Răspuns :
a)
[tex]\bf Al=Aluminiu \implies \begin{cases}\bf Z=13 \\ \\ \bf A=27 \end{cases}[/tex]
[tex]\bf Z=p^+ =e^- = 13 \implies \begin{cases} \bf protoni=electroni=13 \\ \\ \bf neutroni = A-Z=27-13=14 \end{cases}[/tex]
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[tex]\bf Cl=Clor \implies \begin{cases} \bf Z=17 \\ \\ \bf A=35 \end{cases}[/tex]
[tex]\bf Z=p^+ =e^- = 17 \implies \begin{cases} \bf protoni=electroni=17 \\ \\ \bf neutroni = A-Z=35-17=18 \end{cases}[/tex]
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b)
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[tex]\bf Al=Aluminiu \; \rightarrow \; \mathsf{sarcina\; nucleara} = Z = 13[/tex]
[tex]\bf Cl=Clor \; \rightarrow \; \mathsf{sarcina\; nucleara} = Z=17[/tex]
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Învelișurile unui atom sunt K, L, M, N, O, P, Q :
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[tex]\boxed{\bf N_{max} = 2 \cdot n^2}[/tex]
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[tex](1)K \rightarrow 2e^-[/tex]
[tex](2)L \rightarrow 8e^-[/tex]
[tex]3(M) \rightarrow 8e^- \rightarrow 18e^-[/tex]
[tex]4(N) \rightarrow 8e^- \rightarrow 18e^- \rightarrow 32e^-[/tex]
[tex]5(O) \rightarrow 8e^- \rightarrow 18e^- \rightarrow 32e^-[/tex]
[tex]6(P) \rightarrow 8e^- \rightarrow 18e^- \rightarrow 32e^-[/tex]
[tex]7(Q) \rightarrow 8e^- \rightarrow 18e^- \rightarrow 32e^-[/tex]
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Condiție: Pe ultimul strat avem maxim 8 electroni.
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[tex]Al \implies \begin{cases} K=2e^- \\ \\ L=8e^- \\\\ M= 3e^- \end{cases}[/tex]
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[tex]Cl \implies \begin{cases} K=2e^- \\ \\ L=8e^- \\\\ M=7e^- \end{cases}[/tex]
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c) In atașamente.
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d) Aluminiu: perioada 3, grupa III A
Clor: perioada 3, grupa VII A
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