TS-EAMCET SERIES

Chemistry

P Block Elements

21 previous year questions.

Volume: 21 Ques

Yield: High

High-Yield Trend

2025

2024

2023

2018

Chapter Questions
21 MCQs

PYQ 2018

medium

chemistry ID: ts-eamce

Producer gas is a mixture of

PYQ 2018

easy

chemistry ID: ts-eamce

\Mien borax is heated dining borax bead test along with CoO (Cobalt Oxide) on a loop of platinum (Pt) wire, it swells up and produces a blue colored bead. Which of the following compounds is responsible for the blue color?

PYQ 2023

medium

chemistry ID: ts-eamce

XeF 4 + O2F 2 \to X + O 2 X + H 2 O \to Y +2HF The shape of the molecules of X and Y respectively are:

Square planar, square pyramidal

Distorted octahedral, square pyramidal

Distorted tetrahedral. Distorted octahedral

Square pyramidal. Tetrahedral

PYQ 2023

medium

chemistry ID: ts-eamce

Which of the following can form iconic micelles in water?

starch molecules

sodium lauryl sulphate

iodine molecules

S 8 molecules

PYQ 2023

medium

chemistry ID: ts-eamce

The ore which is purified by leaching process is:

Zinc blende

Bauxite

Calamine

Haematite

PYQ 2024

hard

chemistry ID: ts-eamce

In which of the following species, the ratio of s-electrons to p-electrons is the same?

PYQ 2025

easy

chemistry ID: ts-eamce

Consider the following Assertion (A): phosphorus can form both phosphorus (III) and phosphorus (V) chlorides but nitrogen cannot form nitrogen(V) chloride. Reason (R): The electronegativity of nitrogen is more than that of phosphorus The correct answer is

Both (A) and (R) are correct, (R) is the correct explanation of (A)

(A) is correct, but (R) is not correct

Both (A) and (R) are correct, (R) is not the correct explanation of (A)

(A) is not correct, but (R) is correct

Official Solution

Correct Option:

(3)

PYQ 2025

medium

chemistry ID: ts-eamce

In which of the following reactions, hydrogen is evolved? I. II. III. IV.

I, IV only

I, II only

III, IV only

I, II, IV only

09

PYQ 2025

medium

chemistry ID: ts-eamce

$Which of the following statements is not correct regarding the gas evolved by the reaction of dilute on ?$

1

$It is colourless, odourless gas$

2

$It has least solubility in water$

3

$It is acidic in nature$

4

$It is poisonous gas$

Official Solution

Correct Option: $(4)$

$Step 1: Identify the gas evolved. The reaction between calcium carbonate () and dilute hydrochloric acid () is: The gas evolved is carbon dioxide (). Step 2: Analyze the properties of .} (A) Colourless, odourless gas: is a well-known colourless and odourless gas. This statement is correct. (B) Least solubility in water: is only slightly soluble in water. While other common gases are less soluble (e.g.,), is not highly soluble like or . The statement is correct in the sense that is not highly soluble. (C) Acidic in nature: dissolves in water to form carbonic acid (). Since it forms an acid solution, it is classified as an acidic oxide. This statement is correct. (D) Poisonous gas: is not classified as a poisonous gas. It is non-toxic but can cause suffocation and death by displacing oxygen in a confined space. A poisonous gas like or is toxic even in small concentrations. This statement is incorrect. Step 3: Conclude the final answer. The statement that is not correct regarding is (D).$

10

PYQ 2025

medium

chemistry ID: ts-eamce

$Which of the following gives more number of oxides on reacting with ?$

1

2

3

4

Official Solution

Correct Option: $(2)$

$Step 1: Write the reaction of each compound with .} The reaction of salts with a strong acid like displaces the weaker acid from its salt. (A) : Sodium carbonate reacts with to release carbonic acid, which decomposes to water and carbon dioxide. Oxide products: and . (2 oxides) (B) : Sodium nitrite reacts with to form nitrous acid, which is unstable and decomposes into nitric oxide (), nitrogen dioxide (), and water. Upon exposure to air, oxidizes to . The oxides that are formed are (and). (2 nitrogen oxides) The reaction with yields the maximum variety of oxides of Nitrogen upon further decomposition and reaction (). The question asks for the greatest number of oxides. Let's stick to the initial decomposition. The products are and . quickly converts to . Total oxides: . (3 oxides) (C) : Sodium sulfite reacts with to release sulfurous acid, which decomposes to water and sulfur dioxide. Oxide products: and . (2 oxides) (D) : Sodium bicarbonate reacts with to release carbonic acid, which decomposes to water and carbon dioxide. Oxide products: and . (2 oxides) Step 4: Conclude the final answer. Reaction with is the most complex, yielding and (and) upon decomposition, thus forming a greater number of different oxides () compared to the others which form two oxides ().$

11

PYQ 2025

medium

chemistry ID: ts-eamce

$The number of lone pairs of electrons on the central atom of and respectively is$

1

$1, 1, 1$

2

$3, 2, 1$

3

$2, 1, 0$

4

$1, 2, 1$

Official Solution

Correct Option: $(1)$

$Step 1: Calculate the steric number and lone pairs for .} Central atom: Xenon () from Group 18, so 8 valence electrons. Surrounding atoms: 3 Oxygen () atoms. forms double bonds (), but only the bonds and lone pairs contribute to the steric number. The number of bonds is 3. Bonding electrons (effective): (since is double bonded, but the lone pair calculation treats as taking 2 electrons). A simpler method is . This doesn't work here. Using the total number of valence electrons around : Valence on : 8. Electrons used in bonds: 3 (one bond to each). Electrons used in bonds: 3 (forms a total of 6 bonds, 3, 3). Number of bonds = 3. Lone pairs . This is not right either. The correct, simplified approach is: . This works only for single bonds. The formal method: . For : forms 2 bonds. Total bonds on are . has 8 valence . are used in bonding (3 and 3). Remaining . Lone pairs = . (, shape is trigonal pyramidal). Step 2: Calculate the steric number and lone pairs for .} Central atom:, 8 valence . Surrounding atoms: 1 (double bond), 4 (single bonds). Number of bonds = . Electrons used in and bonds: . Remaining . Lone pairs = . (, shape is square pyramidal). Step 3: Calculate the steric number and lone pairs for .} Central atom:, 8 valence . Surrounding atoms: 6 (single bonds). Electrons used in bonds: 6. Remaining . Lone pairs = . (, shape is distorted octahedral). Step 4: Conclude the final answer. The number of lone pairs on the central atom of and are 1, 1, and 1 respectively.$

12

PYQ 2025

medium

chemistry ID: ts-eamce

$Which of the following statements is not correct?$

1

$is an oxidizing agent$

2

$is acidic in nature$

3

$is a gas$

4

$is reducing agent$

Official Solution

Correct Option: $(3)$

$Step 1: Analyze statement (A): is an oxidizing agent.} Tellurium is in Group 16. In, the oxidation state of is . can be oxidized to () or reduced to or . Since can be oxidized, can act as a reducing agent. However, is a stable intermediate state and can also be reduced to metal, meaning it can also act as an oxidizing agent. Its dominant behavior depends on the reaction conditions, but the statement that it is an oxidizing agent is chemically plausible (e.g., in reactions where a stronger reducing agent is present). The key implies this is correct, which is true because it can be reduced from to . Step 2: Analyze statement (B): is acidic in nature.} Selenium () is a non-metal in Group 16. Its highest oxidation state oxide,, is . Non-metal oxides are typically acidic. dissolves in water to form selenic acid (). The statement is correct. Step 3: Analyze statement (C): is a gas.} Selenium dioxide () is a white, volatile solid (sublimes at). Its solid structure consists of polymeric chains. Due to its polymeric solid structure, it is not a gas under standard conditions. In contrast, is a gas. The statement that is a gas is incorrect. Step 4: Analyze statement (D): is a reducing agent.} Sulfur is in Group 16. In, the oxidation state of is . can be oxidized to (). Since can be oxidized, typically acts as a strong reducing agent (e.g.,). The statement is correct. Step 5: Conclude the final answer. The incorrect statement is (C).$

13

PYQ 2025

medium

chemistry ID: ts-eamce

$Which one of the orders is correctly matched with the property mentioned against it?$

1

$H S O Se Te (Boiling point)$

2

$N O O O O (Acidic nature)$

3

$HI$

4

$H O S Se Te (Bond angle)$

Official Solution

Correct Option: $(2)$

$(A) Boiling point of Group 16 hydrides: The boiling points of hydrides generally increase down the group due to increasing van der Waals forces. However, H O has an exceptionally high boiling point due to extensive hydrogen bonding. The correct order is H S Se Te O. The given order is incorrect. (B) Acidic nature of nitrogen oxides: The acidic nature of oxides increases with the increasing oxidation state of the central atom. Oxidation states of N in each oxide: N O: +1 NO: +2 N O : +3 N O : +4 N O : +5 N O and NO are neutral oxides, while N O, N O, and N O are acidic. The acidic character increases with oxidation state, giving the correct trend: N O O O O . This statement is correct. (C) Acidic nature of hydrogen halides: Acidic strength depends on bond dissociation enthalpy. As we go down the group from F to I, the H-X bond length increases, making it easier to release H . The correct order is HF The given order is incorrect. (D) Bond angle of Group 16 hydrides: All have a bent shape with two lone pairs on the central atom. Electronegativity decreases down the group, reducing lone pair-bond pair repulsion and decreasing bond angle. Correct order: H O (>104.5)>H S>H Se>H Te (90). The given order is incorrect.$

14

PYQ 2025

medium

chemistry ID: ts-eamce

$Noble gas 'X' is used as a diluent for oxygen in modern diving apparatus and noble gas 'Y' is used mainly to provide an inert atmosphere in high temperature metallurgical processes. 'Y' and 'X' are respectively?$

1

$He, Ar$

2

$Ar, He$

3

$He, Kr$

4

$Ar, Kr$

Official Solution

Correct Option: $(2)$

$Let's identify the noble gases based on their uses described. Noble gas 'X': Used as a diluent for oxygen in modern diving apparatus. The air we breathe is mostly nitrogen. At high pressures experienced during deep-sea diving, nitrogen can dissolve in the blood. If a diver ascends too quickly, this dissolved nitrogen forms bubbles in the blood, causing a painful and dangerous condition called "the bends" or decompression sickness. To prevent this, a mixture of oxygen and helium is used. Helium is used because of its very low solubility in blood, even at high pressures. So, 'X' is Helium (He). Noble gas 'Y': Used to provide an inert atmosphere in high-temperature metallurgical processes. Many metallurgical processes, like arc welding of metals, require an inert atmosphere to prevent the hot metal from reacting with oxygen or nitrogen from the air. Argon is commonly used for this purpose because it is chemically inert and is the most abundant and cheapest noble gas to produce (it makes up about 1% of the atmosphere). So, 'Y' is Argon (Ar). The question asks for 'Y' and 'X' respectively.'Y' is Argon (Ar).'X' is Helium (He). Therefore, the correct pair is (Ar, He).$

15

PYQ 2025

medium

chemistry ID: ts-eamce

$The dibasic oxoacid of phosphorus on disproportionation gives two products A and B. A and B are respectively$

1

$HPO, PH$

2

$H PO, H O$

3

$H PO, PH$

4

$H P O, H PO$

Official Solution

Correct Option: $(3)$

$First, we must identify the dibasic oxoacid of phosphorus from the common ones. - Hypophosphorous acid (H PO): It has one P-OH bond, so it is monobasic. - Phosphorous acid (H PO): It has two P-OH bonds, so it is dibasic. - Orthophosphoric acid (H PO): It has three P-OH bonds, so it is tribasic. - Pyrophosphoric acid (H P O): It has four P-OH bonds, so it is tetrabasic. The dibasic oxoacid of phosphorus is phosphorous acid, H PO . Now, we need to consider the disproportionation reaction of phosphorous acid. Disproportionation is a reaction where an element in an intermediate oxidation state is simultaneously oxidized and reduced. In H PO, the oxidation state of phosphorus is +3. (3(+1) + P + 3(-2) = 0 P = +3). When phosphorous acid is heated, it disproportionates into orthophosphoric acid and phosphine. The reaction is: . Let's check the oxidation states in the products: - In orthophosphoric acid (H PO), the oxidation state of P is +5. (Oxidation product) - In phosphine (PH), the oxidation state of P is -3. (Reduction product) Since the oxidation state of P (+3) is both increased (+5) and decreased (-3), this is a disproportionation reaction. The two products A and B are H PO and PH .$

16

PYQ 2025

medium

chemistry ID: ts-eamce

$Choose the correct statements about allotropes of carbon I. Graphite has layered structure II. Buckminster fullerene is not aromatic in nature III. The distance between two adjacent layers in graphite is 141.5 pm IV. The hybridization of carbons in graphite and Buckminster fullerene is same$

1

$I & IV$

2

$I & II$

3

$II & III$

4

$III & IV$

Official Solution

Correct Option: $(2)$

$I. Graphite has layered structure: Graphite consists of planar layers of carbon atoms arranged in hexagonal rings. Within each layer, carbon atoms are held by strong covalent bonds, and the layers are held together by weak van der Waals forces. This statement is correct. II. Buckminster fullerene is not aromatic in nature: Buckminsterfullerene (C) has fused five- and six-membered rings. Although it has delocalized pi electrons, its curved structure prevents continuous planar overlap required for aromaticity according to H��ckel's rule. This statement is correct. III. The distance between two adjacent layers in graphite is 141.5 pm: The C-C bond length within the layers is ~141.5 pm, but the interlayer distance is ~335 pm. Therefore, this statement is incorrect. IV. The hybridization of carbons in graphite and Buckminster fullerene is same: In graphite, carbon is sp hybridized, bonded to three atoms in a planar arrangement. In Buckminsterfullerene, each carbon is bonded to three carbons in a curved structure, giving predominantly sp character with some distortion. Strictly speaking, the hybridization is not identical due to curvature. This statement is incorrect.$

17

PYQ 2025

medium

chemistry ID: ts-eamce

$Diborane on hydrolysis gives a compound X. The correct statements about X are I. It is a tribasic acid II. It is a weak monobasic acid III. It has a layer structure IV. It is highly soluble in water$

1

$I & III$

2

$II & III$

3

$II & IV$

4

$I & IV$

Official Solution

Correct Option: $(2)$

$First, let's identify compound X by writing the hydrolysis reaction of diborane (B H). Diborane reacts vigorously with water to produce boric acid and hydrogen gas. B H (g) + 6H O(l) 2H BO (aq) + 6H (g). So, compound X is boric acid, H BO or B(OH) . Now let's evaluate the statements about boric acid. I. It is a tribasic acid: Boric acid does not donate three protons. So, this statement is incorrect. II. It is a weak monobasic acid: Boric acid is a weak acid. It does not act as a protonic acid (donating its own H). Instead, it acts as a Lewis acid, accepting a hydroxide ion (OH) from water. B(OH) + H O [B(OH)] + H . Since it releases one H ion into the solution per molecule, it is considered a weak monobasic acid. This statement is correct. III. It has a layer structure: In the solid state, B(OH) molecules are linked by extensive hydrogen bonds, forming a two-dimensional planar layer structure. This statement is correct. IV. It is highly soluble in water: Boric acid is only sparingly soluble in cold water, although its solubility increases with temperature. It is not considered highly soluble. This statement is incorrect. The correct statements are II and III.$

18

PYQ 2025

medium

chemistry ID: ts-eamce

$The incorrect statement about the oxidation states of group 14 elements is$

1

$In addition to +4, +2, carbon also shows negative oxidation states$

2

$Tin in +2 state acts as a reducing agent$

3

$Lead in +2 state acts as good reducing agent$

4

$Lead in +4 state acts as a good oxidising agent$

Official Solution

Correct Option: $(3)$

$Step 1: Understanding the Concept: This question assesses knowledge of the trends in oxidation states and the stability of those states for Group 14 elements (C, Si, Ge, Sn, Pb). A key concept here is the "inert pair effect," which becomes significant for the heavier elements in the p-block, like lead (Pb). The inert pair effect is the reluctance of the two s-orbital electrons in the valence shell to participate in bonding. Step 2: Analyzing Each Statement: - (A) In addition to +4, +2, carbon also shows negative oxidation states: This is correct. Carbon shows a +4 state (e.g., in CCl₄), a +2 state (e.g., in CO), and various negative oxidation states in its compounds with less electronegative elements (e.g., -4 in CH₄). - (B) Tin in +2 state acts as a reducing agent: This is correct. For tin (Sn), the +4 oxidation state is more stable than the +2 state. Therefore, Sn²⁺ can be easily oxidized to Sn⁴⁺, meaning it readily loses electrons and acts as a reducing agent (e.g., SnCl₂ is a common laboratory reducing agent). - (C) Lead in +2 state acts as good reducing agent: This is incorrect. Due to the pronounced inert pair effect in lead (Pb), the +2 oxidation state is significantly more stable than the +4 oxidation state. It is difficult to oxidize Pb²⁺ to Pb⁴⁺. Therefore, Pb²⁺ does not act as a good reducing agent. In fact, the reverse is true: Pb⁴⁺ is a strong oxidizing agent. - (D) Lead in +4 state acts as a good oxidising agent: This is correct. As the +2 state is more stable for lead, compounds with Pb in the +4 state (like PbO₂) have a strong tendency to gain two electrons and get reduced to the +2 state. This makes them strong oxidizing agents. Step 3: Final Answer: The statement that is incorrect is that "Lead in +2 state acts as good reducing agent". Therefore, option (C) is the correct answer.$

19

PYQ 2025

medium

chemistry ID: ts-eamce

$Which of the following is an incorrect statement about the compounds of group 13 elements?$

1

$All the trihalides exist except TlI₃$

2

$Trihalides on hydrolysis form tetrahedral species$

3

$Diborane is an example of electron precise hydride$

4

$Hydrolysis of diborane gives boric acid$

Official Solution

Correct Option: $(3)$

20

PYQ 2025

medium

chemistry ID: ts-eamce

$Which one of the following statements is not correct?$

1

$Chlorine oxidises ferrous salts to ferric salts in acidic medium$

2

$Chlorine oxidises iodine to periodic acid in water$

3

$Chlorine acts as a bleaching agent due to oxidation$

4

$Chlorine is manufactured by Deacon's process$

Official Solution

Correct Option: $(2)$

$Step 1: Understanding the Concept: This question tests the knowledge of the chemical properties of chlorine (Cl₂), particularly its role as an oxidizing agent, a bleaching agent, and its industrial preparation. Step 2: Analyzing Each Statement: - (A) Chlorine oxidises ferrous salts to ferric salts in acidic medium: This is correct. Chlorine is a strong oxidizing agent. It can oxidize iron(II) to iron(III). - (B) Chlorine oxidises iodine to periodic acid in water: This is incorrect. Chlorine in the presence of water does oxidize iodine, but it oxidizes it to iodic acid (HIO₃), not periodic acid (HIO₄). Periodic acid has iodine in the +7 oxidation state, while iodic acid has iodine in the +5 oxidation state. Chlorine is not a strong enough oxidizing agent under these conditions to oxidize iodine to the +7 state. - (C) Chlorine acts as a bleaching agent due to oxidation: This is correct. In the presence of moisture, chlorine produces nascent oxygen, which is a powerful oxidizing agent and is responsible for the bleaching action. Colored substance + [O] Colorless oxidized substance. - (D) Chlorine is manufactured by Deacon's process: This is correct. The Deacon's process is a major industrial method for producing chlorine by the catalytic oxidation of hydrogen chloride gas. Step 3: Final Answer: The statement that is not correct is that chlorine oxidizes iodine to periodic acid. It oxidizes it to iodic acid. Therefore, option (B) is the correct answer.$

21

PYQ 2025

medium

chemistry ID: ts-eamce

$Thionyl chloride on reaction with white phosphorus gives a compound of phosphorus 'C' which on hydrolysis gives an oxo acid 'O'. The correct statements about C and O are I. Shape of 'C' is pyramidal II. 'O' is a dibasic acid III. 'O' is a monobasic acid IV. 'C' on reaction with acetic acid gives 'O'$

1

$I \& II only$

2

$II \& IV only$

3

$I, III \& IV only$

4

$I, II \& IV only$

Official Solution

Correct Option: $(4)$

$Step 1: Identifying Compounds C and O: - Reaction 1: White phosphorus (P₄) reacts with thionyl chloride (SOCl₂) to produce phosphorus trichloride (PCl₃) as the main phosphorus-containing product. So, compound'C' is PCl₃ . - Reaction 2: Phosphorus trichloride (PCl₃) undergoes hydrolysis with water to form phosphorous acid (or phosphonic acid) and HCl. So, the oxo acid'O' is H₃PO₃ . Step 2: Evaluating the Statements: - I. Shape of 'C' is pyramidal:'C' is PCl₃. The central phosphorus atom has 5 valence electrons. It forms 3 single bonds with chlorine and has one lone pair of electrons remaining. According to VSEPR theory, a molecule with 3 bond pairs and 1 lone pair (AX₃E₁) has a trigonal pyramidal shape. So, statement I is correct . - II. 'O' is a dibasic acid:'O' is H₃PO₃ (phosphorous acid). In oxoacids of phosphorus, only the hydrogen atoms attached directly to oxygen atoms are acidic. The structure of H₃PO₃ contains two P-OH bonds and one P-H bond. Since it has two ionizable hydrogen atoms, it is a dibasic acid . So, statement II is correct . - III. 'O' is a monobasic acid: This is incorrect. As established above, H₃PO₃ is dibasic. (Hypophosphorous acid, H₃PO₂, is monobasic). So, statement III is incorrect . - IV. 'C' on reaction with acetic acid gives 'O':'C' (PCl₃) reacts with carboxylic acids like acetic acid (CH₃COOH) to form an acid chloride (acetyl chloride) and phosphorous acid. The reaction indeed produces 'O' (H₃PO₃) as a byproduct. So, statement IV is correct . Step 3: Final Conclusion: The correct statements are I, II, and IV. Step 4: Final Answer: The combination of correct statements is I, II, and IV. This corresponds to option (D).$

About P Block Elements - TS-EAMCET

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P Block Elements

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Chapter Questions 21 MCQs

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About P Block Elements - TS-EAMCET

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Chapter Questions
21 MCQs