The electronic configuration of the phosphide | vedclass

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

Question

(D) The atomic number of phosphorus $(P)$ is $15$. The phosphide ion $(P^{3-})$ is formed by gaining $3$ electrons,so it has $15 + 3 = 18$ electrons.E

Vedclass · Accepted Answer

Chloride ion $(Cl^-)$

Vedclass · Answer

(D) The atomic number of phosphorus $(P)$ is $15$. The phosphide ion $(P^{3-})$ is formed by gaining $3$ electrons,so it has $15 + 3 = 18$ electrons.Electronic configuration of $P^{3-}$ is $1s^2 2s^2 2p^6 3s^2 3p^6$,which corresponds to the noble gas Argon $(Ar)$.Now,let us check the number of electrons in the options:$A$) Nitride ion $(N^{3-})$: $7 + 3 = 10$ electrons.$B$) Fluoride ion $(F^-)$: $9 + 1 = 10$ electrons.$C$) Sodium ion $(Na^+)$: $11 - 1 = 10$ electrons.$D$) Chloride ion $(Cl^-)$: $17 + 1 = 18$ electrons.Since the chloride ion $(Cl^-)$ has $18$ electrons,its electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^6$,which is the same as the phosphide ion $(P^{3-})$.

The electronic configuration of the phosphide ion $(P^{3-})$ is similar to that of:

Similar Questions

$A$ feasible solution to an $LP$ problem,

In the reaction,the major product formed is:

Blister copper is refined by stirring molten impure metal with green logs of wood because such a wood liberates hydrocarbon gases (like $CH_4$). This process $X$ is called . . . . . . and the metal contains impurities of $Y$ is . . . . . . .

If $z=1+i\sqrt{3}$,then $|\operatorname{Arg} z|+|\operatorname{Arg} \bar{z}|$ is equal to

Which of the following halogens undergoes disproportionation in water?

Vedclass Test Series

Exam Paper Generator

Online Exam Module