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Chapter 22: Problem 16

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

Expert verified
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

01

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.
02

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.
03

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.
04

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.

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