I think the guy who picked b the last time is probably right? He seems to know what hes talking about in every other question I have seen him answer.
Mercury (II) chloride reacts with excess KI to yield _____.
a)Hg2I2
b)HgI4
c)HgI
2006-10-18 22:10:11 · 4 answers · asked by Ruphert J 1 in Science & Mathematics ➔ Chemistry
mercury (ii) iodide is precipitated. HgI2.
excess KI would indicate a redox reaction. mercury (ii) is reduced accordingly to mercury (i). however the answer is not C, because HgI cannot exist.
the actual compound formed is I-Hg-Hg-I, which is technically still a mercury (i) compound. each mercury atom is covalently bonded to aanother Hg atom and ionically bonded to a iodide ion.
the molecular formula of the compound is Hg2I2, (a).
2006-10-18 22:26:52 · answer #1 · answered by Anonymous · 0⤊ 0⤋
I- is a reducing agent, and reduces mercury(II) to mercury(I).
Hg2I2 is called mercury(I) iodide. The excess iodide ions are needed to ensure precipitation.
2006-10-18 22:51:08 · answer #2 · answered by Gervald F 7 · 0⤊ 0⤋
I would go with HgI2
is A really Hg(II)I2 ????
that is Mercuric iodide,
If it changes its valence state, then Hg2I2 can exist
2006-10-18 22:34:36 · answer #3 · answered by Slave to JC 4 · 0⤊ 0⤋
I think its (c). Bcuz mercury has a valency +2 and chlorine -2, so they will cancel each other; I don't think there is any compound like Hg2I2.
2016-05-22 01:32:31 · answer #4 · answered by Anonymous · 0⤊ 0⤋