Question

In the equation \(4 \mathrm{M}+8 \mathrm{CN}^{-}+2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2} \longrightarrow 4\left[\mathrm{M}(\mathrm{CN})_{2}\right]^{-}+\) \(4 \mathrm{OH}^{-}\) identify the metal \(\mathrm{M}\). (a) zinc (b) iron (c) copper (d) silver

Short answer

The metal 'M' is silver (d).

Step-by-step solution

Understanding the Reaction

The equation provided is a balanced chemical reaction where a metal "M" reacts with cyanide ions, water, and oxygen to form a complex metal cyanide and hydroxide ions. Our goal is to determine which metal (zinc, iron, copper, or silver) corresponds to 'M'.

Analyzing Metal Cyanide Complexes

In this equation, the metal 'M' forms a complex with cyanide ions: \[ [M(CN)_2]^- \]. Based on common metal-cyanide complexes, metals like silver (Ag), zinc (Zn), iron (Fe), and copper (Cu) can form such complexes.

Identifying Common Reactions

We need to consider which of these metals typically form stable complexes with cyanide ions, such as the ones mentioned in the equation. Silver typically forms the complex \[ [Ag(CN)_2]^- \], a known silver cyanide complex.

Eliminating Options

Zinc, iron, and copper can also form complexes with cyanide but none conform precisely to the named complex in the equation. Silver is unique in forming a \[ [Ag(CN)_2]^- \] which corresponds directly to the reactants and products in this equation.

Selecting the Metal

Based on the analysis, silver is the metal 'M' that, when reacting with cyanide, forms the complex \[ [M(CN)_2]^- \]. Therefore, 'M' is most likely silver here.

Key concepts

Metal Cyanide Complexes

Metal cyanide complexes are interesting chemistry compounds formed when metals interact with cyanide ions. These complexes comprise a central metal atom bonded to one or more cyanide ions. The general formula in this scenario is \([M(CN)_2]^-\). These complexes are often stable and find use in various industrial processes.

- **Silver** (Ag) is known for forming a stable \([Ag(CN)_2]^-\) complex, making it quite prevalent in silver plating.- **Zinc** (Zn), **iron** (Fe), and **copper** (Cu) can also form metal-cyanide complexes, though they aren't as commonly highlighted in basic chemistry discussions.
The formation of these complexes involves coordination bonds, where the cyanide ions donate electrons to the metal. This creates a complex that varies in its solubility and reactivity based on the metal and the exact configuration.

Understanding these complexes can help predict the behavior and reaction pathways of metals in various environments, especially in industrial contexts like metal extraction and electroplating.

Identifying Metals in Reactions

When identifying metals in a chemical reaction, it’s crucial to consider the nature and characteristics of the metals involved. Each metal has unique properties and tendencies in chemical bonding and reaction formation.

- **Silver (Ag):** Often forms stable complexes like \([Ag(CN)_2]^-\). It is known for its lustrous appearance and excellent conductivity.- **Zinc (Zn):** Forms \[Zn(CN)_4]^{2-}\] complexes. Often used in galvanization.- **Iron (Fe):** Generally forms Fe(CN)₆ complexes. Recognized for its strength and magnetic properties.- **Copper (Cu):** Known for \[Cu(CN)_3]^{2-}\] complexes, often used in electrical wiring and plumbing.
By looking at which metals commonly form specific types of stable complexes, we can deduce which metal might be involved in a particular reaction scheme. For instance, if you see a \([M(CN)_2]^-\) complex, silver is a typical candidate due to its known stability with this formation.

Balancing Chemical Equations

Balancing chemical equations is a fundamental aspect of chemistry that's essential to describe reactions accurately. Each side of a chemistry equation must have the same number of atoms of each element, which reflects the law of conservation of mass.

To balance an equation, follow these steps:- **Identify the most complex substance** - Typically, start with the compound that has the most elements involved.- **Balance the elements one at a time** - Usually, it’s easiest to balance metals first, then non-metals, and leave hydrogen and oxygen for last.
In the given reaction:\[4 \mathrm{M} + 8 \mathrm{CN}^{-} + 2 \mathrm{H}_{2}\mathrm{O} + \mathrm{O}_{2} \longrightarrow 4\left[\mathrm{M}(\mathrm{CN})_{2}\right]^{-} + 4 \mathrm{OH}^{-}\]Each element and ion is balanced:- **M (metal):** 4 atoms on each side.- **CN- (cyanide):** 8 ions on each side.- **H₂O (water) and O₂ (oxygen):** 2 water molecules and 1 oxygen molecule on the reactant side to produce 4 hydroxide ions.
Practicing the balancing of equations helps improve understanding of the concept and prepares students for tackling more complex chemical reactions.