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Post a LessonAnswered on 13 Apr Learn Unit VII: p - Block Elements
Nazia Khanum
Nitrogen is indeed more inert compared to phosphorus, primarily due to differences in their atomic structures and the stability of their compounds.
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.
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.
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.
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.
Answered on 13 Apr Learn Unit VII: p - Block Elements
Nazia Khanum
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.
Answered on 13 Apr Learn Unit VII: p - Block Elements
Nazia Khanum
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.
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Answered on 13 Apr Learn Unit VIII: d and f Block Elements
Nazia Khanum
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.
Answered on 13 Apr Learn Unit VIII: d and f Block Elements
Nazia Khanum
Transition elements exhibit variable oxidation states due to the presence of incompletely filled d orbitals in their atoms. These d orbitals can participate in bonding and can gain or lose electrons to form compounds with different oxidation states.
The number of oxidation states displayed by transition metals is often related to their electronic configurations. Transition metals have multiple incompletely filled d orbitals, which can easily lose or gain electrons to achieve a stable configuration. This flexibility allows them to exhibit a range of oxidation states.
For example, iron (Fe) can form compounds where it has an oxidation state of +2 or +3. In the +2 oxidation state, iron loses two electrons from its 4s orbital, while in the +3 oxidation state, it loses three electrons from both its 4s and 3d orbitals. Similarly, elements like chromium (Cr) can exhibit oxidation states ranging from -2 to +6.
The variability in oxidation states allows transition metals to form a wide variety of compounds with different properties and reactivities, making them essential in many chemical reactions and industrial processes.
Answered on 13 Apr Learn Unit VIII: d and f Block Elements
Nazia Khanum
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.
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Answered on 13 Apr Learn Unit IX: Coordination Compounds
Nazia Khanum
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:
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.
Answered on 13 Apr Learn Unit IX: Coordination Compounds
Nazia Khanum
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.
Answered on 13 Apr Learn Unit IX: Coordination Compounds
Nazia Khanum
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.
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Answered on 13 Apr Learn Unit IX: Coordination Compounds
Nazia Khanum
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