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 $4$.

Column $I$ (Element types)	Column $II$ (Electronic configuration)
$A$. Inert-gas elements	$1. (n-1)d^{1-10} ns^{1-2}$
$B$. Transition elements	$2. ns^2 np^6$
$C$. Inner-transition elements	$3. (n-2)f^{1-14} (n-1)s^2 p^6 d^{0-1} ns^2$

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

Which of the following statements is/are wrong?

Two statements are given below.
Statement $I$: Nitrogen has more ionization enthalpy and electronegativity than beryllium.
Statement $II$: $CrO_3$ and $B_2O_3$ are acidic oxides.
Correct answer is

Identify the incorrect order.

Count the number of properties which have higher value for '$O$' atom than that of '$S$': Ionisation potential, Electron affinity, Covalent radius, Electronegativity.

Vedclass Test Series

Exam Paper Generator

Online Exam Module