Question

question 12 of 18
give the chemical symbol for the element with the ground - state electron configuration ar4s²3d³.
symbol:
determine the quantum numbers n and ℓ and select all possible values for m_ℓ for each subshell of the element.
4s n=
4s ℓ=
the possible values of m_ℓ for the 4s subshell are
-2,-1,0,+1,+2
-1,0,+1
-3,-2,-1,0,+1,+2,+3
0

Explanation:

Step1: Find the element

Argon (\(Ar\)) has \(18\) electrons. The electron configuration \([Ar]4s^{2}3d^{3}\) means the total number of electrons is \(18 + 2+3=23\). The element with atomic number \(23\) is vanadium (\(V\)).

Step2: Determine \(n\) for \(4s\)

The principal quantum number \(n\) for the \(4s\) sub - shell is \(4\) (from the \(4\) in \(4s\)).

Step3: Determine \(\ell\) for \(4s\)

For an \(s\) sub - shell, the angular momentum quantum number \(\ell = 0\) (since \(s\) corresponds to \(\ell=0\), \(p\) to \(\ell = 1\), \(d\) to \(\ell=2\), \(f\) to \(\ell = 3\)).

Step4: Determine \(m_{\ell}\) for \(4s\)

The magnetic quantum number \(m_{\ell}\) has values from \(-\ell\) to \(+\ell\). Since \(\ell = 0\) for \(s\) sub - shell, \(m_{\ell}=0\).

Answer:

symbol: \(V\)
\(4s\) \(n = 4\)
\(4s\) \(\ell=0\)
The possible values of \(m_{\ell}\) for the \(4s\) subshell is \(0\)