Problem 70 - Chemistry The Central Science

Question

Write the formulas for the following compounds, and indicate the oxidation state of the group 4 A element or of boron in each: (a) silicon dioxide, (b) germanium tetra chloride, (c) sodium borohydride, (d) stannous chloride, (e) diborane, (f) boron trichloride.

Step-by-Step Solution

Answer

(a) Silicon dioxide formula is SiO_2, and the oxidation state of Si is +4. (b) Germanium tetra chloride formula is GeCl_4, and the oxidation state of Ge is +4. (c) Sodium borohydride formula is NaBH_4, and the oxidation state of B is -3. (d) Stannous chloride formula is SnCl_2, and the oxidation state of Sn is +2. (e) Diborane formula is B_2H_6, and the oxidation state of both B atoms is +3. (f) Boron trichloride formula is BCl_3, and the oxidation state of B is +3.

Step 1: (a) Silicon dioxide formula and oxidation state

Silicon (Si) is from Group 4A, and oxygen (O) is in Group 6A. The formula for silicon dioxide can be determined by finding the lowest common multiple of their valence electrons to form a neutral compound. Since silicon has 4 valence electrons and oxygen has 6, the compound formula is SiO_2. In this case, the oxidation state of Si is +4 because each oxygen is taking two electrons from silicon.

Step 2: (b) Germanium tetra chloride formula and oxidation state

Germanium (Ge) is from Group 4A, and chlorine (Cl) is in Group 7A. We can determine the compound formula similarly by finding the lowest common number of valence electrons needed to form a neutral compound. Since germanium has 4 valence electrons and chlorine has 7, the compound formula is GeCl_4. The oxidation state of germanium in this compound is +4 because each chlorine contributes a -1 charge.

Step 3: (c) Sodium borohydride formula and oxidation state

Sodium (Na) is from Group 1A, boron (B) is from Group 3A, and hydrogen (H) is in Group 1A. In this case, the negative hydrogen acts as a hydride ion (H^-). Because sodium has a +1 charge and boron requires three electrons from hydrogen to get the lowest common multiple of their valence electrons, the compound forms NaBH_4. The oxidation state of boron in this compound is -3 because it gains three electrons from hydrogen.

Step 4: (d) Stannous chloride formula and oxidation state

"Stannous" indicates that this is a tin(II) compound, which means tin (Sn) has an oxidation state of +2. Chlorine (Cl) has a -1 charge value in the compound. To create a neutral compound, we need two chlorine atoms for each tin atom, resulting in a formula of SnCl_2. The oxidation state of tin is +2, as already mentioned.

Step 5: (e) Diborane formula and oxidation state

Diborane refers to a compound with two boron atoms (B) and six hydrogen atoms (H), therefore the formula is B_2H_6. As it's a unusual compound, both boron atoms have an oxidation state of +3 in diborane. Note, in this case, we need to know the structure of the compound instead of using valence electrons.

Step 6: (f) Boron trichloride formula and oxidation state

Boron (B) is from Group 3A, and chlorine (Cl) is in Group 7A. The lowest common multiple of their valence electrons needed to form a neutral compound is three for boron and one for chlorine. Thus, the compound formula is BCl_3. The oxidation state of boron in this compound is +3 because it loses three electrons to the three chlorine atoms.

Key Concepts

Oxidation State

Understanding the oxidation state of a chemical compound is crucial in learning how elements interact within a compound. Oxidation state, also known as oxidation number, indicates the degree of oxidation of an atom in a chemical compound. It represents the hypothetical charge that an atom would have if all bonds to atoms of different elements were completely ionic.
For instance, in the compound silicon dioxide (SiO\(_2\)), silicon (Si) has an oxidation state of +4. This is because each oxygen (O) atom, being more electronegative, "steals" two electrons from silicon. Therefore, since there are two oxygen atoms, the silicon is considered to have lost four electrons, leading to its +4 oxidation state.

• Positive oxidation states indicate loss of electrons.
• Negative oxidation states indicate gain of electrons.
• In a neutral compound, the sum of the oxidation states equals zero.

Knowing the oxidation state assists in predicting the types of reactions compounds can undergo and their reactivity.

Group 4A Elements

Group 4A elements of the periodic table include carbon (C), silicon (Si), germanium (Ge), tin (Sn), and lead (Pb). These elements belong to the p-block and are characterized by having four valence electrons. The chemistry of these elements, especially silicon and germanium, is quite versatile due to their ability to form a variety of compounds.
Silicon, for example, commonly forms compounds like silicon dioxide (SiO\(_2\)), which is prevalent in the earth's crust as quartz. For silicon, the typical oxidation state in its compounds is +4, reflecting the element's tendency to share its four valence electrons.
Germanium is another Group 4A element and forms compounds like germanium tetrachloride (GeCl\(_4\)), also with an oxidation state of +4. The ability of these elements to achieve different oxidation states allows for varied chemical behavior and practical applications ranging from semiconductors to catalyst roles. Their diverse oxidation states also imply unique properties that are explored in advanced chemistry.

Compound Formulas

Formulating compounds involves understanding the valence electrons of the participating elements to achieve a stable electronic configuration. Valence electrons are the outermost electrons of an atom and are crucial in bond formation.
For instance, the formula for germanium tetrachloride, GeCl\(_4\), illustrates the need for germanium to share its four valence electrons with four chlorine atoms, each needing one additional electron to complete its valence shell. Similarly, the formula for boron trichloride (BCl\(_3\)) arises from boron sharing its three valence electrons with three chlorine atoms.
When composing compound formulas, it's crucial to maintain electrical neutrality. This often involves using subscripts in chemical formulas to show the ratio of elements that achieve such neutrality. The correct knowledge and application of compound formulas enable chemists to accurately describe the structures and properties of various compounds.

• Compounds must balance charges to be neutral.
• Ratios in the formula represent the number of each type of atom present.
• Understanding compound formulas promotes insight into chemical behavior.

Mastering these concepts is foundational for tasks ranging from analytical chemistry to molecular design.