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

Hint: 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$.