Question

which of the following sets of quantum numbers is not allowed?a $n = 2LXB0m_{i}=0$b $n = 3,LXB1m_{i}=0$c $n = 2LXB2m_{i}=2$d $n = 3,LXB3m_{i}=-1$e $n = 4,LXB4m_{i}=1$

Explanation:

Step1: Recall the relationship between quantum numbers

For a given principal quantum number \(n\), the angular - momentum quantum number \(l\) satisfies \(0\leq l\leq n - 1\). For a given \(l\), the magnetic quantum number \(m_{l}\) satisfies \(-l\leq m_{l}\leq l\).

Step2: Analyze option a

Given \(n = 2\), then \(l\) can be \(0\) or \(1\). When \(l = 0\), \(m_{l}=0\). This set of quantum numbers \((n = 2,l = 0,m_{l}=0)\) is allowed.

Step3: Analyze option b

Given \(n = 3\), then \(l\) can be \(0\), \(1\), or \(2\). When \(l = 2\), \(m_{l}\) can be \(- 2,-1,0,1,2\). This set of quantum numbers \((n = 3,l = 2,m_{l}=0)\) is allowed.

Step4: Analyze option c

Given \(n = 2\), then \(l\) can be \(0\) or \(1\) (since \(l\leq n - 1=2 - 1 = 1\)). But here \(l = 2\), which violates the rule \(l\leq n - 1\). So this set of quantum numbers \((n = 2,l = 2,m_{l}=2)\) is not allowed.

Step5: Analyze option d

Given \(n = 3\), then \(l\) can be \(0\), \(1\), or \(2\). When \(l = 1\), \(m_{l}\) can be \(-1,0,1\). This set of quantum numbers \((n = 3,l = 1,m_{l}=-1)\) is allowed.

Step6: Analyze option e

Given \(n = 4\), then \(l\) can be \(0\), \(1\), \(2\), \(3\). When \(l = 2\), \(m_{l}\) can be \(-2,-1,0,1,2\). This set of quantum numbers \((n = 4,l = 2,m_{l}=1)\) is allowed.

Answer:

The set of quantum numbers in option c (\(n = 2\), \(l = 2\), \(m_{l}=2\)) is not allowed.