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
<span>The form of emission that is commonly not written in nuclear equations because they do not affect charges, atomic numbers, or mass numbers is the gamma ray.
Gamma rays have no electrical charge nor atomic mass. It is only used to balance in a nuclear reaction, but gamma ray is considered powerful.
So, letter C. is the answer.
Hope this helps!</span>
The correct answer is A.) dilute
No. This is a double displacement reaction.
This is because the only spectator ion is potassium ion, K+. Sulfuric acid is a strong acid, but only for the first H+ which is produced. The remaining HSO4^- ion is a weak acid, and is only minimally ionized. (*) Potassium hydroxide is a strong base and is completely ionized. As for the products, K2SO4 is completely ionized, but water is not. H+ and OH- combine to make molecular water
H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + 2HOH(l) ….. “molecular” equation
H+ + HSO4^- + 2K+ + 2OH- → 2K+ + SO4^2- + 2HOH(l) …. ionic equation
H+ + HSO4^- + 2OH → SO4^2- + 2HOH(l) ……. net ionic equation
I think its A but im not completely sure.