Problem 11 - Chemistry: The Central Science

Question

A complex is written as \(\mathrm{NiBr}_{2} \cdot 6 \mathrm{NH}_{3}\) (a) What is the oxidation state of the \(\mathrm{Ni}\) atom in this complex? (b) What is the likely coordination number for the complex? (c) If the complex is treated with excess \(\mathrm{AgNO}_{3}(a q)\), how many moles of AgBr will precipitate per mole of complex?

Step-by-Step Solution

Answer

(a) The oxidation state of the \(\mathrm{Ni}\) atom is +2. (b) The likely coordination number for the complex is 6. (c) 2 moles of \(\mathrm{AgBr}\) will precipitate per mole of complex when treated with excess \(\mathrm{AgNO}_{3}(aq)\).

Step 1: Determine the oxidation state of the Ni atom

To find the oxidation state of the \(\mathrm{Ni}\) atom in this complex, we will first analyze the charges of the other components in the complex. In the complex \(\mathrm{NiBr}_{2} \cdot 6 \mathrm{NH}_{3}\), the oxidation state of \(\mathrm{Br}\) is -1, and the \(\mathrm{NH}_{3}\) ligands are neutral, meaning they have an oxidation state of 0. In order to balance the overall charge of the complex, the oxidation state of the \(\mathrm{Ni}\) atom can be calculated using the following equation: Oxidation state of \(\mathrm{Ni}\) = Total charge - (Oxidation state of \(\mathrm{Br}\) x Number of \(\mathrm{Br}\)) - (Oxidation state of \(\mathrm{NH}_{3}\) x Number of \(\mathrm{NH}_{3}\)) The total charge of the complex is 0, as it is a neutral compound. So the oxidation state of \(\mathrm{Ni}\) can be calculated as: Oxidation state of \(\mathrm{Ni}\) = 0 - (-1 x 2) - (0 x 6) = 0 + 2 = +2 Thus, the oxidation state of the \(\mathrm{Ni}\) atom in this complex is +2.

Step 2: Determine the likely coordination number for the complex

The coordination number is defined as the number of coordination bonds (or ligands) attached to the central metal atom/ion. In this complex, the ligands are the \(\mathrm{NH}_{3}\) molecules, which form coordination bonds with the central \(\mathrm{Ni}\) atom. Since there are 6 \(\mathrm{NH}_{3}\) ligands in the complex, the coordination number for this complex is 6.

Step 3: Calculate the moles of AgBr precipitated per mole of complex

When the \(\mathrm{NiBr}_{2} \cdot 6 \mathrm{NH}_{3}\) complex is treated with excess \(\mathrm{AgNO}_{3}(a q)\), the following reaction occurs: \(\mathrm{NiBr}_{2} \cdot 6 \mathrm{NH}_{3 (s)} + 2 \mathrm{AgNO}_{3 (aq)} \rightarrow 2 \mathrm{AgBr (s)} + \mathrm{Ni(NO}_{3)_{2 (aq)} + 6\mathrm{NH}_{3 (aq)}\) From the balanced reaction equation, we can see that 2 moles of \(\mathrm{AgBr}\) are formed for each mole of the complex reacted. Therefore, 2 moles of \(\mathrm{AgBr}\) will precipitate per mole of complex when treated with excess \(\mathrm{AgNO}_{3}(aq)\). To summarize: (a) The oxidation state of the \(\mathrm{Ni}\) atom is +2. (b) The likely coordination number for the complex is 6. (c) 2 moles of \(\mathrm{AgBr}\) will precipitate per mole of complex when treated with excess \(\mathrm{AgNO}_{3}(aq)\).

Key Concepts

Oxidation State

Chemists often need to determine the oxidation state of an element to understand the electron configuration in a compound. This is crucial for predicting the behavior and properties of the compound. In coordination chemistry, the oxidation state is the charge of the metal if all bonds were ionic.

For the \( \mathrm{NiBr}_2 \cdot 6 \mathrm{NH}_3\) complex, nickel's oxidation state can be found by considering the charges of its components. Bromine (\( \mathrm{Br} \)) typically has an oxidation state of -1. With two bromine atoms, we have a total negative charge of -2. The ammonia (\( \mathrm{NH}_3 \)) molecules are neutral, contributing no charge.

• Total charge of the complex: 0 (it's neutral)
• Oxidation state of \( \mathrm{Ni} \): 0 - (2 x -1) = +2

Thus, the nickel in this complex has an oxidation state of +2. By understanding oxidation states, chemists can predict reactions and properties of compounds, which is a fundamental skill in chemistry.

Coordination Number

The coordination number in coordination chemistry is the number of ligand atoms directly bonded to the central metal. It indicates how "packed" the complex ion is. In the nickel complex \(\mathrm{NiBr}_{2} \cdot 6 \mathrm{NH}_{3}\), ammonia molecules act as ligands.

• Ligands are molecules or ions that donate a pair of electrons to the metal.
• Typically, the number of ligands or donor atoms gives the coordination number.

In our example, six \(\mathrm{NH}_{3}\) ligands surround the nickel atom and coordinate with it. Hence, the coordination number for this complex is 6. This sheds light on the structure and potential geometry of the complex, which can affect its reactivity and application.

Precipitation Reaction

Precipitation reactions occur when two solutions react and form an insoluble solid, called the precipitate. In our scenario, treating the nickel complex \(\mathrm{NiBr}_{2} \cdot 6 \mathrm{NH}_{3}\) with \(\mathrm{AgNO}_3\) leads to such a reaction.

• The \(\mathrm{Br}^{-}\) ions from the complex react with \(\mathrm{Ag}^{+}\) ions from \(\mathrm{AgNO}_3\).
• This produces \(\mathrm{AgBr}\), a classic white precipitate.

The chemical equation for this reaction is:\[\mathrm{NiBr}_{2} \cdot 6 \mathrm{NH}_{3(s)} + 2 \mathrm{AgNO}_{3(aq)} \rightarrow 2 \mathrm{AgBr(s)} + \mathrm{Ni(NO}_{3})_{2(aq)} + 6 \mathrm{NH}_{3(aq)}\]According to the equation, 2 moles of \(\mathrm{AgBr}\) precipitate for every mole of the original nickel complex that reacts. Being able to predict and understand precipitation reactions is essential for laboratory work and industrial applications where separating substances is necessary.