Category: Class 12 Chemistry Objective Questions

Chemistry Assessment Questions for Class 12 on “Amines Chemical Reactions – 4”.

1. Which gas is produced when ethanamine reacts with nitrous acid?
a) N₂
b) H₂
c) HCl
d) O₂
Answer: a
Clarification: Nitrogen gas is evolved due to the decomposition of diazonium salts which are formed from the reaction of 1° aliphatic amines with HNO₂. This is used in the estimation of amino acids and proteins.

2. What is the main product of the following reaction?

a) Benzene diazonium chloride
b) Phenol
c) o-Nitrosoaniline
d) p-Nitrosoaniline
Answer: b
Clarification: Primary aromatic amines react with nitrous acid at cold temperatures (273-278K) to form diazonium salts. However, if the temperature is more than 278K, phenol is formed along with the evolution of nitrogen gas.

3. Tertiary amines dissolve in nitrous acid to form corresponding salts.
a) True
b) False
Answer: a
Clarification: Tertiary amines on reaction with cold HNO₂ remain dissolved, forming amine nitrite salts, which decompose to nitrosoamines and alcohol on warming.

4. Which of the following is not a final product of the reaction between propylamine and nitrous acid?
a) CH₃CH₂CH₂N₂Cl
b) CH₃CH₂CH₂OH
c) N₂ gas
d) H₂O
Answer: a
Clarification: CH₃CH₂CH₂N₂Cl is a diazonium salt derivative of propanamine and is first formed when it reacts with HNO₂. This is a very unstable compound and immediately decomposes to give propanol and evolves nitrogen gas.

5. Diethylamine reacts with nitrous acid in the cold to form _______
a) diazonium salt
b) alcohol
c) imine
d) nitrosoamine
Answer: d
Clarification: Secondary amines react slowly with nitrous acid at low temperatures to give yellow oily nitrosoamines. Diethylamine specifically produces N,N-diethylnitrosoamine.

6. Hinsberg’s reagent is _______
a) benzenesulfonic acid
b) benzenesulphonyl chloride
c) p-toluenesulphonyl chloride
d) chlorosulphuric acid
Answer: b
Clarification: Benzenesulphonyl chloride (C₆H₅SO₂Cl), or arylsulphonyl chloride, is known as Hinsberg’s reagent. It is an important compound used in the distinction of different classes of amines.

7. Which of the following amines, on reaction with benzenesulphonyl chloride, will give a sulphonamide that is insoluble in alkali?
a) Ethylamine
b) Ethylmethylamine
c) Trimethylamine
d) Aniline
Answer: b
Clarification: Secondary amines on reaction with Hinsberg’s reagent produce sulphonamides without any hydrogen atom attached to the nitrogen atom. Hence, it is not acidic and therefore insoluble in alkali. The amide produced in this case is N-ethyl-N-methylbenzenesulphonamide.

8. Which of the following amines will form a product that is soluble in KOH, on reaction with Hinsberg’s reagent?
a) Isopropylamine
b) Diethylamine
c) N,N-Dimethylpropylamine
d) N,N-Dimethylaniline
Answer: a
Clarification: The reaction of Hinsberg’s reagent with Isopropylamine gives N-isopropylbenzene sulphonamide. The lone hydrogen attached to the N atom is strongly acidic due to the presence of strong electron withdrawing sulphonyl group. Hence, is it soluble in alkali like KOH.

9. Which of the following reactions/tests does not help in the distinction between ethylamine and diethylamine?
a) Carbylamine test
b) Hinsberg’s test
c) Reaction with HNO₂
d) Reaction with CH₃CH₂Br
Answer: d
Clarification: Both ethylamine and diethylamine on reaction with CH₃CH₂Br eventually gives quaternary ammonium salt. Hence, the alkylation of primary amines cannot be used as a distinction method.

10. Aniline is a _______ directing compound.
a) ortho
b) meta
c) ortho and para
d) ortho, meta and para
Answer: c
Clarification: The ortho and para positions with respect to NH₂ group in aniline are centres of high electron density. This is due to the resonance structures of aniline. Thus, NH₂ is ortho and para directing and a powerful activating group.

11. The activating effect of -NHCOCH₃ group is ______ as compared to -NH₂ group.
a) less
b) same
c) more
d) very high
Answer: a
Clarification: The -NHCOCH₃ group forms two resonance structures, where the lone pair of nitrogen interacts with the lone pairs on oxygen atom. This makes the lone electron pair on N less available for donation to benzene ring.

12. What is the major product formed when aniline reacts with bromine water at room temperature?
a) 2-Bromoaniline
b) 4-Bromoanline
c) 2,6-Dibromoaniline
d) 2,4,6-Tribromoaniline
Answer: d
Clarification: Since amino group is highly ortho and para directing, it substitutes the Br group at the para as well as both ortho positions (2,4,6) to give a white precipitate of a tribromo substituted aniline.

13. What is the order of quantities of all isomers of nitroaniline formed on the reaction of aniline with nitric acid and sulphuric acid at 288K?
a) ortho > meta > para
b) para > ortho > meta
c) para > meta > ortho
d) meta > para > ortho
Answer: c
Clarification: Nitration of aniline produces nitro derivatives. Under controlled conditions, aniline is protonated (by strong acidic medium) to form anilinium ion, which is meta directing. Therefore, apart from p-nitroaniline (51%) and o-nitroaniline (2%), a significant amount of m-nitroaniline (47%) is also formed.

14. It is possible to obtain para isomers of anilines as the major product in electrophilic ring substitution of aniline.
a) True
b) False
Answer: a
Clarification: This is possible by protecting the amino group of aniline by reacting it with acetic anhydride (acetylation) to give acetanilide. This compound on desired substitution reaction followed by hydrolysis gives para substituted aniline as major product.

15. A compound ‘P’ on treating with concentrated H₂SO₄ forms ‘R’. The product ‘R’ on heating at 460K forms a zwitter ionic compound. Identify P, R respectively.
a) Aniline; anilinium hydrogensulphate
b) Aniline; sulphanilic acid
c) Anilinium hydrogensulphate; sulphanilic acid
d) Sulphanilic acid; anilinium hydrogensulphate
Answer: a
Clarification: Aniline (P) reacts with conc. H₂SO₄ to form anilinium hydrogensulphate (R) which on heating at 453-473K produces sulphanilic acid (zwitter ionic compound).

Chemistry Assessment Questions for Class 12,

Chemistry Multiple Choice Questions on “Biomolecules – Hormones”.

1. Which of the following hormone is a polypeptide?
a) Estrogen
b) Insulin
c) Androgen
d) Epinephrine
Answer: b
Clarification: Insulin consists of a chain of 51 amino acids and hence is a polypeptide. Estrogen and androgen are steroids, whereas epinephrine is an amine.

2. Hormones are ______
a) messengers
b) catalysts
c) enzymes
d) inhibitors
Answer: a
Clarification: Hormones are chemical substances produced by the endocrine glands in the human body. They are carried to different parts of the body through the blood stream. Because of the action of hormones as intercellular communicators, they are called chemical messengers.

3. Which of the following is not an amine hormone?
a) Norepinephrine
b) Adrenaline
c) Thyroxine
d) Oxytocin
Answer: d
Clarification: Amine hormones are water-soluble compounds which have amino groups and are structurally derived from amino acids. Oxytocin is a peptide hormone.

4. Identify the hormone that increases the glucose level in blood.
a) Insulin
b) Glucagon
c) Oxytocin
d) Vasopressin
Answer: b
Clarification: Insulin is released in response to the rapid rise in blood glucose level. On the other hand, the hormone glucagon tends to increase the glucose level in blood. These two hormones together regulate the blood glucose level.

5. Which of the following is known as fight or flight hormone?
a) Epinephrine
b) Norepinephrine
c) Insulin
d) Thyroxine
View Answer

Answer: a
Clarification: Epinephrine and norepinephrine are amine hormones that mediate response to external stimuli. Adrenaline plays an important role in fight or flight situations by increasing blood flow to muscles, blood sugar level and pulse rate.

6. Which hormone plays an important role during child birth and post it?
a) Estrogen
b) Progesterone
c) Cortisone
d) Oxytocin
Answer: d
Clarification: Oxytocin controls the contraction of the uterus during child birth and helps in the release of milk from mammary glands. Progesterone is responsible for the preparation of uterus for implantation of fertilised egg.

7. The condition goitre is associated with which hormone?
a) Insulin
b) Thyroxine
c) Adrenaline
d) Cortisone
Answer: b
Clarification: Thyroxine is an iodinated derivative of tyrosine. Abnormally low levels of thyroxine leads to hypothyroidism, which causes enlargement of thyroid gland (goitre). Increases level of thyroxine causes hyperthyroidism.

8. Lack of which component in diet causes hypothyroidism?
a) Potassium
b) Vitamin C
c) Iodine
d) Water
Answer: c
Clarification: Low levels of iodine in the diet may lead to hypothyroidism and enlargement of the thyroid gland. This can be controlled by adding sodium iodide to table salt to form iodised salt.

9. Which of the following does not release steroid hormones?
a) Testes
b) Ovary
c) Adrenal cortex
d) Pancreas
Answer: d
Clarification: Steroid hormones are produced by adrenal cortex and gonads. Hormones released by the adrenal cortex play a very important role in the functions of the bod.

10. Which hormone controls the balance of water and minerals in the body?
a) Vasopressin
b) Mineralocorticoids
c) Testosterone
d) Thyroxine
Answer: b
Clarification: Mineralocorticoids are steroid hormones which control the level of excretion of water and salt by the kidney, thus balancing the water and mineral levels as required.

11. Lack of which hormone causes Addison’s disease?
a) Glucocorticoids
b) Oxytocin
c) Insulin
d) Norepinephrine
Answer: a
Clarification: If adrenal cortex does not function properly then one of the results may be Addison’s disease characterized by hypoglycaemia, weakness and increased susceptibility to stress. This may be fatal unless treated by glucocorticoids and mineralocorticoids.

12. All hormones are proteins.
a) True
b) False
Answer: b
Clarification: Hormones are compounds having varied chemical structures. They may be polypeptide chains or amino acids or contain a steroid nucleus.

13. Estradiol is the main sex hormone in females.
a) True
b) False
Answer: a
Clarification: Estradiol is a steroid hormone which is majorly responsible for the development of secondary sex characteristics in females and participates in the control of menstrual cycle.

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Chemistry MCQs for Schools on “Solid State – Packing Efficiency”.

1. What does the ratio ‘space occupied/total space’ denote?
a) Packing factor
b) Packing efficiency
c) Particle fraction
d) Packing unit
Answer: a
Clarification: Packing factor is a fraction of total space of the unit cell occupied by the constituent particles.

2. What is the dimensional formula of packing fraction?
a) M⁰L³T⁰
b) M⁰L⁰T⁰
c) ML⁰T⁰
d) M⁰L²T⁰
Answer: b
Clarification: Packing fraction is a dimensionless quantity which is the ratio of space occupied to total crystal space available. Since both the quantities have the same units the ratio renders dimensionless.

3. Arrange the types of arrangement in terms of decreasing packing efficiency.
a) BCC b) HCP c) HCP d) CCP Answer: c
Clarification: HCP and CCP have the highest packing efficiency of 74% followed by BCC which is 68%. The simple cubic structure has a packing efficiency of 54%.

4. If the body-centered unit cell is assumed to be a cube of edge length ‘a’ with spherical particles of radius ‘r’ then how is the diameter, d of particle and surface area, S of the cell related?
a) S = 32d⁴/3
b) S = 2d²
c) S = 4d²
d) S = 8d²
Answer: d
Clarification: For BCC unit cell the relation between radius of a particle ‘r’ and edge length of unit cell, a, is r = (frac{sqrt{3}}{4})a.
We know that diameter, d = 2r = (frac{sqrt{3}}{2})a
Implying d² = (frac{3}{4})a²
Therefore, 4d²/3=a²
Multiplying by 6 on both sides gives S = 6a² = 8d², where S is the surface area of the cube = 6a².

5. Which of the following metals would have the highest packing efficiency?
a) Copper
b) Potassium
c) Chromium
d) Polonium
Answer: a
Clarification: Copper metal bears face-centered unit cells in its crystal structure. Potassium and chromium both have body-centered unit cells whereas polonium is the only known metal to bear a simple cubic structure. FCC structure has the highest efficiency.

6. “The packing efficiency can never be 100%”. Is this true or false?
a) False
b) True
Answer: b
Clarification: Packing efficiency can never be 100% because in packing calculations all constituent particles filling up the cubical unit cell are assumed to be spheres.

7. What are the percentages of free space in a CCP and simple cubic lattice?
a) 52% and 74%
b) 48% and 26%
c) 26% and 48%
d) 74% and 52%
Answer: c
Clarification: The packing efficiency in CCP and simple cubic lattice are 74% and 52%, respectively. Hence the corresponding free spaces will be 100% – 74% = 26% and 100% – 52% = 48%.

8. How many atoms surround the central atom present in a unit cell with the least free space available?
a) 4
b) 6
c) 8
d) 12
Answer: d
Clarification: FCC, CCP and HCP are unit cells with least free space available i.e. highest packing efficiency. The coordination number of given cells are 12.

9. If metallic atoms of mass 197 and radius 166 pm are arranged in ABCABC fashion then what is the surface area of each unit cell?
a) 1.32 × 10⁶ pm²
b) 1.32 × 10^-18pm²
c) 2.20 × 10⁵ pm²
d) 2.20 × 10^-19 pm²
Answer: a
Clarification: ABCABC arrangement is found in CCP.
In closed cubic packing, relation between edge length of unit cell, a, and radius of particle, r, is given as a=2(sqrt{2})r.
Surface area (S.A.) = 6a²
From the relationship,
a² = 8r²
S.A. = 6a² = 48r²
When r = 166 pm, S.A. = 48(166pm) = 1.32 x 10⁶ pm².

10. If copper, density = 9.0 g/cm³ and atomic mass 63.5, bears face-centered unit cells then what is the ratio of surface area to volume of each copper atom?
a) 0.0028
b) 0.0235
c) 0.0011
d) 0.0323
Answer: b
Clarification: Density, d of unit cell is given by d = (frac{zM}{a^3N_A})
Given,
Density, d = 9.0 g/cm³
Atomic mass, M = 63.5 g/mole
Edge length = a
N_A = Avogadro’s number = 6.022 x 10²³
z = 4 atoms/cell
On rearranging the equation for density we get a³ = (frac{zM}{dN_A})
Substituting the given values:
a³ = (frac{4 times 63.5}{9 times 6.022 times 10^{23}})
Therefore, a = 360.5 pm
The relation of edge length ‘a’ and radius of particle ‘r’ for FCC packing i.e. a = 2(sqrt{2})r.
On substituting the value of ‘a’ in the given relation, r = (frac{360.5}{2sqrt{2}})=127.46 pm
Now, for spherical particles volume, V = 4πr³/3 and surface area, S = 4πr²
Required ratio = S/V=4πr²/(4πr³/3) = 3/r (after simplifying)
Thus, S/V = 3/127.46 = 0.0235.

Chemistry MCQs for Schools,

Chemistry Assessment Questions for Schools on “Conductance of Electrolytic Solutions”.

1. The voltameter is an instrument in which electrical energy is converted to chemical energy.
a) True
b) False
Answer: a
Clarification: The voltameter is a scientific instrument used to measure the quantity of electricity through electrolytic action. Since it is a type of electrolytic cell, it uses electrical energy in order to perform electrolysis. Therefore, it converts electrical energy to chemical energy.

2. Which of the following solutions cannot conduct electricity?
a) Sugar in water
b) NaCl in water
c) MgCl₂ in water
d) KCl in water
Answer: a
Clarification: For a solution to conduct electricity, it requires the solution to have movable ions. A solution of sugar in water does not conduct electricity as the sugar molecules do not dissociate to form ions. Whereas the salts in the other options dissociate to form ions and help conduct electricity.

3. Which of the following scientists stated that the equivalent conductivity of an electrolyte at infinite dilution is equal to the sum of the conductances of the ions and cations?
a) Svante Arrhenius
b) Friedrich Kohlrausch
c) Hermann Kolbe
d) Antoine Lavoisier
Answer: b
Clarification: Friedrich Kohlrausch was a German scientist who conducted research on the conductive properties of electrolytes. He stated that “at infinite dilution, each ion of an electrolyte contributes a characteristic ionic conductance towards equivalent conductance of electrolyte which is independent of the nature of the other ion present in the solution”.

4. Which of the following statements is correct with respect to electrolytic solutions?
a) Its conductance increases with dilution
b) Its conductance decreases with dilution
c) Its conductivity increases with dilution
d) Its equivalent conductance decreases with dilution
Answer: a
Clarification: Conductance is dependent on the concentration of the electrolytic solution. It is also inversely proportional to the conductivity of the solution. On dilution, the number of ions per unit volume reduces, decreasing conductivity and increasing the conductance. The equivalent conductance increases on dilution ionic mobility increases on dilution.

5. The limiting equivalent conductance for weak electrolytes can be calculated using Kohlrausch’s law.
a) True
b) False
Answer: a
Clarification: According to Kohlrausch’s law, the equivalent conductivity of an electrolyte at infinite dilution is the sum of two values one depending upon the cation and the other upon the anion. Therefore, it can be used to calculate the limiting equivalent conductance for weak electrolytes.

6. Which of the following statements is correct regarding the conductivity of solutions of electrolytes?
a) It is independent of the size of the ions
b) It is independent of the viscosity of the solution
c) It depends on the solvation of ions present in solution
d) It decreases with temperature
Answer: c
Clarification: Conductivity of a solution is inversely proportional to the size of the ions present in it, the viscosity of the solution and the solvation of the ions in the solution. Conductivity increases with temperature as it can cause an increase in the number of ions in solution.

7. Which of the following solutions has theleast value of conductivity?
a) 0.01M Na₂SO₄
b) 0.01M NaCl
c) 0.01M CH₃COOH
d) 0.01M HCl
Answer: c
Clarification: The conductivity of a solution depends on the strength of the ions that make up the solution because strong electrolytes dissociate easily and increase the number of ions, increasing the conductivity of the solution. Since CH₃COOH is not a strong electrolyte, and because all of the solutions have the same concentration, it has the least value of conductivity.

8. What is the value of the cell constant when the conductance of an electrolytic solution is equal to its conductivity?
a) 0
b) 1
c)10
d) 100
Answer: b
Clarification: Given,
Conductance = conductivity
We know that, conductivity = conductance x cell constant
Since the value of conductivity is equal to conductance, the value of the cell constant is equal to 1.

9. Which of the following conditions are satisfied when the electrolytic solutions are infinitely dilute?
a) Electrolyte is 100% dissociated
b) Interionic effects increase
c) Conductance is infinite at infinite dilution
d) Molecules continue to exist in solution
Answer: a
Clarification: At infinite dilution, even weak electrolytes behaves like strong electrolytes and undergo complete ionization. Hence, the electrolyte is 100% dissociated. Also, at infinite dilution, the ions are far apart and interionic effects disappear.

10. Which of the following salts show maximum value of equivalent conductance in their fused state?
a) NaCl
b) KCl
c) RbCl
d) CsCl
Answer: d
Clarification: On moving down a group, the atomic size of the elements increases and so their covalent character also decreases. Thus, CsCl, being the least covalent compound will readily give its ions in its fused state. Therefore, CsCl has the maximum value of equivalent conductance in its fused state.

11. Which of the following is an additive property?
a) Surface tension
b) Viscosity
c) Conductance
d) Volume
Answer: c
Clarification: An additive property is a property which is equal to the sum of corresponding properties of its constituent atoms. In the given list, viscosity, surface tension and volume are not additive properties since it does not depend on the interactions of molecules whereas conductance does. Therefore, conductance is an additive property.

12. Which of the given solutions have an equal value of molar conductivity and equivalent conductivity?
a) 1M BaSO₄
b) 1M KCl
c) 1M BCl₃
d) 1M CaSO₄
Answer: b
Clarification: We know that for electrolytic solutions, ⋀_M = ⋀_E × valency factor
Where ⋀_M = molar conductivity and ⋀_E = equivalent conductivity
For KCl, the valency factor is 1. Therefore, for KCl, the value of molar conductivity is equal to its equivalent conductivity. For other ions, ⋀_M > ⋀_E as their valency factors are > 1.

13. Which of the following is ionic mobility independent of?
a) Size of ion
b) Charge on ion
c) Distance of separation between the electrodes
d) Concentration of electrolyte
Answer: c
Clarification: Ionic mobility is an inherent property of the electrolytic solution and is hence, independent of the distance of separation between the two electrodes. As the charge to size ratio increases, the ionic mobility decreases.

14. Which of the following statements is true regarding ionic speed?
a) It is independent of the size of the ion
b) It depends on the potential difference between the two electrodes
c) It is independent of the concentration of the electrolyte
d) It is independent of the charge of the ion
Answer: b
Clarification: Ionic speed is the speed with which the ion moves in the electrolyte during current passage. It depends on the nature of the ions, potential difference, separation between the electrodes and the concentration of the electrolyte. The ionic speed is directly proportional to the potential difference between the electrodes.

15. Which of the following complex compounds will have minimum conductance in solution?
a) [Co(NH₃)₃Cl₃]
b) [Co(NH₃)₄Cl₂]Cl
c) [Co(NH₃)₆]Cl₃
d) [Co(NH₃)₅Cl]Cl₂
Answer: a
Clarification: Electrolytes that ionise in solution have a higher value of conductance. Among the given options, [Co(NH₃)₃Cl₃] does not ionise, whereas the other complexes ionise. Therefore, [Co(NH₃)₃Cl₃] will have minimum conductance in solution.

Chemistry Assessment Questions for Schools,

Chemistry Online Test for Schools on “Surface Chemistry – Colloids Around Us”.

1. Colloids are responsible for the blue colour of the sky and sea water.
a) True
b) False
Answer: a
Clarification: There are a lot of suspended impurities in the atmosphere and sea water that act as colloids. Due to Tyndall effect and Rayleigh scattering, these colloidal particles scatter blue light and hence, make them appear blue in colour.

2. Which of the following statements is incorrect?
a) Clouds are colloidal systems
b) Fog is a colloidal system
c) Mist is a colloidal system
d) Clouds are not a colloidal system
View Answer

Answer: d
Clarification: When it is cold, the moisture in the air condenses on the surface of dust particles, forming tiny droplets. These droplets are colloidal in nature and form a blanket of fog or mist in the atmosphere. Clouds are colloidal systems as they are aerosols consisting of water droplets suspended in air.

3. Which of the following can be used to cause artificial rain?
a) Silver nitrate
b) Silver iodide
c) Silver chloride
d) Silver sulphate
Answer: b
Clarification: Clouds can be made to cause rain by spraying oppositely charged colloidal dust, sand particles or precipitates of silver iodide over them. This causes a neutralisation reaction between the two colloids and result in the coagulation of water droplets which come down as rain.

4. Presence of colloids helps in sewage disposal.
a) True
b) False
Answer: a
Clarification: Colloidal particles present in sewage such as dirt, mud have an electrical charge. When sewage is through plates at a high potential, the colloidal particles coagulate due to electrophoresis and gets removed.

5. What is the role of Argyrol?
a) It is used to cause artificial rain
b) It is used as a lubricant
c) It is used as an eye lotion
d) It is used as an antacid
Answer: c
Clarification: Argyrol is an antiseptic containing solutions of mild silver proteins. It is made up of a silver sol and is used as an eye lotion majorly to treat gonorrhoeal blindness and other pathogenic infections in the eyes of new-born infants.

6. Which of the following diseases is cured using colloidal antimony?
a) Typhoid
b) Psoriasis
c) Kalaazar
d) Pneumonia
Answer: c
Clarification: Most of the medicines used today are of colloidal forms as they are easy to assimilate in the body. Colloidal antimony is usually used to cure eye diseases and also Kalaazar(also called visceral leishmaniasis).

7. What is “aquadag”?
a) Colloidal solution of graphite in water
b) Colloidal solution of oil in water
c) Colloidal solution of silver in water
d) Colloidal solution of gold in water
Answer: a
Clarification: Aquadag is the trade name of a colloidal solution of graphite in water. It is mainly used as a conductor and a lubricant. On the other hand, a colloidal solution of oil in water is called as “oildag” and a colloidal solution of gold in water is called “Purple of Cassius”.

8. What is the application of a colloidal solution of silver bromide in gelatine?
a) Photography
b) Rubber industry
c) Leather industry
d) Medicine
Answer: a
Clarification: A colloidal solution of silver bromide in gelatine is used in photographical devices. The colloidal solution is applied on glass plates, celluloid films or paper to form sensitive plates in photography.

9. Which of the following can be used in place of tannin for the process of tanning?
a) Zinc salts
b) Copper salts
c) Chromium salts
d) Cobalt salts
Answer: c
Clarification: Tannin, obtained from plants, is a mixture of polyhydroxy benzoic acids. It contains negatively charged colloidal particles which make it useful for tanning as animal hides contain positively charged colloidal particles. Chromium salts can be used in place of tannin.

10. Which of the following is used to cure arsenic poisoning?
a) Colloidal solution of zinc hydroxide
b) Colloidal solution of ferric hydroxide
c) Colloidal solution of chromium hydroxide
d) Colloidal solution of calcium hydroxide
Answer: b
Clarification: Colloidal solution of ferric hydroxide (Fe(OH)₃) is given to a person suffering from arsenic poisoning. It is used as ferric hydroxide adsorbs arsenic and this can be vomited out later.

Chemistry Online Test for Schools,

Chemistry Multiple Choice Questions on “P-Block Elements – Phosphorus Halides”.

1. Phosphorus trichloride reacts violently with water.
a) True
b) False
Answer: a
Clarification: Phosphorus trichloride reacts violently with water to produce phosphorus acid (H₃PO₃) and hydrochloric acid.
PCl₃ + 3H₂O → H₃PO₃ + 3HCl.

2. Which of the following reacts with PCl₃ to form PCl₅?
a) O₂
b) N₂
c) S
d) Cl₂
Answer: d
Clarification: Phosphorus trichloride (PCl₃) reacts with chlorine gas (Cl₂) to form phosphorus pentachloride (PCl₅).
PCl₃(g) + Cl₂(g) → PCl₅(g).

3. Which of the following statement regarding PCl₃ is false?
a) PCl₃ forms metal chlorides on heating with finely divided metals
b) PCl₃ does not react with organic compounds
c) PCl₃ is a colourless pungent smelling liquid
d) PCl₃ boils at 347 K
Answer: b
Clarification: Phosphorus trichloride is widely used as an important reagent in organic chemistry for replacing hydroxyl groups by chlorine atoms in organic reactions. For example, ethanol reacts with phosphorus trichloride to form chloroethane and phosphorous acid.
3CH₃CH₂OH + PCl₃ → 3CH₃CH₂Cl + H₃PO₃.

4. Phosphorus trichloride does not react with Grignard reagents.
a) True
b) False
Answer: b
Clarification: Phosphorus trichloride reacts with Grignard reagents to form substituted phosphines. For example, PCl₃ reacts with phenylmagnesium chloride to produce triphenylphosphate and magnesium chloride.
PCl₃ + 3C₆H₅MgCl → P(C₆H₅)₃ + 3MgCl₂.

5. Which of the following is most stable?
a) AsCl₅
b) SbCl₅
c) PCl₅
d) BiCl₅
Answer: c
Clarification: Phosphorus pentachloride is the most stable. As we move down a group in the periodic table, the stability of the pentahalides decreases due to inert pair effect. So, the correct order of stability is PCl₅ > SbCl₅ > AsCl₅ > BiCl₅.

6. What is the hybridization of phosphorus in PCl₅?
a) sp³
b) sp²
c) sp³d
d) sp³d²
View Answer

Answer: c
Clarification: In PCl₅, phosphorus undergoes sp³d hybridization and has trigonal bipyramidal geometry in gaseous and liquid states. It has three equatorial, P-Cl bonds and two axial, P-Cl bonds. Since two axial P-Cl bonds are repelled by three bond pairs while three equatorial bonds are repelled by two bond pairs, therefore, axial bonds are longer than equatorial bonds.

7. How does PCl₅ exist in its solid state?
a) [PCl₄]⁺[PCl₆]^–
b) [PCl₃]²⁺[PCl₇]^2-
c) [PCl₂]³⁺[PCl₈]^3-
d) [PCl₅][PCl₅]
Answer: a
Clarification: Phosphorus pentachloride is a pale yellow crystalline solid with a characteristic pungent smell. In the solid state, it exists as an ionic solid, [PCl₄]⁺[PCl₆]^– in which the cation, [PCl₄]⁺ is tetrahedral while the anion, [PCl₆]^– is octahedral.

8. What is the liquid product formed on decomposition of PCl₅?
a) Cl₂
b) HCl
c) PCl₃
d) No liquid product is formed
Answer: c
Clarification: Phosphorus pentachloride, when heated, sublimes but decomposes on stronger heating into phosphorus trichloride (a colourless pungent-smelling liquid) and chlorine gas.
PCl₅ ⇌ PCl₃ + Cl₂.

9. Which of the following is not formed on hydrolysis of PCl₅?
a) POCl₃
b) H₃PO₄
c) HCl
d) PCl₃
Answer: d
Clarification: In moist air, phosphorus pentachloride undergoes hydrolysis to first form POCl₃ and then finally phosphoric acid.
PCl₃ + H₂O → POCl₃ + 2HCl
POCl₃ + 3H₂O → H₃PO₄ + 3HCl
However, with excess of water, PCl₅ reacts violently to form H₃PO₄ and HCl.
PCl₅ + 4H₂O (excess) → H₃PO₄ + 5HCl.

10. What type of agent is PCl₅ when it reacts with H₂?
a) Reducing agent
b) Oxidising agent
c) Both reducing and oxidising
d) Neither reducing nor oxidising
Answer: b
Clarification: PCl₅ reacts with H₂ to form PCl₃.
PCl₅ + H₂ → PCl₃ + 2HCl
In this reaction, PCl₅ acts as an oxidising agent and thus oxidises H₂ to HCl.