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 \(\mathrm{AgBr}\) will precipitate per mole of complex?

Short answer

(a) The oxidation state of Ni in the complex is +2. (b) The coordination number for the complex is 6. (c) 2 moles of AgBr will precipitate per mole of complex when treated with excess AgNO3(aq).

Step-by-step solution

Calculate the oxidation state of Ni

In order to find the oxidation state of the Ni atom, we have to look at the other components in the complex. We know that Br has an oxidation state of -1 and NH3 has an oxidation state of 0 since it is a neutral molecule. The formula of the complex is NiBr2 · 6 NH3, which can be written as follows by separating the components: Ni + 2 Br + 6 NH3 Let x be the oxidation state of the Ni atom. Since the compound is neutral, the sum of the oxidation states of all the atoms should be equal to zero. Therefore, the equation for the oxidation state can be written as follows: x + (-1)x2 + (0)x6 = 0 Solving for x: x - 2 = 0 x = 2 So, the oxidation state of Ni in this complex is +2.

Determine the coordination number

Coordination number is the number of ligands that are attached to the central metal ion in a complex. In this complex, the ligands are NH3 molecules. Since there are 6 NH3 molecules in the given complex (NiBr2 · 6 NH3), the coordination number for the complex is 6.

Calculate the moles of AgBr precipitate per mole of complex

In this step, we want to verify how many moles of AgBr will precipitate per mole of complex when the complex is treated with excess AgNO3(aq). Since we have excess AgNO3, this means that Ag ions will react with Br ions present in the given complex. Keep in mind that there are 2 Br ions per 1 mole of the complex in the compound NiBr2 · 6 NH3. The balanced equation for the precipitation reaction between NiBr2 and AgNO3 is: NiBr2(aq) + 2AgNO3(aq) -> Ni(NO3)2(aq) + 2 AgBr(s) From this equation, it is evident that 2 moles of AgBr will form for each mole of NiBr2 that reacts with AgNO3. But we have NiBr2 in excess, so it's safe to say that 2 moles of AgBr will precipitate per mole of complex. To summarize: (a) The oxidation state of Ni in the complex is +2. (b) The coordination number for the complex is 6. (c) 2 moles of AgBr will precipitate per mole of complex when treated with excess AgNO3(aq).

Key concepts

Oxidation State

Understanding the oxidation state is crucial in coordination chemistry as it defines the electron configuration of the central metal atom or ion within a coordination complex. For example, in the complex \(\mathrm{NiBr}_{2} \cdot 6 \mathrm{NH}_{3}\), calculating the oxidation state of nickel (Ni) is done by considering the known oxidation states of the other components. Bromine (Br) typically has an oxidation state of -1, while ammonia (NH3) is neutral, having an oxidation state of 0. The overall charge of the complex is neutral, enabling us to set up an equation that sums the oxidation states to zero and solve for the oxidation state of Ni. In this case, Ni has an oxidation state of +2.

Briefly and simply put, the oxidation state indicates the number of electrons an atom can lose, gain, or share at the time of bond formation. It helps in predicting the stability, reactivity, and magnetic properties of the complex.

Coordination Number

Coordination number plays a pivotal role in specifying the structure and geometry of coordination complexes. It refers to the number of ligand attachment points to the central metal ion or atom. In our textbook problem, the coordination number can be determined by counting the molecules or ions directly bonding with Ni. With six ammonia (NH3) molecules surrounding the Ni ion within the complex \(\mathrm{NiBr}_{2} \cdot 6 \mathrm{NH}_{3}\), Ni displays a coordination number of 6.

This number not only provides information about the shape of the complex—a coordination number of 6 often leads to an octahedral geometry—but also affects the physical and chemical properties of the substance. Coordination numbers can vary widely among different metals and ligands, making them a fundamental aspect to grasp for students exploring transition metal chemistry.

Precipitation Reaction

A precipitation reaction is an essential type of reaction in chemistry, where soluble ions in separate solutions come together to form an insoluble ionic compound that precipitates out of solution. It is a testament to the axiom 'opposites attract', as it involves the attractive forces between cations and anions. The exercise presented involves a precipitation reaction where the complex \(\mathrm{NiBr}_{2} \cdot 6 \mathrm{NH}_{3}\) is treated with an excess of \(\mathrm{AgNO}_{3}(aq)\), leading to the formation of \(\mathrm{AgBr}\) as a precipitate.

In our example, the reaction proceeds until all the bromide (Br-) ions from NiBr2 have reacted with the silver ions (Ag+) from AgNO3 to form the solid AgBr. Since each molecule of NiBr2 provides two bromide ions, and each bromide ion produces one AgBr molecule, two moles of AgBr will precipitate per mole of the complex. This step-by-step approach to analyzing a precipitation reaction assists students in visualizing how stoichiometry governs the amounts of products formed in a chemical reaction.