Chemistry, 19.09.2019 21:30 vickiw

Identify the name of the compound cr2(co3)3.

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Answer from: hwest98

The name of the compound ${\text{C}}{{\text{r}}_{\text{2}}}{\left( {{\text{C}}{{\text{O}}_{\text{3}}}} \right)_3}$ is $\boxed{{\text{chromium }}\left( {{\text{III}}} \right){\text{ carbonate}}}$ .

Further explanation:

An ionic compound is formed when there occurs complete transfer of electrons between the bonded atoms. Atoms that lose electrons form cations whereas that gain electrons forms anions. Cations are positively charged ions and anions are negatively charged ions.ns.

An ionic compound is named by naming its cation first and then by its anion.

Naming of cations:

1. The main group cations are named the same as the name of their corresponding elements. For example, ${\text{N}}{{\text{a}}^ + }$ is named sodium cation.

2. The transition metals cations are named similar to the main group elements. The only difference is that the charge of transition metal cations is written in parenthesis after the name of the cation. For example, ${\text{C}}{{\text{u}}^{2 + }}$ is named copper (II) cation.

3. The polyatomic cation is named by adding suffix “ium”.

Naming of anions:

1. The suffix “ide” is added when a monoatomic anion is formed by an element. For example, ${{\text{H}}^ - }$ is named as a hydride.

2. When two oxyanions are formed by a single element, the oxyanion with fewer oxygen atoms is written with suffix “ite” and that with more oxygen is named with suffix “ate”. For example, ${\text{NO}}_2^ -$ is named nitrite while ${\text{NO}}_3^ -$ is named nitrate.

3. If a series of oxyanions are formed by an element, prefix “hypo” is used when fewer oxygen atoms are present and prefix “per” is used if more oxygen atoms are present. For example, ${\text{Cl}}{{\text{O}}^ - }$ is named hypochlorite while ${\text{ClO}}_4^ -$ is named as perchlorate.

The given compound is ${\text{C}}{{\text{r}}_{\text{2}}}{\left( {{\text{C}}{{\text{O}}_{\text{3}}}} \right)_{\text{3}}}$ . Since Cr is a transition metal so its oxidation state has to be written parantheses.

The expression to calculate the oxidation state in ${\text{C}}{{\text{r}}_{\text{2}}}{\left( {{\text{C}}{{\text{O}}_{\text{3}}}} \right)_{\text{3}}}$ is:

$\left[ {2\left( {{\text{oxidation state of Cr}}} \right) + 3\left( {{\text{oxidation state of C}}} \right) + 9\left( {{\text{oxidation state of O}}} \right)} \right] = 0$ … (1)

Rearrange equation (1) for the oxidation state of Cr.

${\text{Oxidation state of Cr}} = \dfrac{{\left[ { - 3\left( {{\text{oxidation state of C}}} \right) - 9\left( {{\text{oxidation state of O}}} \right)} \right]}}{2}$ …… (2)

Substitute +4 for the oxidation ${\text{state}}$ of C and -2 for the oxidation state of O in equation (2).

$\begin{aligned}{\text{Oxidation state of Cr}} &= \frac{{\left[ { - 3\left( { + {\text{4}}} \right) - 9\left({ - {\text{2}}} \right)} \right]}}{2}\\&= \frac{{ - 12 + 18}}{2}\\&=+ 3\\\end{aligned}$

Therefore the oxidation state of Cr in ${\text{C}}{{\text{r}}_{\text{2}}}{\left( {{\text{C}}{{\text{O}}_{\text{3}}}} \right)_{\text{3}}}$ is +3.

The formula of any compound is formed by crisscross rule. So the charge on the anion will be -2 and the corresponding anion is ${\text{CO}}_3^{2 - }$ whose name is carbonate. Therefore the name of the given compound is chromium (III) carbonate.