Answer:
Explanation:
Formal charge of ICl₂⁻
Formal charge = group no - ( no of non bonding electrons +no of bonds)
In I there are 7 electrons in outermost orbit . If we add one more electrons due to - ve charge on the ion , it becomes eight . This centrally placed iodine forms two single bond with two chlorine atoms on either side.
Each of chlorine atoms also contains 7 valance electrons like iodine.
So formal charge of chlorine
= group no - ( no of non bonding electrons +no of bonds)
= 7 - ( 6 + 1 )
= 0
So formal charge of iodine
= group no - ( no of non bonding electrons +no of bonds)
= 7 - ( 5 + 2 )
=0
Formal charge of ICl₂⁺
In this case , central iodine will have only 6 valence electrons due to absence one electron.
So formal charge of chlorine in ICl₂⁺
= group no - ( no of non bonding electrons +no of bonds)
= 7 - ( 6 + 1 )
= 0
formal charge of iodine in in ICl₂⁺
7 - ( 4 + 2)
= 1
Answer:
Neutral ions
Explanation:
Because they have a neutral charge They can only produce neutrally.
d///////////////////////////////
Answer:
The mole ratio of the cation and the anion in a precipitate is a simple fraction. ( im sorry if this dosent help a lot.)
Explanation:
Answer:
The correct answer is option a.
Explanation:

Equilibrium concentration cadmium ions = ![[Cd^{2+}]=0.0585 M](https://tex.z-dn.net/?f=%5BCd%5E%7B2%2B%7D%5D%3D0.0585%20M)
Equilibrium concentration fluoride ions = ![[F^{-}]=0.117 M](https://tex.z-dn.net/?f=%5BF%5E%7B-%7D%5D%3D0.117%20M)
Molar solubility is the maximum concentration of salt present in water in ionic form beyond that no more salt will exist in its ionic form and will settle down in bottom of the solution.
The molar solubility of the solid cadmium fluoride = 0.0585 M
..[1]

Due to addition of sodium fluoride will increase concentration of fluoride in the solution.And due to common ion effect the equilibrium will shift in backward direction in [1], that is precipitation of more cadmium fluoride.
Hence, decrease in solubility will be observed.