The count comes from adding all elements in Groups 13–18 across periods 2–7. Since period 1 has no p-block, the total is 6 columns × number of periods containing them = 35.

Variable Oxidation States due to available orbitals. Diversity in Physical States —metals, non-metals, metalloids. Allotropy —multiple forms of the same element. Wide Reactivity Range —from highly reactive halogens to inert noble gases.

Variable Oxidation States due to available orbitals. Diversity in Physical States —metals, non-metals, metalloids. Allotropy —multiple forms of the same element. Wide Reactivity Range —from highly reactive halogens to inert noble gases.

Yes. Group 13 elements the boron group—are the first column of the p-block. They include both non-metals (B), metals (Al, Ga, In, Tl), and a radioactive element (Nh). Their outer electron configuration is ns²np¹.

Yes. Group 13 elements the boron group—are the first column of the p-block. They include both non-metals (B), metals (Al, Ga, In, Tl), and a radioactive element (Nh). Their outer electron configuration is ns²np¹.

P-block elements include: Group 13: B, Al, Ga, In, Tl, Nh Group 14: C, Si, Ge, Sn, Pb, Fl Group 15: N, P, As, Sb, Bi, Mc Group 16: O, S, Se, Te, Po, Lv Group 17: F, Cl, Br, I, At, Ts Group 18: He, Ne, Ar, Kr, Xe, Rn, Og

P-block elements include: Group 13: B, Al, Ga, In, Tl, Nh Group 14: C, Si, Ge, Sn, Pb, Fl Group 15: N, P, As, Sb, Bi, Mc Group 16: O, S, Se, Te, Po, Lv Group 17: F, Cl, Br, I, At, Ts Group 18: He, Ne, Ar, Kr, Xe, Rn, Og

If you’re starting p-block in Class 12 for NEET: Read NCERT First: Don’t jump to notes or guides until you’ve read the NCERT text fully. Mark Important Points: Highlight uses, preparation methods, and exceptions directly in the book. Group Order: Study 15 → 16 → 17 → 18, because NEET questions are more frequent here than in 13 & 14 (covered in Class 11). Revise Regularly: Set a 15-minute daily slot for p-block review to keep it fresh. Use Flashcards: For formulas, structures, and names of compounds. This method prevents last-minute cramming and ensures steady retention.

If you’re starting p-block in Class 12 for NEET: Read NCERT First: Don’t jump to notes or guides until you’ve read the NCERT text fully. Mark Important Points: Highlight uses, preparation methods, and exceptions directly in the book. Group Order: Study 15 → 16 → 17 → 18, because NEET questions are more frequent here than in 13 & 14 (covered in Class 11). Revise Regularly: Set a 15-minute daily slot for p-block review to keep it fresh. Use Flashcards: For formulas, structures, and names of compounds. This method prevents last-minute cramming and ensures steady retention.

For JEE Main & Advanced, the p-block is important but requires a deeper approach than NEET because JEE includes conceptual and reaction-based questions. High-Weightage Areas: Group 13 & 14: Boron compounds (borax, boric acid), aluminium properties, carbon allotropes, silicones, silicates. Group 15: Preparation & properties of ammonia, nitric acid, phosphorus allotropes. Group 16: Oxygen allotropes, sulfur compounds like SO₂, H₂SO₄. Group 17: Halogen preparation, interhalogen compounds, oxoacids. Group 18: Xenon compounds (XeF₂, XeF₄, XeO₃). Approach: Understand trends and anomalies (why B is different from Al, why F is most reactive halogen). Practice reaction mechanisms for compound preparation. Apply conceptual reasoning for theoretical questions. For JEE, problem-solving practice is as important as theory—apply knowledge to real chemical situations.

For JEE Main & Advanced, the p-block is important but requires a deeper approach than NEET because JEE includes conceptual and reaction-based questions. High-Weightage Areas: Group 13 & 14: Boron compounds (borax, boric acid), aluminium properties, carbon allotropes, silicones, silicates. Group 15: Preparation & properties of ammonia, nitric acid, phosphorus allotropes. Group 16: Oxygen allotropes, sulfur compounds like SO₂, H₂SO₄. Group 17: Halogen preparation, interhalogen compounds, oxoacids. Group 18: Xenon compounds (XeF₂, XeF₄, XeO₃). Approach: Understand trends and anomalies (why B is different from Al, why F is most reactive halogen). Practice reaction mechanisms for compound preparation. Apply conceptual reasoning for theoretical questions. For JEE, problem-solving practice is as important as theory—apply knowledge to real chemical situations.

In chemistry, the p-block refers to a set of elements in the periodic table whose outermost electrons occupy the p-orbital of their electron configuration. The periodic table is divided into blocks—s, p, d, and f—based on the type of orbital that receives the last electron. The p-block elements are found in Groups 13 to 18 of the periodic table, on the right-hand side. These groups include a wide range of element types, such as metals, metalloids, and non-metals, as well as some of the most reactive and biologically significant elements.The name “p-block” comes directly from the quantum mechanical description of atoms. In the electron configuration notation, the letter “p” refers to the principal orbital type. The p-orbital can hold a maximum of six electrons, meaning there are six columns of elements in this block. This is why periods (rows) that have p-block elements always have six members in this section.One of the defining features of the p-block is its diversity. It contains essential life-supporting elements like oxygen and nitrogen, technologically important materials like aluminium and silicon, and reactive halogens like fluorine and chlorine. It also includes noble gases such as helium, neon, and argon, which are chemically inert under normal conditions. The p-block spans a wide range of oxidation states and chemical behaviours, making it central to understanding both inorganic and organic chemistry.From an academic perspective, studying the p-block helps students grasp periodic trends such as electronegativity, ionization energy, and atomic radius, as these trends are strongly visible across these groups. For competitive exams like NEET and JEE, mastering p-block concepts is crucial due to the high weightage of questions related to its properties, reactions, and applications.

In chemistry, the p-block refers to a set of elements in the periodic table whose outermost electrons occupy the p-orbital of their electron configuration. The periodic table is divided into blocks—s, p, d, and f—based on the type of orbital that receives the last electron. The p-block elements are found in Groups 13 to 18 of the periodic table, on the right-hand side. These groups include a wide range of element types, such as metals, metalloids, and non-metals, as well as some of the most reactive and biologically significant elements.The name “p-block” comes directly from the quantum mechanical description of atoms. In the electron configuration notation, the letter “p” refers to the principal orbital type. The p-orbital can hold a maximum of six electrons, meaning there are six columns of elements in this block. This is why periods (rows) that have p-block elements always have six members in this section.One of the defining features of the p-block is its diversity. It contains essential life-supporting elements like oxygen and nitrogen, technologically important materials like aluminium and silicon, and reactive halogens like fluorine and chlorine. It also includes noble gases such as helium, neon, and argon, which are chemically inert under normal conditions. The p-block spans a wide range of oxidation states and chemical behaviours, making it central to understanding both inorganic and organic chemistry.From an academic perspective, studying the p-block helps students grasp periodic trends such as electronegativity, ionization energy, and atomic radius, as these trends are strongly visible across these groups. For competitive exams like NEET and JEE, mastering p-block concepts is crucial due to the high weightage of questions related to its properties, reactions, and applications.

Studying the p-block can feel overwhelming because it covers many groups, each with different trends, exceptions, and compound types. However, with a structured approach, you can make it manageable—and even interesting. Step 1: Learn Group-Wise Break it down into Groups 13–18 . Focus on one group at a time, understanding physical and chemical properties, oxidation states, and important compounds. Step 2: Master Trends Memorise periodic trends— electronegativity, ionisation energy, melting points, acidic/basic oxides —because these explain most reactions and exceptions. Step 3: Use Mnemonics For remembering element names, oxidation states, or anomalies, mnemonics are powerful. For example, for Group 15: “Naughty People Are Small Babies” (N, P, As, Sb, Bi). Step 4: Link with Real-World Uses Associate each element with something tangible—Aluminium in aircraft, Silicon in chips, Nitrogen in fertilisers. Step 5: Active Recall & Revision Write short notes and keep revisiting them. Attempt MCQs regularly to reinforce concepts.

Studying the p-block can feel overwhelming because it covers many groups, each with different trends, exceptions, and compound types. However, with a structured approach, you can make it manageable—and even interesting. Step 1: Learn Group-Wise Break it down into Groups 13–18 . Focus on one group at a time, understanding physical and chemical properties, oxidation states, and important compounds. Step 2: Master Trends Memorise periodic trends— electronegativity, ionisation energy, melting points, acidic/basic oxides —because these explain most reactions and exceptions. Step 3: Use Mnemonics For remembering element names, oxidation states, or anomalies, mnemonics are powerful. For example, for Group 15: “Naughty People Are Small Babies” (N, P, As, Sb, Bi). Step 4: Link with Real-World Uses Associate each element with something tangible—Aluminium in aircraft, Silicon in chips, Nitrogen in fertilisers. Step 5: Active Recall & Revision Write short notes and keep revisiting them. Attempt MCQs regularly to reinforce concepts.

The 'P' in p-block refers to the p-orbital—one of the four types of atomic orbitals (s, p, d, f) where electrons can be found. In the periodic table, elements are grouped into blocks depending on the type of orbital that receives the last electron in their electron configuration.For p-block elements, the outermost electrons enter the p-subshell. The p-orbital can hold a maximum of six electrons, which explains why there are six columns of elements in this block (Groups 13–18). Each period starts with s-block elements, moves through d-block (in longer periods), and ends with p-block elements before reaching noble gases.The letter ‘p’ itself comes from the word “principal”, as used in the old spectroscopic notation, indicating the type of angular momentum of the orbital (l = 1 for p-orbitals).From a chemical perspective, the filling of the p-orbital influences bonding, reactivity, and oxidation states. For example, oxygen (2p⁴) is highly electronegative and forms covalent bonds, whereas aluminium (3p¹) behaves as a reactive metal.So when we say "p-block", we’re essentially classifying elements by the quantum mechanical property of their electrons, which in turn determines their chemistry.

The 'P' in p-block refers to the p-orbital—one of the four types of atomic orbitals (s, p, d, f) where electrons can be found. In the periodic table, elements are grouped into blocks depending on the type of orbital that receives the last electron in their electron configuration.For p-block elements, the outermost electrons enter the p-subshell. The p-orbital can hold a maximum of six electrons, which explains why there are six columns of elements in this block (Groups 13–18). Each period starts with s-block elements, moves through d-block (in longer periods), and ends with p-block elements before reaching noble gases.The letter ‘p’ itself comes from the word “principal”, as used in the old spectroscopic notation, indicating the type of angular momentum of the orbital (l = 1 for p-orbitals).From a chemical perspective, the filling of the p-orbital influences bonding, reactivity, and oxidation states. For example, oxygen (2p⁴) is highly electronegative and forms covalent bonds, whereas aluminium (3p¹) behaves as a reactive metal.So when we say "p-block", we’re essentially classifying elements by the quantum mechanical property of their electrons, which in turn determines their chemistry.

S-block Elements: Groups 1 and 2 — Alkali metals (Li, Na, K, Rb, Cs, Fr) and Alkaline earth metals (Be, Mg, Ca, Sr, Ba, Ra). P-block Elements: Groups 13 to 18 — Includes boron group, carbon group, nitrogen group, oxygen group, halogens, and noble gases. D-block Elements: Groups 3 to 12 — Transition metals such as Fe, Cu, Zn, Ag, Au, Pt, Ni, Cr, Mn.These divisions are based on which orbital (s, p, d) is being filled with electrons.

S-block Elements: Groups 1 and 2 — Alkali metals (Li, Na, K, Rb, Cs, Fr) and Alkaline earth metals (Be, Mg, Ca, Sr, Ba, Ra). P-block Elements: Groups 13 to 18 — Includes boron group, carbon group, nitrogen group, oxygen group, halogens, and noble gases. D-block Elements: Groups 3 to 12 — Transition metals such as Fe, Cu, Zn, Ag, Au, Pt, Ni, Cr, Mn.These divisions are based on which orbital (s, p, d) is being filled with electrons.

Physical Properties: State: Can be gases (O₂, N₂), liquids (Br₂), or solids (C, S, I₂). Density & Hardness: Varies—carbon (diamond) is extremely hard, while noble gases are light and inert. Conductivity: Non-metals are poor conductors; metals like aluminium conduct electricity well. Allotropes: Many p-block elements exist in multiple forms—carbon (diamond, graphite), phosphorus, sulfur. Chemical Properties: Show variable oxidation states due to availability of d-orbitals in heavier elements. Oxides range from acidic to basic depending on metallic character. Many form covalent compounds, but metals in the block also form ionic compounds. Exhibit reactivity trends—halogens are strong oxidising agents, oxygen is a vital oxidant, and nitrogen forms stable triple bonds. Example of an Interhalogen: An interhalogen compound is formed between two different halogens, e.g., ClF₃ (Chlorine Trifluoride). These are often more reactive than individual halogens and are used in industrial fluorination processes.

Physical Properties: State: Can be gases (O₂, N₂), liquids (Br₂), or solids (C, S, I₂). Density & Hardness: Varies—carbon (diamond) is extremely hard, while noble gases are light and inert. Conductivity: Non-metals are poor conductors; metals like aluminium conduct electricity well. Allotropes: Many p-block elements exist in multiple forms—carbon (diamond, graphite), phosphorus, sulfur. Chemical Properties: Show variable oxidation states due to availability of d-orbitals in heavier elements. Oxides range from acidic to basic depending on metallic character. Many form covalent compounds, but metals in the block also form ionic compounds. Exhibit reactivity trends—halogens are strong oxidising agents, oxygen is a vital oxidant, and nitrogen forms stable triple bonds. Example of an Interhalogen: An interhalogen compound is formed between two different halogens, e.g., ClF₃ (Chlorine Trifluoride). These are often more reactive than individual halogens and are used in industrial fluorination processes.

A comparative study of p-block elements focuses on trends across periods (left to right) and down groups (top to bottom). This helps explain why elements behave differently despite being in the same block. Physical Properties: Metallic Character: Decreases across a period (Boron → Oxygen) and increases down a group (Nitrogen → Bismuth). Electronegativity: Increases across a period and decreases down a group. Fluorine is the most electronegative. Melting and Boiling Points: Vary widely—carbon has one of the highest melting points, while noble gases have extremely low boiling points. Density: Generally increases down the group due to increased atomic mass. Chemical Properties: Acidic/Basic Nature of Oxides: Across a period, oxides change from basic (Al₂O₃) to amphoteric to acidic (SO₂). Down a group, basicity increases. Reactivity: Halogens are highly reactive non-metals, while noble gases are almost inert. Type of Bonding: Covalent bonding dominates in the upper period elements; metallic and ionic bonding become more common down the groups. For example, in Group 15, nitrogen forms strong multiple bonds (N&equiv;N), while bismuth prefers metallic bonding. Understanding these trends allows chemists to predict reactions—critical in designing materials, medicines, and industrial processes.

A comparative study of p-block elements focuses on trends across periods (left to right) and down groups (top to bottom). This helps explain why elements behave differently despite being in the same block. Physical Properties: Metallic Character: Decreases across a period (Boron → Oxygen) and increases down a group (Nitrogen → Bismuth). Electronegativity: Increases across a period and decreases down a group. Fluorine is the most electronegative. Melting and Boiling Points: Vary widely—carbon has one of the highest melting points, while noble gases have extremely low boiling points. Density: Generally increases down the group due to increased atomic mass. Chemical Properties: Acidic/Basic Nature of Oxides: Across a period, oxides change from basic (Al₂O₃) to amphoteric to acidic (SO₂). Down a group, basicity increases. Reactivity: Halogens are highly reactive non-metals, while noble gases are almost inert. Type of Bonding: Covalent bonding dominates in the upper period elements; metallic and ionic bonding become more common down the groups. For example, in Group 15, nitrogen forms strong multiple bonds (N&equiv;N), while bismuth prefers metallic bonding. Understanding these trends allows chemists to predict reactions—critical in designing materials, medicines, and industrial processes.

The chemical properties of p-block elements vary widely because the block contains metals, non-metals, and metalloids across Groups 13–18. However, these elements show predictable patterns based on their position in the periodic table and the number of valence electrons in the p-orbital. 1. Oxidation States: P-block elements exhibit multiple oxidation states. For example, Group 13 elements typically show +3 and sometimes +1 oxidation states due to the inert pair effect , while Group 15 elements range from –3 (e.g., NH₃) to +5 (e.g., P₂O₅). Halogens generally have –1 oxidation states but can show positive states in interhalogen compounds. 2. Reactivity Trends: Non-metals like oxygen, nitrogen, and halogens are highly electronegative and form covalent compounds. Metals such as aluminium and lead form ionic or covalent bonds depending on the partner element. Metalloids like boron and silicon show mixed behaviour. 3. Nature of Compounds: P-block elements form acids (HCl, HNO₃), bases (aluminates), and amphoteric oxides (Al₂O₃). Non-metal oxides are usually acidic, while metal oxides are basic. 4. Allotropes: Many p-block elements exist in multiple allotropes—carbon (diamond, graphite), phosphorus (white, red, black), and sulfur (rhombic, monoclinic). 5. Industrial Importance: They participate in a wide range of industrial reactions—chlorine in water treatment, nitrogen in fertilisers, silicon in electronics, etc.Understanding these properties is key for predicting reactivity in organic, inorganic, and industrial chemistry, which is why p-block mastery is essential for exams like NEET and JEE.

The chemical properties of p-block elements vary widely because the block contains metals, non-metals, and metalloids across Groups 13–18. However, these elements show predictable patterns based on their position in the periodic table and the number of valence electrons in the p-orbital. 1. Oxidation States: P-block elements exhibit multiple oxidation states. For example, Group 13 elements typically show +3 and sometimes +1 oxidation states due to the inert pair effect , while Group 15 elements range from –3 (e.g., NH₃) to +5 (e.g., P₂O₅). Halogens generally have –1 oxidation states but can show positive states in interhalogen compounds. 2. Reactivity Trends: Non-metals like oxygen, nitrogen, and halogens are highly electronegative and form covalent compounds. Metals such as aluminium and lead form ionic or covalent bonds depending on the partner element. Metalloids like boron and silicon show mixed behaviour. 3. Nature of Compounds: P-block elements form acids (HCl, HNO₃), bases (aluminates), and amphoteric oxides (Al₂O₃). Non-metal oxides are usually acidic, while metal oxides are basic. 4. Allotropes: Many p-block elements exist in multiple allotropes—carbon (diamond, graphite), phosphorus (white, red, black), and sulfur (rhombic, monoclinic). 5. Industrial Importance: They participate in a wide range of industrial reactions—chlorine in water treatment, nitrogen in fertilisers, silicon in electronics, etc.Understanding these properties is key for predicting reactivity in organic, inorganic, and industrial chemistry, which is why p-block mastery is essential for exams like NEET and JEE.

Helium has the electron configuration 1s², which means its valence electrons are in the s-orbital . Logically, one might expect it to be placed in the s-block alongside hydrogen. However, in the modern periodic table, helium is placed in Group 18—the noble gases—within the p-block section. This arrangement is based on chemical properties and periodic trends, not just electron configuration.Helium’s position in Group 18 is justified because it shares the same defining property as other noble gases: a completely filled outer shell, which makes it chemically inert under normal conditions. Even though its outer shell is an s-orbital, the “block” classification in the periodic table is often secondary to grouping by similar reactivity. Helium behaves more like neon and argon than hydrogen or lithium, so it is placed alongside them.Another reason for helium’s p-block placement is consistency in the periodic table layout. If helium were placed above beryllium in Group 2 (s-block), it would disrupt the classification of groups by chemical behaviour. Instead, its noble gas characteristics, such as extremely low reactivity, low boiling point, and monatomic gaseous state, align perfectly with the rest of Group 18.This decision reflects the periodic table’s dual purpose: to show electron configurations and chemical similarities. While helium’s configuration is unique, its placement in the p-block ensures it sits with elements that mirror its inert nature, making it easier to study periodic trends and group behaviours.

Helium has the electron configuration 1s², which means its valence electrons are in the s-orbital . Logically, one might expect it to be placed in the s-block alongside hydrogen. However, in the modern periodic table, helium is placed in Group 18—the noble gases—within the p-block section. This arrangement is based on chemical properties and periodic trends, not just electron configuration.Helium’s position in Group 18 is justified because it shares the same defining property as other noble gases: a completely filled outer shell, which makes it chemically inert under normal conditions. Even though its outer shell is an s-orbital, the “block” classification in the periodic table is often secondary to grouping by similar reactivity. Helium behaves more like neon and argon than hydrogen or lithium, so it is placed alongside them.Another reason for helium’s p-block placement is consistency in the periodic table layout. If helium were placed above beryllium in Group 2 (s-block), it would disrupt the classification of groups by chemical behaviour. Instead, its noble gas characteristics, such as extremely low reactivity, low boiling point, and monatomic gaseous state, align perfectly with the rest of Group 18.This decision reflects the periodic table’s dual purpose: to show electron configurations and chemical similarities. While helium’s configuration is unique, its placement in the p-block ensures it sits with elements that mirror its inert nature, making it easier to study periodic trends and group behaviours.

The p-block elements contain all elements in Groups 13, 14, 15, 16, 17, and 18 of the periodic table. These groups are home to an incredible variety of substances, ranging from light gases to heavy metals. Specifically: Group 13 (Boron Group): Boron (B), Aluminium (Al), Gallium (Ga), Indium (In), Thallium (Tl), and Nihonium (Nh). These show a mixture of metallic and non-metallic characteristics, with boron as a metalloid and aluminium as a widely used lightweight metal. Group 14 (Carbon Group): Carbon (C), Silicon (Si), Germanium (Ge), Tin (Sn), Lead (Pb), and Flerovium (Fl). These elements form the backbone of organic chemistry (carbon) and modern electronics (silicon). Group 15 (Nitrogen Group or Pnictogens): Nitrogen (N), Phosphorus (P), Arsenic (As), Antimony (Sb), Bismuth (Bi), and Moscovium (Mc). Known for a range of oxidation states and vital biological roles (e.g., nitrogen in DNA and proteins). Group 16 (Oxygen Group or Chalcogens): Oxygen (O), Sulfur (S), Selenium (Se), Tellurium (Te), Polonium (Po), and Livermorium (Lv). Oxygen is essential for respiration, while sulfur plays a key role in amino acids. Group 17 (Halogens): Fluorine (F), Chlorine (Cl), Bromine (Br), Iodine (I), Astatine (At), and Tennessine (Ts). Extremely reactive non-metals, halogens form salts with metals (e.g., NaCl). Group 18 (Noble Gases): Helium (He), Neon (Ne), Argon (Ar), Krypton (Kr), Xenon (Xe), and Radon (Rn). They are chemically inert due to a complete octet in their outer shell. This diversity makes the p-block unique. It contains gases, solids, metals, non-metals, and metalloids. It also includes elements vital for life (O, C, N, P) and industrial processes (Al, Si, Cl). Understanding the constituents of the p-block is crucial for predicting chemical reactivity, industrial applications, and environmental behaviour of elements.

The p-block elements contain all elements in Groups 13, 14, 15, 16, 17, and 18 of the periodic table. These groups are home to an incredible variety of substances, ranging from light gases to heavy metals. Specifically: Group 13 (Boron Group): Boron (B), Aluminium (Al), Gallium (Ga), Indium (In), Thallium (Tl), and Nihonium (Nh). These show a mixture of metallic and non-metallic characteristics, with boron as a metalloid and aluminium as a widely used lightweight metal. Group 14 (Carbon Group): Carbon (C), Silicon (Si), Germanium (Ge), Tin (Sn), Lead (Pb), and Flerovium (Fl). These elements form the backbone of organic chemistry (carbon) and modern electronics (silicon). Group 15 (Nitrogen Group or Pnictogens): Nitrogen (N), Phosphorus (P), Arsenic (As), Antimony (Sb), Bismuth (Bi), and Moscovium (Mc). Known for a range of oxidation states and vital biological roles (e.g., nitrogen in DNA and proteins). Group 16 (Oxygen Group or Chalcogens): Oxygen (O), Sulfur (S), Selenium (Se), Tellurium (Te), Polonium (Po), and Livermorium (Lv). Oxygen is essential for respiration, while sulfur plays a key role in amino acids. Group 17 (Halogens): Fluorine (F), Chlorine (Cl), Bromine (Br), Iodine (I), Astatine (At), and Tennessine (Ts). Extremely reactive non-metals, halogens form salts with metals (e.g., NaCl). Group 18 (Noble Gases): Helium (He), Neon (Ne), Argon (Ar), Krypton (Kr), Xenon (Xe), and Radon (Rn). They are chemically inert due to a complete octet in their outer shell. This diversity makes the p-block unique. It contains gases, solids, metals, non-metals, and metalloids. It also includes elements vital for life (O, C, N, P) and industrial processes (Al, Si, Cl). Understanding the constituents of the p-block is crucial for predicting chemical reactivity, industrial applications, and environmental behaviour of elements.

P-Block Elements Class 12 Notes – Group 15 to 18 Elements, Properties & Compounds

Q: When preparing p-block for NEET , the key is smart selection and repeated revision rather than trying to memorise everything at once. Step 1: Focus on NCERT Almost all NEET questions are directly from NCERT chemistry. Read the p-block chapters word-for-word, underlining exceptions, preparation methods, and uses. Step 2: Group Prioritisation For NEET, Groups 15, 16, and 17 are high-yield. Questions often focus on preparation, properties, and uses of compounds like HNO₃, H₂SO₄, PCl₅, and halogen oxoacids. Step 3: Diagram & Table Learning NCERT tables (like anomalous properties, trends) are question goldmines. Learn them visually—draw and label trends. Step 4: Previous Years’ Papers Solve last 10 years of NEET chemistry questions for p-block. This reveals patterns—often direct NCERT lines are used. Step 5: Revision Cycles Do 3–4 revisions before the exam. In the last week, only read marked/highlighted parts.This method ensures you hit all NEET-specific points without drowning in unnecessary detail.

When preparing p-block for NEET , the key is smart selection and repeated revision rather than trying to memorise everything at once. Step 1: Focus on NCERT Almost all NEET questions are directly from NCERT chemistry. Read the p-block chapters word-for-word, underlining exceptions, preparation methods, and uses. Step 2: Group Prioritisation For NEET, Groups 15, 16, and 17 are high-yield. Questions often focus on preparation, properties, and uses of compounds like HNO₃, H₂SO₄, PCl₅, and halogen oxoacids. Step 3: Diagram & Table Learning NCERT tables (like anomalous properties, trends) are question goldmines. Learn them visually—draw and label trends. Step 4: Previous Years’ Papers Solve last 10 years of NEET chemistry questions for p-block. This reveals patterns—often direct NCERT lines are used. Step 5: Revision Cycles Do 3–4 revisions before the exam. In the last week, only read marked/highlighted parts.This method ensures you hit all NEET-specific points without drowning in unnecessary detail.

Q: Because the p-orbital has 3 sub-orbitals, each holding 2 electrons—maximum 6 electrons . This creates six columns in the periodic table’s p-block.

Because the p-orbital has 3 sub-orbitals, each holding 2 electrons—maximum 6 electrons . This creates six columns in the periodic table’s p-block.

Group 13 to 18 elements (except helium), in which the last electron enters the p-orbitals, constitute the p-block. In this chapter will be study systematic group wise details of p-block elements.

Elements: B, Al, Ga, In, Tl.

Trends in Chemical Reactivity and Oxidation States

Most common oxidation states are +3 and +1. Stability of +1 oxidation state increases on going down the group from aluminium to Tl. Thus, in aqueous solution Tl⁺ is more stable.

[Reason: Tl⁺ is more stable due to inert pair effect. As we move down the group, size of atom increases and bond energy decreases. Therefore, energy releases due to two extra bond formations is not sufficient to unpair the ns² electron. Thus, stability of elements in +3 oxidation state decreases.]

Chemical Behaviour

Metallic character increases down the group due to decreasing I.E. and easy loss of electrons. Boron having small size cannot easily lose its electron and hence is an exception to metallic character.
Al, Ga, In and Tl having vacant d-orbitals form [M(OH)₄]^-, [M(OH)₄(H₂O)₂]^- and [M(H₂O)₆]³⁺ type of complexes with H₂O.
MH₃ type of hydrides are formed by group 13 elements. These hydrides act as Lewis acids and form adducts with strong Lewis bases. LiAlH₄ is a white crystalline solid obtained as follows:
4LiH + AlCl₃^Et₂O→ LiAlH₄ + 3LiCl
All halides of group 13 elements are known except TlI₃. The halides have halogen bridged dimeric structures. Boron trihalides exist as monomers only as boron is too small to co-ordinate with four large halide ions. In case of BF₃, the energy required to break p–p bond is not released during bridge formation. The ionic character of trihalides increases on moving down the group due to polarization effects according to Fajan's rule.
Reaction with alkalies: Boron dissolves to give borates with evolution of H₂ gas.
2B + 6NaOH ^Fuse→ 2Na₃BO₃ + 3H₂
Al and Ga with conc. alkalies form metaaluminate and gallate respectively.
2Al + 2NaOH + 6H₂O → 2Na[Al(OH)₄] + 3H₂

Sodium metaaluminate

2Ga + 2NaOH + 6H₂O → 2Na[Ga(OH)₄] + 3H₂

Sodium gallate
Oxides: All elements form oxides of formula M₂O₃, except Tl which forms Tl₂O due to inert pair effect.

Boron shows anomalous behaviour due to its small size and high electronegativity. Chemistry of boron is discussed in detail.

BORON

Boron is the only non-metal in its group due to high I.E., it cannot form B³⁺ instead it forms three covalent bonds.

Properties of Boron

It exists in mainly two allotropic forms, i.e. amorphous dark brown powder and crystalline black hard solid.

Reaction with air:

B + N₂ + O₂^973K→ B₂O₃ + BN

Boron nitride

Reaction with metals:

3Mg + 2B → Mg₃B₂

Magnesium boride

Boron does not react with H₂O.

It reacts with oxidizing agents like HNO₃, H₂SO₄ etc.

B + 3HNO₃ → H₃BO₃ + 3NO₂

2B + 3H₂SO₄ → 2H₃BO₃ + 3SO₂

Boron reduces CO₂ to carbon and SiO₂ to Si.

4B + 3CO₂ → 2B₂O₃ + 3C

Compounds of Boron

Borax or sodium tetraborate decahydrate (Na₂B₄O₇.10H₂O)

Borax naturally occurs as tincal, which contains 50% borax and is obtained from mineral colemanite.

Ca₂B₆O₁₁ + 2Na₂CO₃ → Na₂B₄O₇ + 2CaCO₃ + 2NaBO₂

Borax

Its aqueous solution is alkaline due to hydrolysis.

Na₂B₄O₇ + 7H₂O → 2NaOH + 4H₃BO₃

Borax when heated with ethyl alcohol and conc. H₂SO₄ gives volatile vapours of triethyl borate which burns with green edged flame.

Na₂B₄O₇ + H₂SO₄ + 5H₂O → Na₂SO₄ + 4H₃BO₃

H₃BO₃ + 3C₂H₅OH → B(OC₂H₅)₃ + 3H₂O

Triethyl borate

On heating, borax loses water of crystallization and swells up to form fluffy mass. On further heating, transparent glassy bead is formed.

Na₂B₄O₇.10H₂O ^-10H₂O→ Na₂B₄O₇ → 2NaBO₂ + B₂O₃

Sodium metaborate Boric anhydride

Borax bead

Boric acid or orthoboric acid (H₃BO₃) or B(OH)₃:

It can be obtained from borax or from colemanite as follows,

Na₂B₄O₇ + 2HCl + 5H₂O → 2NaCl + 4H₃BO₃

Ca₂B₆O₁₁ + 4SO₂ + 11H₂O → 2Ca(HSO₃)₂ + 6H₃BO₃

It behaves as a weak monobasic acid due to the following reaction in water.

H₃BO₃ + H₂O → [B(OH)₄]^- + H⁺

With ethyl alcohol and conc. H₂SO₄, it gives triethyl borate, which burns with a green edged flame.

H₃BO₃ + 3C₂H₅OH ^{conc.H₂SO₄}→ B(OC₂H₅)₃ + 3H₂O

On heating, boric acid loses water to form B₂O₃ in three stages.

H₃BO₃^370K_-H₂O→ HBO₂ (metaboric acid)

4HBO₂^410K_-H₂O→ H₂B₄O₇ (tetraboric acid)

4HBO₂^{Red heat}_-H₂O→ 2B₂O₃ (Boron trioxide)

Illustration 1: Borax structure contains

(A) two BO₄ groups and two BO₃ groups

(B) four BO₄ groups only

(D) three BO₄ and one BO₃ groups

Solution: (A). Depends on the charge on the anion.

Illustration 2: The order of increasing acidic strength of BF₃, BCl₃ and BBr₃ is

(A) BF₃ > BCl₃ > BBr₃

(B) BF₃ < BCl₃ < BBr₃

(D) BCl₃ > BBr₃ > BF₃

Solution: (B). Moving down the group, size of halogen increases, hence extent of overlap decreases, so back donation from halogen to boron decreases.

Illustration 3: The bonds present in borazole are

(A) 12σ, 3π

(B) 9σ, 6π

(D) 9σ, 9π

Solution: (A). Borazole is B₃N₃H₆.

Exercise 1

Inorganic benzene is
(A) BN (B) BF₄

(C) B₂H₆ (D) B₃N₃H₆
BCl₃ does not exist as dimer but BH₃ exists as dimer (B₂H₆) because
(A) chlorine is more electronegative than hydrogen

(B) there is p-p back bonding in BCl₃ but BH₃ does not contain such multiple bonding

(C) large sized chlorine atoms do not fit in between the smaller boron atoms whereas small sized hydrogen atoms get fitted in between boron atoms

(D) None of the above

COMPOUNDS OF ALUMINIUM

Aluminium chloride (AlCl₃):

It can be prepared by passing dry Cl₂ or HCl gas over heated Al or by heating a mixture of alumina and carbon in a current of dry chlorine.

2Al + 3Cl₂ → 2AlCl₃

2Al + 6HCl → 2AlCl₃ + 3H₂

Al₂O₃ + 3C + 3Cl₂ → 2AlCl₃ + 3CO

It exists as dimer Al₂Cl₆ in which each Al atom is tetrahedrally surrounded by four Cl atoms.

Compound of Aluminium

Anhydrous AlCl₃ fumes in moist air due to formation of HCl.

AlCl₃ + 3H₂O → Al(OH)₃ + 3HCl

It behaves as a Lewis acid and forms adduct with donor molecules. For example,

H₃N → AlCl₃

Alums:

Alums are double salts of formula X₂SO₄.Y₂(SO₄)₃. 24H₂O where X is a monovalent cation and Y is a trivalent cation. For example, K₂SO₄. Al₂(SO₄)₃. 24H₂O is potash alum.

All alums are Isomorphours.
All alums form acidic solutions due to formation of H₂SO₄.

K₂SO₄.Al₂(SO₄)₃.24H₂O ^H₂O→ Al(OH)₃ + 3H₂SO₄

Illustration 4: Al₂O₃ formation involves large quantity of heat evolution, which makes its use in

(A) deoxidizer

(B) confectionary

(D) thermite welding

Solution: (D).

GROUP 14 ELEMENTS

Elements: C, Si, Ge, Sn, Pb

These elements have valence shell electronic configuration ns², np². Common oxidation states observed for group 14 elements are +4 for silicon and +4 and +2 for, Ge, Sn and Pb. Stability of +2 oxidation state increases in the sequence, Ge < Sn < Pb due to inert pair effect.

Trends in Chemical Reactivity

Due to vacant d-orbitals, elements of the group, except carbon, form compounds having co-ordination numbers higher than four like (SiF₅)^-, (SiF₆)^2- and (PbCl₆)^2-.

As we go down the group, the size of atom increases and the interatomic bond strength decreases Thus, the order of catenation decreases as we go down the group.

Tetrahedral, covalent tetrahalides of the type MX₄ are formed by Si, Ge, Si and Pb. Ge, Sn and Pb also form dihalides with increasing stability in the sequence, GeX₂ < SnX₂ < PbX₂ due to increasing stability of divalent state going down the group (inert pair).

Oxides of composition MO₂ are formed by Si, Ge, Sn and Pb. SiO₂ (silica) is a three dimensional network solid of silica and oxygen atoms connected by single covalent bonds with each silicon atom bonded to four oxygen atoms in tetrahedral arrangement. Silica exists both in crystalline and amorphous forms. Monoxides of Sn and Pb are known. Red lead (Pb₃O₄) is a combination of Pb(II) and Pb(IV) oxides, i.e. 2PbO.PbO₂.

Silicates

Silicates are formed by heating metal oxides or metal carbonates with sand, e.g.

Na₂CO₃^{Fused with sand}_SiO₂→ Na₄SiO₄, Na₄(SiO₃)_n, etc.

Silicates have basic unit of SiO₄^4- where each silicon atom is bonded with four oxide ions tetrahedrally.

Type of silicates:

Type	Units	Example
1. Orthosilicates	Single SiO₄^4- unit.	ZrSiO₄, Mg₂SiO₄
2. Pyrosilicates	Two units of SiO₄^4- joined along a corner oxygen. Hence, they are formed of Si₂O₇^6- units.	Sc₂Si₂O₇, Zn₃(Si₂O₇)Zn(OH)₂.H₂O
3. Cyclic silicates	General formula: (SiO₃^2-)_n or (SiO₃)_n^2n-	Si₃O₉^6- in Ca₃Si₃O₉, Si₆O₁₈^12-, Be₃Al₂Si₆O₁₈
4. Chain silicates	(SiO₃)_n^2n- and (Si₄O₁₁)_n^6n-	Li₂SiO₃, MgSiO₃
5. Two dimensional sheet silicates	(Si₂O₅)_n^2n-	Mg(Si₂O₅)₂Mg(OH)₂, Al₂(OH)₄(Si₂O₅)
6. Three dimensional sheet silicates	All 4 oxygen atoms shared with adjacent SiO₄^4- tetrahedral.	Quartz, zeolites, etc.

CARBON

It exists in two or more allotropes having different physical properties but identical chemical properties. Amorphous forms are coal, coke and charcoal etc., while crystalline forms are diamond, graphite and fullerenes.

Among group 14 elements, carbon forms pπ – pπ multiple bond easily with itself (in graphite). Hence, carbon shows pronounced ability to form pπ – pπ multiple bonds.

Compounds of Carbon

Oxides:

Carbon forms two stable oxides, i.e. carbon dioxide (CO₂) and carbon monoxide (CO). Less stable oxides are C₃O₂, C₅O₂ and C₁₂O₉ known as suboxides.

Carbon monoxide:

Can be prepared by heating oxalic acid or from formic acid or potassium ferrocyanide as follows:

H₂C₂O₄^H₂SO₄_-H₂O→ CO + CO₂ + H₂O
HCOOH ^H₂SO₄_373K→ CO(pure) + H₂O
K₄[Fe(CN)₆] + 6H₂SO₄ + 6H₂O → 2K₂SO₄ + FeSO₄ + 3(NH₄)₂SO₄ + 6CO

Carbon monoxide reduces metal oxides to metals.

Fe₂O₃ + 3CO → 2Fe + 3CO₂

CuO + CO → Cu + CO₂

Due to presence of lone pairs of electrons, CO acts as a Lewis base and this is the reason for the formation of metal carbonyls.

Fe + 5CO → Fe(CO)₅

Phosgene, a poisonous gas is formed when CO combines with Cl₂ in the presence of sunlight

CO + Cl₂^hν→ COCl₂ (phosgene)

Carbon dioxide:

Carbon dioxide is prepared from calcium carbonate.

CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

Glucose and fructose fermentation also produces CO₂

C₆H₁₂O₆^Zymase→ 2C₂H₅OH + 2CO₂

CO₂ turns lime water milky due to the formation of insoluble CaCO₃ which dissolves by passing excess of CO₂ due to soluble Ca(HCO₃)₂ formation.

Ca(OH)₂ + CO₂ → CaCO₃ + H₂O

CaCO₃ + H₂O + CO₂ → Ca(HCO₃)₂

In photosynthesis, CO₂ is converted into glucose, starch and O₂.

6CO₂ + 6H₂O ^Chlorophyll_hν → C₆H₁₂O₆ + 6O₂

Glucose

CO₂ is also used in urea formation under a pressure of 220 atm and 453 – 473 K temperature.

2NH₃ + CO₂^453-473K_220atm→ [NH₂COONH₂] → NH₂CONH₂ + H₂O

Urea

Carbon forms other compounds, namely,

carbides, e.g. Be₂C, Al₄C₃
acetylides having (C ≡ C)^2- ions, e.g. MgC₂, CaC₂
allylide having C₃^4- ions, e.g. Mg₂C₃
halides, e,g, CF₄, CCl₄, CBr₄, CFC
sulphides, e.g. CS, CS₂, C₃S₂

GROUP 15 ELEMENTS

Elements: N, P, As, Sb, Bi.

Nitrogen and phosphorus are non-metals. The metallic character increases down the group due to lower I.E. and larger size. Hence, bismuth shows metallic character. The group state electronic configuration is ns², np³.

Trends in Chemical Reactivity

Most common oxidation states shown by group 15 elements are – 3, +3 and +5. The stability of highest oxidation state (+5) decreases down the group.

The covalent character goes on decreasing as we move down the group in the sequence, P > As > Sb > Bi. This is due to increasing size of atom which refers to Fajan's rules.

Sb and Bi are the heavier elements of the group and form M³⁺ cations due to decrease in ionization enthalpy.

In contrast to nitrogen, the size of phosphorus atom forms pπ bonding and forms both cyclic and open chain compounds. Hence, it shows catenation.

NITROGEN

Nitrogen can be obtained from ammonium nitrite, ammonium dichromate, ammonia, urea and barium azide in the laboratory as follows.

(i) NH₄Cl + NaNO₂ → NH₄NO₂ + NaCl
Unstable
NH₄NO₂ → N₂ + 2H₂O

(ii) (NH₄)₂Cr₂O₇ → N₂ + Cr₂O₃ + 4H₂O

(iii) 2NH₃ + 3CuO → N₂ + 3Cu + 3H₂O

(iv) NH₂CONH₂ + 2HNO₂ → 2N₂ + CO₂ + 3H₂O

(v) Ba(N₃)₂ → 3N₂ + Ba

Barium azide

Nitrogen combines with oxygen under high temperature conditions to form nitric oxide

N₂ + 3O₂^3000°C→ 2NO(Nitric oxide)

It combines with H₂ in presence of a catalyst at 200 atm and 400 – 500°C temperature to form ammonia.

N₂ + H₂ ⇌ 2NH₃

With metals it forms nitrides.

6Li + N₂^450°C→ 2Li₃N(Lithium nitride)

3Mg + N₂^450°C→ Mg₃N₂ (Magnesium nitride)

Some non-metals also combine with nitrogen to form nitrides.

2B + N₂ → 2BN(Boron nitride)

3Si + 2N₂ → Si₃N₄ (Silicon nitride)

Compounds of Nitrogen

Ammonia:

It is prepared from ammonium salt, magnesium nitride and by Haber's process as follows.

(i) NH₄Cl + NaOH → NH₃ ↑ + NaCl + H₂O

(ii) Mg₃N₂ + 6H₂O → 3Mg(OH)₂ + 2NH₃

(iii) N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

In water it forms a basic solution.

NH₃ + H₂O ⇌ NH₄⁺ + OH^-

Ammonia reacts with halogens in different stoichiometic ratios.

(i) 8NH₃ + 3Cl₂ → 6NH₄Cl + N₂

(ii) NH₃ + 3Cl₂ → NCl₃ + 3HCl

(iii) 8NH₃ + 3Br₂ → 6NH₄Br + N₂

(iv) NH₃ + 3Br₂ → NBr₃ + 3HBr

(v) 2NH₃ + 3I₂ → NH₃.NI₃ + 3HI

Explodes in dry state

Ammonia acts as a Lewis base in complex formation.

(i) Ag⁺ + 2NH₃ → [Ag(NH₃)₂]⁺

(ii) Cu²⁺ + 4NH₃ → [Cu(NH₃)₄]²⁺

(iii) Cd²⁺ + 4NH₃ → [Cd(NH₃)₄]²⁺

Oxides of nitrogen

Nitrogen forms six different oxides, with oxidation states ranging from +1 to +5. Various oxides with their Lewis formula are listed below:

Oxides of Nitrogen

Preparations of above oxides are as follows:

(a) NH₄NO₃^Δ→ N₂O + 2H₂O

(b) 2NaNO₂ + 2FeSO₄ + 3H₂SO₄ → Fe₂(SO₄)₃ + 2NaHSO₄ + 2H₂O + 2NO

(d) 2Pb(NO₃)₂^673K→ 4NO₂ + 2PbO + O₂

(e) 2NO₂^Cool⇌ N₂O₄^Heat

(f) 2HNO₃ + P₂O₅ → 2HPO₃ + N₂O₅

Chemical reactivity of various oxides can be summarized as follows:

(i) Nitrous oxide is neutral towards litmus and reacts with sulphur, phosphorus, metals and hydrogen as follows:

(a) S + 2N₂O → SO₂ + 2N₂

(b) P₄ + 10N₂O → P₄O₁₀ + 10N₂

(d) 2Na + N₂O → Na₂O + N₂

(e) Cu + N₂O → CuO + N₂

(f) N₂O + H₂ → N₂ + H₂O

(ii) (a) Nitric oxide is a stable oxide and decomposes only when heated at 800°C.

2NO ^800°C→ N₂ + O₂

(b) It dissolves in cold ferrous sulphate solution.

[Fe(H₂O)₆]SO₄ + NO → [Fe(H₂O)₅NO]SO₄ + H₂O

Hydrated nitrosyl
(Brown ring)

FeSO₄ + NO + 5H₂O

(c) It also acts as an oxidizing agent.

SO₂ + 2NO + H₂O → H₂SO₄ + N₂O

H₂S + 2NO + H₂O → S + N₂O

(iii) Nitrogen tetroxide is a reddish brown pungent smelling gas, which associates and dissociates with change in temperature.

(a) It acts both as oxidizing and reducing agent.

NO₂ + 2Na → Na₂O + NO } as oxidizing agent
NO₂ + 2Cu → Cu₂O + NO }

2NO₂ + O₃ → N₂O₅ + O₂ } as reducing agent

(iv) Nitrogen pentoxide (N₂O₅) is a strong oxidizing agent.

I₂ + 5N₂O₅ → I₂O₅ + 10NO₂

(a) It is decomposed by alkali metals.

N₂O₅ + Na → NaNO₃ + NO₂ ↑

(b) With aq. NaCl, the reaction proves that N₂O₅ exists as ionic nitronium nitrate (NO₂⁺NO₃^-).

N₂O₅ + NaCl → NaNO₃ + NO₂Cl

Oxides of nitrogen

Nitrogen forms six different oxides, with oxidation states ranging from +1 to +5. Various oxides with their Lewis formula are listed below:

Formula	Oxidation no. of nitrogen	Lewis formula
N₂O Nitrous oxide	+1	:N≡N→Ö:
NO Nitric oxides	+2	:N=Ö:
N₂O₃ Dinitrogen trioxide	+3	O \ N—N / O
NO₂ Nitrogen dioxide	+4	N / \ O O
N₂O₄ Dinitrogen tetroxide	+4	O O \ / N—N / \ O O
N₂O₅ Dinitrogen pentoxide	+5	O ‖ O N O \ / N / \ O O

Preparations of above oxides are as follows:

(a) NH₄NO₃^Δ→ N₂O + 2H₂O

(b) 2NaNO₂ + 2FeSO₄ + 3H₂SO₄ → Fe₂(SO₄)₃ + 2NaHSO₄ + 2H₂O + 2NO

(d) 2Pb(NO₃)₂^673K→ 4NO₂ + 2PbO + O₂

(e) 2NO₂^Cool⇌ N₂O₄^Heat

(f) 2HNO₃ + P₂O₅ → 2HPO₃ + N₂O₅

Chemical reactivity of various oxides can be summarized as follows:

(i) Nitrous oxide is neutral towards litmus and reacts with sulphur, phosphorus, metals and hydrogen as follows:

(a) S + 2N₂O → SO₂ + 2N₂

(b) P₄ + 10N₂O → P₄O₁₀ + 10N₂

(d) 2Na + N₂O → Na₂O + N₂

(e) Cu + N₂O → CuO + N₂

(f) N₂O + H₂ → N₂ + H₂O

(ii) (a) Nitric oxide is a stable oxide and decomposes only when heated at 800°C.

2NO ^800°C→ N₂ + O₂

(b) It dissolves in cold ferrous sulphate solution.

[Fe(H₂O)₆]SO₄ + NO → [Fe(H₂O)₅NO]SO₄ + H₂O

Hydrated nitrosyl
(Brown ring)

FeSO₄ + NO + 5H₂O

SO₂ + 2NO + H₂O → H₂SO₄ + N₂O

H₂S + 2NO + H₂O → S + N₂O

(iii) Nitrogen tetroxide is a reddish brown pungent smelling gas, which associates and dissociates with change in temperature.

2NO₂^770K⇌ O₂ + 2NO ^410K⇌ 2NO₂
↓
300 K
N₂O₄
Yellow liquid
↓
260 K N₂O₄
Colourless
solid

(a) It acts both as oxidizing and reducing agent.

NO₂ + 2Na → Na₂O + NO } as oxidizing agent
NO₂ + 2Cu → Cu₂O + NO }

2NO₂ + O₃ → N₂O₅ + O₂ } as reducing agent

(iv) Nitrogen pentoxide (N₂O₅) is a strong oxidizing agent.

I₂ + 5N₂O₅ → I₂O₅ + 10NO₂

(a) It is decomposed by alkali metals.

N₂O₅ + Na → NaNO₃ + NO₂ ↑

(b) With aq. NaCl, the reaction proves that N₂O₅ exists as ionic nitronium nitrate (NO₂⁺NO₃^-).

N₂O₅ + NaCl → NaNO₃ + NO₂Cl

Oxyacids of nitrogen:

Out of the five oxyacids of nitrogen, nitric acid (HNO₃) is most important.

It can be prepared in the lab by heating NaNO₃ or KNO₃.

NaNO₃ + H₂SO₄ → NaHSO₄ + HNO₃

It can also be manufactured by Oswald's process.

4NH₃ + 5O₂^Pt/Rh→ 4NO + 6H₂O

2NO + O₂^1120K→ 2NO₂

3NO₂ + H₂O → 2HNO₃ + NO

Reactions of nitric acid are summarized in the following table.

Concentration	Metal	Main Product
Very dilute HNO₃	Mg, Mn	H₂ + M(NO₃)_x
Very dilute HNO₃	Fe, Zn, Sn	NH₄NO₃ + M(NO₃)_x Metal nitrate
Dilute HNO₃	Pb, Cu, Ag, Hg	NO + M(NO₃)_x
Dilute HNO₃	Fe, Zn	N₂O + M(NO₃)_x
Conc. HNO₃	Zn, Fe, Pb, Cu, Ag	NO₂ + M(NO₃)_x
Conc. HNO₃	Sn	NO₂ + H₂SO₄ Metastannic acid

Oxides of phosphorus:

Phosphorus pentoxide (P₄O₁₀) is prepared by burning white phosphorus in excess of air or oxygen.

P₄ + 5O₂ → P₄O₁₀

It acts as a dehydrating agent.

4HNO₃ + P₄O₁₀ → 2N₂O₅ + 4HPO₃

2H₂SO₄ + P₄O₁₀ → 2SO₃ + 4HPO₃

Final product or reaction with water is as follows.

P₄O₁₀^2H₂O→ 4HPO₃^2H₂O→ 2H₄P₂O₇^2H₂O→ 4H₃PO₄

Metaphosphoric Pyrophosphoric Orthophosphoric

Structure: Phosphorus Trioxide

Phosphorous trioxide (P₄O₆) is prepared by burning white phosphorous in limited supply of air.

P₄ + 3O₂ → P₄O₆

It reacts with cold as well as hot water.

P₄O₆ + 6H₂O(cold) → 4H₃PO₃

P₄O₆ + 6H₂O(hot) → 3H₃PO₄ + PH₃

It burns in chlorine to form oxy – chlorides.

P₄O₆ + 4Cl₂ → 2POCl₃ + 2PO₂Cl

Oxyacids of phosphorus are summarized below:

(HPO₃)_n → It exists in polymeric form.

Exercise 2

Which is true among the following?
(A) PH₃ is stronger base than NH₃

(B) Bond angle in PH₃ is more than that of NH₃

(C) PH₃ is stronger reducing agent than NH₃

(D) Boiling point of PH₃ is more than that of NH₃
Orthophosphoric acid on strong heating forms
(A) P₄O₁₀ (B) H₄P₂O₇

(C) (HPO₃)_x (D) P₂O₅

GROUP 16 ELEMENTS

Elements: O, S, Se, Te, Po.

First four elements are called chalcogens. The valence shell electronic configuration is ns², np⁴, which shows each element has two electrons short of the next noble gas configuration.

Trends in Chemical Reactivity and Oxidation States

Important oxidation states of S, Se and Te are -2, +2, +4 and +6. Oxidation states +4 and +6 are more pronounced for sulphur and heavier chalcogens, e.g. SF₆, SeCl₄, Te(OH)₆.
As we go down the group, the electronegativity for S, Se, Te and Po decreases due to increase in size.
Due to easy loss of electron with increasing size, the metallic character increases on descending down the group. Hence, O, S are non-metals; Se, Te are metalloids whereas Po shows metallic character.
Tendency for catenation decreases with increasing size as we move down the group.
Thermal stability of hydrides decreases in the order: H₂O > H₂S > H₂Se > H₂Te > H₂Po.
Tendency to form multiple bonds decreases as we go down the group. Thus, S = C = S is quite stable as compared to Se = C = Se, which is unstable and decomposes readily while Te = C = Te is unknown.

OXYGEN

Oxygen can be prepared from various oxides and salts as follows:

(a) 2HgO ^450°C→ 2Hg + O₂

(b) 3MnO₂^Δ→ Mn₃O₄ + O₂

(d) 4K₂Cr₂O₇^400°C→ 4K₂CrO₄ + 2Cr₂O₃ + 3O₂

(e) 2KClO₃^400°C→ 2KCl + 3O₂

Ozone:

In the laboratory ozone can be prepared by passing oxygen through a strong electric field.

3O₂ ⇌ 2O₃

Being a metastable allotrope, it always has a tendency to convert back to oxygen.

2O₃ → 3O₂ ΔG = -163J/mole

It acts as a powerful oxidizing agent.

6NO₂ + O₃ → 3N₂O₅

O₃ + 2KI + H₂O → I₂ + 2KOH + O₂

The amount of ozone in a gas mixture can be determined by passing it through KI solution (at constant pH 9.2). The iodine liberated is titrated with sodium thiosulphate (hypo) solution.

O₃ + 2I^- + H₂O → I₂ + 2OH^- + O₂

I₂ + 2S₂O₃^2- → 2I^- + S₄O₆^2-

O₃ layer in the upper atmosphere is destroyed by exhaust gases containing nitrogen oxides. Halogens also damage O₃ layer.

(a) NO + O₃ → NO₂ + O₂

(b) O₃ + hν → O₂ + O

Also,

(d) Cl + O₃ → ClO + O₂

(e) O₃ + hν → O₂ + O

(d) ClO + O → Cl + O₂

SULPHUR

Sulphur can be extracted from underground deposits by the Frasch process. It can also be prepared from H₂S gas as follows:

H₂S + ³⁄₂O₂ → SO₂ + H₂O

2H₂S(g) + SO₂(g) — Fe₂O₃ (catalyst), 673 K —→³⁄₈S₈(g) + 2H₂O(g)

Allotropes of Sulphur

Sulphur exists in three allotropic forms:

Rhombic sulphur (octahedral sulphur) which exists as S₈ molecules. It is a bright yellow solid, soluble in CS₂.
Monoclinic sulphur or β-sulphur is dull yellow and soluble in CS₂. Below 369 K it slowly changes to rhombic sulphur. At 369 K (the transition temperature) both forms can coexist.
Plastic sulphur (amorphous sulphur or γ-sulphur) is a rubber-like, transparent yellow material, insoluble in CS₂ and H₂O. It is regarded as a super-cooled liquid that exists in long, random, inverted chains of sulphur atoms.

Compounds of Sulphur

Sulphur Dioxide

Among the several oxides of sulphur, SO₂ (sulphur dioxide) and SO₃ (sulphur trioxide) are the most important.

Preparation

S₈ + 8O₂ → 8SO₂
4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂

SO₂ shows both oxidizing and reducing properties.

Reactions illustrating its dual behaviour

2H₂S + SO₂ → 2H₂O + 3S ↓ (oxidizing property of SO₂)
SO₂ + 2Mg → 2MgO + S ↓ (oxidizing property of SO₂)
2KMnO₄ + 5SO₂ + 2H₂O → K₂SO₄ + 2MnSO₄ + 2H₂SO₄ (SO₂ acts as a reducing agent)
2FeCl₃ + SO₂ + 2H₂O → H₂SO₄ + 2FeCl₂ + 2HCl (SO₂ acts as a reducing agent)

Sulphur Trioxide, Sulphuric Acid & Sodium Thiosulphate

Sulphur trioxide

(SO₃) is prepared by catalytic oxidation of sulphur dioxide.

2SO₂(g) + O₂(g) — V₂O₅ catalyst, ~723 K —→ 2SO₃(g)

It exists in three allotropic forms: α-SO₃, β-SO₃, γ-SO₃.

SO₃ is an acidic oxide:

SO₃ + H₂O → H₂SO₄ + heat

It forms oleum with concentrated sulphuric acid:

H₂SO₄ + SO₃ → H₂S₂O₇ (oleum)

Sulphuric acid

Sulphur forms many oxyacids; one important acid is sulphuric acid (H₂SO₄), also called oil of vitriol. It is prepared by the contact process and the lead chamber process.

Contact process (outline)

2SO₂ + O₂— catalyst —→ 2SO₃

H₂SO₄ + SO₃ → H₂S₂O₇ (Oleum)

H₂S₂O₇ + H₂O → 2H₂SO₄

Properties

Sulphuric acid is a low-volatile liquid, a strong acid with a strong affinity for water. It removes water from wet gases that do not react with the acid and dehydrates many organic compounds (e.g., carbohydrates), causing charring.

C₁₂H₂₂O₁₁ + 11 H₂SO₄ → 12C + 11 H₂SO₄ + 11 H₂O

Both metals and non-metals are oxidized by concentrated sulphuric acid; the acid itself is reduced to SO₂.

C + 2H₂SO₄ → CO₂ + 2SO₂ + 2H₂O

Cu + 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O

Sodium thiosulphate

Preparation

Prepared by boiling aqueous solutions of metal sulphites with elemental sulphur.

Na₂SO₃ + ¹⁄₈S₈— H₂O, ~378 K —→ Na₂S₂O₃

It can also be manufactured by Spring’s reaction, treating sodium sulphide and sodium sulphite with iodine:

Na₂S + Na₂SO₃ + I₂ → Na₂S₂O₃ + 2NaI

Reaction with acids

Na₂S₂O₃ + 2HCl → 2NaCl + H₂O + S ↓ + SO₂

Group 17 (Halogens) – Notes & MCQs

Illustrations (MCQs)

Illustration 5: Oxide of nitrogen used as a catalyst in the lead chamber process for the manufacture of H₂SO₄ is

(A) NO (B) N₂O (C) N₂O₃ (D) N₂O₅

Solution: (A).

Illustration 6: SO₃ is a powerful oxidizing agent and it oxidizes P to

(A) PH₃ (B) H₃PO₄ (C) P₂O₃ (D) P₄O₁₀

Sol.: (D). Preparation of phosphorus oxides.

Illustration 7: P₄O₁₀ is not used to dry NH₃ gas because

(A) it is very costly
(B) it is not a dehydrating agent
(C) it reacts with moisture to give an acid which reacts with NH₃ (base)
(D) it is basic while NH₃ is acidic

Solution: (C). H₃PO₄ formed reacts with NH₃ to give (NH₄)₃PO₄.

GROUP 17 ELEMENTS (HALOGENS)

Elements: F, Cl, Br, I, At. Astatine is radioactive. The melting and boiling points of halogens increase down the group due to increasing atomic size and hence stronger van der Waals forces. Thus, fluorine and chlorine are pale-yellow and greenish-yellow gases, bromine is a deep reddish-brown liquid, and iodine is a lustrous greyish-black crystalline solid.

Bond energy (X–X) order: F–F < Cl–Cl > Br–Br > I–I. The smaller bond energy of F–F is due to strong lone-pair repulsions in F₂. From Cl₂ to I₂, increasing size reduces effective overlap and the bond strength decreases.

Trends in Oxidation States and Chemical Reactivity

Fluorine, the most electronegative element, shows oxidation state −1 in all compounds and easily accepts electrons. Halogens act as oxidizing agents and can oxidize halide ions of halogens below them in the group.

Halogens react with metals and non-metals to form halides. Fluorine forms compounds with all elements except He, Ar and Ne.

Halogens form metal halides, M–X, whose ionic character decreases in the order M–F > M–Cl > M–Br > M–I. With increasing halide ion size, polarizability increases, covalent character rises, and ionic character falls. For a metal with multiple oxidation states, the halide of the higher oxidation state is more covalent due to the smaller cation size. Thus, SnCl₄, PbCl₄ and SbCl₅ are more covalent than SnCl₂, PbCl₂ and SbCl₃, respectively.

Hydrogen halides (HF, HCl, HBr, HI) are hydrohalic acids. Only HF is a liquid due to H-bonding. In aqueous solution their acid strength increases in the order HF < HCl < HBr < HI, following the decreasing H–X bond strength.

Other Comparative Properties

Dipole moment: HI < HBr < HCl < HF.
Bond length: HF < HCl < HBr < HI.
Bond strength: HI < HBr < HCl < HF.
Thermal stability: HI < HBr < HCl < HF.
Acid strength: HF < HCl < HBr < HI.
Reducing power: HF < HCl < HBr < HI (HI strongest).

Oxyacids

Halogens form various oxyacids. Fluorine does not form oxyacids. Important oxyacids of chlorine are listed below.

Oxyacids of Chlorine

Formula	Name	Oxidation no. of Cl
HClO	Hypochlorous acid	+1
HClO₂	Chlorous acid	+3
HClO₃	Chloric acid	+5
HClO₄	Perchloric acid	+7

Acidic strength: Increases with oxidation number of the halogen: HClO₄ > HClO₃ > HClO₂ > HClO.

Group 17 (Halogens) – Notes & MCQs

Illustrations (MCQs)

Illustration 5: Oxide of nitrogen used as a catalyst in the lead chamber process for the manufacture of H₂SO₄ is

(A) NO (B) N₂O (C) N₂O₃ (D) N₂O₅

Solution: (A).

Illustration 6: SO₃ is a powerful oxidizing agent and it oxidizes P to

(A) PH₃ (B) H₃PO₄ (C) P₂O₃ (D) P₄O₁₀

Sol.: (D). Preparation of phosphorus oxides.

Illustration 7: P₄O₁₀ is not used to dry NH₃ gas because

(A) it is very costly
(B) it is not a dehydrating agent
(C) it reacts with moisture to give an acid which reacts with NH₃ (base)
(D) it is basic while NH₃ is acidic

Solution: (C). H₃PO₄ formed reacts with NH₃ to give (NH₄)₃PO₄.

GROUP 17 ELEMENTS (HALOGENS)

Trends in Oxidation States and Chemical Reactivity

Fluorine, the most electronegative element, shows oxidation state −1 in all compounds and easily accepts electrons. Halogens act as oxidizing agents and can oxidize halide ions of halogens below them in the group.

Halogens react with metals and non-metals to form halides. Fluorine forms compounds with all elements except He, Ar and Ne.

Other Comparative Properties

Dipole moment: HI < HBr < HCl < HF.
Bond length: HF < HCl < HBr < HI.
Bond strength: HI < HBr < HCl < HF.
Thermal stability: HI < HBr < HCl < HF.
Acid strength: HF < HCl < HBr < HI.
Reducing power: HF < HCl < HBr < HI (HI strongest).

Oxyacids

Halogens form various oxyacids. Fluorine does not form oxyacids. Important oxyacids of chlorine are listed below.

Oxyacids of Chlorine

Formula	Name	Oxidation no. of Cl
HClO	Hypochlorous acid	+1
HClO₂	Chlorous acid	+3
HClO₃	Chloric acid	+5
HClO₄	Perchloric acid	+7

Acidic strength: Increases with oxidation number of the halogen: HClO₄ > HClO₃ > HClO₂ > HClO.

Interhalogens & Noble Gases – Notes and MCQs

Reason: This is because the loss of H⁺ ion from each oxyacid forms the conjugate base as (oxo-anion). The greater the number of oxygen atoms in the ion, the greater is the dispersal of negative charge and hence the higher is the stability of the resulting ion. Hence, the ease of formation of the resulting ions increases accordingly.

Interhalogen Compounds

Halogens can also form interhalogen compounds by reacting with each other. General formula: AB_n where n = 1, 3, 5, 7; A = heavier halogen and B = lighter halogen; e.g., ICl₃, IF₇, etc. All interhalogen compounds are covalent and are generally more reactive than the individual halogens.

The stability of interhalogen compounds increases with the size of the central atom. Shapes of some interhalogen compounds are given below:

Interhalogen compound	Shape	Hybridization
Type AB (n = 1) — BrF, BrCl, IBr, IF	Linear	sp
Type AB₃ (n = 3) — BrF₃, ICl₃, IF₃	T-shape	sp³d
Type AB₅ (n = 5) — BrF₅, ICl₅, IF₅	Square pyramidal	sp³d²
Type AB₇ (n = 7) — IF₇	Pentagonal bipyramidal	sp³d³

Pseudo Halide Ions and Pseudo Halogen

Ions consisting of two or more atoms, of which at least one is nitrogen, and having properties similar to halide ions are called pseudohalide ions. Some of these pseudohalide ions can be oxidized to form covalent dimers comparable to halogens. Such dimers are called pseudohalogens; e.g., CN⁻ is a pseudohalide ion and (CN)₂ is called cyanogen. SCN⁻ is a pseudohalide ion (thiocyanate) and (SCN)₂ is thiocyanogen.

Illustrations (MCQs)

Illustration 8: Which radical can bring about the highest oxidation state of a transition metal?
(A) F⁻ (B) Cl⁻ (C) Br⁻ (D) I⁻
Solution: (A). F⁻ because it has the smallest size and highest electronegativity.

Illustration 9: Which forms an “acid salt” with base?
(A) HCl (B) HBr (C) HI (D) HF
Sol.: (D). HF exists as H₂F₂ and forms KHF₂ (acid salt) and KF (normal salt).

Illustration 10: Among the fluorides given below which will further react with F₂?
(A) NaF (B) CaF₂ (C) SF₆ (D) IF₅
Solution: (D).

Illustration 11: The following acids have been arranged in the order of decreasing acidic strength. Identify the correct order.
(A) ClOH > BrOH > IOH (B) BrOH > ClOH > IOH (C) IOH > BrOH > ClOH (D) ClOH > IOH > BrOH
Solution: (A). Electronegativity of halogen is inversely proportional to basic strength and directly proportional to acidic strength.

Exercise 3

When Cl₂ water is added to an aqueous solution of potassium halide in the presence of CHCl₃, a violet colour is obtained. On adding more Cl₂ water, the violet colour disappears and a colourless solution is obtained. This test confirms the presence of which in the aqueous solution?
(A) Iodide (B) Bromide (C) Chloride (D) Iodide and Bromide
Which of the following species is linear?
(A) I₃⁻ (B) XeF₂ (C) ICl₂⁻ (D) All of the above

GROUP 18 ELEMENTS

Elements: He, Ne, Ar, Kr, Xe, Rn. These are noble gases. Their valence shells are fully occupied (ns² np⁶), so gain or loss of electrons is difficult and they are chemically least reactive. All noble gases are monoatomic.

The natural abundance of noble gases in dry air is around 1%, of which argon is the major component. The main commercial source of He is natural gas. Radon is obtained as the decay product of the Radium-226 isotope.

Isolations: Rare gases are isolated from air by Ramsay and Rayleigh’s method. Several xenon compounds with very electronegative elements like fluorine and oxygen have been synthesized, while no true compounds of He, Ne and Ar are known. Krypton forms compounds rarely (notably KrF₂). This trend is explained by increasing atomic size down the group, decreasing ionization energy, and the availability of vacant d-orbitals.

Xenon Fluoride and Xenon Oxygen Compounds

XeF₂, XeF₄ and XeF₆ are the main fluorides of xenon.

XeF_x are stable only in Ni containers. They are strong fluorinating agents and are hydrolysed even by traces of water. Xenon fluorides react with fluoride-ion acceptors to form cationic species and with fluoride-ion donors to form fluoroanions.

Hydrolysis of xenon fluorides produces xenon oxygen compounds. XeO₃ with aqueous alkali forms hydrogen xenate ion.

Structure of xenon compounds

Compound	Structure	Hybridization
XeF₂	Linear	sp³d
XeF₄	Square planar	sp³d²
XeF₆	Distorted octahedron	sp³d³
XeO₃	Pyramidal	sp³
XeOF₄	Square pyramidal	sp³d²
XeO₄	Tetrahedral	sp³
XeO₂F₂	Distorted octahedron	sp³d

Separation of Rare Gases

Rare gases are separated individually from their mixture by Dewar’s charcoal adsorption method. After removing O₂, CO₂ and N₂ from dry air, the remaining mixture of rare gases is adsorbed by coconut-charcoal at about −100 °C in a Dewar flask.

Illustration 12 :

Geometry of XeOF₄ molecule is

(A) square planar (B) square pyramidal
(C) triangular bipyramidal (D) distorted octahedron

Solution. (B). It involves sp³d² hybridization with one lone pair.

Exercise 4

i) The case of liquefaction of noble gases decreases in the order

(A) He > Ne > Ar > Kr > Xe (B) Xe > Kr > Ar > Ne > He
(C) Kr > Xe > He > Ar > Me (D) Ar > Kr > Xe > He > Me

ii) The gaseous mixture used by divers for respiration is

(A) N₂ + O₂ mixture (B) He + O₂ mixture
(C) Ar + O₂ mixture (D) Ne + O₂ mixture

ANSWER TO EXERCISE

Exercise 1. 1) D 2) C
Exercise 2. 1) C 2) D
Exercise 3. 1) A 2) D
Exercise 4. 1) B 2) B

SOLVED EXAMPLES

1. Alumina on heating with carbon in nitrogen atmosphere gives

(A) Al + CO (B) Al + CO₂
(C) AlN + CO (D) Al + CO + N₂

Sol. (C).

2. Thallium shows different oxidation states

(A) as it is a transition metal (B) due to inert pair effect
(C) because of its amphoteric character (D) due to its high reactivity

Sol. (B).

3. Aluminum vessels should not be washed with materials containing washing soda because

(A) washing soda is expensive
(B) washing soda is easily decomposed
(C) washing soda reacts with aluminum to form soluble aluminates
(D) washing soda reacts with aluminum to form insoluble aluminum oxide

Sol. (C).

4. Which of the following is pseudoalum?

(A) (NH₄)₂SO₄·Fe₂(SO₄)₃·24H₂O (B) K₂SO₄·Al₂(SO₄)₃·24H₂O
(C) MnSO₄·Al₂(SO₄)₃·24H₂O (D) None of the above

Sol. (C).

5. Which is used in high temperature thermometry?

(A) Na (B) Tl (C) Ga (D) Hg

Sol. (C).

6. The main factor responsible for weak acidic nature of B–F bonds in BF₃ is

(A) large electronegativity of fluorine
(B) three centred two electron bonds in BF₃
(C) pπ–dπ back bonding
(D) pπ–pπ back bonding

Sol. (D).

7. Which of the following is methanide?

(A) Be₂C (B) Al₄C₃ (C) Mg₂C₃ (D) Both (A) and (C)

Sol. (D). Both Be₂C and Al₄C₃ are called methanides because they react with water to give methane.

8. Which of the following nitrate will produce laughing gas on heating?

(A) NaNO₃ (B) Ca(NO₃)₂ (C) Hg(NO₃)₂ (D) NH₄NO₃

Sol. (D).

9. Among the following oxides, least acidic is

(A) P₄O₆ (B) P₄O₁₀ (C) As₄O₆ (D) As₄O₁₀

Sol. (C). Trioxides are less acidic than pentoxides. Acidic character decreases down the group.

10. The equivalent weight of phosphoric acid (H₃PO₄) in the reaction — is

(A) 25 (B) 49 (C) 59 (D) 98

Sol. (D).

ASSIGNMENT PROBLEMS

1. Which of the following has the minimum heat of dissociation?
(A) (B) (C) (D)

2.The role of fluorspar (CaF₂) which is added in small quantities in the electrolytic reduction of alumina dissolved in fused cryolite (Na₃AlF₆) is
(A) as a catalyst
(B) to make the fused mixture very conducting
(C) to increase the temperature of the melt
(D) to decrease the rate of oxidation of carbon at the anode

3. The dissociation of Al(OH)₃ by a solution of NaOH results in the formation of
(A) (B) (C) (D)

4. B–H–B bridge in B₂H₆ is formed by the sharing of
(A) 2 electrons (B) 4 electrons (C) 1 electron (D) 3 electrons

5. Fluorine is more electronegative than both boron and phosphorus. What conclusion can be drawn from the fact that BF₃ has no dipole moment but PF₃ does?
(A) BF₃ is not spherically symmetrical
(B) BF₃ must be triangular planar
(C) BF₃ must be linear
(D) Atomic radius of P is larger than atomic radius of B

6. Identify the reagent I and II respectively in the following sequence:
[sequence]
(A) Acid, Al (B) Acid, C (C) Acid, Fe (D) Acid, Mg

7. The structure of quartz, mica and asbestos have same common units of
(A) (SiO₄)⁴⁻ (B) (SiO₃)²⁻ (C) (SiO₄)²⁻ (D) SiO₂

8. Antidote for CO poisoning is
(A) Carborundum (B) Pure CO₂ (C) Carbogen (D) Carbonyl chloride

9. R₃SiCl on hydrolysis forms
(A) R₃SiOH (B) R₃Si–O–SiR₃ (C) R₃Si=O (D) None of the above

10. The thermal stability order for group 14 halides is
(A) CX₄ > SiX₄ > GeX₄ > SnX₄
(B) SnX₄ > GeX₄ > SiX₄ > CX₄
(C) SiX₄ > CX₄ > GeX₄ > SnX₄
(D) None of the above

11. Molecular shapes of SF₄, CF₄ and XeF₄ are
(A) same with 2, 0 and 1 lone pair of electrons respectively
(B) same with 1, 1 and 1 lone pair of electrons respectively
(C) different with 0, 1 and 2 lone pair of electrons respectively
(D) different with 1, 0 and 2 lone pair of electrons respectively

12. The number of moles of water required to completely hydrolyze one mole of P₄O₁₀ is
(A) two (B) three (C) four (D) six

13. Ammonium dichromate is used in some fireworks. The green coloured powder blown during explosion is
(A) CrO₃ (B) Cr₂O₃ (C) Cr (D) CrO(O₂)

14. Antichlor is a compound
(A) which absorbs chlorine
(B) which removes chlorine from a material
(C) which liberates Cl₂ from bleaching powder
(D) which acts as a catalyst in the manufacture of chlorine

15. Which of the following is not a fertilizer?
(A) 4Ca(H₂PO₄)₂
(B) 3Ca(H₂PO₄)₂ + 7CaSO₄ + 2HF
(C) Ca(NO₃)₂ + NH₄NO₃
(D) CaF₂ + (NH₄)₃PO₄

ANSWERS TO ASSIGNMENT PROBLEMS

1. D 2. B 3. B 4. A 5. B 6. D
7. A 8. C 9. B 10. A 11. D 12. D
13. B 14. B 15. B

Frequently Asked Questions

When preparing p-block for NEET, the key is smart selection and repeated revision rather than trying to memorise everything at once.

Step 1: Focus on NCERT
Almost all NEET questions are directly from NCERT chemistry. Read the p-block chapters word-for-word, underlining exceptions, preparation methods, and uses.

Step 2: Group Prioritisation
For NEET, Groups 15, 16, and 17 are high-yield. Questions often focus on preparation, properties, and uses of compounds like HNO₃, H₂SO₄, PCl₅, and halogen oxoacids.

Step 3: Diagram & Table Learning
NCERT tables (like anomalous properties, trends) are question goldmines. Learn them visually—draw and label trends.

Step 4: Previous Years’ Papers
Solve last 10 years of NEET chemistry questions for p-block. This reveals patterns—often direct NCERT lines are used.

Step 5: Revision Cycles
Do 3–4 revisions before the exam. In the last week, only read marked/highlighted parts.

This method ensures you hit all NEET-specific points without drowning in unnecessary detail.

Period 1 has only hydrogen and helium, filling only the 1s orbital. The p-orbital starts from period 2, so there are no p-block elements in the first period.

Because the p-orbital has 3 sub-orbitals, each holding 2 electrons—maximum 6 electrons. This creates six columns in the periodic table’s p-block.

S.no	Formulas List
1.	Liquid Solutions
2.	Electrochemistry
3.	Chemical Kinetics
4.	Ores and Metallurge
5.	P-Block Elements
6.	D-Block and Co-ordination Compounds
7.	Chemistry of Heavy Metals
8.	Alkyl and Aryl Halides
9.	Alcohol phenol and ether
10.	Aldehyde and Ketone

P-Block Elements

Trends in Chemical Reactivity and Oxidation States

Chemical Behaviour

BORON

Properties of Boron

Reaction with air:

Reaction with metals:

Compounds of Boron

Borax or sodium tetraborate decahydrate (Na2B4O7.10H2O)

Boric acid or orthoboric acid (H3BO3) or B(OH)3:

Illustration 1: Borax structure contains

Illustration 2: The order of increasing acidic strength of BF3, BCl3 and BBr3 is

Illustration 3: The bonds present in borazole are

COMPOUNDS OF ALUMINIUM

Aluminium chloride (AlCl3):

Alums:

Illustration 4: Al2O3 formation involves large quantity of heat evolution, which makes its use in

GROUP 14 ELEMENTS

Trends in Chemical Reactivity

Silicates

Type of silicates:

CARBON

Compounds of Carbon

Oxides:

Carbon monoxide:

Carbon dioxide:

GROUP 15 ELEMENTS

Trends in Chemical Reactivity

NITROGEN

Compounds of Nitrogen

Ammonia:

Oxides of nitrogen

Chemical reactivity of various oxides can be summarized as follows:

Oxides of nitrogen

Preparations of above oxides are as follows:

Chemical reactivity of various oxides can be summarized as follows:

Oxyacids of nitrogen:

Reactions of nitric acid are summarized in the following table.

Oxides of phosphorus:

Exercise 2

GROUP 16 ELEMENTS

Trends in Chemical Reactivity and Oxidation States

OXYGEN

Ozone:

SULPHUR

Allotropes of Sulphur

Compounds of Sulphur

Sulphur Dioxide

Preparation

Reactions illustrating its dual behaviour

Sulphur trioxide

Sulphuric acid

Contact process (outline)

Properties

Sodium thiosulphate

Preparation

Reaction with acids

Illustrations (MCQs)

GROUP 17 ELEMENTS (HALOGENS)

Trends in Oxidation States and Chemical Reactivity

Other Comparative Properties

Oxyacids

Oxyacids of Chlorine

Illustrations (MCQs)

GROUP 17 ELEMENTS (HALOGENS)

Trends in Oxidation States and Chemical Reactivity

Other Comparative Properties

Oxyacids

Oxyacids of Chlorine

Interhalogen Compounds

Pseudo Halide Ions and Pseudo Halogen

Illustrations (MCQs)

Exercise 3

GROUP 18 ELEMENTS

Xenon Fluoride and Xenon Oxygen Compounds

Structure of xenon compounds

Separation of Rare Gases

Illustration 12 :

Exercise 4

ANSWER TO EXERCISE

Borax or sodium tetraborate decahydrate (Na₂B₄O₇.10H₂O)

Boric acid or orthoboric acid (H₃BO₃) or B(OH)₃:

Illustration 2: The order of increasing acidic strength of BF₃, BCl₃ and BBr₃ is

Aluminium chloride (AlCl₃):

Illustration 4: Al₂O₃ formation involves large quantity of heat evolution, which makes its use in