Question

Calomel is:
A. $H{g_2}C{l_2}$
B. $HgC{l_2}$
C. $H{g_2}C{l_2} + Hg$
D. $Hg + HgC{l_2}$

Answer 1

The chemical name of Calomel is mercurous chloride. Calomel is the principle and most significant example of a mercury (I) compound. Calomel is a diamagnetic compound with a metal-metal bond between 2 mercurous ions.

Complete step by step answer:
Now, we will look at the compounds given in the options:
The first compound is $H{g_2}C{l_2}$
$H{g_2}C{l_2}$is the chemical formula of mercurous chloride i.e. calomel. Here, mercury exists as a mercurous ion i.e. in the form of $H{g_2}^{2 + }$ with oxidation state $+ 1$. The $+ 1$ oxidation state of mercury is however unstable with unpaired electrons in 6s orbital. A metal-metal bond is formed between these two mercuric ions.
The second compound is $HgC{l_2}$
The chemical name of $HgC{l_2}$ is mercuric chloride. In this compound, mercury exists as a mercuric ion i.e. in the form of $H{g^{2 + }}$ with oxidation state $+ 2$. Mercuric chloride is a highly toxic compound.
The third compound is $H{g_2}C{l_2} + Hg$
This is the combination of mercurous chloride and elemental mercury. Here one mercury exist as mercurous ion($H{g_2}^{2 + }$ with oxidation state $+ 1$) and another in elemental state i.e. $H{g^0}$ with oxidation state 0.
The fourth compound is $Hg + HgC{l_2}$
The above compound is a combination of elemental mercury ($Hg$) and mercuric chloride $HgC{l_2}$. In this compound, one mercury exists in elemental form with oxidation state 0 and the other one exist as mercuric ion i.e. $H{g^{2 + }}$ with oxidation as $+ 2$. Thus, option A is the correct option because the first compound has $+ 1$ oxidation state and its chemical name is mercurous chloride.

So, the correct answer is Option A.

Note: The oxidation state of the free element is always zero. For example: $H{g^0}$
The oxidation state of the electronegative element of group 17 (example - chlorine) in binary compounds is $- 1$.
Thus, the oxidation state of $H{g_2}C{l_2}$can be found out as:
$2(x) + 2( - 1) = 0$
Here, $x$ is the oxidation state of $Hg$
By solving this equation, we get the oxidation state of mercury as $+ 1$.