a)
Theo đề có: \(n+l+m_l+m_s=2,5\)
=> X phải khác H, He => \(n\ge2\)
TH1: \(n=2,m_s=+\dfrac{1}{2}\) => \(l+m_l=0\)
+) \(l=m_l=0\Rightarrow...2s^1\Rightarrow X:Li\)
+) \(l=1\Rightarrow m_l=-1\Rightarrow....2p^1\Rightarrow X:B\)
Vị trí:
\(Li\left(Z=3\right):1s^22s^1\Rightarrow\left\{{}\begin{matrix}STT:3\\CK:2\\nhóm.IA\end{matrix}\right.\)
\(B\left(Z=5\right):1s^22s^22p^1\Rightarrow\left\{{}\begin{matrix}STT:5\\CK:2\\nhóm.IIIA\end{matrix}\right.\)
TH2: \(n=2,m_s=-\dfrac{1}{2}\Rightarrow l+m_l=1\Rightarrow l=1,m_l=0\Rightarrow...2p^5\Rightarrow X:F\)
Vị trí:
\(F\left(Z=9\right):1s^22s^22p^5\Rightarrow\left\{{}\begin{matrix}STT:9\\CK:2\\nhóm:VIIA\end{matrix}\right.\)
TH3: \(n=3,m_S=-\dfrac{1}{2}\Rightarrow l+m_l=0\)
+) \(l=0\Rightarrow m_l=0\Rightarrow...3s^2\Rightarrow X:Mg\)
+) \(l=1\Rightarrow m_l=-1\Rightarrow....3p^4\Rightarrow X:S\)
+) \(l=2\Rightarrow m_l=-2\Rightarrow....3d^6\Rightarrow X:Fe\)
Vị trí:
\(Mg\left(Z=12\right):\left[Ne\right]3s^2\Rightarrow\left\{{}\begin{matrix}STT:12\\CK:3\\nhóm:IIA\end{matrix}\right.\\ S\left(Z=16\right):\left[Ne\right]3s^23p^4\Rightarrow\left\{{}\begin{matrix}STT:16\\CK:3\\nhóm.VIA\end{matrix}\right.\\ Fe\left(Z=26\right):\left[Ar\right]3d^64s^2\Rightarrow\left\{{}\begin{matrix}STT:26\\CK:4\\nhóm.VIIIB\end{matrix}\right.\)
b)
Có điều kiện \(n+l=3\Rightarrow n\ge2;m_l+m_s=+\dfrac{1}{2}\), ta có 2 trường hợp.
TH1: \(n=3\Rightarrow l=0\Rightarrow m_l=0\Rightarrow m_s=+\dfrac{1}{2}\Rightarrow....3s^1\)
=> Na
TH2: \(n=2\Rightarrow l=1\Rightarrow m_l=-1;0;+1\)
Từ điều kiện \(m_l+m_s=+\dfrac{1}{2}\) có:
+) \(m_l=0\Rightarrow m_s=+\dfrac{1}{2}\Rightarrow...2p^2\left(C\right)\)
+) \(m_l=+1\Rightarrow m_s=-\dfrac{1}{2}\Rightarrow...2p^4\left(O\right)\)
