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

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.

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.

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.

To answer this question, it's important to understand what an isotone is. Isotones are nuclides (a specific type of atom) that have the same number of neutrons but different numbers of protons. The term isotone comes from the root "iso-" meaning same and "-tone" referring to neutrons, similar to how isotope refers to atoms with the same number of protons but different numbers of neutrons.

First, let's figure out the number of neutrons in the given isotope of Germanium $_{32}^{76}Ge$. You can calculate the number of neutrons by subtracting the atomic number (the number of protons, denoted as Z) from the mass number (the sum of protons and neutrons, denoted as A). This gives us:

$\text{Number of Neutrons} = A - Z$

For $_{32}^{76}Ge$, this calculation would be:

$76 - 32 = 44$

So, $_{32}^{76}Ge$ has 44 neutrons. An isotone of $_{32}^{76}Ge$ would therefore also need to have 44 neutrons. Now, we will calculate the number of neutrons in each option to see which ones match this criteria.

Option A: $_{32}^{77}Ge$

$\text{Neutrons} = 77 - 32 = 45 \text{ (not an isotone)}$

Option B: $_{33}^{77}As$ (since the atomic number 33 corresponds to Arsenic, not Germanium)

$\text{Neutrons} = 77 - 33 = 44 \text{ (is an isotone)}$

Option C: $_{34}^{77}Se$ (since the atomic number 34 corresponds to Selenium, not Germanium)

$\text{Neutrons} = 77 - 34 = 43 \text{ (not an isotone)}$

Option D: $_{34}^{78}Se$ (once again, atomic number 34 is for Selenium)

$\text{Neutrons} = 78 - 34 = 44 \text{ (is an isotone)}$