Question

$\huge\underline\mathbb\color{plum}{QuEStiON}$
✨ Explanation needed
✨ Please don't post spam and copied answers​

Answer 1

Option C

__________________________

Explanation :

Along the period Atomi size will decrease.

Down the group size will increase.

So

Nitrogen and oxygen and Fluorine is along the period.

Size will decreases.

Here Nitrogen is longest radii than oxygen.

So

171 > 140 > 137

__________________________

Hope u understand

Tag me as brainliest

Answer 2

$\sf \bold {Given\::}$

$\sf N^{-3}$ & $\sf O^{-2}$ & $\sf F^{-1}$ are isoelectronic

$\sf \bold {To\:Find\::}$

Order of ionic radius

$\sf \bold {Solution\::}$

Number of electrons in $\sf N^{-3}$ = (7)+(3) = 10electron

Number of electrons in $\sf O^{-2}$ = (8)+(2) = 10electron

Number of electrons in $\sf F^{-1}$ = (9)+(1) = 10electron

Number of Protons in $\sf N^{-3}$ = 7

Number of Protons in $\sf O^{-2}$ = 8

Number of Protons in $\sf F^{-1}$ = 9

All three have same number of electrons that is 10 , but different number of Protons

Ionic radius depends upon Z/e , { here Z is number of proton & e is number of electrons}

As Z/e increases size of ion decrease & as size of Z/e decrease size of ion increase

Therefore , as we can see Z/e for $\sf F^{-1}$ is maximum followed by $\sf O^{-2}$ , and least for $\sf N^{-3}$

This gives that , size of ion and hence radius of ion increase from $\sf N^{-3}$ (maximum) to $\sf F^{-1}$ (least) with $\sf O^{-2}$ as intermediate

$\sf \bold {ANSWER\::}$

Ionic radius =>

$\sf N^{-3}$ = 171 pm

$\sf O^{-2}$ = 140 pm

$\sf F^{-1}$ = 136 pm