A possible set of values for the four quantum numbers for the highest-energy electron in V4+ (Z = 23) is:

Question

BlackTom AI · Accepted Answer

To determine the four quantum numbers for the highest-energy electron in V4+ (Z = 23), we first establish the electron configuration.
- Vanadium has 23 electrons in its neutral state. In V4+, four electrons are removed, giving 19 electrons total.
- The typical ground-state configuration for V4+ ends with a single electron in the 3d subshell, i.e., [Ar] 3d1. This means the highest-energy (outermost) electron resides in a 3d orbital with principal quantum number n = 3 and azimuthal quantum number l = 2.
Now, evaluate each option:
Option 1: n = 4, l = 2, ml = –2, ms = –1/2
- This would place the electron in the 4d subshell (n = 4, l = 2). Since the highest-energy electron is in the 3d shell (n = 3) for V4+ with configuration 3d1, this option places the electron in a higher principal quantum number than correct. It also conflicts with the actual valence shell for the highest-energy electron in this ion.
Option 2: n = 3, l = 0, ml = 0, ms = –1/2
- Here l = 0 corresponds to an s orbital (3s), not a d orbital. The highest-energy electron in V4+ occupies a d orbital (3d), so this option is inconsistent with the correct orbital type.
Option 3: n = 3, l = 1, ml = 0, ms = +1/2
- This describes a 3p electron (l = 1). But the outermost electron for V4+ is in 3d (l = 2), not 3p, making this option incorrect.
Option 4: n = 3, l = 2, ml = +1, ms = +1/2
- This exactly places the electron in the 3d subshell (n = 3, l = 2) with a permissible magnetic quantum number ml = +1 from the set {–2, –1, 0, +1, +2}, and an up-spin (ms = +1/2). Since there is a single 3d electron in V4+, this combination is consistent with the highest-energy electron being in a 3d orbital with a valid ml and ms.
Option 5: n = 4, l = 0, ml = 0, ms = –1/2
- This would correspond to a 4s electron. In V4+, the valence is 3d, not 4s, so this does not describe the highest-energy electron.
In summary, options 1, 2, 3, and 5 misplace either the principal quantum number, the subshell type, or both, relative to the known ground-state configuration of V4+ (3d1). Only option 4 correctly identifies the electron as residing in the 3d subshell with valid ml and ms values for the highest-energy electron in this ion.

A possible set of values for the four quantum numbers for the highest-energy electron in V4+ (Z = 23) is:Single choice

Similar Questions

There are [Fill in the blank] orbitals in a 5p subshell. The angular momentum quantum number (l) of these orbitals is [Fill in the blank] . (Input numbers only.)

Identify the quantum numbers n [Fill in the blank] and l [Fill in the blank] and the number of radial [Fill in the blank] and angular [Fill in the blank] nodes for the following simulated orbital cross-section:

Identify the quantum numbers n [Fill in the blank] and l [Fill in the blank] and the number of radial [Fill in the blank] and angular [Fill in the blank] nodes for the following simulated orbital cross-section:

Orbitals with the same principal quantum number (n) are said to be in the same shell.

Which quantum number must be the same for the electrons in the three orbitals shown below? 17A

Which of the following represents a possible set of quantum numbers for an electron in an atom?

Match the quantum numbers needed for each of the following 1: determination of the energy of an electron in a many electron atom 2: determination of the size of an orbital 3: determination of the orientation of an orbital 4: determination of the general shape of an orbital

More Practical Tools for Students Powered by AI Study Helper

Homework AI Solver

Stylized AI Paper Writer

Plagiarism Checker Assistant

Citation AI Academic Writing Tool

In-Class Translation Assistant

AI Note Generator

AI Quiz Answers

Past Exam Questions from University Test Bank

Smart Practice Assistant

Adaptive Practice

Making Your Study Simpler