Learn Science with Free Lessons & Tips

Nitrogen is indeed more inert compared to phosphorus, primarily due to differences in their atomic structures and the stability of their compounds.

1. Bond Strength: Nitrogen forms a very strong triple bond (N≡N) in molecular nitrogen (N2), which is difficult to break. This makes nitrogen gas quite unreactive under normal conditions. Phosphorus, on the other hand, tends to form weaker single bonds (P-P) in its elemental form (P4), making it more reactive.

2. Electronegativity: Nitrogen has a higher electronegativity compared to phosphorus. This means that nitrogen atoms attract electrons more strongly, which stabilizes the molecules they form and makes them less prone to reacting with other substances.

3. Size of Atom: Nitrogen atoms are smaller than phosphorus atoms, which affects their ability to form stable bonds. Nitrogen's smaller size allows for stronger overlap of atomic orbitals in the formation of multiple bonds, contributing to the stability of nitrogen compounds.

4. Hybridization: Nitrogen often undergoes sp2 hybridization, leading to planar geometry in many of its compounds. This geometric arrangement can enhance the stability of nitrogen compounds. Phosphorus, however, can exhibit various hybridizations and geometries, which may render its compounds more reactive.

These factors collectively contribute to the relative inertness of nitrogen compared to phosphorus. However, despite nitrogen's inertness in its diatomic form, it can react vigorously under certain conditions to form a wide variety of compounds, especially when it reacts with highly reactive elements or under specific catalytic conditions.

Of the two ions you mentioned, PCl₄⁻ (tetrahedral tetrachlorophosphate ion) is more likely to exist than PCl₄⁺ (tetrahedral tetrachlorophosphonium ion). This is because phosphorus typically forms covalent bonds with other atoms, such as chlorine in this case, rather than losing or gaining electrons to form ions.

In PCl₄⁻, phosphorus has a valence electron configuration of 3s²3p³. By accepting four electrons from chlorine atoms, phosphorus completes its octet, achieving a more stable electron configuration. This is consistent with the tendency of elements to gain electrons to achieve a noble gas configuration.

However, for PCl₄⁺ to exist, phosphorus would need to lose its lone pair of electrons, which is less energetically favorable due to the electronegativity difference between phosphorus and chlorine. Additionally, the formation of positively charged phosphorus is less common because phosphorus typically forms covalent bonds rather than losing electrons.

Therefore, PCl₄⁻ is more likely to exist than PCl₄⁺ due to the stability gained through electron gain rather than electron loss.

Between PH3 (phosphine) and H2S (hydrogen sulfide), H2S is more acidic.

Acidity is typically measured by the ease with which a compound donates a proton (H⁺ ion) in solution. In both PH3 and H2S, the central atom (phosphorus in PH3 and sulfur in H2S) is bonded to three hydrogen atoms. However, the central atoms in these molecules differ in electronegativity.

Sulfur is more electronegative than phosphorus, meaning it has a stronger pull on the shared electrons in the hydrogen-sulfur bonds compared to phosphorus in the hydrogen-phosphorus bonds. This results in the hydrogen-sulfur bond being more polarized, with a partial positive charge on the hydrogen atom.

Consequently, the hydrogen atom in H2S is more easily ionizable (loses a proton) compared to the hydrogen atom in PH3. Therefore, H2S is considered a stronger acid compared to PH3.

Sure, let's break down each of these statements:

(i) Transition metals generally form colored compounds: Reason: The color of transition metal compounds arises due to the presence of partially filled d orbitals in the transition metal ions. When light interacts with these compounds, electrons in the d orbitals can absorb certain wavelengths of light, causing them to transition to higher energy levels. The absorbed wavelengths correspond to the complementary color of the one observed, resulting in the compound appearing colored. This phenomenon is known as d-d transition. The energy gap between the d orbitals varies depending on the metal ion and its oxidation state, leading to a wide range of colors observed in transition metal compounds.

(ii) Manganese exhibits the highest oxidation state of +7 among the 3d series of transition elements: Reason: Manganese, being a member of the 3d transition metal series, can exhibit multiple oxidation states due to the availability of its d orbitals for electron transfer. However, among the 3d series elements, manganese has the highest number of unpaired electrons available in its 3d orbitals, which allows it to achieve its highest oxidation state of +7. This occurs in compounds like potassium permanganate (KMnO4), where manganese is in the +7 oxidation state. The ability of manganese to access this high oxidation state is attributed to its electron configuration and its position within the periodic table.

These are interesting observations that can be explained by considering the electronic configurations and trends in oxidation states across transition metals.

(i) Cr2+ is reducing in nature while Mn3+ is an oxidizing agent: This can be explained by looking at the electronic configurations of Cr2+ and Mn3+.

• Cr2+ has an electronic configuration of [Ar] 3d4, where it has a half-filled d orbital. Half-filled orbitals have lower energy due to greater exchange energy, making it energetically favorable for Cr2+ to lose electrons and become Cr3+ in order to achieve a stable half-filled d orbital, thus acting as a reducing agent.

• On the other hand, Mn3+ has an electronic configuration of [Ar] 3d4, which is one electron short of achieving a stable half-filled d orbital. So, Mn3+ tends to gain an electron to achieve a stable half-filled d orbital, making it an oxidizing agent as it oxidizes other species by accepting electrons.

(ii) In a transition series of metals, the metal which exhibits the greatest number of oxidation states occurs in the middle of the series: This observation can be explained by considering the trends in the filling of d orbitals across the transition series.

• At the beginning of the transition series, elements have fewer d electrons available for oxidation, limiting the number of oxidation states they can exhibit.

• Toward the middle of the series, there's a peak in the number of oxidation states exhibited. This is because these elements have a balance between gaining and losing electrons, allowing them to exhibit a wider range of oxidation states.

• Towards the end of the series, the number of oxidation states generally decreases as elements have a higher tendency to gain electrons rather than lose them, leading to fewer oxidation states.

So, the middle of the transition series tends to have elements that can exhibit the greatest number of oxidation states due to the balance between gaining and losing electrons facilitated by their electronic configurations.