States of Matter Class 11 NEET Questions

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    States of Matter Class 11 NEET Questions. We covered all the States of Matter Class 11 NEET Questions in this post for free so that you can practice well for the exam.

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    mcq on States of Matter for NEET Students

    Which of the following accurately describes the structure of the P₄ Molecule found in white phosphorus?

    (A) It contains three P–P single bonds

    (B) It has four electron lone pairs

    (C) It features five P–P single bonds

    (D) It holds six lone pairs of electrons

    Explanation:
    The question focuses on the Molecular arrangement and Bonding behavior in elemental phosphorus when it exists in its most reactive Solid form. In this form, phosphorus atoms do not remain isolated but instead connect in a compact three-dimensional cluster. The structure is highly strained due to the way atoms are forced into a small closed geometry, leading to deviation from ideal bond angles expected in standard tetrahedral arrangements.

    Each Atom in the cluster forms covalent connections with the other surrounding atoms, creating a cage-like structure. Along with Bonding interactions, each Atom also retains a pair of non-Bonding electrons. These lone electron pairs contribute to electron repulsion within the structure, increasing internal strain and influencing reactivity. Because of this arrangement, the Molecule is far more reactive compared to other elemental forms of phosphorus.

    The structure is important in understanding how atomic Bonding and lone pair interactions together determine stability and shape in polyatomic elemental molecules. The combination of single covalent Bonding and localized electron pairs leads to a unique geometry that explains the physical and chemical behavior of white phosphorus in reactions and storage conditions.

    Option b – It has four electron lone pairs

    A hydrocarbon ‘Z’ reacts with warm dilute H₂SO₄ in the presence of HgSO₄ at 333 K to produce compound X, which then rearranges to form propanone. Identify ‘Z’.

    (A) Propanol

    (B) Propanal

    (C) Propyne

    (D) Propene

    Explanation:
    The question deals with the transformation of an unsaturated hydrocarbon into a carbonyl compound through Acid-catalyzed hydration. In the presence of dilute Acid and a mercury Salt catalyst, water adds across multiple bonds of unsaturated Hydrocarbons. This process often follows an orientation rule where the hydrogen and hydroxyl group attach in a specific manner, leading to the formation of an intermediate that may not be the most stable structure initially.

    When the starting compound contains a carbon-carbon multiple bond, hydration leads to the formation of an unstable enol intermediate. This intermediate does not remain in its original form and undergoes rapid rearrangement through tautomerism. The rearrangement converts the enol into a more stable carbonyl compound, specifically a ketone in this case. The final product mentioned indicates that the rearrangement leads to a methyl ketone structure, which is characteristic of such hydration reactions involving terminal unsaturated Hydrocarbons.

    This type of reaction is widely used in Organic synthesis to convert unsaturated Hydrocarbons into oxygen-containing functional groups and demonstrates how reaction conditions and intermediates determine the final product formed in solution Chemistry.

    Option c – Propyne

    Which of the following statements is INCORRECT regarding d- and l-forms of a compound?

    (A) They are non-superimposable mirror images

    (B) These are also referred to as optical antipodes

    (C) They rotate plane-polarized Light in opposite directions

    (D) They differ in chemical properties but react at the same rate

    Explanation:
    The question is based on stereoisomerism, specifically optical isomerism, where molecules exist in forms that are mirror images of each other. These mirror-image forms arise due to the presence of chiral centers, where a carbon Atom is attached to four different substituents, creating non-superimposable structures. Such forms are commonly referred to as enantiomers and they exhibit identical physical properties in most cases, except in their interaction with plane-polarized Light and other chiral environments.

    These enantiomers rotate plane-polarized Light in equal magnitude but opposite directions, which is a key characteristic used to distinguish them experimentally. They are also often called optical antipodes due to their opposite optical behavior. However, their chemical behavior is generally very similar in non-chiral environments because their Bonding framework remains the same.

    In biochemical and stereochemical contexts, differences may arise in reactions involving chiral catalysts or biological systems, but under standard conditions, their reaction rates and chemical properties are usually comparable. The concept is crucial in understanding how Molecular spatial arrangement affects optical activity and interaction with polarized Light in Organic Chemistry systems.

    Option d – They differ in chemical properties but react at the same rate

    A hydrocarbon has 60% carbon content. What Mass of CO₂ will be formed when 0.6 g of this compound undergoes complete combustion?

    (A) 2.64 g

    (B) 1.32 g

    (C) 0.65 g

    (D) 0.6 g

    Explanation:
    The question involves stoichiometric analysis of combustion reactions, where a hydrocarbon reacts completely with oxygen to form carbon dioxide and water. The key idea is that all carbon present in the hydrocarbon is ultimately converted into carbon dioxide during complete combustion. The percentage composition of carbon helps determine how much carbon is present in the given sample.

    Once the Mass of carbon in the compound is determined, it is converted into carbon dioxide using the Molecular relationship between carbon and carbon dioxide. Each Atom of carbon contributes to one Molecule of carbon dioxide, so the Mass conversion depends on molar Mass ratios. This approach avoids needing the exact Molecular formula and relies purely on proportional relationships between elements.

    Such calculations are commonly used in analytical Chemistry to determine product yield based on elemental composition. The concept highlights conservation of Mass and the direct transformation of carbon atoms into gaseous carbon dioxide during combustion reactions, making it a standard method in quantitative chemical analysis.

    Option b – 1.32 g

    How many lone electron pairs are present in the N₂O₃ Molecule?

    (A) 2

    (B) 6

    (C) 7

    (D) 8

    Explanation:
    The question deals with electron distribution in a covalently bonded nitrogen oxide compound where atoms share electrons to achieve stable configurations. In such molecules, both Bonding and non-Bonding electron pairs must be considered to understand the overall electronic structure. Nitrogen and oxygen atoms follow octet considerations, but oxygen typically carries more non-Bonding electron pairs due to its higher electronegativity.

    The structure of the Molecule includes nitrogen atoms connected through oxygen bridges, forming a mixed Bonding Network. Each oxygen Atom tends to retain lone pairs after forming single or double bonds with nitrogen. These lone pairs contribute to electron density regions that affect Molecular shape and polarity. The total count of lone pairs is determined by summing non-bonding electrons across all atoms after distributing valence electrons according to bonding requirements.

    Understanding lone pair distribution is essential in predicting Molecular geometry using electron pair repulsion theory. The presence of multiple oxygen atoms significantly increases the number of non-bonding electron pairs, which influences bond angles and Molecular polarity in such nitrogen oxides.

    Option d – 8

    Which of the following statements about alkaline Earth metal carbonates is INCORRECT?

    (A) Barium carbonate dissolves less in water than magnesium carbonate

    (B) Beryllium carbonate is the most thermally stable of the group

    (C) Heating any of these carbonates produces metal oxide and CO₂

    (D) Solubility in water decreases from top to bottom in the group

    Explanation:
    The question is based on Periodic trends and thermal stability of carbonates formed by Group 2 elements. These carbonates exhibit systematic changes in properties as one moves down the group due to increasing atomic size and decreasing charge density of the metal ions. These factors influence solubility, thermal decomposition behavior, and lattice stability.

    Generally, as the size of the metal ion increases, the stability of the carbonate decreases, making decomposition easier upon heating. Conversely, smaller ions tend to stabilize the carbonate lattice more effectively due to stronger electrostatic interactions. Solubility trends are also influenced by hydration energy and lattice energy balance, which changes across the group.

    Thermal decomposition of these carbonates produces corresponding metal oxides and carbon dioxide gas. The ease of decomposition and solubility trends are important for understanding chemical reactivity and stability of alkaline Earth compounds. The statement that contradicts these systematic Periodic trends is considered incorrect because it violates expected group behavior based on ionic size and lattice energy considerations.

    Option b – Beryllium carbonate is the most thermally stable of the group

    What is the bond order of the superoxide ion (O₂⁻)?

    (A) 2.5

    (B) 2

    (C) 1.5

    (D) 1

    Explanation:
    The question is based on Molecular orbital theory, which explains bonding in terms of electron distribution across bonding and antibonding orbitals rather than simple shared pairs. Oxygen molecules and their ions are analyzed using Molecular orbital energy diagrams where electrons fill orbitals in order of increasing energy.

    In oxygen-based species, the addition or removal of electrons affects the occupancy of antibonding orbitals, which directly influences bond strength and stability. The bond order is calculated using the difference between electrons in bonding and antibonding orbitals, divided by two. A decrease in bond order indicates weaker bonding and increased bond length.

    For the superoxide ion, an extra electron occupies an antibonding orbital, reducing overall bond strength compared to Molecular oxygen. This results in a fractional bond order, indicating partial double bond character. Molecular orbital theory is essential in explaining paramagnetism and bond variations in oxygen species, which cannot be explained adequately by simple valence bond theory.

    Option c – 1.5

    Which compound provides the nitrogen source in Gabriel phthalimide synthesis of amines?

    (A) Sodium azide (NaN₃)

    (B) Sodium nitrite (NaNO₂)

    (C) Potassium cyanide (KCN)

    (D) Potassium phthalimide

    Explanation:
    The question relates to a synthetic Organic Chemistry method used to prepare primary amines selectively. The Gabriel synthesis involves the use of a nitrogen-containing compound that can be converted into an amine through a sequence of nucleophilic substitution and hydrolysis reactions. The key idea is to introduce nitrogen in a controlled manner without forming secondary or tertiary amines.

    The nitrogen source in this reaction is a cyclic imide derivative that contains a nitrogen Atom bonded within a stable ring system. This compound reacts with alkyl halides to form intermediate products, which upon hydrolysis release primary amines. The stability of the intermediate prevents over-alkylation, making the process highly selective.

    This method is widely used because it allows controlled synthesis of primary amines without side products that commonly occur in direct alkylation reactions. The nitrogen-containing starting compound plays a central role in ensuring selectivity and efficiency in amine synthesis.

    Option d – Potassium phthalimide

    Zone refining separates Metals based on the principle that:

    (A) Metals with lower boiling points can be distilled off

    (B) Impurities are more soluble in molten form than in Solid state

    (C) Components adsorb differently onto an adsorbent

    (D) Volatile metal vapors decompose into pure metal

    Explanation:
    The question is based on purification techniques used in Metallurgy to obtain highly pure Metals. Zone refining relies on the difference in solubility of impurities between Solid and molten states of a metal. A narrow region of a metal rod is melted and slowly moved along its length, allowing impurities to migrate with the molten zone.

    Impurities generally dissolve more readily in the liquid phase than in the Solid phase. As the molten zone moves, impurities are carried along and concentrated at one end of the metal rod. This repeated process gradually increases the purity of the remaining Solid metal. The technique is especially useful in producing semiconductors where extremely high purity is required.

    The method is based on phase distribution principles and differences in solubility behavior rather than chemical reactivity or volatility. It is a physical separation process that exploits equilibrium differences between Solid and liquid phases to achieve purification.

    Option b – Impurities are more soluble in molten form than in Solid state

    Which compound is a synthetic derivative of progesterone?

    (A) Novestrol

    (B) Norethindrone

    (C) Chloroxylenol

    (D) Cholesterol

    Explanation:
    The question relates to steroid Chemistry and structural modification of naturally occurring hormones. Progesterone is a steroid hormone that plays a key role in reproductive Biology, and synthetic derivatives are often designed to mimic or modify its biological activity. These derivatives are commonly used in Pharmaceutical applications such as oral contraceptives and hormonal therapies.

    Synthetic modification typically involves changes to functional groups or side chains of the steroid framework, resulting in compounds with altered stability, potency, or receptor affinity. These changes can enhance oral activity or prolong biological effects compared to the natural hormone.

    Such derivatives are important in medicinal Chemistry because they allow controlled regulation of hormonal pathways. Understanding structural relationships between natural hormones and their synthetic analogs helps in predicting biological activity and therapeutic applications.

    Option b – Norethindrone

    Bakelite is synthesized by the polymerization of:

    (A) Urea and formaldehyde

    (B) Phenol and formaldehyde

    (C) Ethylene glycol and phthalic Acid

    (D) 1,3-butadiene and acrylonitrile

    Explanation:
    The question is based on condensation polymerization, a process where small Organic molecules join together while eliminating simple molecules like water. In this type of reaction, the formation of a rigid, highly cross-linked polymer depends on the presence of functional groups capable of repeated bonding. The resulting material is thermosetting in nature, meaning it cannot be reshaped once SET due to extensive cross-linking between chains.

    The process involves aromatic compounds reacting with aldehydes under controlled conditions, forming a three-dimensional Network structure. This Network provides high mechanical strength, Heat resistance, and electrical insulating properties. The reaction proceeds stepwise, forming intermediate linkages that eventually build a rigid polymer matrix.

    Such Polymers are widely used in electrical switches, handles, and industrial components due to their durability and resistance to Heat. The structure formed is highly cross-linked, making it distinct from linear or branched Polymers. The concept highlights how multifunctional monomers lead to rigid polymer networks through repeated condensation reactions.

    Option b – Phenol and formaldehyde

    An orange dye is formed when benzene diazonium chloride couples with Phenol. What is the identity of this dye?

    (A) p-Hydroxyazobenzene

    (B) p-Nitrophenol

    (C) m-Hydroxyazobenzene

    (D) p-Aminoazobenzene

    Explanation:
    The question involves azo coupling reactions, which are important in dye Chemistry. These reactions occur when a diazonium Salt reacts with an activated aromatic compound, typically containing electron-donating groups such as hydroxyl groups. The coupling takes place at positions where electron density is higher, usually ortho or para relative to activating groups.

    In this reaction, the diazonium compound acts as an electrophile and forms a nitrogen-nitrogen double bond linkage with the aromatic ring of Phenol. This creates an extended conjugated system, which is responsible for the color of the dye. The intensity and wavelength of absorbed Light depend on the extent of conjugation in the Molecule.

    Azo compounds are widely used in textile and pigment industries because of their vivid colors and stability. The formation of such dyes demonstrates how electronic effects in aromatic compounds control substitution patterns and optical properties.

    Option a – p-Hydroxyazobenzene

    Osmosis can be reversed by:

    (A) Applying pressure higher than osmotic pressure on the concentrated solution side

    (B) Applying pressure greater than osmotic pressure on the dilute solution side

    (C) Applying lower pressure on the concentrated solution side

    (D) Applying equal pressure on both sides of the membrane

    Explanation:
    The question is based on the concept of osmotic pressure and membrane Transport phenomena. Osmosis refers to the movement of solvent molecules from a region of lower solute concentration to higher solute concentration through a semipermeable membrane. This movement continues until equilibrium is reached.

    When external pressure is applied on the concentrated solution side, it can oppose the natural flow of solvent molecules. If the applied pressure exceeds the osmotic pressure, the direction of solvent movement is reversed. This process is known as reverse osmosis and is widely used in water purification and desalination technologies.

    The principle relies on overcoming the natural chemical potential difference across the membrane. By applying sufficient external force, solvent molecules are forced to move against the concentration gradient. This demonstrates how physical pressure can control molecular movement in solution systems.

    Option a – Applying pressure higher than osmotic pressure on the concentrated solution side

    When a 20 g mixture of ethanol and acetaldehyde is heated with Fehling’s solution, it produces 14.3 g of a red precipitate. What is the percentage of acetaldehyde in the mixture?

    (A) 30%

    (B) 15%

    (C) 22%

    (D) 11%

    Explanation:
    The question is based on quantitative analysis using redox reactions in Organic Chemistry. Fehling’s solution is used to detect aldehydes, which reduce copper(II) ions to form a red precipitate of copper(I) oxide. Alcohols like ethanol do not participate in this reaction under the same conditions, making the test selective for aldehydes.

    The amount of precipitate formed is directly related to the amount of aldehyde present in the mixture. By using stoichiometric relationships, the Mass of the aldehyde can be calculated from the Mass of copper oxide formed. This allows determination of the composition of the mixture.

    Such analytical techniques are important in identifying functional groups and quantifying components in mixtures. The reaction provides a clear relationship between chemical reactivity and measurable product formation, allowing indirect calculation of composition through product yield.

    Option c – 22%

    Identify the INCORRECT statement related to Molecular Orbital Theory (MOT):

    (A) Electrons in molecules are found in molecular orbitals

    (B) Atomic orbitals with similar energies and symmetry can combine

    (C) Molecular orbitals are spread over more than one nucleus

    (D) Bonding molecular orbitals possess higher energy than antibonding orbitals

    Explanation:
    The question is based on molecular orbital theory, which describes bonding in molecules through the combination of atomic orbitals to form molecular orbitals. These orbitals extend over the entire Molecule and are classified as bonding or antibonding depending on their energy and electron distribution.

    Bonding molecular orbitals result from constructive overlap and have lower energy than the original atomic orbitals, leading to stabilization. Antibonding orbitals arise from destructive overlap and have higher energy, which destabilizes the Molecule when occupied. Electrons fill these orbitals according to energy principles, influencing bond strength and magnetic properties.

    Understanding the energy ordering of these orbitals is essential for predicting molecular stability. Incorrect statements usually contradict the fundamental rule that bonding orbitals are lower in energy than antibonding orbitals. Molecular orbital theory provides a more accurate description of bonding than simple valence bond models.

    Option d – Bonding molecular orbitals possess higher energy than antibonding orbitals

    Which colloidal silver preparation is used as an eye lotion?

    (A) Colloidal gold

    (B) Colloidal antimony

    (C) Argyrol

    (D) Milk of magnesia

    Explanation:
    The question is based on colloidal Chemistry and medicinal applications of metal sols. Colloidal solutions consist of finely dispersed particles that remain suspended in a medium without settling. Silver-based colloids are known for their antimicrobial properties and have been historically used in medical formulations.

    These preparations are effective in treating infections due to the ability of silver particles to inhibit microbial growth. In ophthalmic applications, certain stabilized silver colloids are used as antiseptic solutions for eye treatment. The effectiveness depends on particle size, stability, and dispersion medium.

    Colloidal systems are widely used in pharmaceuticals because they provide controlled release and improved surface interaction. The use of silver in eye preparations highlights the intersection of nanomaterials and medical Chemistry in therapeutic applications.

    Option c – Argyrol

    Which reagent is used to oxidize CH₃CH₂CH₂CH₂CH₂OH into CH₃CH₂CH₂CH₂COOH?

    (A) Tollen’s reagent

    (B) Jones reagent

    (C) Fehling’s reagent

    (D) Schiff’s reagent

    Explanation:
    The question involves oxidation of primary Alcohols to carboxylic Acids in Organic Chemistry. Primary Alcohols undergo stepwise oxidation, first forming aldehydes and then further oxidizing to carboxylic Acids under strong oxidizing conditions. The choice of reagent determines the extent of oxidation achieved.

    Strong oxidizing agents supply oxygen or remove hydrogen from the Alcohol functional group, facilitating conversion into a carboxylic Acid. These reagents are commonly used in laboratory synthesis to achieve complete oxidation of Alcohols without stopping at the aldehyde stage.

    The process involves breaking and reforming bonds around the carbon attached to the hydroxyl group. Controlled oxidation is important in synthetic chemistry for preparing Acids from Alcohol precursors in a predictable manner.

    Option b – Jones reagent

    What is the compound formed when diborane reacts with trimethylamine?

    (A) B₂H₆·2NH₃

    (B) B₃N₃H₆

    (C) BH₃·NMe₃

    (D) B₂H₆·2NMe₃

    Explanation:
    The question is based on Lewis Acid-Base interactions involving electron-deficient and electron-rich species. Diborane is electron-deficient because boron atoms do not have complete octets, making them capable of accepting electron pairs. Trimethylamine, on the other hand, contains a lone pair on nitrogen and acts as a Lewis Base.

    When these two species react, the lone pair on nitrogen is donated to boron, forming a coordinate bond. This stabilizes the electron-deficient boron compound. The resulting product is an adduct where electron deficiency is reduced through donor-acceptor interaction.

    Such reactions are important in understanding coordination chemistry and the behavior of electron-deficient compounds. The formation of adducts demonstrates how Lewis Acid-Base theory explains bonding beyond simple covalent interactions.

    Option c – BH₃·NMe₃

    Choose the correct order of first ionization energies:

    (A) C < Si < Ge

    (B) C > Si > Ge

    (C) Ge < C < Si

    (D) Si > C > Ge

    Explanation:
    The question is based on Periodic trends in ionization energy, which is the energy required to remove the outermost electron from an isolated gaseous Atom. Ionization energy depends on atomic size, nuclear charge, and shielding effect. As we move across a period, ionization energy generally increases due to increasing nuclear attraction.

    Down a group, ionization energy decreases because atomic size increases and outer electrons are further from the nucleus, experiencing more shielding. This makes electron removal easier. These trends help compare elements across periods and groups.

    Understanding ionization energy trends is essential in predicting chemical reactivity and metallic character. Elements with higher ionization energy are less likely to lose electrons and form positive ions.

    Option b – C > Si > Ge

    Which of the following is an electron-precise hydride?

    (A) NH₃

    (B) CH₄

    (C) H₂O

    (D) B₂H₆

    Explanation:
    The question is based on classification of hydrides according to electron count and bonding behavior. Electron-precise hydrides are compounds in which atoms achieve stable electron configurations following standard covalent bonding rules. In these molecules, all atoms satisfy their typical valence requirements without electron deficiency or excess.

    Such hydrides follow normal covalent bonding patterns and do not require additional electrons or exhibit unusual bonding like electron-deficient or electron-rich hydrides. They are generally stable and have well-defined molecular structures consistent with octet rule compliance.

    Understanding hydride classification helps in studying bonding variations across different chemical compounds. Electron-precise hydrides represent the most stable and conventional bonding situation in covalent chemistry.

    Option b – CH₄

    Identify the catalyst used in the contact process for manufacturing sulfuric Acid:

    (A) Iron

    (B) Vanadium(V) oxide

    (C) Raney nickel

    (D) Anhydrous aluminium chloride

    Explanation:
    The question is based on the industrial production of sulfuric Acid, where sulfur dioxide is converted into sulfur trioxide before absorption into water to form the final Acid. This conversion step is crucial because it determines the efficiency and yield of the entire process. Since the reaction is reversible and exothermic, a catalyst is required to increase the rate without affecting the equilibrium position.

    A Solid catalyst is used to enhance the oxidation of sulfur dioxide in the presence of oxygen. This catalyst provides an alternative pathway with lower activation energy, allowing the reaction to proceed efficiently at industrially feasible temperatures. The catalyst remains unchanged after the reaction and can be reused, making it suitable for large-scale production.

    The process is widely used in chemical industries due to its efficiency and economic viability. The choice of catalyst ensures high conversion rates while maintaining stability under high-temperature conditions required for the reaction. It plays a key role in optimizing sulfuric Acid production in the contact process.

    Option b – Vanadium(V) oxide

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