Which of the following statements are true? (a) Si can form an ion with six fluorine atoms, \(\mathrm{SiF}_{6}^{2-}\), whereas carbon cannot. (b) Si can form three stable compounds containing two \(\mathrm{Si}\) 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.

We can examine the possibility of Si forming an ion with six fluorine atoms by checking for the available valence electrons. Silicon, a group 4A element, has four valence electrons. However, it can accommodate an expanded valence shell to form \(\mathrm{SiF}_{6}^{2-}\). Carbon, also a group 4A element, cannot accommodate an expanded valence shell since it doesn't have available d-orbitals to have more than four covalent bonds. Therefore, the statement is true.

We need to analyze whether Si can form stable compounds containing two Si atoms and varying numbers of H atoms: 1. \(\mathrm{Si}_{2}\mathrm{H}_{2}\): There are not enough H atoms to create a stable compound with two Si atoms. Each Si requires at least two H atoms to fulfill their valence requirements; hence, this compound is not stable. 2. \(\mathrm{Si}_{2}\mathrm{H}_{4}\): This compound is stable and exists as disilene. Each Si atom forms two single bonds with two H atoms and a double bond with the other Si atom. 3. \(\mathrm{Si}_{2}\mathrm{H}_{6}\): This compound is also stable, known as disilane. Each Si creates four single bonds with H atoms (three of them) and the other Si atom. The statement is incorrect; there are only two stable compounds containing two Si atoms.

We need to calculate the oxidation states of the central atoms in the following compounds: 1. \(\mathrm{HNO}_{3}\): N is the central atom, and its oxidation state can be found as \( x - (3) = -1 \), where x is the oxidation state of N. So, \( x = +5 \). 2. \(\mathrm{H}_{3}\mathrm{PO}_{4}\): P is the central atom, and its oxidation state can be calculated as \( x + (3) - (4) = 0 \), where x is the oxidation state of P. Thus, \( x = +5 \) The oxidation states of the central atoms, N and P, in \(\mathrm{HNO}_{3}\) and \(\mathrm{H}_{3}\mathrm{PO}_{4}\) are the same (+5), making the statement false.

We must investigate the electronegativities of sulfur (S) and selenium (Se). Both elements belong to Group 6A of the periodic table. As we go down the group, the electronegativity generally decreases due to the increasing atomic radii. Consequently, sulfur has a higher electronegativity (2.58) than selenium (2.55). The statement is true. In conclusion, statements (a) and (d) are true, while statements (b) and (c) are false.