The electronic configuration 1s^{2}2s^{2}2p^{ | vedclass

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

Question

(D) The given electronic configuration is $1s^{2}2s^{2}2p^{6}3s^{1}$.Counting the total number of electrons: $2 + 2 + 6 + 1 = 11$ electrons.For $Al^{3

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) The given electronic configuration is $1s^{2}2s^{2}2p^{6}3s^{1}$.Counting the total number of electrons: $2 + 2 + 6 + 1 = 11$ electrons.For $Al^{3+}$ (atomic number $13$),the number of electrons is $13 - 3 = 10$. Configuration: $1s^{2}2s^{2}2p^{6}$.For $Ne$ (atomic number $10$),the ground state configuration is $1s^{2}2s^{2}2p^{6}$. An excited state would involve promotion of an electron to a higher orbital,such as $1s^{2}2s^{2}2p^{5}3s^{1}$.For $Mg^{+}$ (atomic number $12$),the number of electrons is $12 - 1 = 11$. The ground state configuration is $1s^{2}2s^{2}2p^{6}3s^{1}$.Since the configuration $1s^{2}2s^{2}2p^{6}3s^{1}$ represents the ground state of $Mg^{+}$,and none of the options correctly identify this as a ground state species,the correct choice is $D$.

The electronic configuration $1s^{2}2s^{2}2p^{6}3s^{1}$ corresponds to which of the following species?

Similar Questions

Nickel atom can lose two electrons to form $Ni^{2+}$ ion. The atomic number of nickel is $28$. From which orbital will nickel lose two electrons?

Statement $A$: In a potassium atom,the $19^{th}$ electron enters the $4s$-orbital instead of the $3d$-orbital.
Reason $R$: The $(n + l)$ rule is followed to determine the minimum energy of an orbital.

The azimuthal quantum number for the last electron of the $Na$ atom is:

If the value of $(n+\ell)$ is not greater than $3$,which of the following subshells is not possible $-$

Which orbital is closer to the nucleus: $2p$ or $3p$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module