Ch 1Basic Concepts of Chemistry and Chemical Calculations
5-Mark Questions
The equivalent mass of ferrous oxalate is (a) \(\frac{\text { molar mass of ferrous oxalate }}{1}\) (b) \(\frac{\text { molar mass of ferrous oxalate }}{2}\) (c) \(\frac{\text { molar mass of ferrous oxalate }}{3}\) (d) none of these
(c) \(\frac{\text { molar mass of ferrous oxalate }}{3}\) Converting grams to moles calculator is as simple as multiplying the total mass by the moles per unit of mass.
Two 22.4 liter containers A and B contains 8 g of O 2 and 8 g of SO 2 respectively at 273 K and 1 atm pressure, then (a) Number of molecules in A and B are the same (b) Number of molecules in B is more than that in A. (c) The ratio between the number of molecules in A to number of molecules in B is 2: 1 (d) Number of molecules in B is three times greater than the number of molecules in A.
(c) The ratio between the number of molecules in A to number of molecules in B is 2: 1 No. of moles of oxygen = 8 g/32 g = 0.25 moles of oxygen No. of moles of sulphur dioxide = 8 g/64 g = 0.125 moles of sulphur dioxide Ratio between the no. of molecules = 0.25: 0.125 = 2: 1
2-Mark Questions
The correct increasing order of the oxidation state of sulphur in the anions SO 4 2-, SO 3 2-, S 2 O 4 2-, S 2 O 6 2- is (a) SO 3 2- < SO 4 2- < S 2 O 4 2- < S 2 O 6 2- (b) SO 4 2- < S 2 O 4 2- < S 2 O 6 2- < SO 3 2- (c) S 2 O 4 2- < SO 3 2- < S 2 O 6 2- < SO 4 2- (d) S 2 O 6 2- < S 2 O 4 2- < SO 4 2- < SO 3 2-
(c) S 2 O 4 2- < SO 3 2- < S 2 O 6 2- < SO 4 2-
If Avogadro number were changed from 6.022 × 10 23 to 6.022 × 10 20, this would change (a) the ratio of chemical species to each other in a balanced equation (b) the ratio of elements to each other in a compound (c) the definition of mass in units of grams (d) the mass of one mole of carbon
(d) the mass of one mole of carbon
Choose the disproportionation reaction among the following redox reactions. (a) 3Mg (s) + N 2(g) → Mg 3 N 2(s) (b) P 4(s) + 3NaOH + 3H 2 O → PH 3(g) + 3NaH 2 PO 2(aq) (c) Cl 2(g) + 2KI (aq) → 2KCl (aq) + I 2(s) (d) Cr 2 O 3(s) + 2Al (s) → Al 2 O 3(s) + 2Cr (s)
b) P 4(s) + 3NaOH + 3H 2 O → PH 3(g) + 3NaH 2 PO 2(aq)
1-Mark Questions (MCQ)
The equivalent mass of potassium permanganate in an alkaline medium is MnO 4 + 2H 2 O + 3e → MnO 2 + 4OH –
(b) 52.7 The reduction reaction of the oxidising agent (Mn0 4 ) involves the gain of 3 electrons. Hence the equivalent mass = (Molar mass of KMnO 4 )/3 = 158.1/3 = 52.7
Ch 2Quantum Mechanical Model of Atom
5-Mark Questions
Based on equation E = -2.178 × 10 -18 J(z 2 /n 2 ), certain conclusions are written. Which of them is not correct? (a) Equation can be used to calculate the change in energy when the electron changes orbit (b) For n – 1, the electron has a more negative energy then it does for n = 6 which means that the eiectron is more loosely bound in the smallest allowed orbit (c) The negative sign in equation simply means that the energy of electron bound to the nucleus is lower than it would be if the electrons were at the infinite distance ffome nucleus. (d) Larger the value of n, the larger is the orbit radius.
(b) For n – 1, the electron has a more negative energy then it does for n = 6 which means that the eiectron is more loosely bound in the smallest allowed orbit.
How many orbitals are possible for n = 4?
When n = 0, l = 0, 1,2 and 3. Hence, there are four subshells namely, s, p, d and f l = 0, m 1 = 0; one 4s orbital, l = 1, m = -1, 0, +1; three 4p orbitals, l = 2, m 1 = – 2, -1, 0, +1, +2; five 4d orbitals and l = 3, m 1 = – 3, -2, -1, 0, 1, 2, 3; seven 4f orbitals. Hence, the number of possible orbitals when n = 4 are sixteen.
2-Mark Questions
The electronic configuration of Eu (Atomic no, 63), Gd (Atomic no. 64), and Tb (Atomic no. 65) are (a) [Xe] 4f 6 5d 1 6s 2, [Xe] 4f 7 5d 1 6s 2 and [Xe] 4f 8 5d 1 6s 2 (b) [Xe] 4f 7 6s 2, [Xe] 4f 1 5d 1 6s 2 and [Xe] 4f 9 6s 2 (c) [Xe] 4f 7 6s 2, [Xe] 4f 8 6s 2 and [Xe] 4f 8 5d 1 6s 2 (d) [Xe] 4f 6 5d 1 6s 2, [Xe] 4f 1 5d 1 6s 2 and [Xe] 4f 9 6s 2
(b) [Xe] 4f 7 6s 2, [Xe] 4f 1 5d 1 6s 2 and [Xe] 4f 9 6s 2
If n = 6, the sequence for filling electrons will be, (a) ns → (n – 2)f → (n – 1)d → np (b) ns → (n – 1 )d → (n – 2)f → np (c) ns → {n – 2)f → np → (n – 1 )d (d) none of these are correct
(a) ns → (n – 2)f → (n – 1)d → np
Which quantum number reveals information about the shape, energy, orientation, and size of orbitals?
Magnetic quantum numbers reveal information about the shape, energy, orientation, and size of orbitals.
1-Mark Questions (MCQ)
Electronic configuration of species M 2+ is 1s 2 2s 2 2p 6 3s 2 3p 6 3d 6 and its atomic weight is 56. The number of neutrons in the nucleus of species M is
(c) 30
Ch 3Periodic Classification of Elements
5-Mark Questions
What are isoelectronic ions? Give examples.
Two ions having the same number of electrons are called isoelectronic ions. Example: Na + (1s 2 2s 2 2p 6 ) and F – (1s 2 2s 2 2p 6 ). Both these ions contain eight electrons.
What is an effective nuclear charge?
The net nuclear charge experienced by valence electrons in the outermost shell is called the effective nuclear charge. Z eff = Z – S Where Z = Atomic number S = Screening constant calculated by using Slater’s rules.
2-Mark Questions
Identify the wrong statement. (a) Amongst the isoelectronic species, smaller the positive charge on cation, smaller is the ionic radius (b) Amongst isoelectric species greater the negative charge on the anion, larger is the ionic radius (c) Atomic radius of the elements increases as one moves down the first group of the periodic table (d) Atomic radius of the elements decreases as one moves across from left to right in the 2nd period of the periodic table.
(a) Amongst the isoelectronic species, smaller the positive charge on cation, smaller is the ionic radius
Define modern periodic law.
The modem periodic law states that “The physical and chemical properties of the elements are a periodic function of their atomic numbers.”
Define electronegativity.
Electronegativity is defined as the relative tendency of an element present in a covalently bonded molecule, to attract the shared pair of electrons towards itself.
1-Mark Questions (MCQ)
What would be the IUPAC name for an element with atomic number 222?
(d) bibibium
Ch 4Hydrogen
5-Mark Questions
Explain why hydrogen is not placed with the halogen in the periodic table.
Hydrogen resembles alkali metals as well as halogens. Hydrogen resembles more alkali metals than halogens. The electron affinity of hydrogen is much less than that of halogen atoms. Hence the tendency to form hydride ion is low compared to that of halogens. In most of its compounds hydrogen exists in a +1 oxidation state. Therefore it is reasonable to place the hydrogen in group 1 along with alkali metals as shown in the latest periodic table published by IUPAC.
A cube at 0°C is placed in some liquid water at 0°C, the ice cube sinks – Why?
In ice, each atom is surrounded tetrahedrally by four water molecules through hydrogen bonds. That is, the presence of two hydrogen atoms and two lone pairs of electrons on oxygen atoms in each water molecule allows the formation of a three-dimensional structure. This arrangement creates an open structure, which accounts for the lower density of ice compared with water at 0°C. While in liquid water, unlike ice where hydrogen bonding occurs over a long-range, the strong hydrogen bonding prevails only in a short-range and therefore the denser packing. Hence, the ice cube sinks in water.
1-Mark Questions (MCQ)
Which of the following statements about hydrogen is incorrect?
(c) Hydrogen has three isotopes of which tritium is the most common.
Ch 5Alkali and Alkaline Earth Metals
5-Mark Questions
Why sodium hydroxide is much more water-soluble than chloride?
NaOH + H 2 O ⇌ Na + + OH – 1. This reaction is an exothermic reaction. Sodium hydroxide is a strong base, completely dissociated in an aqueous medium. The heat evolved increases the stability. This phenomenon is strong enough to prove that sodium hydroxide crystals are readily dissolved in water. 2. NaCl is geologically stable. If kept dry, it will remain a free-flowing solid for years. Water can dissolve NaCl because the Na + ions are attracted by OH – in water and Cl – ions are attracted by H + in water. The solubility of NaCl does not increase the temperature. …
Explain what to mean by efflorescence.
* Efflorescence is a process of losing water of hydration from hydrate. * Sodium carbonate crystallises as decahydrate which is white in colour. * Upon heating, it loses the water of crystallization to form a monohydrate. * Monohydrate (Na 2 CO 3.H 2 O) is formed as a result of efflorescence. Na 2 CO 3.10H 2 O → Na 2 CO 3.H 2 O + 9H 2 O
2-Mark Questions
Find the wrong statement (a) sodium metal is used in organic qualitative analysis (b) sodium carbonate is soluble in water and it is used in inorganic qualitative analysis (c) potassium carbonate can be prepared by solvay process (d) potassium bicarbonate is acidic salt
(c) potassium carbonate can be prepared by solvay process
Match the flame colours of the alkali and alkaline earth metal salts in the bunsen burner (p) Sodium (1) Brick red (q) Calcium (2) Yellow (r) Barium (3) Violet (s) Strontium (4) Apple green (t) Cesium (5) Crimson red (u) Potassium (6) Blue (a) p – 2, q – 1, r – 4, s – 5, t – 6, u – 3 (b) p – 1, q – 2, r – 4, s – 5, t – 6, u – 3 (c) p – 4, q – 1, r – 2, s – 3, t – 5, u – 6 (d) p – 6, q – 5, r – 4, s – 3, t – 1, u – 2
(a) p – 2, q – 1, r – 4, s – 5, t – 6, u – 3
Which is the correct sequence of solubility of carbonates of alkaline earth metals ? (a) BaCO 3 > SrCO 3 > CaCO 3 > MgCO 3 (b) MgCO 3 > CaCO 3 > SrCO 3 > BaCO 3 (c) CaCO 3 > BaCO 3 > SrCO 3 > MgCO 3 (d) BaCO 3 > CaCO 3 > SrCO 3 > MgCO 3
(b) MgCO 3 > CaCO 3 > SrCO 3 > BaCO 3
1-Mark Questions (MCQ)
For alkali metals, which one of the following trends is incorrect?
(c) Density: Li < Na < K < Rb
Ch 6Gaseous State
5-Mark Questions
State Boyle’s law.
Boyle’s law states that at a given temperature. the volume occupied by a fixed mass of a gas is inversely proportional to its pressure. V α \(\frac {1}{P}\); where T and n are fixed or PV = Constant = k
A balloon filled with air at room temperature and cooled to a much lower temperature can be used as a model for Charle’s law.
Yes, a balloon filled with air at room temperature and cooled to a much lower temperature can be used as a model for Charle’s law. The volume of balloon decreases when the temperature reduced from room temperature to low temperature. When cooled, the kinetic energy of the gas molecules decreases, so that the volume of the balloon also decreases.
2-Mark Questions
Would it be easier to drink water with a straw on the top of Mount Everest?
It would be harder on the top of the mountain; because the external pressure pushing on the liquid to force it up the straw is less.
It takes 192 sec for an unknown gas to diffuse through a porous wall and 84 sec for N2 gas to effuse at the same temperature and pressure. What Is the molar mass of the unknown gas?
A gas’s partial pressure formula is the gas pressure exerted if that gas were alone.
1-Mark Questions (MCQ)
Gases deviate from ideal behavior at high pressure. Which of the following statement(s) is correct for non-ideality? (a) at high pressure the collision between the gas molecule become enormous (b) at high pressure the gas molecules move only in one direction (c) at high pressure, the volume of gas become insignificant (d) at high pressure the intermolecular interactions become significant
(d) at high pressure the intermolecular interactions become significant
Ch 7Thermodynamics
5-Mark Questions
State the first law of thermodynamics.
The first law of thermodynamics states that “the total energy of an isolated system remains constant though it may change from one form to another” (or) Energy can neither be created nor destroyed but may be converted from one form to another.
Define Hess’s law of constant heat summation.
The heat changes in chemical reactions are equal to the difference in internal energy (∆U) or heat content (∆H) of the products and reactants, depending upon whether the reaction is studied at constant volume or constant pressure. Since ∆U and ∆H are functions of the state of the system, the heat evolved or absorbed in a given reaction depends only on the initial state and final state of the system and not on the path or the steps by which the change takes place.
2-Mark Questions
∆S is expected to be maximum for the reaction (a) Ca(S) + 1/2 O 2 (g) → CaO(S) (b) C(S) + O 2 (g) → CO 2 (g) (c) N 2 (g) + O 2 (g) → 2NO(g) (d) CaCO 3 (S) → CaO(S) + CO 2 (g)
(d) CaCO 3 (S) → CaO(S) + CO 2 (g)
Explain intensive properties with two examples.
The property that is independent of the mass or size of the system is called an intensive property. e.g., Refractive index and surface tension.
Give Kelvin a statement of the second law of thermodynamics.
Kelvin-Planck statement: It is impossible to take heat from a hotter reservoir and convert a cyclic process heat to a cooler reservoir.
1-Mark Questions (MCQ)
The amount of heat exchanged with the surrounding at constant temperature and pressure is given by the quantity
(b) ∆H
Ch 8Physical and Chemical Equilibrium
5-Mark Questions
If there is no change in concentration, why is the equilibrium state considered dynamic?
At chemical equilibrium the rate of two opposing reactions are equal and the concentration of. reactants and products do not change with time. This condition is not static and is dynamic because both the forward and reverse reactions are still occurring at the same rate and no macroscopic change is observed. So the chemical equilibrium is in a state of dynamic equilibrium.
What is the relation between K p and K c ? Given one example for which K p is equal to K c.
The relation between K p and K c is K p = K C (RT) ∆n g K p = equilibrium constant is terms of partial pressure. K c = equilibrium constant is terms of concentration. R = gas constant T = Temperature. ∆n g = Difference between the sum of the number of moles of products and the sum of number of moles of reactants in the gas phase. When ∆n g = 0 K p = K C (RT) 0 = K C i.e., K p = K C Example: H 2 (g) + I 2 (g) \(\rightleftharpoons\) 2HI(g) ∆n g = 2 – 2 = 0 ∴ K p = K C for the synthesis of HI.
2-Mark Questions
Match the equilibria with the corresponding conditions: i) Liquid ⇌ Vapour ii) Solid ⇌ Liquid iii) Solid ⇌ Vapour iv) Solute(s) ⇌ Solute (Solution) 1) Melting point 2) Saturated solution 3) Boiling point 4) Sublimation point 5) Unsaturated solution
b) 3 1 4 2
The equilibrium constants of the following reactions are: N 2 + 3H 2 ⇌ 2NH 3; K 1 N 2 + O 2 ⇌ 2NO; K 2 H 2 + 1/2O 2 ⇌ H 2 O; K 3 The equilibrium constant (K) for the reaction; 2NH 3 + 5/2 O 2 ⇌ 2NO + 3H 2 O, will be a) K 2 3 K 3 /K 1 b) K 1 K 3 3 /K 2 c) K 2 K 3 3 /K 1 d) K 2 K 3 /K 1
c) K 2 K 3 3 /K 1
A 20 litre container at 400 K contains CO 2 (g) at pressure 0.4 atm and an excess (neglect the volume of solid SrO). The volume of the container is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when the pressure of CO 2 attains its maximum value will be: Given that: SrCO 3 (S) ≅ SrO + CO 2 (g) [K p = 1.6 atm] a) 2 litre b) 5 litre c) 10 litre d) 4 litre
b) 5 litre II. Write brief answer to the following questions:
1-Mark Questions (MCQ)
If K b and K f for a reversible reaction are 0.8 × 10 -5 and 1.6 × 10 -4 respectively, the value of the equilibrium constant is, a) 20 b) 0.2 × 10 -1 c) 0.05 d) None of these
a) 20
Ch 9Solutions
5-Mark Questions
At same temperature, which pair of the following solutions are isotonic? a) 0.2 M BaCl 2 and 0.2 M urea b) 0.1 M glucose and 0.2 M urea c) 0.1 M NaCl and 0.1 M K 2 SO 4 d) 0.1 M Ba(NO 3 ) 2 and 0.1 M Na 2 SO 4
d) 0.1 M Ba(NO 3 ) 2 and 0.1 M Na 2 SO 4 Formula of normality · In the case of acid-base chemistry, normality is used to express the concentration of hydronium ions (H3O+).
The observed depression in freezing point of water for a particular solution is 0.093°C calculate the concentration of the solution in molality. Given that molal depression constant for water is 1.86 K Kg mol -1.
∆T f = 0.093°C = 0.093 K, m = ? K f = 1.86 K Kg mol -1 ∆T f = K f.m ∴ m = \(\frac{\Delta \mathrm{T}_{\mathrm{f}}}{\mathrm{K}_{\mathrm{f}}}=\frac{0.093 \mathrm{~K}}{1.86 \mathrm{~K} \mathrm{Kg} \mathrm{mol}^{-1}}\) = 0.05 mol Kg -1 = 0.05 m
2-Mark Questions
P 1 and P 2 are the vapour pressures of pure liquid components, 1 and 2 respectively of an ideal binary solution If x 1 represents the mole fraction of component 1, the total pressure of the solution formed by 1 and 2 will be a) P 1 + x 1 (P 2 – P 1 ) b) P 2 – x 1 (P 2 + P 1 ) c) P 1 – x 2 (P 1 – P 2 ) d) P 1 + x 2 (P 1 – P 2 )
c) P 1 – x 2 (P 1 – P 2 )
Two liquids X and Y on mixing gives a warm solution. The solution is a) ideal b) non-ideal and shows positive deviation from Raoult’s law c) ideal and shows negative deviation from Raoult’s Law d) non-ideal and shows negative deviation from Raoult’s Law
d) non-ideal and shows negative deviation from Raoult’s Law
What is osmosis?
Osmosis is a spontaneous process by which the solvent molecules pass through a semipermeable membrane from a solution of lower concentration to the solution of higher concentration.
1-Mark Questions (MCQ)
The molality of a solution containing 1.8 g of glucose dissolved in 250 g of water is a) 0.2 M b) 0.01 M c) 0.02 M d) 0.04 M
d) 0.04 M
Ch 10Chemical Bonding
5-Mark Questions
Pick out the incorrect statement from the following: a) sp 3 hybrid orbitals are equivalent and are at an angle of 109°28’ with each other b) dsp 2 hybrid orbitals are equivalent and bond angle between any two of them is 90° c) All five sp 3 d hybrid orbitals are not equivalent out of these five sp 3 d hybrid orbitals, three are at an angle of 120° remaining two are perpendicular to the plane containing the other three d) none of these
c) All five sp 3 d hybrid orbitals are not equivalent out of these five sp 3 d hybrid orbitals, three are at an angle of 120° remaining two are perpendicular to the plane containing the other three
Define the following: i) Bond order ii) Hybridisation iii) σ – bond
i) Bond order: The number of bonds formed between the two bonded atoms in a molecule is called the bond order. ii) Hybridisation: Hybridisation is the process of mixing of atomic orbitals of the same atom with comparable energy to form an equal number of new equivalent orbitals with the same energy. iii) σ – bond: When two atomic orbitals overlap linearly along the axis, the resultant bond is called a sigma (σ) bond.
2-Mark Questions
In ClF 3, NF 3 and BF 3 molecules the chlorine, nitrogen and boron atoms are a) sp 3 hybridised b) sp 3, sp 3 and sp 2 respectively c) sp 3 hybridised d) sp 3 d, sp 3 and sp hybridised respectively
d) sp 3 d, sp 3 and sp hybridised respectively
When one s and three p orbitals hybridise, a) four equivalent orbitals at 90° to each other will be formed b) four equivalent orbitals at 109°28’ to each other will be formed c) four equivalent orbitals, that are lying the same plane will be formed d) none of these
b) four equivalent orbitals at 109°28’ to each other will be formed
Which of these represents the correct order of their increasing bond order. a) C 2 + < C 2 2- < O 2 2- < O 2 b) C 2 2- < C 2 + < O 2 < O 2 2- c) O 2 2- < O 2 < C 2 2- < C 2 + d) O 2 2- < C 2 + < O 2 < C 2 2-
d) O 2 2- < C 2 + < O 2 < C 2 2-
1-Mark Questions (MCQ)
In which of the following compound does the central atom obey the octet rule? a) XeF 4 b) AlCl 3 c) SF 6 d) SCl 2
d) SCl 2
Ch 11Fundamentals of Organic Chemistry
5-Mark Questions
Give the general characteristics of organic compounds.
All organic compounds have the following characteristic properties. * They are covalent compounds of carbon and generally insoluble in water and readily soluble in organic solvent such as benzene, toluene, ether, chloroform etc… * Many of the organic compounds are inflammable (except CCl 4 ). They possess low boiling and melting points due to their covalent nature. * Organic compounds are characterized by functional groups. A functional group is an atom or a specific combination of bonded atoms that react in a characteristic way, irrespective of the organic molecule in which it is present. …
Write a note on homologous series.
Homologous series: A series of organic compounds each containing a characteric functional group and the successive members differ from each other in molecular formula by a CH 2 group is called homologous series. Example: Alkanes: Methane (CH 4 ), Ethane (C 2 H 6 ), Propane (C 3 H 8 ) etc.. Alcohols: Methanol (CH 3 OH), Ethanol (C 2 H 5 OH) Propanol (C 3 H 7 OH) etc…
2-Mark Questions
Select the molecule which has only one π bond. a) CH 3 – CH = CH – CH 3 b) CH 3 – CH = CH – CHO c) CH 3 – CH = CH – COOH d) All of these
a) CH 3 – CH = CH – CH 3
In the hydrocarbon the state of hybridization of carbon 1, 2, 3, 4 and 7 are in the following sequence. a) sp, sp, sp 3, sp 2, sp 3 b) sp 2, sp, sp 3, sp 2, sp 3 c) sp, sp, sp 2, sp, sp 3 d) none of these
a) sp, sp, sp 3, sp 2, sp 3
The IUPAC name of the compound CH 3 – CH = CH – C ≡ CH is a) Pent – 4- yn – 2 – ene b) Pent – 3- en – 1- yne c) Pent – 2 – en – 4 – yne d) Pent – 1 yn – 3 – ene
b) Pent – 3- en – 1- yne
1-Mark Questions (MCQ)
The general formula for alkadiene is a) C n H 2n b) C n H 2n – 1 c) C n H 2n – 2 d) C n H n – 2
c) C n H 2n – 2
Ch 12Basic Concepts of Organic Reactions
5-Mark Questions
Write short notes on a) Resonance b) Hyper Conjugation
a) Resonance (or) Mesomeric effect: The resonance is a chemical phenomenon which is observed in certain organic compounds possessing double bonds at a suitable position. Certain organic compounds can be represented by more than one structure and they differ only in the position of bonding and lone pair of electrons. Such structures are called resonance structures (canonical structures) and this phenomenon is called resonance. This phenomenon is also called mesomerism or mesomeric effect. …
What are electrophiles and nucleophiles? Give suitable examples for each.
Electrophiles: Electrophiles are reagents that are attracted towards negative charge or electron-rich center. They are either positively charged ions or electron-deficient neutral molecules. Example: CO 2, AlCl 3, BF 3, FeCl 3, NO +, NO + 2,, etc. Nucleophiles: Nucleophiles are reagents that has high affinity for electropositive centers. They possess an atom that has an unshared pair of electrons. They are usually negatively charged ions or electron-rich neutral molecules. Example. NH 3, R-NH 2, R-SH, H 2 O, R-OH, CN –. OH – etc.
2-Mark Questions
Decreasing order of nucleophilicity is a) OH – > NH 2 – > -OCH 3 > RNH 2 b) NH 2 – > OH – > -OCH3 > RNH 2 c) NH 2 – > CH 3 O – > OH – > RNH 2 d) CH 3 O – > NH 2 – > OH – > RNH 2
b) NH 2 – > OH – > -OCH3 > RNH 2
Which of the group has highest + I effect? a) CH 3 – b) CH 3 – CH 2 – c) (CH 3 ) 2 – CH- d) (CH 3 ) 3 – C –
d) (CH 3 ) 3 – C –
The geometrical shape of carbocation is a) Linear b) tetrahedral c) Planar d) Pyramidal
c) Planar II. Write brief answer to the following questions:
1-Mark Questions (MCQ)
For the following reactions (A) CH 3 CH 2 CH 2 Br + KOH → CH 3 – CH = CH 2 + KBr + H 2 O (B) (CH 3 ) 3 CBr + KOH → (CH 3 ) 3 COH + KBr (C) Which of the following statement is correct? a) (A) is elimination, (B) and (C) are substitution b) (A) is substitution, (B) and (C) are elimination c) (A) and (B) are elimination and (C) is addition reaction d) (A) is elimination, (B) is substitution and (C) is addition reaction
d) (A) is elimination, (B) is substitution and (C) is addition reaction
Ch 13Hydrocarbons
5-Mark Questions
Write short notes on ortho, para directors in aromatic electrophilic substitution reactions.
All the activating groups are ‘ortho-para’ directors. Example: – OH, – NH 2, -NHR, -NHCOCH 3, -OCH 3 -CH 3 – C 2 H 5 etc. Let us consider the directive influences of phenolic (-OH) group. Phenol is the resonance hybrid of following structures. In these resonance structures, the (-) charge residue is present on ortho and para position of ring structure. It is quite evident that the lone pair of electron on the atom which is attached to the ring involves in resonance and makes the ring more electron rich than benzene. …
Describe the mechanism of Nitration of benzene.
It is prepared by the action of a mixture of con. HNO 3 and con. H 2 SO 4 (nitrating mixture) on benzene maintaining the temperature below 333 K. Nitration: Sulphuric acid generates the electrophile – NO 2 +, nitronium ion-from nitric acid. This is an example of aromatic electrophilic substitution reaction. The generation of nitronium ion H 2 SO 4 + HONO 2 → + HSO 4 – To the nitronium ion (being an electron deficient species) the π bond of benzene, donates a pair of electrons forming a 6-bond. A species with a + ve charge is formed as an intermediate. …
2-Mark Questions
The correct statement regarding the comparison of staggered and eclipsed conformations of ethane, is a) The eclipsed conformation of ethane is more stable than staggered conformation even though the eclipsed conformation has torsional strain. b) The staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain. c) The staggered conformation of ethane is less stable than eclipsed conformation, because staggered conformation has torsional strain. d) The staggered conformation of ethane is less stable than eclipsed conformation, because staggered conformation has no torsional strain.
b) The staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain.
The compound formed at anode in the electrolysis of an aqueous solution of potassium acetate are a) CH 4 and H 2 b) CH 4 and CO 2 c) C 2 H 6 and CO 2 d) C 2 H 4 and Cl 2
c) C 2 H 6 and CO 2
Which among the following alkenes on reductive ozonolysis produces only propanone? a) 2 – Methyl propene b) 2 – Methyl but – 2 – ene c) 2, 3 – Dimethyl but – 1- ene d) 2, 3 – Dimethyl but – 2 – ene
d) 2, 3 – Dimethyl but – 2 – ene
1-Mark Questions (MCQ)
C 2 H 5 Br + 2Na C 4 H 10 + 2NaBr. The above reaction is an example of which of the following a) Reimer Tiemann reaction b) Wurtz reaction c) Aldol condensation d) Hoffmann reaction
b) Wurtz reaction
Ch 14Haloalkanes and Haloarenes
5-Mark Questions
Classify the following compounds in the form of alkyl, allylic, vinyl, benzylic halides. i) CH 3 – CH = CH – Cl ii) C 6 H 5 CH 2 I iii) iv) CH 2 = CH – Cl
i) CH 3 – CH = CH – Cl = Allylic halide ii) C 6 H 5 CH 2 I = Benzylic halide iii) = Alkyl halide iv) CH 2 = CH – Cl = Vinyl halide
Why chlorination of methane is not possible in dark?
Chlorination of methane is a free radical substitution reaction. Before chlorine reacts with methane, the Cl-Cl single bond must break to form free radicals and this can only be done in the presence of ultraviolet light. In dark, chlorine-free radicals formation is not possible and so chlorination of methane is not possible in dark. The ultraviolet light is a source of energy and is being used to break of Cl-Cl and produce Cl free radical Free radicals which can attack methane. in dark, this is not possible.
2-Mark Questions
The IUPAC name of is a) 2 – Bromo pent – 3 – ene b) 4 – Bromo pent – 2 – ene c) 2 – Bromo pent – 4 – ene d) 4 – Bromo pent – 1 – ene
b) 4 – Bromo pent – 2 – ene
Match the compounds given in Column I with suitable items given in Column II: Column I (Compound) Column II (Uses) A. Iodoform 1. Fire extinguisher B. Carbon tetra chloride 2. Insecticide C. CFC 3. Antiseptic D. DDT 4. Refrigerants Code a) A → 2 B → 4 C → 1 D → 3 b) A → 3 B → 2 C → 4 D → 1 c) A → 1 B → 2 C → 3 D → 4 d) A → 3 B → 1 C → 4 D → 2
d) A → 3 B → 1 C → 4 D → 2
The major products obtained when chlorobenzene is nitrated with HNO 3 and con H 2 SO 4 a) 1 – chloro – 4 – nitrobenzene b) 1 – chloro – 2 – nitrobenzene c) 1 – chloro – 3 – nitrobenzene d) 1 – chloro – 1 – nitrobenzene
a) 1 – chloro – 4 – nitrobenzene
1-Mark Questions (MCQ)
Of the following compounds, which has the highest boiling point? a) n – Butyl chloride b) Isobutyl chloride c) t – Butyl chloride d) n – Propyl chloride
a) n – Butyl chloride
Ch 15Environmental Chemistry
5-Mark Questions
Dissolved oxygen in water is responsible for aquatic life. What processes are responsible for the reduction in dissolved oxygen in water?
Organic matter such as leaves, grass, trash can pollute water. Microorganisms present in water can decompose this organic matter and consume dissolved oxygen in water. Eutrophication is a process by which water bodies receive excess nutrients that stimulates excessive plant growth. This enhanced plant growth in water bodies is called algal bloom. The growth of algae in extreme abundance covers the water surface and reduces the oxygen concentration in water. Thus, bloom-infeded water inhibits the growth of others. living organisms in the water body. …
What would happen, if the greenhouse gases were totally missing in the earth’s atmosphere?
The solar energy radiated back from the earth’s surface is absorbed by the greenhouse gases. (CO 2, CH 4, O 3, CFCs) are present near the earth’s surface. They heat up the atmosphere near the earth’s surface and keep it warm. As a result of these, there is the growth of vegetation that supports life. In the absence of this effect, there will be no life of both plant and animal on the surface of the earth.
2-Mark Questions
Identify the wrong statement in the following. (a) The clean water would have a BOD value of less than 5 ppm (b) Greenhouse effect is also called Global warming (c) Minute solid particles in air are known as particulate pollutants (d) Biosphere is the protective blanket of gases surrounding the earth
(c) Minute solid particles in air are known as particulate pollutants
Match the List I with List-II and select the correct answer using the code given below the lists: List I List II A. Depletion of the ozone layer 1. CO 2 B. Acid rain 2. NO C. Photochemical smog 3. SO 2 D. Greenhouse effect 4. CFC Code:
(a)
Match List I with List-II and select the correct answer using the code given below the lists. List I List II A. Stone leprosy 1. CO B. Biological magnification 2. Greenhouse gases C. Global warming 3. Acid rain D. Combination with hemoglobin 4. DDT Code:
(d)
1-Mark Questions (MCQ)
The gaseous envelope around the earth is known as atmosphere. The region lying between an altitudes of 11 – 50 km is _______.
(d) Stratosphere
Frequently asked questions
- The equivalent mass of ferrous oxalate is (a) \(\frac{\text { molar mass of ferrous oxalate }}{1}\) (b) \(\frac{\text { molar mass of ferrous oxalate }}{2}\) (c) \(\frac{\text { molar mass of ferrous oxalate }}{3}\) (d) none of these
- (c) \(\frac{\text { molar mass of ferrous oxalate }}{3}\) Converting grams to moles calculator is as simple as multiplying the total mass by the moles per unit of mass.
- Two 22.4 liter containers A and B contains 8 g of O 2 and 8 g of SO 2 respectively at 273 K and 1 atm pressure, then (a) Number of molecules in A and B are the same (b) Number of molecules in B is more than that in A. (c) The ratio between the number of molecules in A to number of molecules in B is 2: 1 (d) Number of molecules in B is three times greater than the number of molecules in A.
- (c) The ratio between the number of molecules in A to number of molecules in B is 2: 1 No. of moles of oxygen = 8 g/32 g = 0.25 moles of oxygen No. of moles of sulphur dioxide = 8 g/64 g = 0.125 moles of sulphur dioxide Ratio between the no. of molecules = 0.25: 0.125 = 2: 1
- The correct increasing order of the oxidation state of sulphur in the anions SO 4 2-, SO 3 2-, S 2 O 4 2-, S 2 O 6 2- is (a) SO 3 2- < SO 4 2- < S 2 O 4 2- < S 2 O 6 2- (b) SO 4 2- < S 2 O 4 2- < S 2 O 6 2- < SO 3 2- (c) S 2 O 4 2- < SO 3 2- < S 2 O 6 2- < SO 4 2- (d) S 2 O 6 2- < S 2 O 4 2- < SO 4 2- < SO 3 2-
- (c) S 2 O 4 2- < SO 3 2- < S 2 O 6 2- < SO 4 2-
- If Avogadro number were changed from 6.022 × 10 23 to 6.022 × 10 20, this would change (a) the ratio of chemical species to each other in a balanced equation (b) the ratio of elements to each other in a compound (c) the definition of mass in units of grams (d) the mass of one mole of carbon
- (d) the mass of one mole of carbon
These important questions are selected from the Samacheer Kalvi Class 11 Chemistry textbook book-back exercises to help you revise the most useful questions. Mark weightage (5/2/1) follows the usual exam pattern and may vary by exam — always check your latest syllabus and question pattern. Open each chapter for the complete set of questions and answers.