Question

Which of the following statements are true? (a) Si can form an ion with six fluorine atoms, SiF \(_{6}^{2-}\) ,whereas carbon cannot.(b) Si can form three stable compounds containing two Si atoms each, \(\mathrm{Si}_{2} \mathrm{H}_{2}, \mathrm{Si}_{2} \mathrm{H}_{4},\) and \(\mathrm{Si}_{2} \mathrm{H}_{6}\) (c) In \(\mathrm{HNO}_{3}\) and \(\mathrm{H}_{3} \mathrm{PO}_{4}\) the central atoms, \(\mathrm{N}\) and \(\mathrm{P},\) have different oxidation states. (d) \(\mathrm{S}\) is more electronegative than Se.

Short answer

Statements (a) and (d) are true, while statements (b) and (c) are false. This is because Si can form SiF\(_{6}^{2-}\) due to access to its 3d orbitals, carbon cannot form a hexafluoride ion, there is no known stable compound with the formula \(\mathrm{Si}_{2} \mathrm{H}_{2}\), and sulfur is more electronegative than selenium.

Step-by-step solution

Statement (a)

We will analyze if SiF \(_{6}^{2-}\) is possible, and why carbon cannot form a similar ion. Silicon (Si) and carbon (C) belong to the same group, Group 14, in the periodic table and share some similarities. However, Si can form stable covalent hexafluorides with an expanded octet, like in the case of SiF\(_{6}^{2-}\), while carbon cannot. This is because Si has access to its empty 3d orbitals and can accommodate more than eight electrons in its valence shell. Carbon (C), on the other hand, lacks access to vacant d orbitals and can only accommodate a maximum of eight electrons around it, not allowing it to form a hexafluoride ion. Hence, statement (a) is true.

Statement (b)

We will analyze if \(\mathrm{Si}_{2} \mathrm{H}_{2}, \mathrm{Si}_{2} \mathrm{H}_{4},\) and \(\mathrm{Si}_{2} \mathrm{H}_{6}\) are stable compounds. All three compounds mentioned are silicon hydrides and are also known as silanes. These compounds follow the general formula \(\mathrm{Si}_{n} \mathrm{H}_{2n+2}\), similar to hydrocarbons. Disilane (\(\mathrm{Si}_{2} \mathrm{H}_{6}\)) and other higher-order silanes are known to exist and are stable. However, there is no known stable compound with the formula \(\mathrm{Si}_{2} \mathrm{H}_{2}\). Therefore, statement (b) is false.

Statement (c)

We will determine the oxidation states of nitrogen (N) and phosphorus (P) in \(\mathrm{HNO}_{3}\) and \(\mathrm{H}_{3} \mathrm{PO}_{4}\). In \(\mathrm{HNO}_{3}\), the oxidation state of nitrogen (N) is +5. This can be determined by assigning the oxidation numbers to each element in the compound and equating the sum to 0, considering the compound is neutral. In \(\mathrm{H}_{3} \mathrm{PO}_{4}\), the oxidation state of phosphorus (P) is also +5. As both N and P have the same oxidation state in their respective compounds, statement (c) is false.

Statement (d)

We will compare the electronegativity of sulfur (S) and selenium (Se). When we consult the periodic table, we find that sulfur (S) and selenium (Se) belong to the same group, Group 16. As we move down within the same group, electronegativity generally decreases. Since sulfur is above selenium in the periodic table, sulfur is more electronegative than selenium. Therefore, statement (d) is true. In conclusion, statements (a) and (d) are true, while statements (b) and (c) are false.

Key concepts

Silicon-fluorine compounds

Silicon can form unique compounds with fluorine due to its ability to expand its valence shell. A well-known example is the hexafluorosilicate ion, SiF \(_{6}^{2-}\). This structure is possible because silicon can utilize its vacant 3d orbitals to accommodate more than eight electrons.
A key difference here is that carbon in the same group cannot do this. Carbon lacks 3d orbitals, restricting it to a maximum of eight valence electrons. This is why carbon cannot form a similar six-fluoride ion.
This ability of silicon to expand its octet and form complex anions like SiF \(_{6}^{2-}\) demonstrates its unique chemistry compared to other Group 14 elements like carbon.

Oxidation states in chemistry

Oxidation states are a way we keep track of electron transfer in chemical reactions. The oxidation state tells us how many electrons an atom gains, loses, or shares when it forms chemical bonds.
For example, in the nitric acid molecule (HNO\(_3\)), nitrogen has an oxidation state of +5. This means nitrogen is sharing a significant amount of its electron cloud with the very electronegative oxygen atoms. Similarly, phosphorus in phosphoric acid (H\(_3\)PO\(_4\)) is also +5. Although different elements, both nitrogen and phosphorus exhibit the same oxidation state in these compounds due to their electron configurations and bonding environment.
Understanding oxidation states is essential, as it helps predict how elements will react with each other.

Electronegativity trends

Electronegativity is a measure of how strongly an atom can attract electrons in a chemical bond. It is key to understanding why atoms form certain types of bonds and how these bonds affect compound properties.
Generally, electronegativity decreases as you move down a group on the periodic table. This trend is due to increasing atomic size; electrons are farther away from the nucleus and are less strongly attracted. For example, sulfur is more electronegative than selenium because it is higher up in Group 16. Sulfur can pull electrons more effectively towards itself compared to selenium.
These trends help us understand the behavior and properties of elements in chemical reactions and compound formation.

Periodic table group 14

Group 14 of the periodic table contains elements like carbon, silicon, germanium, tin, and lead. These elements share certain properties but also exhibit unique traits, particularly in their ability to form different types of bonds and compounds.
Carbon is well-known for forming long carbon chains and the vast array of organic compounds. Silicon, meanwhile, relies on its ability to form extended networks in minerals and structures like tetrahedral silicates. These compound structures are vital in things like electronics and ceramics.
While carbon focuses more on forming stable covalent bonds with itself, silicon can form additional bonds due to its ability to access empty d orbitals, contributing to its versatility in forming compounds like silanes and silicones.