hydrogen atom ground state energy is:

$E_{S_1} = -\frac{Z^2}{n^2} E_0 = -\frac{3^2}{2^2}E_0 = -\frac{9}{4}E_0$

Now, to express this in units of the hydrogen atom ground state energy ($-E_0$):

$\frac{E_{S_1}}{E_0} = -\frac{9}{4} = -2.25$

So, considering the provided options and the fact that energy levels are usually considered in positive values when comparing magnitudes, the correct answer is:

Option C: 2.25.

To identify the orbital angular momentum quantum number, $l$, for the state $S_2$ of a hydrogen-like species such as $Li^{2+}$, we can refer to the given information and the known equations for the energy levels of hydrogen-like atoms. Hydrogen-like atoms or ions have only one electron and their energy in a particular state is given by the equation:

$E_n = -\frac{Z^2}{n^2} E_0$

where:

• $E_n$ is the energy of the electron in the nth energy level,
• $Z$ is the atomic number of the species (for $Li$, $Z = 3$),
• $n$ is the principal quantum number,
• $E_0$ is the energy of the ground state of hydrogen ($-13.6$ eV).

The problem states that $S_2$ has energy equal to the ground state energy of the hydrogen atom. The ground state of hydrogen corresponds to $n = 1$ and $E_0 = -13.6$ eV. However, for $Li^{2+}$, with $Z = 3$, the same energy level could be attained at a different value of $n$ since the $Z^2$ factor magnifies the energy levels with increasing atomic number. Let's calculate the principal quantum number, $n$, for the $Li^{2+}$ ion that would give it an energy equal to $E_0$:

For $Li^{2+}$ ion to have the ground state energy of a hydrogen atom, we can set the energies equal and solve for $n$:

$-\frac{Z^2}{n^2}E_0 = E_0$

Substituting $Z = 3$ and simplifying:

$-\frac{9}{n^2} = 1$

From which we find, $n^2 = 9$ and thus $n = 3$.

Furthermore, the problem states that $S_1$ has one radial node and upon absorbing light, it transitions to $S_2$ which also has one radial node. Radial nodes are related to the principal quantum number $n$ and the angular quantum number $l$ by the formula:

Number of radial nodes = $n - l - 1$

Given that $S_2$ has one radial node, we can plug $n = 3$ into the radial node formula to solve for $l$:

$1 = 3 - l - 1$

This simplifies to:

$l = 3 - 2$

$l = 1$

Therefore, the orbital angular momentum quantum number of the state $S_2$ is 1, which corresponds to option B.

(A) Orbital angular momentum of the electron in a hydrogen-like atomic orbital is represented by azimuthal quantum number.

(B) A hydrogen-like one-electron wave function obeying Pauli principle correspond to Electron spin quantum number.

(C) Shape, size and orientation of hydrogen-like atomic orbitals are determined by Azimuthal, principal and Magnetic quantum number.

(D) Probability density of electron at the nucleus in hydrogen-like atom is determined by Principal quantum number, Azimuthal quantum number and Magnetic quantum number.

With increase in the atomic number, the proton-proton electrostatic repulsion begins to overcome attractive forces involving proton and neutrons in heavier nuclides. The stability relationship can be represented by a line with a slope of 45$^\circ$, i.e., the maximum stability is attained when N = Z. Right of the curve a radioactive nuclide would be neutron rich and would decay by $\beta$-emission to produce a daughter nucleus with a lower n/p ratio. For heavier nuclides, p-p repulsions start to offset the attractive forces and an excess of neutrons over protons, is required for stability.

Given:

A positron is emitted from:

$ {}^{23}_{11}\text{Na} $

Step 1: Recall what happens in positron emission.

A positron is a $\beta^+$ particle, i.e. $ {}^0_{+1}e $.

In positron emission:

$ p \rightarrow n + e^+ + \nu_e $

So inside the nucleus:

• Number of protons decreases by 1,

• Number of neutrons increases by 1,

• The mass number (A) remains the same.

Step 2: Apply this to sodium-23.

$ {}^{23}_{11}\text{Na} \rightarrow {}^{23}_{10}\text{X} + e^+ + \nu_e $

So the new nucleus has:

• Atomic number $Z = 10$

• Mass number $A = 23$

The element with $Z = 10$ is Neon (Ne).

Hence, the daughter nucleus is $ {}^{23}_{10}\text{Ne} $.

Step 3: Find the required ratio.

They ask for the ratio of atomic mass (mass number) to atomic number of the resulting nuclide:

$ \frac{A}{Z} = \frac{23}{10} $

✅ Final Answer:

$ \boxed{\frac{23}{10}} $

Option C: $\dfrac{23}{10}$

(A) The $\mathrm{V}_n$ and $\mathrm{K}_n$ denotes potential energy and kinetic energy of an electron.

The expression for kinetic theory and potential energy are given by Bohr atomic theory as shown below.

$ \begin{array}{rlrl} & \mathrm{V}_n =-\frac{\mathrm{Ze}^2}{r} \text { and } \mathrm{K}_n=\frac{\mathrm{Ze}^2}{2 r} \\ \text { Hence, } & \frac{\mathrm{V}_n}{\mathrm{~K}_n} =\frac{-\frac{\mathrm{Ze}^2}{r}}{\frac{\mathrm{Ze}^2}{2 r}} \\ & \therefore \frac{\mathrm{~V}_n}{\mathrm{~K}_n} =-2 \end{array} $

Hence, option A in column I matches with option R in column II.

(B) The $\mathrm{E}_n$ denoted the total energy of an electron which is the sum of potential and kinetic energy of an electron.

The radius of $n^{\text {th }}$ orbit of an atom is denoted by $r_n$.

Where $n$ denoted the number of an orbit.

The radius of $n^{\text {th }}$ orbit is directly proportional to the square of $n$ and the total energy of an electron is inversely proportional to that of square of $n$.

$ r_n \propto n^2 \text { and } \mathrm{E}_n \propto \frac{1}{n^2} $

Therefore,

$ \mathrm{E}_n \propto \frac{1}{r_n} \text { or } r_n \propto \mathrm{E}_n^{-1} $

The value of $x$ will be,

$ x=-1 $

Hence, option B in column I matches with option Q in column II.

(C) According to the Bohr theory, angular momentum for an orbital is given as,

$ \omega=\sqrt{l(l+1)} \frac{h}{2 \pi} $

Here, $\omega$ is the angular momentum for an orbital, $l$ is the azimuthal quantum number and $h$ is the Planck's constant.

The lowest atomic orbital in any atom is $1 s$, which has least energy in the atom.

For $1 s$ orbital, the azimuthal quantum number is 0 as the orbital possess spherical shape.

Substituting the value of $l$ in above equation for 1s orbital,

$ \begin{aligned} & \omega=\sqrt{0(0+1)} \frac{h}{2 \pi} \\ & \omega=0 \end{aligned} $

Hence, the value of angular momentum for lowest orbital will be zero.

Hence, option C in column I matches with option P in column II.

(D) The Z denotes the atomic number of an element.

The radius of an orbit and atomic number of an element are related by following relation.

$ \frac{1}{r_n} \propto Z_n $

If $\frac{1}{r_n} \propto Z^y$, then $y=1$

Hence, option D in column I matches with option $S$ in column II.

To find the number of radial nodes in an orbital, we use the formula:

$\text{Number of radial nodes} = n - l - 1$

where $n$ is the principal quantum number, and $l$ is the azimuthal quantum number (also known as the angular momentum quantum number).

For the 3s orbital:

$n = 3$ (since it is the 3rd energy level) and $l = 0$ (s orbital corresponds to $l = 0$).

Plugging these values into the formula gives:

$\text{Number of radial nodes for 3s} = 3 - 0 - 1 = 2$

For the 2p orbital:

$n = 2$ (since it is the 2nd energy level) and $l = 1$ (p orbital corresponds to $l = 1$).

Therefore, the number of radial nodes for the 2p orbital is:

$\text{Number of radial nodes for 2p} = 2 - 1 - 1 = 0$

Thus, the number of radial nodes of 3s and 2p orbitals are respectively 2 and 0.

Option A is correct: 2, 0

To compare the radius of these different orbits to the first Bohr's orbit of the hydrogen atom, we can use the formula for the radius of the nth orbit in the hydrogen-like atom, which is given by:

$ r_n = \frac{n^2h^2}{4\pi^2kme^2}Z^{-1} $

Where:

• $r_n$ is the radius of the nth orbit,
• $n$ is the principal quantum number,
• $h$ is Planck's constant,
• $k$ is Coulomb's constant,
• $m$ is the mass of the electron,
• $e$ is the charge of the electron,
• $Z$ is the atomic number or the number of protons in the nucleus (which also equals the charge of the nucleus in a hydrogen-like atom).

For the first Bohr orbit of the hydrogen atom ($n=1$ and $Z=1$), the radius is:

$ r_1 = \frac{1^2h^2}{4\pi^2kme^2}(1)^{-1} $

We must compare this with the options provided. The formula simplifies to identifying where the $n^2/Z$ ratio for each option is equal to 1, as it is for hydrogen's first orbit.

• Option A: He⁺ ($Z=2$, $n=2$)

  $\frac{n^2}{Z} = \frac{2^2}{2} = 2$

• Option B: Li²⁺ ($Z=3$, $n=2$)

  $\frac{n^2}{Z} = \frac{2^2}{3} \approx 1.33$

• Option C: Li²⁺ ($Z=3$, $n=3$)

  $\frac{n^2}{Z} = \frac{3^2}{3} = 3$

• Option D: Be³⁺ ($Z=4$, $n=2$)

  $\frac{n^2}{Z} = \frac{2^2}{4} = 1$

Thus, the radius is the same as that of the first Bohr's orbit of hydrogen atom only for Option D: Be³⁺ ($n=2$), because in this case, the ratio $n^2/Z = 1$, matching the ratio for hydrogen's first orbit.

Isoelectronic species are those that have the same number of electrons. Isostructural species have the same geometric arrangement of atoms.

First, let's determine if the given pairs are isoelectronic.

• $NO_3^-$: Nitrogen has 7 electrons, and each oxygen has 8 electrons. The negative charge adds one more electron. Therefore, $NO_3^-$ has $7 + 3 \cdot 8 + 1 = 32$ electrons.
• $CO_3^{2-}$: Carbon has 6 electrons, each oxygen has 8 electrons, and the ${2-}$ charge adds 2 more electrons. Therefore, $CO_3^{2-}$ has $6 + 3 \cdot 8 + 2 = 32$ electrons.
• $ClO_3^-$: Chlorine has 17 electrons, each oxygen has 8 electrons, and the negative charge adds 1 more electron. Therefore, $ClO_3^-$ has $17 + 3 \cdot 8 + 1 = 42$ electrons.
• $SO_3$: Sulfur has 16 electrons, and each oxygen has 8 electrons. Therefore, $SO_3$ has $16 + 3 \cdot 8 = 40$ electrons.

Based on the number of electrons, $NO_3^-$ and $CO_3^{2-}$ are isoelectronic.

Now, let's determine if these species are isostructural. Both $NO_3^-$ and $CO_3^{2-}$ have a trigonal planar structure.

Therefore, the correct option is:

Option A: $NO_3^-$, $CO_3^{2-}$

The correct option is Option D: helium nuclei, which impinged on a metal foil and got scattered.

Rutherford's experiment is a landmark in the study of atomic structure. In this experiment, a beam of alpha particles, which are essentially helium nuclei (comprising two protons and two neutrons, and thus positively charged), was directed at a thin gold foil. Most of the alpha particles passed through the foil with little to no deflection, which suggested that atoms are mostly empty space. However, a small fraction of the alpha particles were deflected at large angles, and a very few even bounced back towards the source. This observation was unexpected based on the prevailing model of the atom at the time, which was J.J. Thomson's "plum pudding" model where the atom was thought to be a uniform sphere of positive charge with electrons embedded within it.

The results of Rutherford's experiment led him to propose a new model of the atom, known as the nuclear model. In this model, an atom consists of a small, dense nucleus that contains most of the atom's mass and all of its positive charge, surrounded by electrons that occupy the remaining vast space of the atom. The nucleus is tiny compared to the overall size of the atom, explaining why most alpha particles passed through the foil unimpeded. The large deflections occurred when an alpha particle came close to the nucleus, being repelled by its positive charge. This discovery was fundamental in moving away from the idea of an atom as a homogenous sphere to an atom with a dense central nucleus around which electrons orbit.

The correct answer is Option C: Pauli exclusion principle.

The Pauli exclusion principle states that no two electrons can have the same set of four quantum numbers. In simpler terms, it means that each orbital can hold a maximum of two electrons, and these two electrons must have opposite spins. The quantum numbers that define an electron's state in an atom include the principal quantum number (n), the angular momentum quantum number (l), the magnetic quantum number (m_l), and the spin quantum number (m_s).

The electronic configuration given as 1s⁷ would imply trying to place seven electrons into the 1s orbital. However, an s orbital can accommodate a maximum of two electrons, provided they have opposite spins. This arrangement means that trying to place more than two electrons in the 1s orbital would indeed violate the Pauli exclusion principle because it would require at least some of these electrons to have the same set of quantum numbers, which is not possible.

Let's briefly address why the other options do not fit:

Option A: Heisenberg uncertainty principle concerns itself with the limit to which the position and momentum (or velocity) of an electron can be known simultaneously. This principle does not directly relate to the issue of how many electrons can occupy a given orbital or energy level.

Option B: Hund's rule deals with the distribution of electrons across orbitals of the same energy (degeneracy). It states that electrons will fill each degenerate orbital singly before any orbital gets a second electron, and all singly occupied orbitals will have electrons with the same spin. This rule does not apply to the violation seen in the 1s⁷ configuration.

Option D: Bohr's postulate of stationary orbits pertains to the quantum condition that electrons orbit the nucleus in certain stable orbits without radiating energy. This principle is more historical and foundational to quantum theory and does not address the occupancy limits of orbitals or the exclusion principle directly.

Therefore, the concept violated by having a 1s⁷ configuration is the Pauli exclusion principle (Option C), as this configuration suggests more than two electrons attempting to occupy the same quantum state within the 1s orbital, which is not permissible.

To calculate the wavelength associated with a moving object, we use the de Broglie wavelength formula, which is given as:

$\lambda = \frac{h}{mv}$

where:

• $\lambda$ is the wavelength,


• $h$ is the Planck constant ($6.626 \times 10^{-34} \, \text{m}^2 \text{kg} / \text{s}$),


• $m$ is the mass of the object (in kg),


• $v$ is the velocity of the object (in m/s).

Given that the golf ball weighs 200 g (which is 0.2 kg) and is moving at a speed of 5 m/h. First, let's convert the speed into m/s since the units need to be consistent:

$ 5 \, \text{m/h} \times \left(\frac{1 \, \text{h}}{3600 \, \text{s}}\right) = \frac{5}{3600} \, \text{m/s} = \frac{1}{720} \, \text{m/s} $

Now, we can substitute the values into the de Broglie equation:

$ \lambda = \frac{6.626 \times 10^{-34} \, \text{m}^2\text{kg}/\text{s}}{0.2 \, \text{kg} \times \frac{1}{720} \, \text{m/s}} $

$ \lambda = \frac{6.626 \times 10^{-34}}{0.2 \times \frac{1}{720}} = \frac{6.626 \times 10^{-34}}{\frac{1}{3600}} $

$ \lambda = 6.626 \times 10^{-34} \times 3600 $

$ \lambda = 2.38536 \times 10^{-30} \, \text{m} $

The calculated wavelength is therefore of the order of $10^{-30}$ m.

Correct Answer: Option C, $10^{-30}$ m.

The correct option is D: two quantum mechanical spin states which have no classical analogue. The concept of electron spin with quantum numbers +1/2 and -1/2 indeed represents two intrinsic angular momentum states of an electron, but it doesn't directly correspond to the physical picture of the electron spinning around its own axis like a planet, either clockwise or anticlockwise. This description of spin is a quantum mechanical property that has no classical analogue, meaning that it cannot be accurately described using classical physics concepts. These two states can be associated with the magnetic moment of the electron pointing in two opposite directions, often referred to simplistically as "up" and "down" in the context of its orientation in an external magnetic field, but it's important to understand this is a representation that helps visualize their quantum mechanical behavior regarding their interaction with magnetic fields, rather than a literal physical spinning.

The number of nodal planes in an atomic orbital refers to the planes through the nucleus where the probability of finding an electron is zero. For a p_x orbital, which is one of the three p orbitals (px, py, pz), the characteristic feature is its dumbbell-shaped electron cloud which is symmetric about the nucleus and is oriented along the x-axis (assuming px for this case).

A nodal plane is essentially a region that divides the electron cloud of an orbital into two or more parts where the probability of finding an electron is zero. In the case of p orbitals, there is a central nodal plane that passes through the nucleus and is perpendicular to the axis along which the orbital is oriented. For the p_x orbital, this central nodal plane is oriented such that it divides the orbital into two lobes along the y and z axes. This means there is one nodal plane in the p_x orbital, where the probability of finding an electron is zero.

Therefore, the correct answer is:

Option A: one

The electronic configuration provided is: 1s², 2s² 2p⁶, 3s² 3p⁶ 3d⁵, 4s¹. This configuration gives us a total electron count of $2 + 2 + 6 + 2 + 6 + 5 + 1 = 24$ electrons, which is precisely the number of electrons that chromium (Cr), an element with atomic number 24, possesses in its neutral state.

However, the standard ground state electron configuration for chromium is typically written as: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1$ This is due to the increased stability offered by having a half-filled d-subshell, which is an exception to the expected order where the 4s orbital is filled before the 3d orbitals begin to fill. Despite this seeming irregularity, this electron configuration does not represent an excited state; rather, it's a common anomaly observed due to subshell energy levels and the stability provided by half-filled or fully filled subshells.

Given the options:

• Option A - Excited State: This is not correct because the configuration provided does match chromium’s recognized ground state configuration, due to the stability preferences of electron arrangements.
• Option B - Ground State: This is correct because this configuration corresponds to the stable configuration for chromium, taking into account the exception for electron distribution for enhanced stability due to half-filled d-subshells.
• Option C - Cationic Form: This is not correct because a cationic form would suggest the loss of electrons, which is not indicated in this configuration.
• Option D - Anionic Form: This is not correct because an anionic form would imply the addition of electrons, which is also not indicated in this configuration.

Therefore, the correct answer is Option B - Ground State.

To place the electrons in order of increasing energy based on their quantum numbers, we consider the principles that dictate electron energy levels in atoms. The energy of an electron in an atom is primarily determined by its principal quantum number ($n$) and azimuthal (or angular momentum) quantum number ($l$). Generally, the energy of an electron increases with an increase in the $n$ value. However, for electrons within the same $n$ level, those with higher $l$ values have slightly higher energies due to electron penetration and shielding effects. In multi-electron atoms, due to electron-electron repulsion and orbital shapes, orbitals with the same $n$ but different $l$ values do not have the same energy.

Given the quantum numbers:

• (i) $n = 4, l = 1$
• (ii) $n = 4, l = 0$
• (iii) $n = 3, l = 2$
• (iv) $n = 3, l = 1$

We'll arrange them in order of increasing energy.

The principal quantum number, $n$, primarily determines the energy level. Hence, states with $n = 3$ would generally have lower energy compared to states with $n = 4$. Within the same $n$ level, the energy increases with $l$. Thus, a higher $l$ value indicates a slightly higher energy within the same $n$ level.

Based on these guidelines, we can place the given states in the following order:

1. (iv) $n = 3, l = 1$
2. (iii) $n = 3, l = 2$
3. (ii) $n = 4, l = 0$
4. (i) $n = 4, l = 1$

So, the electrons in order of increasing energy, from the lowest to highest, are Option A (iv)<(iii)<(ii)<(i). This ranking reflects that within the same principal quantum number, $n$, the electron energy increases with the azimuthal quantum number, $l$, and electrons with a lower principal quantum number typically have lower energy than those with a higher $n$.

Let's analyze each option one by one:

Option A: The electronic configuration of Cr is [Ar] 3d⁵4s¹. (Atomic number of Cr = 24)

This statement is correct. Chromium (Cr) has an atomic number of 24, which means it has 24 electrons. The expected configuration, following the Aufbau principle, would be [Ar] 3d⁴4s². However, chromium is an exception to the expected order. To achieve a more stable and lower energy configuration, one electron from the 4s orbital moves to the 3d orbital, resulting in the configuration [Ar] 3d⁵4s¹. This configuration is more stable due to the half-filled 3d subshell, providing extra stability to the atom.

Option B: The magnetic quantum number may have negative value

This statement is correct. The magnetic quantum number, denoted as \( m_l \), describes the orientation of the orbital in space and can have integer values ranging from \(-l\) to \(+l\), where \(l\) is the azimuthal quantum number (also known as the orbital angular momentum quantum number). For example, if \(l = 2\) (d orbitals), \(m_l\) can have the values of -2, -1, 0, +1, +2, indicating that there are five different orientations for d orbitals in space.

Option C: In a silver atom, 23 electrons have a spin of one type and 24 of the opposite type. (Atomic number of Ag = 47)

This statement is correct. Silver (Ag) has an atomic number of 47, indicating it has 47 electrons. Electrons are fermions, and they follow the Pauli exclusion principle, which states that no two electrons can have the same set of quantum numbers within an atom. This implies that electrons fill orbitals in pairs, with opposite spins. In the case of silver, with 47 electrons, this means there will be an unpaired electron. Thus, there will be 23 electrons with one spin direction and 24 electrons with the opposite spin, making the statement correct.

Option D: The oxidation state of Nitrogen in HN₃ is -3

This statement is incorrect. HN₃, also known as hydrazoic acid, contains nitrogen in a relatively unusual oxidation state. To determine the oxidation state of nitrogen in HN₃, let's assign oxidation numbers based on the usual oxidation states hydrogen (H) as \(+1\) and nitrogen (N), which will be determined. The sum of the oxidation states in a neutral compound is zero. In HN₃, with one H atom, hydrogen contributes \(+1\) to the total. If the nitrogen's oxidation state were \(-3\), the total would not balance to 0 with three nitrogen atoms. Instead, the structure and electronegativity of nitrogen suggest a different setup, where one nitrogen is more positive and the other two are more negative, but overall, nitrogen typically has positive oxidation states in HN₃. A common way to rationalize the states is to consider the central nitrogen in a covalent triple bond with a negative oxidation state, but overall, the nitrogen connected with hydrogen tends to have a positive oxidation state due to the high electronegativity of nitrogen compared to hydrogen. The correct total oxidation state of nitrogen in HN₃ depends on how the bonding is considered but is certainly not -3 overall for all nitrogen atoms.

Therefore, the correct statements are Option A and Option B, and Option C, while Option D is incorrect.