NEB Grade 12 Chemistry Important Question Collection with Solution [Physical + Inorganic]

NEB Grade 12 Chemistry Important Question with Solution
Recommended:Organic Chemistry Important Question Collection with Solution

Physical Chemistry

  1. Give the defination:
    1. Normality:

      It is a measure of the concentration of a solution expressed as the number of equivalent weights of solute per liter of solution.

    2. Formality:

      It is a measure of the concentration of a solution expressed as the number of formula weights of solute per liter of solution.

    3. Molality:

      It is a measure of the concentration of a solution expressed as the number of moles of solute per kilogram of solvent.

    4. Molarity:

      It is a measure of the concentration of a solution expressed as the number of moles of solute per liter of solution.

    5. End point:

      It is the point during a titration where the indicator changes color, indicating that the reaction is complete.

    6. Equivalent point:

      It is the point during a titration where the stoichiometrically equivalent amounts of reactants have been mixed.

    7. Semi-normal solution:

      It is a solution containing one-half of a normal solution, or 0.5N.

    8. Decinormal solution:

      It is a solution containing one-tenth of a normal solution, or 0.1N.

    9. Indicator:

      It is a substance that is added to a solution to detect the presence or absence of another substance by producing a visible change, such as a change in color.

    10. Electrochemical equivalent (ECE):

      It is the amount of a substance that is deposited or liberated at an electrode during an electrochemical reaction, per unit of electrical charge that passes through the electrode.

    11. Standard electrode potential:

      It is the potential difference between an electrode and a standard reference electrode under standard conditions, which is used to measure the tendency of a half-reaction to occur.

    12. One Faraday:

      It is the amount of electrical charge carried by one mole of electrons, which is equal to 96,485.34 coulombs.

    13. pH:

      It is a measure of the acidity or basicity of a solution, defined as the negative logarithm of the hydrogen ion concentration.

    14. pOH:

      It is a measure of the acidity or basicity of a solution, defined as the negative logarithm of the hydroxide ion concentration.

    15. Enthalpy of Formation:

      It is the heat released or absorbed when one mole of a compound is formed from its constituent elements in their standard states.

    16. Enthalpy of Combustion:

      It is the heat released when one mole of a substance is completely burned in oxygen under standard conditions.

  2. Difference Between:
    1. Primary and Secondary Standard Solution with examples.
      Primary Standard Solution Secondary Standard Solution
      A solution of known purity and exact concentration. A solution that is standardized against a primary standard solution.
      Typically used in analytical chemistry, where accuracy is crucial. Typically used when the primary standard is not easily obtainable or is unstable.
      The concentration is determined by direct measurement or by gravimetric analysis. The concentration is determined by titration against a primary standard solution.
      Examples: Sodium chloride (NaCl) for chloride ion determination, Potassium hydrogen phthalate (KHP) for acid-base titrations. Examples: Sodium carbonate (Na2CO3) for standardizing hydrochloric acid (HCl), Potassium permanganate (KMnO4) for standardizing hydrogen peroxide (H2O2).
      Expensive and difficult to prepare but offer high accuracy and precision. Less expensive and easier to prepare but offer lower accuracy and precision.
    2. Intensive and Extensive property with example.
      Intensive Property Extensive Property
      A property that does not depend on the amount of substance present. A property that depends on the amount of substance present.
      Can be used to identify substances and determine their quality. Cannot be used to identify substances or determine their quality.
      Independent of the size of the system. Dependent on the size of the system.
      Does not change with physical changes (e.g., cutting, crushing, melting). Changes with physical changes (e.g., cutting, crushing, melting).
      Examples: Density, Melting point, Boiling point, Color. Examples: Mass, Volume, Length, Energy.
    3. Order of Reaction and Molecularity
      Order of Reaction Molecularity
      It is the sum of the powers to which the concentration terms are raised in the rate law equation. It is the number of molecules or atoms that participate in the rate-determining step of a chemical reaction.
      It is a measure of the reaction rate with respect to the reactant concentrations. It is a characteristic of an elementary reaction that can be determined experimentally.
      The order of a reaction can be fractional or zero, but molecularity is always a whole number. Molecularity is always an integer and is only applicable to elementary reactions.
      Can be determined experimentally by the method of initial rates or graphical analysis. Can be determined experimentally by studying the dependence of the reaction rate on the concentration of each reactant.
      Useful in chemical kinetics and reaction engineering. Useful in understanding the mechanism of a reaction and designing new reactions.
  3. Relationship between molarity and normality.
    Answer:

    We Have
    $\begin{array}{l}Molarity(M){\rm{ = }}\frac{{gm/l}}{{Molecular\,Mass}} \times 1000 - - - - - - (i)\\Normality(N){\rm{ = }}\frac{{gm/l}}{{Equivalent{\rm{ Weight}}}} \times 1000 - - - - (ii)\end{array}$ Dividing above Equation and we get
    $\frac{{Normality}}{{Molarity}} = \frac{{Equivalent{\rm{ Weight}}}}{{Molecular\,Mass}}$

  4. State Ostwald’s dilution law. What is the limitation of this law?
    Answer:

    Ostwald's Dilution Law states that the degree of dissociation of a weak electrolyte is directly proportional to the square root of its dilution.
    Mathematically, it can be expressed as α = K √C
    where α is the degree of dissociation, K is a constant characteristic of the electrolyte, and C is the concentration.

    Here are some limitations of Ostwald's Dilution Law in point form:
    • It is only applicable to weak electrolytes and not to strong electrolytes.
    • It assumes that the dissociation of a weak electrolyte is a reversible process, which may not be true in all cases.
    • It assumes that there are no interactions between the ions, which may not be true for some electrolytes.
    • It assumes that the solvent is ideal and does not affect the dissociation of the electrolyte, which may not be true.
  5. Write the application of Common ion effect.
    Answer: The Common Ion Effect has many applications in chemistry, including:
    1. Buffers: The common ion effect can be used to make buffers, which are solutions that resist changes in pH when an acid or base is added. By adding a strong acid or base to a buffer solution, the common ion effect will decrease the dissociation of the weak acid or base, thereby minimizing changes in pH.
    2. Solubility Equilibria: The common ion effect can be used to predict and control the solubility of a salt in a solution. By adding a common ion to a solution, the solubility of the salt will decrease due to the shift in the equilibrium of the dissolution reaction.
    3. Precipitation Reactions: The common ion effect can be used to control the precipitation of a salt from a solution. By adding a common ion to the solution, the solubility of the salt will decrease, causing it to precipitate out of the solution.
    4. Acid-Base Equilibria: The common ion effect can be used to shift the equilibrium of an acid-base reaction. By adding a common ion to a solution, the dissociation of a weak acid or base will decrease, thereby shifting the equilibrium towards the undissociated form.
    5. Chromatography: The common ion effect can be used in chromatography to separate and identify ions in a mixture. By adding a common ion to the eluent, the separation of the ions can be improved, leading to better resolution and identification.
  6. What will happen when HCl gas is passed over a saturated solution of NaCl. Explain the principle involved?

    When HCl gas is passed through a saturated NaCl solution, the number of Cl−1 ions increases. The increased number of Cl−1 interact with Na+1 ions to form crystallized NaCl. Thus, the solubility of NaCl decreases and it is recovered in crystalline form. This phenomenon is called "common ion effect".

  7. Derive relationship between pH and pOH.
    Answer:

    pH is defined as the negative logarithm taken for hydronium ion (H+) concentration i.e.

    pH = -log[H+] or -log[H3O+] -------(i)

    Similarly, pOH is defined as the negative logarithm talen for hydroxy ion (OH-) concentration i.e.,

    pOH = -log[OH-] ------(ii)

    We know that, [H3O+][OH-] = 10-14 at 25oC

    Taking log on both sides, log{[H3O+][OH-]} =log(10-14)

    Or, log([H3O+]) + log([OH-]) = -14 log 10

    [Note log 10= 1]

    Or, log([H3O+]) + log([OH-]) = -14

    From (i) and (ii)

    Or, - pH - pOH = -14

    Thus, pH + pOH = 14

  8. Mention one important application of standard hydrogen electrode giving example.
    Answer:

    Example: The standard hydrogen electrode is to determine the standard potentials of other electrodes.
    Ans: Example: To determine the standard potential of Zn2+ (1 M) | Zn(s) electrode, it is combined with SHE to a galvanic cell.
    Zn(s)|Zn2+(1M)||H+(1M)|H2(g,1atm)|Pt Single Hydrogen Potential: SHE
    The standard cell potential, E0cell, is measured.

    E0cell = E0H2 - E0Zn = -E0Zn, because E0H2 is zero.

    Thus, the measured emf of the cell is equal to standard potential of Zn2+(1M)|Zn(s) electrode.
  9. State first law of thermodynamics and point out its limitation.
    Answer:

    The First Law of Thermodynamics, also known as the Law of Conservation of Energy, states that energy cannot be created or destroyed, only transferred, or converted from one form to another. Limitations of the First Law of Thermodynamics:

    1. The First Law does not predict the direction of a process.
    2. It does not account for the quality or usefulness of energy.
    3. It does not consider the irreversibility of certain processes.
    4. It assumes that the internal energy of the system can be uniquely defined and measured.
  10. How does catalyst increase the rate of reaction? Describe based on energy profile diagram for catalyzed and uncatalyzed reactions.
    Answer:
    1. Catalyst lowers activation energy and increases reaction rate.
    2. Catalyst provides an alternate pathway with lower activation energy for both forward and backward reactions.
    3. Lowering activation energy makes it easier for reactants to cross the energy barrier, increasing the reaction rate.
    Energy Profile Diagram
  11. State second law of thermodynamics. Explain this law based on entropy change.
    Answer:

    The Second Law of Thermodynamics states that the entropy of an isolated system always increases over time. The entropy of the universe (system + surrounding) always increases during every spontaneous (natural) change. ΔSuniverse>0. Energy of universe is conserved whereas entropy of universe always increases in any natural or spontaneous process.

  12. What is meant by effective collision of reacting species? What are the essential conditions for the effective collision of reacting species?
    Answer:

    Effective collision of reacting species refers to a collision between two or more reactant molecules that results in a chemical reaction. Not all collisions between reactant molecules lead to a reaction, as the reactants must collide with sufficient energy and in the correct orientation to break the existing bonds and form new ones. The essential conditions for effective collision of reacting species are:

    1. Sufficient energy: The colliding particles must have sufficient kinetic energy to overcome the activation energy barrier and break the existing bonds.
    2. Correct orientation: The colliding particles must have the correct orientation so that the existing bonds can be broken, and new ones can be formed.
    3. Proper collision frequency: The number of collisions between the reactant molecules must be high enough to ensure that there are sufficient effective collisions leading to the formation of products.
  13. Find the unit of rate constant of zero, first, second and third order reaction.
    Answer: The unit of rate constant (k) for different order reactions are as follows:

    The unit of rate constant (k) for different order reactions are as follows:

    1. Zero order reaction: mol L-1 s-1
    2. First order reaction: s-1
    3. Second order reaction: L mol-1s-1
    4. Third order reaction: L2 mol-2 s-1

    The unit of rate constant for any order reaction can be derived by substituting the units of concentration and time in the rate law equation for that particular order.

    For example, consider a third order reaction with the rate law:

    Rate = k[A]2[B]

    Where [A] and [B] are the concentrations of reactants A and B, respectively, and k is the rate constant. The units of rate are mol L-1 s-1, and the units of concentration are mol L-1.

    Substituting these units in the rate law equation gives:

    mol L-1 s-1 = k x (mol L-1)2 x (mol L-1)

    Simplifying this equation, we get:

    k = mol-2 L4 s-1

    Therefore, the unit of rate constant for a third order reaction is L2 mol-2 s-1, as mentioned in the previous answer.

Inorganic Chemistry

Transistion Metal

  1. Cu2+ complexes are colored while those of Zn2+ are colorless. Explain the reason.
    Answer: Electronic Configuration of

    Cu+2= [Ar]3d9
    Zn+2−[Ar]3d10

    Copper has an unpaired electron which acts as a F center (F-centers are point defects and can be readily formed in alkaline halides) and allows electron transition in visible region importing color while Zn+2 is having no unpaired electrons hence colorless.

  2. Why Fe3+ is more stable than Fe2+?
    Answer:Electronic Configuration of:

    Fe2+: [Ar] 3d6
    Fe3+: [Ar] 3d5

    Among Fe3+ and Fe2+, Fe3+ is more stable due to half-filled d-orbital. This can be explained by the Aufbau principle. Half-filled and completely filled d-orbitals are more stable than partially filled d-orbitals. So Fe3+ is more stable than Fe2+.
  3. Sc3+, Ti4+, V5+ complexes are white even though all are transition elements. Give reason.
    Answer:Electronic configurations of Sc3+ and Ti4+ are: Sc3+: [Ar] 3d0 and Ti4+: [Ar] 3d0, respectively.

    The ions Sc3+ and Ti4+ have completely empty d-orbitals i.e., no unpaired electrons are present. Thus, their salts are colourless, as d-d transitions are not possible.
  4. What are transition metals? Why are they called so? Why is zinc not considered as transition element?
    Answer: Transition metals are a group of metallic elements located in the central block of the periodic table, between the groups 2 and 13.

    The transition metals are called so because they exhibit properties that are intermediate between those of metals and nonmetals, and they also form a bridge between the highly reactive alkali and alkaline earth metals and the less reactive metals in the later groups.
  5. What are transition metals? Why are they called so? Why is zinc not considered as transition element?
    Answer: Zinc is not considered a transition metal because it does not have partly filled (i.e., incomplete) d-orbital. It has 3d-orbital fulfilled, [Ar]3d10. It is located in the same period as transition metals but does not exhibit the characteristic properties of transition metals.
  6. What is lithopone? How is it prepared?
    Answer: Lithopone is a white pigment that is commonly used in a variety of applications such as paints, coatings, plastics, and rubber.
    The preparation of lithopone involves the following reaction: ZnSO4(aq) + BaS(aq) → ZnS(s) + BaSO4(s)

Non Metal

  1. Describe the process for Extraction of with reaction:
    Answer:
    1. (Blister) copper from copper pyrites.
    2. Steel (Mainly Open-Hearth process)
    3. Iron from iron pyrites.
    4. Zinc from zinc blend (sulphide ore).
    5. Mercury from cinnabar (HgS) ore.
  2. Write the Chemistry of:
    Answer:
    1. Blue vitriol (CuSO4.5H2O)
    2. White vitriol (ZnSO4.7H2O)
    3. Zinc white (ZnO)
    4. Calomel (Hg2Cl2)
    5. Corrosive sublimate (HgCl2)
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