Bài 2. Hạt nhân nguyên tử. Nguyên tố hóa học. Đồng vị

HH

Tổng số electron trong anion (AB3)2- là 42 . Trong hạt nhân của A và B số proton = số notron. Tính số khối A và B.

Giúp mình nha. Cảm ơn nhiều.

VT
17 tháng 8 2016 lúc 14:30

Gọi \(x;y\) là số proton trong hạt nhân \(A,B\)

Theo giả thiết : \(x+3y+2=42\leftrightarrow y=\frac{40-x}{3};\) hay \(y< \frac{40}{3}=13,3\)

Vì \(B\) là phi kim ( tạo anion ) và có \(Z< 13,3\) nên \(B\) là \(F,O,N\)

\(A\) có \(Z=13\leftrightarrow A\) là \(Al\)

Công thức anion \(AB\frac{2-}{3}\) là \(AlF\frac{2-}{3}\leftrightarrow Al^++3F^-\) , vô lí không  có  \(Al^+\)

Nếu B là O ( Z = 8 ) \(\rightarrow x=42-2-3.8=16\)

A có \(Z=16\rightarrow A\) là S . Công thức anion \(SO\frac{2-}{3}\) ( phù hợp )

Nếu B là N ( Z = 7 ) . Công thức ainon \(KN\frac{2-}{3}\rightarrow K^{7+}+3N^{3-}\) vô lí .

Vậy A : S số khối là \(16.2=32,B\) là O số khối là \(8.2=16\)

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LH
17 tháng 8 2016 lúc 14:31

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