Electronic Configuration

To determine the number of unpaired electrons, we must first write the electron configuration of the ions: - V (Vanadium ): The electron configuration of is . For , it loses two electrons, so the configuration is , meaning there are 3 unpaired electrons. - Ni (Nickel ): The electron configuration of is . For , it loses two electrons, so the configuration is , meaning there are 2 unpaired electrons. - Cr (Chromium ): The electron configuration of is . For , it loses two electrons, so the configuration is , meaning there are 4 unpaired electrons. - Mn (Manganese ): The electron configuration of is . For , it loses two electrons, so the configuration is , meaning there are 5 unpaired electrons. - Fe (Iron ): The electron configuration of is . For , it loses two electrons, so the configuration is , meaning there are 4 unpaired electrons. - Sc (Scandium ): The electron configuration of is . For , it loses two electrons, so the configuration is , meaning there is 1 unpaired electron. - Mn (Manganese ): The electron configuration of is . For , it loses three electrons, so the configuration is , meaning there are 4 unpaired electrons. - Fe (Iron ): The electron configuration for was previously calculated as , and there are 4 unpaired electrons. Comparing the number of unpaired electrons: - Cr (4 unpaired electrons) and Mn (5 unpaired electrons) do not match in the number of unpaired electrons. Thus, the correct pair with equal unpaired electrons is option (2) and .

Step 1: Understanding the Concept:
The Aufbau principle states that electrons fill orbitals in order of increasing energy. The order is: 1s, 2s, 2p, 3s, 3p, 4s, 3d, etc. We must count to the 19th electron following this sequence.

Step 2: Key Formula or Approach:
1. Electron 1-2: 1s orbital 2. Electron 3-4: 2s orbital 3. Electron 5-10: 2p orbitals 4. Electron 11-12: 3s orbital 5. Electron 13-18: 3p orbitals (Ar configuration) 6. Electron 19: 4s orbital (starts the K shell/4th period)

Step 3: Detailed Explanation:
1. The 18th electron completes the subshell. 2. According to the rule, the next orbital to be filled is ( ), because is less than (for ). 3. Therefore, the 19th electron enters the orbital. 4. Quantum numbers for 4s: , , , .

Step 4: Final Answer:
The correct set is .

Step 1: Understanding the Concept:
This question tests fundamental principles of atomic structure, specifically the rule for orbital energy, orbital sizes, and the calculation of radial nodes.
Step 2: Detailed Explanation:
Statement A: This is the Aufbau principle's rule. If two orbitals have the same value, the one with the lower principal quantum number ( ) is lower in energy (e.g., 3d has and 4p has , but 3d is filled first). This is True.

Statement B: As the atomic number ( ) increases, the effective nuclear charge increases. This causes the orbital to be attracted more strongly toward the nucleus, which actually decreases the energy (makes it more negative). Thus, the statement is False.

Statement C: Orbital size is primarily determined by the principal quantum number . Higher means the electron is likely further from the nucleus. Since for and for , is larger. This is True.

Statement D: Radial nodes are calculated as .
5f: .
6s: .
4d: .
5p: .
5d: .
Since the 5d orbital does have 2 radial nodes, the statement "none of the orbitals have 2 radial nodes" is False.
Step 3: Final Answer:
Statements A and C are correct. Therefore, the answer is Option (B).

Step 1: Understanding the Concept:
Molecular orbitals form by the overlap of atomic orbitals. - orbitals form by lateral (sideways) overlap. Bonding orbitals have constructive interference, while antibonding orbitals have destructive interference, resulting in a nodal plane between the two nuclei.

Step 2: Key Formula or Approach:
Identify the characteristics of MO:
1. Lateral overlap of p-orbitals.
2. Destructive overlap results in a nodal plane perpendicular to the internuclear axis.
3. Opposite phases are adjacent across the vertical node.

Step 3: Detailed Explanation:
Looking at the provided diagrams:
- Diagram (A) shows side-on constructive overlap (bonding ).
- Diagram (B) shows end-on destructive overlap (antibonding ).
- Diagram (C) shows lateral p-orbitals with opposite phases ( and ) adjacent to each other across a nodal plane between the atoms. This is the hallmark of a orbital.
- Diagram (D) shows a different sigma overlap.

Step 4: Final Answer:
Diagram (C) correctly represents the molecular orbital.