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In H3PO2, also known as hypophosphorous acid, the oxidation number of hydrogen (H) is typically +1.

The sum of the oxidation numbers in a neutral molecule must equal zero. Since there are three hydrogen atoms, each with an oxidation number of +1, their total contribution is +3.

For oxygen (O), the typical oxidation number is -2, except in peroxides and when it's bonded to fluorine. In H3PO2, oxygen's oxidation number is -1.

Given that the overall charge of the molecule is zero, and knowing the oxidation numbers of hydrogen and oxygen, you can calculate the oxidation number of phosphorus (P).

Let's denote the oxidation number of phosphorus as xx:

(+1×3)+(−1×2)+x=0(+1×3)+(−1×2)+x=0

3−2+x=03−2+x=0

1+x=01+x=0

x=−1x=−1

So, in H3PO2, the oxidation number of phosphorus is -1.

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.

(i) Copper (I) ion is not known in aqueous solution primarily because copper tends to exist in the +2 oxidation state in aqueous solutions. This is due to the relative stability of the Cu(II) oxidation state compared to Cu(I) in aqueous environments. The standard reduction potential for the Cu(II)/Cu(I) couple is higher than that for many other metal ions, making the Cu(II) state more stable in water. Additionally, Cu(II) ions readily hydrolyze in water, forming insoluble Cu(OH)₂, further reducing the concentration of Cu(I) ions in solution.

(ii) Actinoids exhibit a greater range of oxidation states than lanthanoids due to the presence of f-orbitals in their electron configurations. Actinoid elements have more extended series of f-orbitals available for electron configuration, leading to a greater variety of possible oxidation states. The lanthanoid series, on the other hand, have electrons filling 4f orbitals, which are relatively shielded from the outer environment by the 5s and 5p orbitals. As a result, lanthanoid elements generally exhibit fewer accessible oxidation states compared to actinoids. Additionally, the actinoid series is longer than the lanthanoid series, providing more elements with a greater variety of electron configurations and oxidation states.

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.

Linkage isomerism occurs in coordination compounds where the ligand can coordinate through different atoms. One common example is the linkage isomerism in nitro and nitrito complexes.

Consider the complex [Co(NH₃)₅(NO₂)]²⁺. Here, the nitro ligand (NO₂) can bind to the central cobalt atom either through the nitrogen atom (forming a nitro ligand) or through one of the oxygen atoms (forming a nitrito ligand).

So, the two possible linkage isomers of this complex are:

1. [Co(NH₃)₅(NO₂)]²⁺ (Nitro linkage isomer)
2. [Co(NH₃)₅(ONO)]²⁺ (Nitrito linkage isomer)

In the nitro linkage isomer, the nitrogen atom of the NO₂ ligand is coordinated to the cobalt atom, while in the nitrito linkage isomer, one of the oxygen atoms of the NO₂ ligand is coordinated to the cobalt atom.

Coordination isomerism occurs when both cation and anion in a complex ion are exchanged with each other. Here's an example:

Consider the coordination compounds [Co(NH₃)₅Cl]Cl₂ and [CoCl₅(NH₃)]²⁻.

In the first compound, [Co(NH₃)₅Cl]Cl₂, the cobalt ion is surrounded by five ammonia ligands and one chloride ion. The counter ion is another chloride ion.

In the second compound, [CoCl₅(NH₃)]²⁻, the cobalt ion is surrounded by five chloride ions and one ammonia ligand. The counter ion is a neutral ammonia molecule.

In both cases, the coordination sphere of cobalt is different, but the overall formula of the compounds remains the same. This is an example of coordination isomerism.