Test: Main Group - 1 - Chemistry MCQ

Diborane contains a unique type of bonding known as three-centre two-electron (3c–2e) bonds. These bonds are formed by the interaction of two boron atoms and one hydrogen atom.

In diborane, there are:

• Two boron atoms.
• Four hydrogen atoms.

Each boron atom forms:

• Two 3c–2e bonds with hydrogen atoms.
• This results in a total of two 3c–2e bonds in the molecule.

Thus, the total number of 3c–2e bonds present in diborane is 2.

The decomposition temperature of carbonates varies among different compounds. Here are some key points regarding their thermal stability:

• Barium carbonate (BaCO₃) has the highest decomposition temperature.
• Strontium carbonate (SrCO₃) decomposes at a lower temperature than barium carbonate but higher than calcium carbonate.
• Calcium carbonate (CaCO₃) follows, with its decomposition occurring at a lower temperature than strontium carbonate.
• Magnesium carbonate (MgCO₃) has the lowest decomposition temperature among the four.

Thus, the order of decomposition temperatures from highest to lowest is:

• Barium carbonate (BaCO₃)
• Strontium carbonate (SrCO₃)
• Calcium carbonate (CaCO₃)
• Magnesium carbonate (MgCO₃)

This order is crucial for understanding the thermal properties and stability of these carbonates in various applications.

• Rb⁺ (Rubidium ion): When Rubidium (Rb) loses one electron to form Rb⁺, it has the electron configuration of Kr (Krypton).
• Kr (Krypton): Kr is a noble gas with a stable electron configuration.
• Br⁻ (Bromide ion): When Bromine (Br) gains one electron to form Br⁻, it also achieves the electron configuration of Kr.

Since all three species (Rb⁺, Kr, and Br⁻) have the same electron configuration (that of Krypton), they are isoelectronic. They also each have the same number of valence electrons, as they all possess 36 electrons.

Why the other options are not correct:

• B: Have same magnitude of effective nuclear charge: This is not true. The effective nuclear charge (Z_eff) depends on both the nuclear charge and the shielding effect of other electrons. The number of protons in Rb⁺, Kr, and Br⁻ are different, so the effective nuclear charge will be different for each.
• C: Have same magnitude of first ionization potential: This is also incorrect because the ionization potential depends on the atomic size and effective nuclear charge. Since Rb⁺, Kr, and Br⁻ have different atomic sizes and nuclear charges, their ionization potentials will differ.

Therefore, the correct answer is A and D.

The blue solution formed by dissolving sodium in liquid ammonia exhibits the following characteristics:

1. Paramagnetic (B): The solvated electrons (unpaired) contribute to paramagnetism.

2. Reducing in nature (C): The free electrons act as strong reducing agents.

3. Conducts electricity (D): Mobile solvated electrons enable electrical conductivity.

Key Points:

• Diamagnetic (A) is incorrect as the solution contains unpaired electrons.

• The properties align with the presence of solvated electrons, confirming B, C, D as correct.

Due to better guarding of s electrons by p if we directly remove an s electron without removing the p electrons, it would take more energy had we removed all outer electrons before.

Hence A is correct.

Correct Answer :- b

Explanation : Fluoroantimonic acid is the strongest superacid based on the measured value of its Hammett acidity function (H0), which has been determined for different ratios of HF:SbF5. While the H0 of pure HF is −15, addition of just 1 mol % of SbF5 lowers it to around −20.

The nature of silicon–oxygen bonds in silica (SiO₂) is primarily characterized by the following features:

• Polar covalent bonds are formed between silicon and oxygen due to the difference in electronegativity between the two elements.
• The oxygen atom is more electronegative than silicon, which leads to an uneven distribution of electron density.
• This results in a slight negative charge on the oxygen and a slight positive charge on the silicon, creating a dipole.
• These polar covalent bonds contribute to the strong structural integrity of silica.
• Silica's properties, such as its high melting point and hardness, can be attributed to the strength of these bonds.

The melting point of lithium and sodium differs due to several factors:

• Metallic bonding strength: The bonding in lithium is stronger than in sodium. This is due to the smaller size of lithium ions, which leads to a stronger attraction between the delocalised electrons and the metal cations.
• Attraction to metal cations: In lithium, the delocalised electrons are more strongly attracted to the lithium cation compared to sodium. This increased attraction contributes to a higher melting point.
• Charge density: Lithium cations (Li⁺) possess a greater charge density than sodium cations (Na⁺). A higher charge density indicates a stronger electrostatic attraction, which is crucial for bonding.
• Ion size: The size of Li⁺ is smaller than that of Na⁺. This smaller size allows lithium ions to pack more closely, enhancing the strength of the metallic bonds.

These factors combined explain why lithium has a higher melting point than sodium.

The hydrolysis of NCl₃ (compared to NF₃) occurs because chlorine (Cl) can expand its octet by utilizing d-orbitals, enabling the formation of intermediates during the reaction. Fluorine (F), being in the second period, lacks available d-orbitals, making NF₃ unable to undergo similar hydrolysis. This difference in orbital availability is the key reason for the distinct reactivity of the two compounds.

Key Points:

1. Octet Expansion: Chlorine (third-period element) has vacant 3d orbitals, allowing it to form hypervalent intermediates during hydrolysis.

2. Fluorine's Limitation: Fluorine (second-period element) lacks d-orbitals, restricting NF₃ from expanding its octet and reacting with water.

The correct answer is D, as the ability of chlorine to use d-orbitals facilitates hydrolysis in NCl₃, while NF₃ remains inert due to fluorine's inability to expand its octet.

The Lewis acidity of BF₃ is less than that of BCl₃ despite fluorine's greater electronegativity compared to chlorine. This difference can be attributed to several factors:

• Stronger 2p(B)–2p(F) σ-bonding in BF₃ allows for enhanced stability, reducing its acidity.

• The π-bonding interactions in BF₃ are also significant, as they further strengthen the bond between boron and fluorine.

• In contrast, BCl₃ exhibits weaker 2p(B)–2p(Cl) σ-bonding, which contributes to its higher acidity.

• The corresponding π-bonding in BCl₃ is less effective than in BF₃, making BF₃ a weaker Lewis acid overall.

In pyroxenes, there is a total of 3 oxygen atoms per silicon atom, so the simplest repeating unit is 

Hexachlorophosphazene gives Sn2 reaction with NaF and heating hexachlorophosphazene to  250 °C induces polymerization.

Rapid hydrolysis refers to the quick breakdown of molecules when they react with water. Certain molecules are more susceptible to this process due to their chemical structure. The following points explain why some molecules undergo rapid hydrolysis:

• BCl₃ is a strong Lewis acid, which readily reacts with water, leading to hydrolysis.
• SiCl₄ also hydrolyses quickly in the presence of water, forming silicic acid.
• PCl₅ is known to hydrolyse rapidly to produce phosphoric acid and other by-products.

In contrast, molecules like SF₆ and Al₂Cl₆ are more stable and do not hydrolyse as rapidly due to their stronger bonds and less reactive nature. Thus, the group of molecules that undergoes rapid hydrolysis consists of:

• BCl₃
• SiCl₄
• PCl₅

This understanding of hydrolysis is crucial in various fields, including chemistry and materials science, where the behaviour of these molecules can affect reactions and product stability.

The reaction of solid XeF₂ with AsF₅ in a 1:1 ratio produces:

• [XeF]⁺[AsF₆]^–

This reaction forms a complex ion, where:

• [XeF]⁺ represents the cation from xenon fluoride.
• [AsF₆}]^– is the anion derived from arsenic fluoride.

The formation of these ions indicates the stability and reactivity of xenon and arsenic fluorides under the given conditions.

For Et₂AlX (where X can be PPh₂^–, Ph^–, Cl^{– –}, and H^–), the tendency to form a dimeric structure varies as follows:

• PPh₂^– >Cl^– >H^– >Ph^–
• Cl^– >PPh₂^– >H^– >Ph^–
• Ph^– >H^– >Cl^– >PPh₂^–
• H^– >Ph^– >PPh₂^– >Cl^–

This indicates that the presence of PPh₂^– significantly enhances the likelihood of dimer formation, followed by Cl^– and H^–. In contrast, Ph^– demonstrates the least tendency to form dimers.

2Na₂HPo₄ + 3NaH₂Po₄ --->Na₇P₅o₁₆
 

Correct Answer :- a

Explanation : Sodium tripolyphosphate is produced by heating a stoichiometric mixture of disodium phosphate, Na₂HPO₄, and monosodium phosphate, NaH2PO4, under carefully controlled conditions.

2Na₂HPO₄ + NaH₂PO₄ → Na₅P₃O₁₀ + 2 H₂O

STPP is a colourless salt, which exists both in anhydrous form and as the hexahydrate. The anion can be described as the pentanionic chain [O₃POP(O)2OPO₃]^5-

To determine which substituted silane forms a cross-linked silicone polymer upon hydrolysis, consider the number of reactive sites (Cl atoms) available for hydrolysis and subsequent condensation:

• A: (CH₃)₄Si
  No Cl atoms. Hydrolysis does not occur; no polymer forms.

• B: CH₃SiCl₃
  Three Cl atoms. Hydrolysis produces three silanol (Si–OH) groups, enabling 3D cross-linking through condensation. This forms a rigid, cross-linked polymer.

• C: (CH₃)₂SiCl₂
  Two Cl atoms. Hydrolysis yields two silanol groups, forming linear or lightly branched chains (e.g., silicone oils).

• D: (CH₃)₃SiCl
  One Cl atom. Hydrolysis produces one silanol group, leading to no polymerization (only dimerization possible).

Only CH₃SiCl₃ (Option B) provides sufficient reactive sites (3 Cl atoms) to form a cross-linked silicone polymer.

Among the following donors, the one that forms the most stable adduct with the Lewis acid B(CH₃)₃ is:

• 4-Methylpyridine is a moderately strong Lewis base. It has a nitrogen atom that donates electrons effectively, but steric hindrance is present.

• 2,6-Dimethylpyridine is bulkier due to its two methyl groups, which can hinder the interaction with the Lewis acid, making it less stable.

• 4-Nitropyridine contains a nitro group, which is electron-withdrawing, reducing its ability to donate electrons and form a stable adduct.

• 2,6-Di- tert-butylpyridine has significant steric hindrance from its bulky tert-butyl groups, which can severely limit its interaction with the Lewis acid.

The most stable adduct is likely formed by 4-Methylpyridine due to its balanced electron-donating ability and lower steric hindrance compared to the other options.

The IUPAC nomenclature of Na[PCl₆] is:

• Sodium hexachlorophosphate (V)

• Sodium hexachlorophosphine (V)

• Sodium hexachlorophosphine

• Sodium hexachlorophosphite (V)

The correct designation is Sodium hexachlorophosphate (V).

When a mixture of NaCl, concentrated H₂SO₄, and K₂Cr₂O₇ is heated in a dry test tube, a red vapour is produced. This vapour dissolves in aqueous NaOH, resulting in a yellow solution. When treated with AgNO₃, this yellow solution forms a red solid.

• The red vapour is identified as CrO₂Cl₂.
• The yellow solution formed after dissolving the vapour in NaOH is a chromate.
• When this chromate reacts with AgNO₃, it produces Ag₂CrO₄, a red solid.

In summary:

• P (the red vapour) is CrO₂Cl₂.
• Q (the red solid) is Ag₂CrO₄.

NH₄Cl + NaBH₄ ------------> B₃N₃H₆ + H₂ + NaCl

When heating a mixture of ammonium chloride (NH₄Cl) and sodium tetrahydridoborate (NaBH₄), one of the resulting products is a liquid, referred to as compound X. The reaction can be represented as follows:

NH₄Cl + NaBH₄ → X + H₂ + NaCl

Key Points:

• The reaction produces hydrogen gas (H₂) and sodium chloride (NaCl) along with compound X.
• Compound X is identified based on its structural properties.
• It can exhibit characteristics of various chemical classes, but in this context, it is particularly relevant to heterocycles.

Based on the properties of compound X, we can categorise it as an:

• Unsaturated heterocycle, which is a cyclic compound containing at least one unsaturated bond (double or triple) within its ring structure.

Nido-boranes are a class of boron hydrides with a distinctive cage-like structure. The most common example of a nido-borane is:

• B₆H₁₀ - This is a well-known nido-borane, featuring a cage structure with six boron atoms.
• Other structures, like B₄H₁₀, B₆H₁₂, and B₈H₁₄, do not fit the nido classification.

In summary, the primary nido-borane is represented by B₆H₁₀, characterised by its unique arrangement of boron atoms.

To determine the number of S–S bonds in H₂S₅O₆, let's first analyze the structure of the compound:

Structure of H₂S₅O₆:

The compound H₂S₅O₆ consists of sulfur (S), hydrogen (H), and oxygen (O) atoms. A typical structure for such a compound is likely to have S–S linkages (bonds between sulfur atoms) due to the high oxidation state of sulfur.

The structure can be represented as:

• There are 5 sulfur atoms and they are likely to be connected in a chain, with some of them being connected through S–S bonds.
• The compound may be a disulfide structure or a more complex polysulfide.

Number of S–S bonds:

Given the number of sulfur atoms (5), we can assume that there are 4 S–S bonds. This is because sulfur atoms typically form S–S bonds in chain-like or ring structures, and with 5 sulfur atoms, you would expect 4 bonds between them.

Thus, the number of S–S bonds in H₂S₅O₆ is 4.

The ease of formation of the adduct NH₃·BX₃ depends on the Lewis acidity of BX₃. Boron trihalides (BX₃) act as Lewis acids, and their acidity increases with the size of the halogen atom (F < Cl < Br). Larger halogens (Br) have weaker B–X bonds and reduced back-bonding, making the boron atom more electron-deficient. This trend results in stronger Lewis acidity for BBr₃ compared to BCl₃ and BF₃. Consequently, the adduct forms most readily with BBr₃ and least with BF₃.

Order of Lewis acidity: BBr₃ > BCl₃ > BF₃
Order of adduct formation ease: BF₃ < BCl₃ < BBr₃

Ammonolysis of S₂Cl₂ in an inert solvent results in the formation of several compounds. The key products include:

• S₂N₂: This compound consists of sulphur and nitrogen.

• S₂N₂Cl: A chlorinated derivative of the previous compound.

• S₂N₂H₄: This compound includes hydrogen alongside sulphur and nitrogen.

• S₄N₄: A more complex structure with increased quantities of both sulphur and nitrogen.

Among these, the most significant product is S₄N₄, which indicates a higher level of reaction between the components.

Wade's rule helps predict the structure of boron hydrides, including B₅H₉. This compound can be classified based on its structure type, which is related to the number of boron atoms and the bonding configurations.

• Closo: This structure has a closed, cage-like formation with all vertices occupied by atoms.
• Nido: This structure resembles a "nest" with one vertex missing, indicating a more open form than closo.
• Arachno: This structure resembles a "spider web," with two vertices missing, providing an even less compact arrangement.
• Scorpionato: This is a more complex structure, often involving multiple bonding arrangements.

The structure of B₅H₉ aligns with the Nido type, characterised by its less compact formation compared to closo structures.

A superoxide is a chemical compound containing the superoxide ion  which is an oxygen molecule (O₂) with an extra electron.

• KO₂ (Potassium superoxide) contains the superoxide ion  When potassium reacts with oxygen, it forms the superoxide ion in the compound.

Other compounds listed:

• Na₂O₂ is sodium peroxide, which contains the peroxide ion  not the superoxide ion.
• K₂O is potassium oxide, and it does not contain superoxide or peroxide ions.
• Fe₃O₄ is iron(II,III) oxide, which is a mixed oxide, and does not contain the superoxide ion.

Therefore, the correct superoxide is KO₂.

Aqua regia is an extremely potent oxidising agent, composed of a mixture of hydrochloric acid and nitric acid. This unique combination allows it to dissolve noble metals like gold and platinum. The strength of aqua regia can be attributed to the following components:

• Chlorine: The presence of chlorine plays a crucial role in its oxidising properties.
• Nitric Acid: This acid contributes to the formation of nitrosyl chloride, enhancing its reactivity.
• Oxidation Process: Aqua regia facilitates the oxidation of metals, enabling their dissolution.

In summary, aqua regia's effectiveness as an oxidising agent arises from the interaction of its components, which create a highly reactive environment suitable for breaking down noble metals.