The electronic configuration of a dipositive | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$30$

Vedclass · Answer

(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 $n = A - Z$. $n = 56 - 26 = 30$.

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

Similar Questions

$A$ sodium cation $(Na^+)$ has a different number of electrons from which of the following species?

The number of protons,neutrons and electrons in ${ }_{71}^{175} Lu$,respectively,are:

The spectrum of $He$ is expected to be similar to

Which of the following species is isotonic with $_{37}Rb^{86}$?

Which of the following species are isoelectronic (i.e.,have the same number of electrons)?
$Na^{+}, K^{+}, Mg^{2+}, Ca^{2+}, S^{2-}, Ar$

Vedclass Test Series

Exam Paper Generator

Online Exam Module