Chapter 5, Problem 37A

Interpretation Introduction

Interpretation: The invalid orbital designation needs to be identified.

Concept Introduction: Quantum numbers are defined as a set of numbers required to explain the position of an electron in an atom. The four quantum numbers are principal quantum number ( n ), azimuthal quantum number ( l ), magnetic quantum number ( m ), and spin quantum number (s). For any orbital of a known principal quantum number, the azimuthal quantum number is up to $n-1$ . Also, for s orbital the value of an azimuthal quantum number is 0, for p orbital, the value is 1 and for d orbital, the value is 2.

Answer to Problem 37A

The correct answer is option (b) and option (c).

Explanation of Solution

Reason for correct option/s:

In option (b), the given orbital is 2d . Thus, the principal quantum number is 2. Now, there will be up to $n-1$ or $2-1=1$ azimuthal quantum numbers, that is, 0 and 1.

Here, 0 indicates s orbital and 1 indicates p orbital. Since the d orbital is not possible for principal quantum number 2, the 2d orbital is an invalid designation.

In option (c), the given orbital is 3f . Thus, the principal quantum number is 3. Now, there will be up to $n-1$ or $3-1=2$ azimuthal quantum numbers, that is, 0, 1, and 2.

Here, 0 indicates s orbital, 1 indicates p orbital and 2 indicates d orbital. Since the f orbital is not possible for principal quantum number 3, the 3f orbital is an invalid designation.

Therefore, options (b) and (c) are correct.

Conclusion

Reason for incorrect option/s:

In option (a), the given orbital is 4s . Thus, the principal quantum number is 4. Now, there will be up to $n-1$ or $4-1=3$ azimuthal quantum numbers that are 0, 1, 2, and 3.

Here, 0 indicates s orbital, 1 indicates p orbital, 2 indicates d orbital and 3 indicates f orbital. Thus, 4s orbital is a valid designation.

In option (d), the given orbital is 3p . Thus, the principal quantum number is 3. Now, there will be up to $n-1$ or $3-1=2$ azimuthal quantum numbers that is 0, 1, and 2.

Here, 0 indicates s orbital, 1 indicates p orbital and 2 indicates d orbital. Thus, 3p orbital is a valid designation.

Therefore, options (a) and (d) are incorrect.