Question

question 2
10 points
each of the following sets of quantum numbers is supposed to specify an orbital. choose the one set of quantum numbers that does not contain an error.
a n = 2, l = 2, ml =+1
b n = 4, l = 3, ml =-2
c n = 3, l = 2, ml =+3
d n = 2, l = -1, ml =0
question 3
10 points

Explanation:

Step1: Recall quantum - number rules

The principal quantum number \(n\) can take positive integer values (\(n = 1,2,3,\cdots\)). The angular - momentum quantum number \(l\) ranges from \(0\) to \(n - 1\). The magnetic quantum number \(m_l\) ranges from \(-l\) to \(+l\).

Step2: Analyze option A

For \(n = 2\), \(l\) can be \(0\) or \(1\) (since \(l\) ranges from \(0\) to \(n - 1\)). Here \(l = 2\) is incorrect as \(2\geq2\) is not allowed.

Step3: Analyze option B

For \(n = 4\), \(l\) can be \(0,1,2,3\) (because \(l\) ranges from \(0\) to \(n - 1\)). Since \(l = 3\) is within this range, and for \(l = 3\), \(m_l\) can range from \(- 3\) to \(+3\). Here \(m_l=-2\) is within the range \(-l\leq m_l\leq l\). So this set is correct.

Step4: Analyze option C

For \(n = 3\), \(l\) can be \(0,1,2\). Since \(l = 2\), \(m_l\) can range from \(-2\) to \(+2\). Here \(m_l = + 3\) is outside this range, so it is incorrect.

Step5: Analyze option D

The angular - momentum quantum number \(l\) cannot be negative. Here \(l=-1\) is incorrect.

Answer:

B. \(n = 4, l = 3, m_l=-2\)