Question

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.

Short answer

Statements (a) and (d) are true, while statements (b) and (c) are false.

Step-by-step solution

Statement (a) Investigation

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.

Statement (b) Investigation

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.

Statement (c) Investigation

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.

Statement (d) Investigation

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.

Key concepts

Oxidation States

Oxidation states help describe the degree of oxidation, or loss of electrons, of an atom in a chemical compound. They are essential for understanding how atoms interact in various chemical reactions.
To determine the oxidation state of an atom, we consider the hypothetical charge it would have if all bonds were 100% ionic. Generally, more electronegative atoms take on negative oxidation states by gaining electrons, while less electronegative atoms have positive oxidation states by losing electrons.
In compounds, the sum of the oxidation states must equal the overall charge of the compound. For example, in

• ℍ𝓝𝖮₃, nitrogen (N) has an oxidation state of \(+5\), considering the three oxygen atoms each have an oxidation state of \(-2\), contributing to overall -6, while the hydrogen atom has \(+1\). So, the nitrogen must be \(+5\) to make the compound neutral.
• Similarly, in ℍ₃𝖯𝖮₄, phosphorus (P) also has an oxidation state of \(+5\) to balance the compound to neutrality.

Electronegativity

Electronegativity is the tendency of an atom to attract electrons in a chemical bond. It explains why atoms form bonds with others that have similar or different electronegativity.
In the periodic table, electronegativity generally increases from left to right across a period and decreases down a group. Elements with high electronegativity, such as fluorine, have a strong tendency to attract electrons. Conversely, elements with low electronegativity, like cesium, are likely to lose electrons.
For example, sulfur and selenium are in the same group. Sulfur, with an electronegativity of 2.58, is slightly more electronegative than selenium, which has an electronegativity of 2.55. This trend is because sulfur is smaller and its valence electrons are closer to the nucleus.

Covalent Bonds

Covalent bonds form when atoms share pairs of electrons. They are a fundamental type of bond in molecules. Covalent bonds occur in two primary forms: single and multiple bonds.

• Single bonds involve the sharing of one pair of electrons, as in \(\displaystyle{\mathrm{H}_{2}}\). Each hydrogen atom shares one electron, forming a strong bond.
• Multiple bonds, like double or triple bonds, involve sharing more than one pair of electrons. In the example of disilene \(\displaystyle{\mathrm{Si}_{2}H_{4}}\), each silicon atom forms a double bond with the other silicon, leading to stronger covalent interactions.

Covalent bonds are often found between nonmetals with similar electronegativities, facilitating the sharing of electrons.

Periodic Table

The periodic table organizes elements based on atomic number and similar properties, providing a framework to understand chemical behaviors.
Elements in the same group, or column, often have similar chemical properties and valence electron configurations. For instance, silicon and carbon are both in Group 4A, which accounts for their ability (or inability) to form certain compounds.
As you move across the table, periods represent increasing atomic number. Polonium, another Group 6A element, shows that the electronegativity decreases as you move down the group, due to the increase in atomic radii and the addition of electron shells.
Moreover, the table highlights trends like ionization energy, atomic radius, and electron affinity beyond just electronegativity, allowing for predicting chemical reactions and bond formations.