Question

1. give 4 quantum numbers for the following elements:

a. cs n=
l=
m=
m_s=
b. i n=
l=
m=
m_s=

Explanation:

Part a: Cesium (Cs)

Step 1: Determine the principal quantum number (\(n\))

Cesium (Cs) has an atomic number of 55. Its electron configuration is \([Xe] 6s^1\). The outermost electron is in the \(n = 6\) shell.
\(n = 6\)

Step 2: Determine the azimuthal quantum number (\(l\))

For an \(s\)-orbital (the outermost electron of Cs is in an \(s\)-orbital), the value of \(l\) for \(s\)-orbitals is \(0\) (since \(l = 0\) for \(s\), \(l = 1\) for \(p\), \(l = 2\) for \(d\), \(l = 3\) for \(f\), etc.).
\(l = 0\)

Step 3: Determine the magnetic quantum number (\(m\))

For \(l = 0\), the magnetic quantum number \(m\) can only have the value \(0\) (because \(m\) ranges from \(-l\) to \(+l\), so when \(l = 0\), \(m = 0\)).
\(m = 0\)

Step 4: Determine the spin quantum number (\(m_s\))

The spin quantum number can be either \(+\frac{1}{2}\) or \(-\frac{1}{2}\). For the single electron in the \(6s\) orbital of Cs, we can take \(m_s = +\frac{1}{2}\) (the choice between \(+\frac{1}{2}\) and \(-\frac{1}{2}\) is arbitrary here, but we'll use \(+\frac{1}{2}\) as a common choice).
\(m_s = +\frac{1}{2}\)

Step 1: Determine the principal quantum number (\(n\))

Iodine (I) has an atomic number of 53. Its electron configuration is \([Kr] 4d^{10} 5s^2 5p^5\). The outermost electrons are in the \(n = 5\) shell (specifically, the \(5p\) subshell for the valence electrons we can consider for quantum numbers; we'll take a valence electron in the \(5p\) subshell).
\(n = 5\)

Step 2: Determine the azimuthal quantum number (\(l\))

For a \(p\)-orbital (the outermost valence electrons of I are in \(p\)-orbitals), the value of \(l\) for \(p\)-orbitals is \(1\) (since \(l = 0\) for \(s\), \(l = 1\) for \(p\), etc.).
\(l = 1\)

Step 3: Determine the magnetic quantum number (\(m\))

For \(l = 1\), the magnetic quantum number \(m\) can range from \(-1\), \(0\), to \(+1\). For a \(p\)-orbital electron, we can choose one of these values. Let's take \(m = +1\) (the choice is arbitrary among the possible values for \(l = 1\); other valid choices are \(-1\) or \(0\)).
\(m = +1\) (or \(-1\) or \(0\))

Step 4: Determine the spin quantum number (\(m_s\))

The spin quantum number can be either \(+\frac{1}{2}\) or \(-\frac{1}{2}\). For an electron in the \(5p\) orbital of I, we can take \(m_s = +\frac{1}{2}\) (or \(-\frac{1}{2}\); the choice is arbitrary).
\(m_s = +\frac{1}{2}\) (or \(-\frac{1}{2}\))

Answer:

(for Cs):
\(n = 6\), \(l = 0\), \(m = 0\), \(m_s = +\frac{1}{2}\) (or \(m_s = -\frac{1}{2}\))

Part b: Iodine (I)