The electronic configuration of a dipositive | vedclass

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(A) The electronic configuration of the dipositive ion $M^{2+}$ is given as $2, 8, 14$. The total number of electrons in $M^{2+}$ is $2 + 8 + 14 = 24$

Vedclass · Accepted Answer

$30$

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(A) The electronic configuration of the dipositive ion $M^{2+}$ is given as $2, 8, 14$. The total number of electrons in $M^{2+}$ is $2 + 8 + 14 = 24$. Since the ion has a $+2$ charge,the neutral metal atom $M$ must have $24 + 2 = 26$ electrons. Therefore,the atomic number $(Z)$ of the metal is $26$. The atomic weight $(A)$ is given as $56 \ a.m.u.$ The number of neutrons $(n)$ is calculated using the formula $n = A - Z$. $n = 56 - 26 = 30$.

Species	Number of Electrons
$(i) Be^{2+}$	$(a) 0$
$(ii) H^{+}$	$(b) 10$
$(iii) Na^{+}$	$(c) 2$
$(iv) Mg^{+}$	$(d) 11$

The electronic configuration of a dipositive metal $M^{2+}$ is $2, 8, 14$ and its atomic weight is $56 \ a.m.u.$ The number of neutrons in its nuclei would be

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