Question

Arrange in increasing irder of ionic radi f-, o2-,mg2+, na+, al3+

Answer 1



 
0

Home

»

Forum

»

Physical Chemistry

»

Consider the following species : N3–, O2–,...

Consider the following species : N3–, O2–, F–, Na+, Mg2+ and Al3+ (a) What is common in them? (b) Arrange them in the order of increasing ionic radii.

4 years ago

Answers : (2)

Hello student
-ve sign means more of electrons and +ve sign means lack of electrons.
Here N3- with atomic no. 7 has 7+3= 10 electrons
O2- witha tomic no. 8 has 8+2 = 10 electrons
F- with atomic no. 9 has 9+1= 10 electrons
Na+ with atomcatomic no. 11 has 9-1= 10 electrons
Mg2+with atomcatomic no. 12 has 12-2= 10 electrons
Al3+with atomcatomic no. 13 has 13-3= 10 electrons
Since all these ions have 10 electrons in their shell therefore these areisoelectronic speicies
The more + the charge, the smaller the ionic radius. Remember that - means adding electrons. These electrons go in the outermost shells. Also, when an atom loses electrons, it clings ever more tightly to the ones it has left, further reducing the ionic radius. therefore the order of ionic radii will be →

AL3+ Mg2+ Na+ F- O2- N3- (increasing order)


I HOPE YOU LIKE IT

Answer 2

$F^{-}$ ⇒ No of elecrons = 9 +1 =10

$O^{2-}$ ⇒ No of electrons = 8 +2 = 10

$Mg^{2+}$ ⇒ No of Electrons = 12-2=10

$Na^{+}$ ⇒ No of Electrons = 11 - 1= 10

$Al^{3+}$ ⇒ No of Electrons = 13 - 3 = 10

All has same number of Electrons !!!

So lets go to Protons ""

As number of protons increases , the ionic radius decreases

So your Answer is

$Al^{3-}$ < $Mg^{2+}$ < $Na^{+}$ < $F^{-}$<$O^{2-}$