Learn Science with Free Lessons & Tips

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.

The "lanthanoid contraction" refers to a phenomenon observed in the periodic table involving the contraction in atomic and ionic radii as you move across the lanthanide series (also known as the rare earth elements) from left to right.

This contraction occurs due to the poor shielding effect of f-electrons in the lanthanoid series. As electrons are added to the f-orbitals, they are not very effective at shielding the increasing nuclear charge from the outermost s- and p-electrons. As a result, the effective nuclear charge experienced by the outer electrons increases, leading to a contraction in the size of the atoms and ions as you move across the lanthanide series.

The lanthanoid contraction has significant consequences in various chemical properties, including ionization energy, atomic and ionic radii, and complex formation.

(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.

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.

Sure! Ionization isomerism is a type of structural isomerism where the composition of ions within a complex compound changes.

An example of ionization isomerism is seen in the coordination compound [Co(NH3)5SO4]Br and [Co(NH3)5Br]SO4.

In the first compound, [Co(NH3)5SO4]Br, the sulfate ion (SO4) is coordinated to the cobalt ion (Co) while the bromide ion (Br) is outside the coordination sphere.

In the second compound, [Co(NH3)5Br]SO4, the bromide ion (Br) is coordinated to the cobalt ion (Co) while the sulfate ion (SO4) is outside the coordination sphere.

So, in these two compounds, the ions are arranged differently around the central cobalt ion, leading to ionization isomerism.