The p-Block Elements
Multiple Choice Questions (MCQs)
1. on addition of conc.
(a)
(b)
(c) HI gets oxidised to
(d)
Answer
(c) Hydrogen iodide
-
(a)
reduces to : This statement is incorrect because it is actually the that reduces to , not the other way around. is oxidized to in the process. -
(b)
is of violet colour: This statement is incorrect because itself is not violet in color. It is a colorless gas. The violet color is due to the formation of . -
(d)
changes to : This statement is incorrect because does not change to in this reaction. Instead, is oxidized to while reducing to .
2. In qualitative analysis when
(a) deep blue precipitate of
(b) deep blue solution of
(c) deep blue solution of
(d) deep blue solution of
Answer
(b) In qualitative analysis when
On boiling CuS with dil.
-
(a) deep blue precipitate of
: This option is incorrect because the addition of excess aqueous ammonia to the blue solution of does not result in the formation of a precipitate of . Instead, ammonia acts as a ligand and forms a complex ion with copper, leading to the formation of a deep blue solution of . -
(c) deep blue solution of
: This option is incorrect because itself does not form a deep blue solution. The deep blue color is specifically due to the formation of the complex ion when excess ammonia is added to the solution containing ions. -
(d) deep blue solution of
: This option is incorrect because is not a known compound that forms a deep blue solution. The deep blue color is due to the formation of the tetraammine copper(II) complex, , and not due to any compound like .
3. In a cyclotrimetaphosphoric acid molecule, how many single and double bonds are present?
(a) 3 double bonds; 9 single bonds
(b) 6 double bonds; 6 single bonds
(c) 3 double bonds; 12 single bonds
(d) Zero double bond; 12 single bonds
Answer
(c) Cyclotrimetaphosphoric acid contains three double bonds and 9 single bonds as shown below
-
Option (a) 3 double bonds; 9 single bonds: This option is incorrect because, while it correctly identifies the presence of 3 double bonds, it underestimates the number of single bonds. Cyclotrimetaphosphoric acid actually contains 12 single bonds, not 9.
-
Option (b) 6 double bonds; 6 single bonds: This option is incorrect because it overestimates the number of double bonds and underestimates the number of single bonds. Cyclotrimetaphosphoric acid has only 3 double bonds and 12 single bonds, not 6 double bonds and 6 single bonds.
-
Option (d) Zero double bond; 12 single bonds: This option is incorrect because it fails to recognize the presence of any double bonds. Cyclotrimetaphosphoric acid contains 3 double bonds in addition to the 12 single bonds.
4. Which of the following elements can be involved in
(a) Carbon
(b) Nitrogen
(c) Phosphorus
(d) Boron
Answer
(c) Among given four elements i.e., carbon, nitrogen, phosphorus and boron. Only phosphorus has vacant
- Carbon: Carbon does not have vacant
-orbitals, so it cannot participate in bonding. - Nitrogen: Nitrogen also lacks vacant
-orbitals, making it unable to form bonds. - Boron: Boron does not possess vacant
-orbitals, thus it cannot engage in bonding.
5. Which of the following pairs of ions are isoelectronic and isostructural?
(a)
(b)
(c)
(d)
Answer
(a) Compounds having same value of total number of electrons are known as isoelectronic.
For
Total number of electrons
Total number of electrons
Hence,
Both have triangular planar structure as in both the species carbon and nitrogen are
-
(b)
:- These ions are not isoelectronic.
- For
:- Total number of electrons = 17 (Cl) + 3 × 8 (O) + 1 (charge) = 17 + 24 + 1 = 42 electrons.
- For
:- Total number of electrons = 6 (C) + 3 × 8 (O) + 2 (charge) = 6 + 24 + 2 = 32 electrons.
- Since they do not have the same number of electrons, they are not isoelectronic and thus cannot be isostructural.
-
(c)
:- These ions are not isoelectronic.
- For
:- Total number of electrons = 16 (S) + 3 × 8 (O) + 2 (charge) = 16 + 24 + 2 = 42 electrons.
- For
:- Total number of electrons = 7 (N) + 3 × 8 (O) + 1 (charge) = 7 + 24 + 1 = 32 electrons.
- Since they do not have the same number of electrons, they are not isoelectronic and thus cannot be isostructural.
-
(d)
:- These ions are not isoelectronic.
- For
:- Total number of electrons = 17 (Cl) + 3 × 8 (O) + 1 (charge) = 17 + 24 + 1 = 42 electrons.
- For
:- Total number of electrons = 16 (S) + 3 × 8 (O) + 2 (charge) = 16 + 24 + 2 = 42 electrons.
- Although they have the same number of electrons and are isoelectronic, they are not isostructural.
has a trigonal pyramidal structure due to the presence of a lone pair on chlorine, while has a trigonal planar structure.
6. Affinity for hydrogen decreases in the group from fluorine to iodine. Which of the halogen acids should have highest bond dissociation enthalpy?
(a)
(b)
(c)
(d)
Answer
(a)
-
(b) HCl: The bond dissociation enthalpy of HCl is lower than that of HF because the bond between hydrogen and chlorine is weaker than the bond between hydrogen and fluorine. This is due to the larger atomic radius of chlorine compared to fluorine, which results in a longer and weaker H-Cl bond.
-
(c) HBr: The bond dissociation enthalpy of HBr is lower than that of HF because the bond between hydrogen and bromine is even weaker than the bond between hydrogen and chlorine. Bromine has a larger atomic radius than chlorine, leading to an even longer and weaker H-Br bond.
-
(d) HI: The bond dissociation enthalpy of HI is the lowest among the given options because iodine has the largest atomic radius of all the halogens listed. This results in the longest and weakest H-I bond, making it easier to dissociate compared to HF, HCl, and HBr.
7. Bond dissociation enthalpy of
Compound | ||||
---|---|---|---|---|
389 | 322 | 297 | 255 |
(a)
(b)
(c)
(d)
Answer
(d) On moving top to bottom, size of central atom increases. Bond length of
Hence,
-
(a)
: The bond dissociation enthalpy of is the highest among the given compounds (389 kJ/mol). This means that the bond is the strongest and requires the most energy to break. As a result, is the least likely to donate electrons and act as a reducing agent. -
(b)
: The bond dissociation enthalpy of is 322 kJ/mol, which is lower than that of but still relatively high compared to and . This indicates that the bond is stronger than the and bonds, making a weaker reducing agent than and . -
(c)
: The bond dissociation enthalpy of is 297 kJ/mol, which is lower than that of and but higher than that of . This means that the bond is weaker than the and bonds but stronger than the bond. Therefore, is a stronger reducing agent than and , but not as strong as .
8. On heating with concentrated
(a) It is highly poisonous and has smell like rotten fish
(b) It’s solution in water decomposes in the presence of light
(c) It is more basic than
(d) It is less basic than
Answer
(c) White phosphorous on reaction with
-
(a) This statement is correct. Phosphine (PH₃) is indeed highly poisonous and has a smell similar to rotten fish.
-
(b) This statement is correct. The solution of phosphine in water decomposes in the presence of light.
-
(d) This statement is incorrect. Phosphine (PH₃) is less basic than ammonia (NH₃), which aligns with the correct answer provided.
9. Which of the following acids forms three series of salts?
(a)
(b)
(c)
(d)
Answer
(c) Structure of
-
(a)
: This acid, also known as hypophosphorous acid, has the structure . It contains only one ionizable hydrogen atom, which means it can only form one series of salts. -
(b)
: This acid, also known as boric acid, has the structure . It behaves as a monobasic acid in water, meaning it can only donate one proton and thus forms only one series of salts. -
(d)
: This acid, also known as phosphorous acid, has the structure . It contains two ionizable hydrogen atoms, which means it can form only two series of salts.
10. Strong reducing behaviour of
(a) low oxidation state of phosphorus
(b) presence of two
(c) presence of one
(d) high electron gain enthalpy of phosphorus
Answer
(c) Strong reducing behaviour of
Hypophosphorous following (Monobasic)
-
(a) Low oxidation state of phosphorus: While the low oxidation state of phosphorus in
does contribute to its reducing nature, it is not the primary reason for its strong reducing behavior. The presence of bonds is more directly responsible for this property. -
(b) Presence of two
groups and one bond: This option is incorrect because actually has two bonds and only one group. The presence of two bonds is crucial for its strong reducing behavior. -
(d) High electron gain enthalpy of phosphorus: This option is incorrect because the electron gain enthalpy of phosphorus is not the primary factor influencing the reducing behavior of
. The reducing behavior is mainly due to the presence of bonds, which can donate electrons easily.
11. On heating lead nitrate forms oxides of nitrogen and lead. The oxides formed are…… .
(a)
(b)
(c)
(d)
Answer
(b) On heating lead nitrate it produces brown coloured nitrogen dioxide
-
Option (a)
: This option is incorrect because heating lead nitrate does not produce nitrous oxide ( ). The thermal decomposition of lead nitrate specifically produces nitrogen dioxide ( ), not . -
Option (c)
: This option is incorrect because heating lead nitrate does not produce nitric oxide ( ). The correct nitrogen oxide produced is nitrogen dioxide ( ), not . -
Option (d)
: This option is incorrect because heating lead nitrate does not produce nitric oxide ( ) or lead dioxide ( ). The correct products are nitrogen dioxide ( ) and lead (II) oxide ( ).
12. Which of the following elements does not show allotropy?
(a) Nitrogen
(b) Bismuth
(c) Antimony
(d) Arsenic
Answer
(a) Nitrogen does not show allotropy due to its weak
-
Bismuth: Bismuth does not exhibit allotropy because it exists in only one stable form under normal conditions. It has a rhombohedral crystal structure and does not form different structural modifications.
-
Antimony: Antimony exhibits allotropy, with several allotropes including metallic antimony and amorphous antimony. Therefore, it is incorrect to say that antimony does not show allotropy.
-
Arsenic: Arsenic also exhibits allotropy, with several allotropes including gray arsenic (metallic), yellow arsenic, and black arsenic. Therefore, it is incorrect to say that arsenic does not show allotropy.
13. Maximum covalency of nitrogen is…… .
(a) 3
(b) 5
(c) 4
(d) 6
Answer
(c) Maximum covalency of nitrogen is 4 in which one electron is made available by
-
(a) 3: This option is incorrect because while nitrogen commonly forms three bonds (as in NH₃), it can also form a fourth bond by utilizing its lone pair, leading to a covalency of 4.
-
(b) 5: This option is incorrect because nitrogen does not have available d-orbitals in its valence shell to expand its covalency beyond 4. Therefore, it cannot form five bonds.
-
(d) 6: This option is incorrect because nitrogen lacks the necessary d-orbitals in its valence shell to accommodate six bonds. Its maximum covalency is limited to 4.
14. Which of the following statements is wrong?
(a) Single
(b)
(c)
(d) Covalency of nitrogen in
Answer
(a) True statement is that single
(i)
(ii)
(iii) Covalency of nitrogen in
-
(b) Incorrect Reason:
can indeed act as a ligand in the formation of coordination compounds with transition elements due to the presence of a lone pair of electrons on the phosphorus atom. -
(c) Incorrect Reason:
is paramagnetic in nature because it has an odd number of electrons, resulting in one unpaired electron. -
(d) Incorrect Reason: The covalency of nitrogen in
is indeed four, as each nitrogen forms four covalent bonds in the structure of .
15.
(a)
(b)
(c)
(d)
Answer
(a) When freshly prepared solution of
-
(b)
: This compound is not formed during the brown ring test. The brown ring test specifically involves the formation of a complex between and , not . -
(c)
: This complex does not form in the brown ring test. The correct complex formed is , which involves only one molecule coordinated to the iron center. -
(d)
: This compound is not relevant to the brown ring test. The test involves the formation of a complex ion, not a simple mixture or compound of and .
16. Elements of group- 15 form compounds in +5 oxidation state. However, bismuth forms only one well characterised compound in +5 oxidation state. The compound is
(a)
(b)
(c)
(d)
Answer
(b) Stability of +5 oxidation state decreases top to bottom and +3 oxidation state increases top to bottom due to inert pair effect. Meanwhile compound having +5 oxidation state of
: Bismuth does not form a stable oxide in the +5 oxidation state due to the inert pair effect and the relatively larger size and lower electronegativity of oxygen compared to fluorine. : Bismuth does not form a stable chloride in the +5 oxidation state because chlorine is less electronegative than fluorine and cannot stabilize the +5 oxidation state of bismuth effectively. : Bismuth does not form a stable sulfide in the +5 oxidation state due to the much larger size and lower electronegativity of sulfur, which makes it unable to stabilize the +5 oxidation state of bismuth.
17. On heating ammonium dichromate and barium azide separately we get
(a)
(b)
(c)
(d)
Answer
(a) On heating ammonium dichromate and barium azide it produces
-
Option (b) is incorrect because ammonium dichromate does not produce
; it produces instead. Additionally, barium azide does not produce ; it produces . -
Option (c) is incorrect because ammonium dichromate does not produce
; it produces . However, barium azide correctly produces . -
Option (d) is incorrect because ammonium dichromate does not produce
; it produces . Additionally, barium azide does not produce ; it produces .
18. In the preparation of
(a) 2
(b) 3
(c) 4
(d) 6
Answer
(a) Two moles of
-
Option (b) is incorrect because the balanced chemical equation shows that 4 moles of
produce 4 moles of . Therefore, 2 moles of would produce 2 moles of , not 3 moles. -
Option (c) is incorrect because the balanced chemical equation shows that 4 moles of
produce 4 moles of . Therefore, 2 moles of would produce 2 moles of , not 4 moles. -
Option (d) is incorrect because the balanced chemical equation shows that 4 moles of
produce 4 moles of . Therefore, 2 moles of would produce 2 moles of , not 6 moles.
19. The oxidation state of central atom in the anion of compound
(a) +3
(b) +5
(c) +1
Answer
(c) Let oxidation state of
-
Option (a) +3: This is incorrect because the calculation of the oxidation state of phosphorus (P) in
does not result in +3. The correct calculation shows that the oxidation state of P is +1. -
Option (b) +5: This is incorrect because the calculation of the oxidation state of phosphorus (P) in
does not result in +5. The correct calculation shows that the oxidation state of P is +1. -
Option (d) -3: This is incorrect because the calculation of the oxidation state of phosphorus (P) in
does not result in -3. The correct calculation shows that the oxidation state of P is +1.
20. Which of the following is not tetrahedral in shape?
(a)
(b)
(c)
(d)
Answer
(c)
It has trigonal bipyramidal geometry having
-
(a)
: The ammonium ion ( ) has a tetrahedral shape because it has four hydrogen atoms symmetrically arranged around the central nitrogen atom. The nitrogen atom undergoes hybridization, resulting in a tetrahedral geometry. -
(b)
: Silicon tetrachloride ( ) has a tetrahedral shape because the silicon atom is surrounded by four chlorine atoms. The silicon atom undergoes hybridization, leading to a tetrahedral geometry. -
(d)
: The sulfate ion ( ) has a tetrahedral shape because the sulfur atom is surrounded by four oxygen atoms. The sulfur atom undergoes hybridization, resulting in a tetrahedral geometry.
21. Which of the following are peroxoacids of sulphur?
(a)
(b)
(c)
(d)
Answer
(a) Peroxoacids of sulphur must contain one-
-
Option (b)
and : (disulfuric acid) does not contain a peroxo bond ( bond). It is not a peroxoacid of sulfur. -
Option (c)
and : (disulfuric acid) does not contain a peroxo bond ( bond). It is not a peroxoacid of sulfur. -
Option (d)
and : Neither (dithionic acid) nor (disulfuric acid) contain a peroxo bond ( bond). They are not peroxoacids of sulfur.
22. Hot conc.
(a)
(b)
(c)
(d)
Answer
(c)
While carbon on oxidation with
-
(a)
: Copper ( ) reacts with hot concentrated sulfuric acid ( ) to produce copper sulfate ( ), sulfur dioxide ( ), and water ( ). Only one gaseous product, , is formed in this reaction. -
(b)
: Sulfur ( ) reacts with hot concentrated sulfuric acid ( ) to produce sulfur dioxide ( ) and water ( ). Only one gaseous product, , is formed in this reaction. -
(d)
: Zinc ( ) reacts with hot concentrated sulfuric acid ( ) to produce zinc sulfate ( ), sulfur dioxide ( ), and water ( ). Only one gaseous product, , is formed in this reaction.
23. A black compound of manganese reacts with a halogen acid to give greenish yellow gas. When excess of this gas reacts with
(a) -3 to +3
(b) -3 to 0
(c) -3 to +5
(d) 0 to -3
Answer
(a) Black coloured compound
-
Option (b) -3 to 0:
- This option is incorrect because in the reaction between
and , the nitrogen in changes its oxidation state from -3 to +3, not to 0. There is no intermediate step where nitrogen reaches an oxidation state of 0 in this reaction.
- This option is incorrect because in the reaction between
-
Option (c) -3 to +5:
- This option is incorrect because the reaction between
and results in the formation of , where nitrogen has an oxidation state of +3. There is no formation of a compound where nitrogen has an oxidation state of +5 in this reaction.
- This option is incorrect because the reaction between
-
Option (d) 0 to -3:
- This option is incorrect because the initial oxidation state of nitrogen in
is -3, not 0. The reaction involves the change of nitrogen’s oxidation state from -3 to +3, not from 0 to -3.
- This option is incorrect because the initial oxidation state of nitrogen in
24. In the preparation of compounds of
(a) both
(b) both
(c) both
(d) both
Answer
(c) Bertlett had taken
- (a) Both
and do not have the same size. is a diatomic molecule with a smaller size compared to the monatomic noble gas . - (b) Both
and do not have the same electron gain enthalpy. has a higher tendency to gain electrons compared to , which is a noble gas with a stable electronic configuration and very low electron affinity. - (d) While it is true that both
and are gases, this is not the reason Bartlett chose as a base compound. The key factor was the similarity in ionisation enthalpy, not their physical state.
25. In solid state
(a) covalent solid
(b) octahedral structure
(c) ionic solid with
(d) ionic solid with
Answer
(d) In solid state
-
(a) covalent solid: This option is incorrect because in the solid state,
does not exist as a covalent solid. Instead, it dissociates into ionic species, forming and ions. -
(b) octahedral structure: This option is incorrect because
in the solid state does not form a simple octahedral structure. Instead, it forms an ionic lattice with ions having a tetrahedral geometry and ions having an octahedral geometry. -
(c) ionic solid with
octahedral and tetrahedral: This option is incorrect because the charges and geometries of the ions are reversed. In the solid state, forms ions with a tetrahedral geometry and ions with an octahedral geometry, not the other way around.
26. Reduction potentials of some ions are given below. Arrange them in decreasing order of oxidising power.
Ion | |||
---|---|---|---|
Reduction potential |
(a)
(b)
(c)
(d)
Thinking Process
This problem is based on concept of standard reduction potential of species and oxidising property.
Answer
(c) Greater the SRP value of species higher will be its oxidising power.
Here,
-
Option (a)
: This option is incorrect because it suggests that has the highest oxidizing power, followed by and then . However, the standard reduction potentials indicate that has the highest value (1.74 V), followed by (1.65 V), and then (1.19 V). Therefore, should have the lowest oxidizing power, not the highest. -
Option (b)
: This option is incorrect because it suggests that has a higher oxidizing power than . However, the standard reduction potentials indicate that (1.74 V) has a higher value than (1.65 V), meaning should have a higher oxidizing power than . -
Option (d)
: This option is incorrect because it suggests that has a higher oxidizing power than . However, the standard reduction potentials indicate that (1.65 V) has a higher value than (1.19 V), meaning should have a higher oxidizing power than .
27. Which of the following is isoelectronic pair?
(a)
(b)
(c)
(d)
Answer
(b) Isoelectronic pair have same number of electrons
Total number of electrons |
Hence, (b) is the correct choice, while in another cases this value is not equal.
Hence, only (b) is the correct choice.
-
For option (a)
and : : electrons : electrons- The total number of electrons is not equal, hence they are not isoelectronic.
-
For option (c)
and : : electrons : electrons- The total number of electrons is not equal, hence they are not isoelectronic.
-
For option (d)
and : : electrons : electrons- The total number of electrons is not equal, hence they are not isoelectronic.
Multiple Choice Questions (More Than One Options)
28. If chlorine gas is passed through hot
(a) 0 to +5
(b) 0 to +3
(c) 0 to -1
(d) 0 to +1
Answer
When chlorine gas is passed through hot
Oxidation state varies from 0 to -1 and 0 to +5 .
Hence, (a) and (c) are correct choices.
-
Option (b) 0 to +3 is incorrect because chlorine does not exhibit an oxidation state of +3 in the reaction between chlorine gas and hot NaOH solution. The oxidation states involved are 0, -1, and +5.
-
Option (d) 0 to +1 is incorrect because chlorine does not exhibit an oxidation state of +1 in the reaction between chlorine gas and hot NaOH solution. The oxidation states involved are 0, -1, and +5.
29. Which of the following options are not in accordance with the property mentioned against them?
(a)
(b)
(c)
(d)
Answer
This is the incorrect order of ionic character of metal halide.
Correct order can be written as
As electronegativity difference between metal and halogen increases ionic character increases.
This is incorrect order of bond dissociation energy. Correct order is
-
For option (c)
Bond dissociation enthalpy: The given order is incorrect because the bond dissociation enthalpy does not follow this trend. The correct order is . This is due to the electronic repulsion among lone pairs in the molecule, which makes the bond weaker than the bond. Additionally, the bond length increases from to , making the bonds easier to break. -
For option (d)
Hydrogen-halogen bond strength: The given order is incorrect because the bond strength of hydrogen-halogen bonds does not follow this trend. The correct order is . This is because the bond strength decreases as the size of the halogen atom increases, making the bond longer and weaker. Hence, the bond is the strongest due to the small size and high electronegativity of fluorine, while the bond is the weakest due to the large size of iodine.
30. Which of the following is correct for
(a) It has 6 lone pairs of electrons
(b) It has six
(c) It has three
(d) It has four lone pairs of electrons
Answer
(
Structure of
It has total four lone pairs of electrons situated at each P-atom.
It has six
-
Option (a) is incorrect because the
molecule has a total of 4 lone pairs of electrons, not 6. Each phosphorus atom in the molecule has one lone pair, resulting in a total of 4 lone pairs. -
Option (c) is incorrect because the
molecule has six single bonds, not three. Each phosphorus atom forms three single bonds with three other phosphorus atoms, resulting in a total of six single bonds.
31. Which of the following statements are correct?
(a) Among halogens, radius ratio between iodine and fluorine is maximum.
(b) Leaving
(c) Among interhalogen compounds maximum number of atoms are present in iodine fluoride.
(d) Interhalogen compounds are more reactive than halogen compounds.
Answer
(a) Among halogens, radius ratio between iodine and fluorine is maximum because iodine has maximum radius and fluorine has minimum radius.
(b) It can be correctly stated as in general interhalogen compounds are more reactive than halogen. This is because
(c) Among interhalogen compounds maximum number of atoms are present in iodine fluoride because radius ratio of iodine and fluorine has maximum value.
(d) Interhalogen compounds are more reactive than halogen due to weaker
- (b) The statement is incorrect because it implies that all halogens have weaker
bonds than bonds in interhalogens, except for the bond. However, this is not universally true for all halogens. The correct statement should be that interhalogen compounds are generally more reactive than halogen compounds because the bond in interhalogens is typically weaker than the bond in halogens, with the exception of the bond.
32. Which of the following statements are correct for
(a) It acts as bleaching agent in moist conditions.
(b) Its molecule has linear geometry.
(c) Its dilute solution is used as disinfectant.
(d) It can be prepared by the reaction of dilute
Answer
(a, c)
(a) In moist condition
e.g., it converts
(b) is incorrect it has bent structure.
(c) Its dilute solution is used as a disinfectant.
(d) It can be prepared by the reaction of
while metal on treatment with
Hence, options (a) and (c) are correct choices.
-
(b) is incorrect because the
molecule has a bent structure, not a linear geometry. -
(d) is incorrect because
is not prepared by the reaction of dilute with metal sulphide. Instead, this reaction produces . can be prepared by the reaction of with sulphide ore.
33. Which of the following statements are correct?
(a) All the three
(b) All
(c)
(d)
Answer
(c,
(a) All the three
(b) All
(c)
(d)
-
(a) All the three
bond lengths in are not equal because has one double bond and two single bonds, leading to different bond lengths. -
(b) All
bond lengths in molecule in gaseous state are not equal because the molecule has a trigonal bipyramidal structure, where the axial bonds are longer than the equatorial bonds.
34. Which of the following orders are correct as per the properties mentioned against each?
Answer
(a)
(b) Correct order is
(b) Correct order is
(c)
(d)
-
(b) The given order is incorrect because the enthalpy of vaporization generally decreases as we move down the group due to the increase in molecular size and weaker intermolecular forces. The correct order should be ( \text{AsH}_3 > \text{PH}_3 > \text{NH}_3 ).
-
(c) The given order is incorrect because the electron gain enthalpy becomes more negative as we move from left to right across a period and less negative as we move down a group. The correct order should be ( \text{F} > \text{Cl} > \text{O} > \text{S} ).
35. Which of the following statements are correct?
(a)
(b) In peroxosulphuric acid
(c) Iron powder along with
(d) Change in enthalpy is positive for the preparation of
Answer
(
(a) Structure of
It contains one
(b) In peroxosulphuric acid
Structure of
Let oxidation state of
(c) During preparation of ammonia, iron oxide with small amount of
(d) Change in enthalpy is negative for preparation of
-
(c) During the preparation of ammonia by the Haber process, iron oxide (Fe₂O₃) is used as the primary catalyst, with small amounts of K₂O and Al₂O₃ added as promoters to increase the efficiency of the catalyst. Iron powder is not used as the catalyst.
-
(d) The preparation of SO₃ by the catalytic oxidation of SO₂ is an exothermic reaction, meaning that the change in enthalpy (ΔH) is negative, not positive.
36. In which of the following reactions conc.
(a)
(b)
(c)
(d)
Answer
In the above given four reactions, (b) and (c) represent oxidising behaviour of
Here,
-
In reaction (a)
, conc. is acting as an acid, not as an oxidizing agent. There is no change in the oxidation state of sulfur. -
In reaction (d)
, conc. is also acting as an acid, not as an oxidizing agent. There is no change in the oxidation state of sulfur.
37. Which of the following statements are true?
(a) Only type of interactions between particles of noble gases are due to weak dispersion forces.
(b) lonisation enthalpy of molecular oxygen is very close to that of xenon.
(c) Hydrolysis of
(d) Xenon fluorides are not reactive.
Answer
(a,
(a) Only one type of interactions between particles of noble gases are due to weak dispersion forces.
(b) Ionisation enthalpy of molecular oxygen is very close to that of xenon. This is the reason for the formation of xenon oxides.
(c) Hydrolysis of
(d) Xenon fluorides are highly reactive hydrolysis readily even by traces of water.
-
(c) Hydrolysis of
is not a redox reaction because there is no change in the oxidation states of xenon and fluorine during the reaction. The oxidation state of xenon remains +6 and that of fluorine remains -1. -
(d) Xenon fluorides are highly reactive and hydrolyze readily even in the presence of traces of water.
Short Answer Type Questions
38. In the preparation of
Answer
In Contact process
Note
~~
39. Write a balanced chemical equation for the reaction showing catalytic oxidation of
Answer
Ammonia
Balanced chemical reaction can be written as
~~ 40. Write the structure of pyrophosphoric acid.
Answer
Molecular formula of pyrophosphoric acid is
Pyrophosphoric acid
~~
41.
Answer
Dissolution of
~~
42. In
Answer
It has trigonal bipyramidal geometry, in which two
Thus, difference in bond length is due to fact that axial bond pairs suffer more repulsion as compared to equatorial bond pairs.
~~ 43. Why is nitric oxide paramagnetic in gaseous state but the solid obtained on cooling it is diamagnetic?
Answer
In gaseous state,
f0bd3a00/public)
~~
44. Give one reason to explain why
Answer
Existance of
~~
45. Out of
Answer
Bond angle of
~~
46.
Answer
Fluorine atom is smaller in size so, six
~~
47. on reaction with
Answer
Phosphorus on reaction with
When ’
When ‘A’ and ‘B’ are hydrolysed
~~
48. In the ring test of
Answer
This test is known as brown ring test of nitrates generally used to identify the presence of nitrate ion in given solution.
~~ 49. Explain why the stability of oxoacids of chlorine increases in the order given below.
Answer
Oxygen is more electronegative than chlorine, therefore dispersal of negative charge present on chlorine increases from
Due to increase in stability of conjugate base, acidic strength of corresponding acid increases in the same order
~~ 50. Explain why ozone is thermodynamically less stable than oxygen?
Answer
Ozone is thermodynamically less stable than oxygen because its decomposition into oxygen results in the liberation of heat (
~~
51.
Thinking Process
This problem includes conceptual mixing of chemical properties of oxides of phosphorus, mole concept and stoichiometry.
Answer
Neutralisation
Adding Eqs. (i) and (ii)
Number of moles of
(Molar mass of
Given,
~~
52. White phosphorus reacts with chlorine and the product hydrolyses in the presence of water. Calculate the mass of
Thinking Process
This problem is based on concept of chemical reaction of phosphorus and stoichiometry. Write balanced chemical reaction and then calculate the amount of
Answer
Equations for the reactions
~~ 53. Name three oxoacids of nitrogen. Write the disproportionation reaction of that oxoacid of nitrogen in which nitrogen is in +3 oxidation state.
Answer
Three oxoacids of nitrogen having oxidation state +3 are
(a)
(b)
(c) Hyponitrous acid,
In
Disproportionation reaction
~~
54. Nitric acid forms an oxide of nitrogen on reaction with
Answer
Resonating structures of
~~ 55. (i) white phosphorus (ii) red phosphorus and (iii) black phosphorus. Write the difference between white red and black phosphorus on the basis of their structure and reactivity.
Phosphorus has three allotropic forms -
Answer
White phosphorus | Red phosphorus | Black phosphorus |
---|---|---|
It is less stable form of P | More stable than white P. | It is most stable form of P |
It is highly reactive | Less reactive than white P. | It is least reactive |
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It has regular tetrahedron structure | It has polymeric structure | It has a layered structure. |
~~ 56. Give an example to show the effect of concentration of nitric acid on the formation of oxidation product.
Answer
Effect of concentration of nitric acid on the formation of oxidation product can be understood by its reaction with conc
~~
57.
Write the reactions involved to explain what happens.
Answer
~~ 58. Phosphorus forms a number of oxoacids. Out of these oxoacids, phosphinic acid has strong reducing property. Write its structure and also write a reaction showing its reducing behaviour.
Answer
Among various forms of oxoacids, phosphinic acid has stronger reducing property.
Structure of phosphinic acid
Reaction showing reducing behaviour of phosphinic acid is as follows
~~
Matching The Columns
59. Match the compounds given in Column I with the hybridisation and shape given in Column II and mark the correct option.
Column I | Column II | |||
---|---|---|---|---|
A. | 1. | |||
B. | 2. | |||
C. | 3. | |||
D. | 4. |
Codes
A | B | C | D | |
---|---|---|---|---|
(a) | 1 | 3 | 4 | 2 |
(b) | 1 | 2 | 4 | 3 |
(c) | 4 | 3 | 1 | 2 |
(d) | 4 | 1 | 2 | 3 |
Answer
(a) A.
B.
C.
D.
~~ 60. Match the formulas of oxides given in Column I with the type of oxide given in Column II and mark the correct option.
Column I | Column II | ||
---|---|---|---|
A. |
1. | Neutral oxide | |
B. |
2. | Acidic oxide | |
C. |
3. | Basic oxide | |
D. |
4. | Mixed oxide |
Codes
A | B | C | D | |
---|---|---|---|---|
(a) | 1 | 2 | 3 | 4 |
(b) | 4 | 1 | 2 | 3 |
(c) | 3 | 2 | 4 | 1 |
(d) | 4 | 3 | 1 | 2 |
Answer
(b) A.
B.
C.
D.
Formulas of the compound | Type of oxide | |
---|---|---|
A. | Mixed oxide | |
B. | Neutral oxide | |
C. | Acidic oxide | |
D. | Basic oxide |
~~ 61. Match the items of Columns I and II and mark the correct option.
Column I | Column II | ||
---|---|---|---|
A. | 1. | Highest electron gain enthalpy | |
B. | 2. | Chalcogen | |
C. | 3. | Tear gas | |
D. | Sulphur | 4. | Storage batteries |
Codes
A | B | C | D | |
---|---|---|---|---|
(a) | 4 | 3 | 1 | 2 |
(b) | 3 | 4 | 1 | 2 |
(c) | 4 | 1 | 2 | 3 |
(d) | 2 | 1 | 3 | 4 |
Answer
(a) A.
B.
C.
D.
A.
B.
C.
D. Sulphur is a member of chalcogen i.e., ore producing elements.
~~ 62. Match the species given in Column I with the shape given in Column II and mark the correct option.
Column I | Column II | ||
---|---|---|---|
A. | 1. | Tetrahedral | |
B. | 2. | Pyramidal | |
C. | 3. | Sea-saw shaped | |
D. | 4. | Bent T-shaped |
Codes
A | B | C | D | |
---|---|---|---|---|
(a) | 3 | 2 | 1 | 4 |
(b) | 3 | 4 | 2 | 1 |
(c) | 1 | 2 | 3 | 4 |
(d) | 1 | 4 | 3 | 2 |
Answer
(b) A.
B.
C.
D.
~~ 63. Match the items of Columns I and II and mark the correct option.
Column I | Column II | ||
---|---|---|---|
A. | Its partial hydrolysis does not change oxidation state of central atom. | 1. | |
B. | It is used in modern diving apparatus. | 2. | |
C. | It is used to provide inert atmosphere for filling electrical bulbs. | 3. | |
D. | Its central atom is in |
4. |
Codes
A | B | C | D | ||
---|---|---|---|---|---|
(a) | 1 | 4 | 2 | 3 | |
(b) | 1 | 2 | 3 | 4 | |
(c) | 2 | 1 | 4 | 3 | |
(d) | 1 | 3 | 2 | 4 |
Answer
(c) A.
B.
C.
D.
(A) Partial hydrolysis of
(B) He is used in modern diving apparatus.
(C) Ar is used to provide inert atmosphere for filling electrical bulbs
(D) Central atom
~~
Assertion and Reason
In the following questions a statement of Assertion (A) followed by a statement of Reason (
(a) Both Assertion and Reason are correct statements, and Reason is the correct explanation of the Assertion.
(b) Both Assertion and Reason are correct statements, and Reason is not the correct explanation of the Assertion.
(c) Assertion is correct, but Reason is wrong statement.
(d) Assertion is wrong but Reason is correct statement.
(e) Both Assertion and Reason are wrong statements.
64. Assertion (A)
Reason (R) Nitrogen has more electron gain enthalpy than phosphorus.
Answer
(c) Assertion is true, but reason is false.
~~
65. Assertion (A)
Reason (R)
Answer
(c) Assertion is true, but reason is false.
~~
66. Assertion (A) HI cannot be prepared by the reaction of KI with concentrated
Reason (R) HI has lowest
Answer
(b) Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion.
~~
67. Assertion (A) Both rhombic and monoclinic sulphur exist as
Reason (R) Oxygen forms
Answer
(a) Both assertion and reason are correct statements, and reason is the correct explanation of the assertion.
Both rhombic and monoclinic sulphur exist as
Structure of
~~
68. Assertion (A)
Reason (R)
Answer
(a) Both assertion and reason are correct statements, and reason is the correct explanation of the assertion.
But on adding
~~
69. Assertion (A)
Reason (R) Six F-atoms in
Answer
(a) Assertion and reason both are true and reason is the correct explanation of assertion.
~~
Long Answer Type Questions
70. An amorphous solid "
Thinking Process
This problem is based on concept of properties of sulphur and its oxide.
Amorphous solid A gives B is a gas which turns lime water milky and also produced as a by product during roasting of sulphide ore. This gas decolourises acidified aqueous
Answer
Since, the by-product of roasting of sulphide ore is
(i)
(ii)
(iii)
(iv)
~~
71. On heating lead (II) nitrate gives a brown gas "
Thinking Process
This problem is based on preparation and properties of
Answer
Structures
(i)
(ii)
~~
72. On heating compound (A) gives a gas (B) which is a constituent of air. This gas when treated with 3 moles of hydrogen
Answer
The main constituents of air are nitrogen (78%) and oxygen (21%). Only
Reactions involved can be given, as
(i)
(ii)
(iii)
(Iv)
(v)