91Ó°ÊÓ

The molecule permanganate, represented as \( \mathrm{MnO}_4^- \), contains manganese (Mn) and oxygen (O). Oxygen typically has an oxidation state of \(-2\). In a compound like this, the sum of oxidation states must equal the overall charge of the ion, which is \(-1\).
The formula to set for manganese in this scenario looks like this: \( x + 4(-2) = -1 \), where \( x \) is defined as the oxidation state of manganese. Solving this equation:

• The total contribution from oxygen is \( 4 \times (-2) = -8 \).
• Thus, \( x - 8 = -1 \).
• Solving for \( x \) gives \( x = +7 \).

Therefore, manganese in \( \mathrm{MnO}_4^- \) is in the \(+7\) oxidation state. This means that manganese atoms are highly oxidized in this compound, lending the ion significant oxidizing properties.

Chromium hexacyanide, or \( \mathrm{Cr(CN)}_6^{3-} \), consists of chromium (Cr) and cyanide ions (CN). Each cyanide ion, \( \mathrm{CN}^- \), carries a charge of \(-1\). Since there are six cyanide ions, their total charge amounts to \(-6\).
For the entire complex, which has a charge of \(-3\), the equation looks like this: \( y + 6(-1) = -3 \), with \( y \) representing the oxidation state of chromium. Solving the equation involves:

• The contribution from cyanide is \( 6 \times (-1) = -6 \).
• Thus, \( y - 6 = -3 \).
• Solving for \( y \) gives \( y = +3 \).

So, chromium is in a \(+3\) oxidation state in this complex. This intermediate oxidation state allows chromium in \( \mathrm{Cr(CN)}_6^{3-} \) to be both a reducing and an oxidizing agent under different conditions.

Nickel hexafluoride, denoted as \( \mathrm{NiF}_6^{2-} \), comprises nickel (Ni) and fluorine (F). Fluorine generally has an oxidation state of \(-1\). With six fluorine atoms, their collective contribution to the charge adds up to \(-6\).
For the whole ion, which has a \(-2\) charge, the equation can be developed as follows: \( z + 6(-1) = -2 \), where \( z \) is the oxidation state of nickel. Solving this equation involves:

• The total contribution from fluorine is \( 6 \times (-1) = -6 \).
• Thus, \( z - 6 = -2 \).
• Solving for \( z \) results in \( z = +4 \).

Therefore, nickel is in a \(+4\) oxidation state within this complex. In this state, nickel can engage in various chemical reactions, exhibiting different catalytic properties due to its unusual oxidation state.

The compound \( \mathrm{CrO}_2\mathrm{Cl}_2 \) includes chromium (Cr), oxygen (O), and chlorine (Cl). Here, oxygen typically possesses an oxidation state of \(-2\), resulting in a total of \(-4\) for the two oxygen atoms. Chlorine generally has an oxidation state of \(-1\), contributing a total of \(-2\) for the two chlorine atoms.
For a neutral compound, the equation becomes \( w + 2(-2) + 2(-1) = 0 \), with \( w \) as the oxidation state of chromium. To solve:

• The contribution from oxygen is \( 2 \times (-2) = -4 \).
• The contribution from chlorine is \( 2 \times (-1) = -2 \).
• Thus, the equation is \( w - 4 - 2 = 0 \).
• Solving for \( w \) results in \( w = +6 \).

As a result, chromium in \( \mathrm{CrO}_2\mathrm{Cl}_2 \) is in the \(+6\) oxidation state, showcasing the versatile oxidation chemistry of chromium. This high state indicates strong oxidizing behavior and significant reactivity in various chemical processes.