If the electronic configuration of A^{-} is [ | vedclass

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

Question

(C) The electronic configuration of $A^{-}$ is $[Ar] \ 4s^2 \ 3d^{10} \ 4p^1$. This corresponds to a total of $18 + 2 + 10 + 1 = 31$ electrons. Since

Vedclass · Accepted Answer

Group $- 12$,Period $- 4$

Vedclass · Answer

(C) The electronic configuration of $A^{-}$ is $[Ar] \ 4s^2 \ 3d^{10} \ 4p^1$. This corresponds to a total of $18 + 2 + 10 + 1 = 31$ electrons. Since $A^{-}$ has $31$ electrons,the neutral atom $A$ has $30$ electrons,meaning its atomic number $Z = 30$. The element with $Z = 30$ is Zinc $(Zn)$,which belongs to Group $12$ and Period $4$. Removing electrons to form $A^{+2}$ (i.e.,$Zn^{+2}$) does not change the identity of the element or its position in the periodic table. Therefore,$A^{+2}$ remains in Group $12$ and Period $4$.

If the electronic configuration of $A^{-}$ is $[Ar] \ 4s^2 \ 3d^{10} \ 4p^1$,then $A^{+2}$ will belong to which group and period?

Similar Questions

The electronic configurations of four elements are:
$I. [Xe] \, 6s^1$
$II. [Xe] \, 4f^{14} \, 5d^1 \, 6s^2$
$III. [Ar] \, 4s^2 \, 4p^5$
$IV. [Ar] \, 3d^7 \, 4s^2$
Select the incorrect statement about these elements.

In what properties do the first elements of a group differ from the subsequent elements of the same group?

Which of the following statement$(s)$ is/are correct?

In which of the following sets,elements are not correctly arranged with the property shown in brackets?

Choose the incorrect order.

Vedclass Test Series

Exam Paper Generator

Online Exam Module