The electronic configuration of a divalent me | vedclass

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(B) The electronic configuration of $M^{2+}$ is $2, 8, 14$. Total number of electrons in $M^{2+} = 2 + 8 + 14 = 24$. Since $M^{2+}$ is formed by the l

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$32$

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(B) The electronic configuration of $M^{2+}$ is $2, 8, 14$. Total number of electrons in $M^{2+} = 2 + 8 + 14 = 24$. Since $M^{2+}$ is formed by the loss of $2$ electrons,the number of electrons in the neutral atom $M$ is $24 + 2 = 26$. Therefore,the atomic number $(Z)$ of the metal $M$ is $26$. The atomic mass $(A)$ is given as $58 \, a.m.u.$ The number of neutrons $(N)$ is calculated as $N = A - Z$. $N = 58 - 26 = 32$.

The electronic configuration of a divalent metal ion $M^{2+}$ is $2, 8, 14$ and its atomic mass is $58 \, a.m.u.$ What is the number of neutrons in its nucleus?

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