Subshell Electronic Configuration

To understand the given statements, let's analyze both the assertion and the reason:

1. Assertion A: The energy of the 2s orbital of a hydrogen atom is greater than that of the 2s orbital of lithium.
   • The energy levels of orbitals within an atom depend on both the principal quantum number ( ) and the effective nuclear charge experienced by the electrons.
   • In a hydrogen atom, only one electron is present, and it experiences the full nuclear charge, as there is no shielding effect from other electrons.
   • In a lithium atom, which has a higher atomic number, there are other electrons that cause repulsion, and the effective nuclear charge felt by the 2s electron is less than the actual nuclear charge. Thus, the electron is more tightly bound due to the higher nuclear charge, leading to lower energy of the orbital.
2. Reason R: Energies of the orbitals in the same subshell decrease with an increase in the atomic number.
   • The Reason R states a general principle in atomic structure: as the atomic number increases, more protons are present, increasing the nuclear charge and making the electrons in the same subshell more tightly bound.
   • This results in a decrease in the energy of the orbitals because the electrons are held closer to the nucleus due to the increased attraction.

Conclusion:

• Both Assertion A and Reason R are true. As we move from hydrogen to lithium, the increasing atomic number results in the 2s orbital of lithium having a lower energy compared to hydrogen, as explained by Reason R.
• Therefore, Reason R correctly explains why the energy of the 2s orbital in lithium is less than that in hydrogen.

Thus, the correct answer is: Both A and R are true and R is the correct explanation of A.

To solve this question, we need to examine each statement individually and determine their validity.

1. Statement I: "In the Lanthanoids, the formation of Ce⁺⁴ is favoured by its noble gas configuration."
   • Cerium (Ce) is a lanthanide with atomic number 58. In its elemental form, it has the electron configuration: .
   • When Ce is oxidized to Ce⁴⁺, it loses four electrons, resulting in the configuration: , which is a noble gas configuration.
   • The stability of the noble gas configuration explains why Ce⁴⁺ can form under certain conditions, supporting statement I as true.
2. Statement II: "Ce⁺⁴ is a strong oxidant reverting to the common +3 state."
   • Ce⁴⁺ is indeed a strong oxidizing agent and has a tendency to revert to the more stable Ce³⁺ state.
   • Ce³⁺ is the more common oxidation state in lanthanides due to the stability of the partially filled f-orbitals.
   • This reversion process is due to the high tendency of Ce⁴⁺ to gain an electron to become Ce³⁺, making it a strong oxidant, therefore statement II is true.

Given both explanations, we conclude that both Statement I and Statement II are true. Thus, the correct option is: Both Statement I and Statement II are true.

To solve this question, we need to determine the number of unpaired electrons in each of the given elements and then arrange them in increasing order. The unpaired electron count is determined by the electronic configuration of each element.

1. Find the atomic numbers and electron configurations:
   • Scandium (Sc): Atomic number = 21, Configuration = [Ar] 3d¹ 4s²
   • Chromium (Cr): Atomic number = 24, Configuration = [Ar] 3d⁵ 4s¹
   • Vanadium (V): Atomic number = 23, Configuration = [Ar] 3d³ 4s²
   • Titanium (Ti): Atomic number = 22, Configuration = [Ar] 3d² 4s²
   • Manganese (Mn): Atomic number = 25, Configuration = [Ar] 3d⁵ 4s²
2. Count the number of unpaired electrons:
   • Sc (3d¹): 1 unpaired electron
   • Cr (3d⁵): 6 unpaired electrons (5 in 3d and 1 in 4s)
   • V (3d³): 3 unpaired electrons
   • Ti (3d²): 2 unpaired electrons
   • Mn (3d⁵): 5 unpaired electrons
3. Arrange the elements in increasing order based on their unpaired electrons:
   • Sc (1 unpaired electron), Ti (2 unpaired electrons), V (3 unpaired electrons), Mn (5 unpaired electrons), Cr (6 unpaired electrons)

Thus, the increasing order of elements based on the number of unpaired electrons is:

Sc (A) < Ti (D) < V (C) < Mn (E) < Cr (B)

Correct Answer: (A) < (D) < (C) < (E) < (B)

The problem asks to determine the total number of species from the given list where the central atom is hybridized.

Concept Used:

The hybridization of a central atom in a molecule can be determined using the Steric Number (SN) method. The steric number is the sum of the number of sigma ( ) bonds formed by the central atom and the number of lone pairs of electrons on it.

The hybridization is then determined based on the steric number:

• If SN = 2, the hybridization is .
• If SN = 3, the hybridization is .
• If SN = 4, the hybridization is .

For molecules with multiple "central" atoms (like hydrocarbons), the hybridization of each of those atoms is considered.

Step-by-Step Solution:

Step 1: Analyze the hybridization of each species in the list.

: The central atom is Nitrogen (N). It forms 3 bonds with H atoms and has 1 lone pair.

: The central atom is Sulfur (S). It forms 2 bonds with O atoms and has 1 lone pair.

: Silicon dioxide has a giant covalent network structure where each Silicon (Si) atom is the central atom bonded to 4 Oxygen atoms tetrahedrally.

: The central atom is Beryllium (Be). It forms 2 bonds with Cl atoms and has 0 lone pairs.

: In acetylene (H-C≡C-H), each Carbon (C) atom forms 2 bonds (one with H, one with C) and has 0 lone pairs.

: In ethene ( ), each Carbon (C) atom forms 3 bonds (two with H, one with C) and has 0 lone pairs.

: The central atom is Boron (B). It forms 3 bonds with Cl atoms and has 0 lone pairs.

: In formaldehyde, the central atom is Carbon (C). It forms 3 bonds (two with H, one with O) and has 0 lone pairs.

: In benzene, each of the 6 Carbon (C) atoms in the ring forms 3 bonds (one with H, two with adjacent C atoms) and has 0 lone pairs.

: The central atom is Boron (B). It forms 3 bonds with F atoms and has 0 lone pairs.

: In 1,2-dichloroethane ( ), each Carbon (C) atom forms 4 bonds and has 0 lone pairs.

Final Computation & Result:

Step 2: Count the number of species with hybridized central atoms.

The species from the list with hybridization are:

The total number of species with a central atom utilising hybrid orbitals is 6.

The question asks for the electronic configuration of Neodymium (Nd), which has an atomic number of 60. To find its electronic configuration, we need to consider how electrons are distributed in its atomic orbitals based on their energy levels.

Neodymium is a lanthanide, and its electrons fill the orbitals in the following order:

• The core electrons are represented with xenon ([Xe]) because neodymium follows xenon in the periodic table.
• The lanthanides involve the filling of the 4f orbitals. In this case, we need to consider the partially filled 4f and 6s orbitals.

Therefore, the electronic configuration for Neodymium is:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f⁴

The condensed form of this configuration, which uses the noble gas shorthand, would be:

Let's evaluate the given options to confirm the answer:

• – This matches our derived configuration.
• – Incorrect, as the 5f orbitals are not involved in the filling sequence for Nd.
• – Incorrect, this suggests an incorrect count of electrons in 4f orbitals.
• – Incorrect, both 5f and 4d are incorrectly used for Neodymium.

Thus, the correct electronic configuration for Neodymium is indeed .