Maharashtra Board Class 11 Chemistry Chapter 9 Elements of Group 13 14 and 15 Solutions

Chemistry Class 11 Chapter 9 Exercise Solutions

1. Choose Correct Option.

Question A. Which of the following is not an allotrope of carbon ?
(a) buckyball
(b) diamond
(c) graphite
(d) emerald
Answer: (d) emerald
In simple words: Carbon has several allotropic forms like diamond, graphite, and buckyballs, which are different structural arrangements of carbon atoms. Emerald is a gemstone, not an allotrope of carbon.

🎯 Exam Tip: Understanding allotropes helps differentiate between elemental forms and compounds.

Question B. .......... is inorganic graphite.
(a) borax
(b) diborane
(c) boron nitride
(d) colemanite
Answer: (c) boron nitride
In simple words: Boron nitride has a layered structure similar to graphite, making it often referred to as "inorganic graphite."

🎯 Exam Tip: Recognizing specific names and their properties is key in inorganic chemistry.

Question C. Haber's process is used for preparation of ..........
(a) HNO\(_{3}\)
(b) NH\(_{3}\)
(c) NH\(_{2}\)COΝΗ\(_{2}\)
(d) NH\(_{4}\)OH
Answer: (b) NH\(_{3}\)
In simple words: The Haber process is an industrial method used to synthesize ammonia (NH\(_{3}\)) from nitrogen and hydrogen gases.

🎯 Exam Tip: Knowledge of industrial processes and their products is important for competitive exams.

Question D. Thallium shows different oxidation state because ..........
(a) of inert pair effect
(b) it is inner transition element
(c) it is metal
(d) of its high electronegativity
Answer: (a) of inert pair effect
In simple words: Thallium, being a heavy element in Group 13, exhibits the inert pair effect, where its outermost s-electrons are reluctant to participate in bonding, leading to multiple oxidation states.

🎯 Exam Tip: The inert pair effect is a crucial concept for understanding the chemistry of heavier p-block elements.

Question E. Which of the following shows most prominent inert pair effect ?
(a) C
(b) Si
(c) Ge
(d) Pb
Answer: (d) Pb
In simple words: The inert pair effect becomes more prominent as you go down a group, especially for heavier elements like Lead (Pb) in Group 14.

🎯 Exam Tip: The inert pair effect's trend within a group is a common evaluation point.

Question 2. Identify the group 14 element that best fits each of the following description.
(A) Non-metallic element
(B) Form the most acidic oxide
(C) They prefer +2 oxidation state.
(D) Forms strong \( \pi \) bonds.
Answer:
(i) Carbon (C)
(ii) Carbon
(iii) Tin (Sn) and lead (Pb)
(iv) Carbon
In simple words: Carbon is a non-metal and forms acidic oxides, and its small size allows it to form strong pi bonds. Tin and Lead, being heavier elements in Group 14, exhibit a more stable +2 oxidation state due to the inert pair effect.

🎯 Exam Tip: Understanding the trends in metallic character, acidity of oxides, preferred oxidation states, and pi bond formation across Group 14 is crucial.

3. Give Reasons.

Question A. Ga\(^{3+}\) salts are better reducing agent while Tl\(^{3+}\) salts are better oxidising agent.
Answer:
(i) Both gallium (Ga) and thallium (Tl) belong to group 13.
(ii) Ga is lighter element compared to thallium Tl. Therefore, its +3 oxidation state is stable. Thus, Ga\(^{+}\) loses two electrons and get oxidized to Ga\(^{3+}\). Hence, Ga\(^{+}\) salts are better reducing agent.
(iii) Thallium is a heavy element. Therefore, due to the inert pair effect, Tl forms stable compounds in +1 oxidation state. Thus, Tl\(^{3+}\) salts get easily reduced to Tl\(^{1+}\) by accepting two electrons. Hence, Tl\(^{3+}\) salts are better oxidizing agent.
[Note: This question is modified so as to apply the appropriate textual concept.]
In simple words: Due to the inert pair effect, the +1 oxidation state is more stable for heavier elements like Thallium (Tl), making Tl\(^{3+}\) an oxidizing agent. For lighter elements like Gallium (Ga), the +3 state is more stable, making Ga\(^{+}\) a reducing agent.

🎯 Exam Tip: The inert pair effect explains the varying stability of oxidation states for heavier p-block elements, influencing their redox properties.

Question B. PbCl\(_{4}\) is less stable than PbCl\(_{2}\)
Answer:
(i) Pb has electronic configuration [Xe] 4f\(^{14}\) 5d\(^{10}\) 6s\(^2\) 6p\(^2\).
(ii) Due to poor shielding of 6s\(^2\) electrons by inner d and f electrons, it is difficult to remove 6s\(^2\) electrons (inert pair).
(iii) Thus, due to inert pair effect, +2 oxidation state is more stable than +4 oxidation state.
Hence, PbCl\(_{4}\) is less stable than PbCl\(_{2}\).
In simple words: Lead (Pb) exhibits the inert pair effect, making its +2 oxidation state much more stable than its +4 oxidation state. This is why compounds like PbCl\(_{2}\) (Pb in +2 state) are more stable than PbCl\(_{4}\) (Pb in +4 state).

🎯 Exam Tip: The stability of oxidation states for heavier elements is directly linked to the inert pair effect, which is critical for understanding their chemical behavior.

4. Give The Formula Of A Compound In Which Carbon Exhibit An Oxidation State Of

Question A. +4
Answer: CCl\(_{4}\)
In simple words: In carbon tetrachloride (CCl\(_{4}\)), carbon forms four bonds with chlorine atoms, resulting in a +4 oxidation state.

🎯 Exam Tip: Calculating oxidation states is a fundamental skill in chemistry, especially for covalent compounds where electronegativity differences determine electron distribution.

Question B. +2
Answer: CO
In simple words: In carbon monoxide (CO), carbon forms a double bond and a coordinate bond with oxygen, resulting in a +2 oxidation state.

🎯 Exam Tip: Remember that oxidation states can vary for a given element, especially in compounds with different bonding arrangements, like carbon in CO vs CO\(_{2}\).

Question C. -4
Answer: CH\(_{4}\)
In simple words: In methane (CH\(_{4}\)), carbon is bonded to four hydrogen atoms, which are less electronegative, thus carbon exhibits a -4 oxidation state.

🎯 Exam Tip: The oxidation state of carbon often depends on the atoms it is bonded to, particularly hydrogen and oxygen.

5. Explain The Trend Of The Following In Group 13 Elements :

Question A. atomic radii
Answer:
(i) In group 13, on moving down the group, the atomic radii increases from B to Al.
(ii) However, there is an anomaly observed in the atomic radius of gallium due to the presence of 3d electrons. These inner 3d electrons offer poor shielding effect and thus, valence shell electrons of Ga experience greater nuclear attraction. As a result, atomic radius of gallium is less than that of aluminium.
(iii) However, the atomic radii again increases from Ga to Tl.
(iv) Therefore, the atomic radii of the group 13 elements varies in the following order:
\( B < Al > Ga < In < Tl \)
In simple words: Atomic radii generally increase down a group, but in Group 13, gallium is an exception because its inner 3d electrons provide poor shielding, causing its atomic radius to be smaller than aluminium despite being below it.

🎯 Exam Tip: The anomalous trend in atomic radii for Group 13 elements, especially the Al-Ga contraction, is a frequently tested concept related to shielding effects.

Question B. Ionization enthalpy:
Answer:
(i) Ionization enthalpies show irregular trend in the group 13 elements.
(ii) As we move down the group, effective nuclear charge decreases due to addition of new shells in the atom of the elements which leads to increased screening effect. Thus, it becomes easier to remove valence shell electrons and hence, ionization enthalpy decreases from B to Al as expected.
(iii) However, there is a marginal difference in the ionization enthalpy from Al to Tl.
(iv) The ionization enthalpy increases slightly for Ga but decreases from Ga to In.
(v) In case of Ga, there are 10 d-electrons in its inner electronic configuration which shield the nuclear charge less effectively than the s and p-electrons and therefore, the outer electron is held fairly strongly by the nucleus. As a result, the ionization enthalpy increases slightly.
(vi) Number of d electrons and extent of screening effect in indium is same as that in gallium. However, the atomic size increases from Ga to In. Due to this, the first ionization enthalpy of In decreases.
(vii) The last element Tl has 10 d-electrons and 14 f-electrons in its inner electronic configuration which exert still smaller shielding effect on the outer electrons. Consequently, its first ionization enthalpy increases considerably.
In simple words: Ionization enthalpy in Group 13 shows an irregular trend. While it generally decreases down the group, the poor shielding by d and f electrons in gallium and thallium causes an unexpected increase in their ionization enthalpies compared to the expected trend.

🎯 Exam Tip: The irregular trend in ionization enthalpy for Group 13 elements, particularly involving Ga and Tl, is a key concept influenced by d and f-block contraction and inert pair effect.

Question C. Electron affinity:
Answer:
(a) Electron affinity shows irregular trend. It first increases from B to Al and then decreases. The less electron affinity of boron is due to its smaller size. Adding an electron to the 2p orbital in boron leads to a greater repulsion than adding an electron to the larger 3p orbital of aluminium.
(b) From Al to Tl, electron affinity decreases. This is because, nuclear charge increases but simultaneously the number of shells in the atoms also increases. As a result, the effective nuclear charge decreases down the group resulting in increased atomic size and thus, it becomes difficult to add an electron to a larger atom. The electron affinity of Ga and In is same.
Note: Electron affinity of group 13 elements:

Element	B	Al	Ga	In	Tl
Electron affinity (kJ mol\(^{-1}\))	27	43	29	29	20

In simple words: Electron affinity in Group 13 initially increases from Boron to Aluminium due to Al's larger orbital size reducing electron-electron repulsion, then generally decreases for heavier elements as increasing atomic size makes it harder to attract an additional electron.

🎯 Exam Tip: Anomalies in electron affinity, particularly for small elements like boron, are often attributed to high electron density and repulsion in small orbitals.

6. Answer The Following

Question A. What is hybridization of Al in AlCl\(_{3}\)?
Answer: Al is sp\(^2\) hybridized in AlCl\(_{3}\).
In simple words: In AlCl\(_{3}\), the central aluminum atom forms three bonds with chlorine atoms, requiring three hybrid orbitals, which correspond to sp\(^2\) hybridization.

🎯 Exam Tip: Hybridization helps predict the geometry of molecules; sp\(^2\) hybridization implies a trigonal planar geometry.

Question B. Name a molecule having banana bond.
Answer: Diborane (B\(_{2}\)H\(_{6}\))
In simple words: Diborane contains special three-center-two-electron bonds, often called banana bonds, which are crucial for its structure and bonding.

🎯 Exam Tip: Banana bonds are a characteristic feature of electron-deficient compounds like boranes, important for understanding their unique bonding.

7. Draw The Structure Of The Following

Question A. Orthophosphoric acid
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह ऑर्थोफॉस्फोरिक एसिड (H\(_{3}\)PO\(_{4}\)) की संरचना है। इसमें एक केंद्रीय फॉस्फोरस परमाणु होता है जो एक ऑक्सीजन परमाणु के साथ दोहरा बंधन (P=O) बनाता है और तीन अन्य ऑक्सीजन परमाणुओं से एकल बंधन (P-OH) से जुड़ा होता है, जिससे कुल चार ऑक्सीजन परमाणु जुड़े होते हैं। तीन हाइड्रोजन परमाणु ऑक्सीजन परमाणुओं से जुड़े होते हैं।
In simple words: Orthophosphoric acid (H\(_{3}\)PO\(_{4}\)) has a central phosphorus atom double-bonded to one oxygen and single-bonded to three -OH groups, giving it a tetrahedral geometry.

🎯 Exam Tip: Knowing the structures of common oxyacids, including bond types and hybridization, is essential for understanding their chemical properties.

Question B. Resonance structure of nitric acid
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह नाइट्रिक एसिड (HNO\(_{3}\)) की अनुनादी संरचनाओं को दर्शाता है। केंद्रीय नाइट्रोजन परमाणु एक ऑक्सीजन परमाणु के साथ दोहरा बंधन और दूसरे ऑक्सीजन परमाणु के साथ एकल बंधन बनाता है, जो आगे एक हाइड्रोजन परमाणु से जुड़ा होता है। तीसरी ऑक्सीजन परमाणु नाइट्रोजन से एक सहसंयोजक बंधन के माध्यम से जुड़ी होती है, जिसमें नाइट्रोजन पर धनात्मक आवेश और सहसंयोजक ऑक्सीजन पर ऋणात्मक आवेश होता है। ये संरचनाएँ अनुनाद संकर को दर्शाती हैं जहाँ पाई इलेक्ट्रॉन और ऋण आवेश पूरी अणु में फैले होते हैं।
In simple words: Nitric acid (HNO\(_{3}\)) exhibits resonance, meaning its actual structure is a hybrid of several contributing Lewis structures, where the pi electrons are delocalized over the nitrogen and oxygen atoms.

🎯 Exam Tip: Resonance structures are crucial for accurately representing the bonding and stability of molecules like nitric acid, where electron delocalization occurs.

8. Find Out The Difference Between

Question A. Diamond and Graphite
Answer:
Diamond:
(1) It has a three-dimensional network structure.
(2) In diamond, each carbon atom is sp\(^3\) hybridized.
(3) Each carbon atom in diamond is linked to four other carbon atoms.
(4) Diamond is poor conductor of electricity due to absence of free electrons.
(5) Diamond is the hardest known natural substance.
Graphite:
(1) It has a two-dimensional hexagonal layered structure.
(2) In graphite, each carbon atom is sp\(^2\) hybridized.
(3) Each carbon atom in graphite is linked to three other carbon atoms.
(4) Graphite is good conductor of electricity due to presence of free electrons in its structure.
(5) Graphite is soft and slippery.
In simple words: Diamond and graphite are allotropes of carbon with vastly different properties due to their distinct atomic structures and hybridization; diamond is sp\(^3\) hybridized and forms a hard 3D network, while graphite is sp\(^2\) hybridized and forms soft 2D layers with delocalized electrons.

🎯 Exam Tip: Understanding the structural differences (hybridization, bonding) between allotropes is key to explaining their varied physical and chemical properties.

Question B. White phosphorus and Red phosphorus
Answer:
White phosphorus:
(1) It consists of discrete tetrahedral P\(_{4}\) molecules.
(2) It is less stable and more reactive.
(3) It exhibits chemiluminescence.
(4) It is poisonous.
Red phosphorus:
(1) It consists chains of P\(_{4}\) molecules linked together by covalent bonds.
(2) It is stable and less reactive.
(3) It does not exhibit chemiluminescence.
(4) It is nonpoisonous.
In simple words: White phosphorus exists as discrete P\(_{4}\) tetrahedra, making it highly reactive and poisonous, whereas red phosphorus is a polymeric form with chained P\(_{4}\) units, resulting in greater stability and non-toxicity.

🎯 Exam Tip: The structural differences between allotropes directly impact their stability, reactivity, and other physical properties, which is a frequent examination point.

9. What Are Silicones? Where Are They Used?

Question 9. What are silicones? Where are they used?
Answer:
(i) (a) Silicones are organosilicon polymers having R\(_{2}\)SiO (where, R = CH\(_{3}\) or C\(_{6}\)H\(_{5}\) group) as a repeating unit held together by siloxane (-Si-O-Si-) linkage.
(b) Since the empirical formula R\(_{2}\)SiO (where R = CH\(_{3}\) or C\(_{6}\)H\(_{5}\) group) is similar to that of ketones (R\(_{2}\)CO), these compounds are named as silicones.
(ii) Applications: They are used as
(a) insulating material for electrical appliances.
(b) water proofing of fabrics.
(c) sealant.
(d) high temperature lubricants.
(e) for mixing in paints and enamels to make them resistant to high temperature, sunlight and chemicals.
In simple words: Silicones are polymers containing repeating siloxane (Si-O-Si) units with organic groups attached to silicon, known for their excellent thermal stability, water repellency, and insulating properties, making them useful in many applications from sealants to lubricants.

🎯 Exam Tip: Understanding the basic structure (siloxane linkage, organic groups) and diverse applications of silicones is crucial for polymer chemistry questions.

10. Explain The Trend In Oxidation State Of Elements From Nitrogen To Bismuth.

Question 10. Explain the trend in oxidation state of elements from nitrogen to bismuth.
Answer:
(i) Group 15 elements have five valence electrons (ns\(^2\) np\(^3\)). Common oxidation states are -3, +3 and +5. The range of oxidation state is from -3 to +5.
(ii) Group 15 elements exhibit positive oxidation states such as +3 and +5. Due to inert pair effect, the stability of +5 oxidation state decreases and +3 oxidation state increases on moving down the group.
(iii) Group 15 elements show tendency to donate electron pairs in -3 oxidation state. This tendency is maximum for nitrogen.
(iv) The group 15 elements achieve +5 oxidation state only through covalent bonding.
(v) e. g. NH\(_{3}\), PH\(_{3}\), AsH\(_{3}\), SbH\(_{3}\), and BiH\(_{3}\) contain 3 covalent bonds. PCl\(_{5}\) and PF\(_{5}\) contain 5 covalent bonds.
In simple words: In Group 15, elements can show -3, +3, and +5 oxidation states. Down the group, due to the inert pair effect, the stability of the +5 oxidation state decreases while the +3 oxidation state becomes more stable, and the tendency for nitrogen to show -3 oxidation state is highest.

🎯 Exam Tip: The inert pair effect is a critical concept for explaining the varying stability of oxidation states for heavier elements in p-block groups, especially Group 15.

11. Give The Test That Is Used To Detect Borate Radical Is Qualitative Analysis.

Question 11. Give the test that is used to detect borate radical is qualitative analysis.
Answer:
(i) Borax when heated with ethyl alcohol and concentrated H\(_{2}\)SO\(_{4}\), produces volatile vapours of triethyl borate, which burn with green edged flame.
(a) \( Na_{2}B_{4}O_{7} + H_{2}SO_{4} + 5H_{2}O \xrightarrow{\Delta} Na_{2}SO_{4} + 4H_{3}BO_{3} \)
Borax (conc.) Sodium Orthoboric
sulphate acid

(b) \( H_{3}BO_{3} + 3C_{2}H_{5}OH \)
\( \implies \) \( B(OC_{2}H_{5})_{3} + 3H_{2}O \)
Orthoboric Ethyl Triethyl borate
acid alcohol
(ii) The above reaction is Used as a test for the detection and removal of borate radical (BO\(_{3}\)\(^{-}\)) in qualitative analysis.
In simple words: The borate radical can be detected by the green flame test, where borax reacts with sulfuric acid and ethyl alcohol to form volatile triethyl borate, which burns with a characteristic green flame.

🎯 Exam Tip: Flame tests are important qualitative analysis techniques; remember specific colors associated with different elements, like the green flame for borates.

12. Explain Structure And Bonding Of Diborane.

Question 12. Explain structure and bonding of diborane.
Answer:
(i) Electronic configuration of boron is 1s\(^2\) 2s\(^2\) 2p\(^1\). Thus, it has only three valence electrons.
(ii) In diborane, each boron atom is sp\(^3\) hybridized. Three of such hybrid orbitals are half filled while the fourth sp\(^3\) hybrid orbital remains vacant.
(iii) The two half-filled sp\(^3\) hybrid orbitals of each B atom overlap with 1s orbitals of two terminal H atoms and form four B – H covalent bonds. These bonds are also known as two-centred-two-electron (2c-2e) bonds.
(iv) When '1s' orbital of each of the remaining two H atoms simultaneously overlap with half-filled hybrid orbital of one B atom and the vacant hybrid orbital of the other B atom, it produces two three-centred-two- electron bonds (3c-2e) or banana bonds.
(v) Hydrogen atoms involved in (3c-2e) bonds are the bridging H atoms i.e., H atoms in two B-H-B bonds.
(vi) In diborane, two B atoms and four terminal H atoms lie in one plane, while the two bridging H atoms lie symmetrically above and below this plane.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह डाइबोरेन की बंधन और संरचना को दर्शाता है। इसमें दो बोरॉन परमाणु होते हैं जो चार टर्मिनल हाइड्रोजन परमाणुओं से सीधे जुड़े होते हैं, और दो ब्रिजिंग हाइड्रोजन परमाणुओं से जुड़े होते हैं जो दो बोरॉन परमाणुओं के बीच 3-केंद्र-2-इलेक्ट्रॉन "बनाना बांड" बनाते हैं। ब्रिजिंग बॉन्ड में 97° का कोण और 134 pm की दूरी होती है, जबकि टर्मिनल बॉन्ड में 120° का कोण और 119 pm की दूरी होती है।
In simple words: Diborane (B\(_{2}\)H\(_{6}\)) features sp\(^3\) hybridized boron atoms. It contains four terminal 2-center-2-electron B-H bonds and two unique 3-center-2-electron "banana bonds" where a hydrogen atom bridges two boron atoms.

🎯 Exam Tip: The concept of 3-center-2-electron bonds (banana bonds) is crucial for explaining the electron-deficient nature and structure of diborane, a common topic in boron chemistry.

13. A Compound Is Prepared From The Mineral Colemanite By Boiling It With A Solution Of Sodium Carbonate. It Is White Crystalline Solid And Used For Inorganic Qualitative Analysis.

Question 13. A compound is prepared from the mineral colemanite by boiling it with a solution of sodium carbonate. It is white crystalline solid and used for inorganic qualitative analysis.
(a) Name the compound produced.
(b) Write the reaction that explains its formation.
Answer:
(a) Borax
(b) Borax is obtained from its mineral colemanite by boiling it with a solution of sodium carbonate.
\( Ca_{2}B_{6}O_{11} + 2Na_{2}CO_{3} \xrightarrow{boil} Na_{2}B_{4}O_{7} + 2NaBO_{2} + 2CaCO_{3}\downarrow \)
Colemanite Sodium Borax Sodium Calcium
carbonate metaborate carbonate
In simple words: Borax, a white crystalline solid used in qualitative analysis, is industrially prepared by boiling colemanite mineral with a solution of sodium carbonate, producing borax along with sodium metaborate and calcium carbonate precipitate.

🎯 Exam Tip: Knowledge of mineral processing and industrial preparation of common compounds like borax, including the balanced chemical equations, is important.

14. Ammonia Is A Good Complexing Agent. Explain.

Question 14. Ammonia is a good complexing agent. Explain.
Answer:
(i) The lone pair of electrons on nitrogen atom facilitates complexation of ammonia with transition metal ions. Thus, ammonia is a good complexing agent as it forms complex by donating its lone pair of electrons.
(a) \( Cu^{2+}(aq) + 4NH_{3}(aq) \)
\( \implies \) \( [Cu(NH_{3})_{4}]^{2+}(aq) \)
(Blue) (Deep blue)

(b) \( AgCl(s) + 2NH_{3}(aq) \)
\( \implies \) \( [Ag(NH_{3})_{2}]Cl(aq) \)
(White) (Colourless)
(ii) This reaction is used for the detection of metal ions such as Cu\(^{2+}\) and Ag\(^{+}\).
In simple words: Ammonia acts as a good complexing agent because its nitrogen atom possesses a lone pair of electrons, which it can donate to transition metal ions to form stable coordination complexes, often resulting in characteristic color changes used in qualitative analysis.

🎯 Exam Tip: Ligands with available lone pairs, like ammonia, are common in complex formation; understanding this principle is fundamental to coordination chemistry and qualitative analysis.

15. State True Or False. Correct The False Statement.

Question A. The acidic nature of oxides of group 13 increases down the graph.
Answer: False
The acidic nature of oxides of group 13 decreases down the group. It changes from acidic through amphoteric to basic.
In simple words: In Group 13, as you move down, the metallic character increases, leading to a decrease in the acidic nature of their oxides, transitioning from acidic to amphoteric and then to basic.

🎯 Exam Tip: Remember the trend that acidic character of oxides decreases and basic character increases down a group as metallic character increases.

Question B. The tendency for catenation is much higher for C than for Si.
Answer: True
In simple words: Carbon has a much stronger ability to form long chains with itself (catenation) compared to silicon, due to stronger C-C bonds and optimal atomic size.

🎯 Exam Tip: Catenation is a key property of carbon; comparing its extent with other group members like silicon helps understand bond strengths and stability.

16. Match The Pairs From Column A And B.

Question 16. Match the pairs from column A and B.

	Column A		Column B
i.	BCl\(_{3}\)	a.	Angular molecule
ii.	SiO\(_{2}\)	b.	Linear covalent molecule
iii.	CO\(_{2}\)	c.	Tetrahedral molecule
		d.	Planar trigonal molecule

Answer:
(i) – (d)
(ii) – (c)
(iii) – (b)
In simple words: BCl\(_{3}\) has a planar trigonal geometry, SiO\(_{2}\) forms a tetrahedral network, and CO\(_{2}\) is a linear molecule.

🎯 Exam Tip: VSEPR theory is essential for predicting the shapes and geometries of molecules based on their central atom's electron pairs and bonds.

17. Give The Reactions Supporting Basic Nature Of Ammonia.

Question 17. Give the reactions supporting basic nature of ammonia.
Answer:
In the following reactions ammonia reacts with acids to form the corresponding ammonium salts which indicates basic nature of ammonia.
(i) \( NH_{3} + HCl \)
\( \implies \) \( NH_{4}Cl \)
Ammonia Hydrochloric Ammonium
acid chloride

(ii) \( 2NH_{3} + H_{2}SO_{4} \)
\( \implies \) \( (NH_{4})_{2}SO_{4} \)
Ammonia Sulphuric Ammonium
acid sulphate
In simple words: Ammonia acts as a base by accepting protons from acids (like HCl or H\(_{2}\)SO\(_{4}\)) to form ammonium salts, demonstrating its basic nature.

🎯 Exam Tip: The ability of ammonia to accept a proton (Brønsted-Lowry base) or donate a lone pair (Lewis base) is fundamental to its chemistry.

18. Shravani Was Performing Inorganic Qualitative Analysis Of A Salt. To An Aqueous Solution Of That Salt, She Added Silver Nitrate. When A White Precipitate Was Formed. On Adding Ammonium Hydroxide To This, She Obtained A Clear Solution. Comment On Her Observations And Write The Chemical Reactions Involved.

Question 18. Shravani was performing inorganic qualitative analysis of a salt. To an aqueous solution of that salt, she added silver nitrate. When a white precipitate was formed. On adding ammonium hydroxide to this, she obtained a clear solution. Comment on her observations and write the chemical reactions involved.
Answer:
(i) When silver nitrate (AgNO\(_{3}\)) is added to an aqueous solution of salt sodium chloride (NaCl), a white precipitate of silver chloride (AgCl) is formed.
\( NaCl(aq) + AgNO_{3}(aq) \)
\( \implies \) \( AgCl(s) + NaNO_{3}(aq) \)
Sodium Silver Silver Sodium
chloride nitrate chloride nitrate
(White ppt)
(ii) On adding ammonium hydroxide (NH\(_{4}\)OH) to this, the white precipitate of silver chloride gets dissolved and thus, a clear solution is obtained.
\( AgCl(s) + 2NH_{4}OH(aq) \)
\( \implies \) \( Ag(NH_{3})_{2}Cl(aq) + 2H_{2}O \)
White ppt Ammonium Diamine
hydroxide silver chloride
(Colourless)
In simple words: The formation of a white precipitate with silver nitrate suggests the presence of chloride ions, and its subsequent dissolution in ammonium hydroxide confirms the presence of silver ions, forming a soluble diamminesilver(I) complex.

🎯 Exam Tip: This is a classic qualitative analysis test for chloride and silver ions; remembering the observations and the complex formation is key.

11th Chemistry Digest Chapter 9 Elements Of Group 13, 14 And 15 Intext Questions And Answers

Can You Recall? (Textbook Page No. 123)

Question 1. If the valence shell electronic configuration of an element is 3s\(^2\) 3p\(^1\), in which block of the periodic table is it placed?
Answer: The element having valence shell electronic configuration 3s\(^2\) 3p\(^1\) must be placed in the p-block of the periodic table as its last electron enters in p-subshell (3p).
In simple words: An element with a valence shell configuration ending in p\(^1\) (like 3p\(^1\)) is always located in the p-block of the periodic table because its last electron occupies a p-orbital.

🎯 Exam Tip: The block of an element in the periodic table is determined by the subshell in which its last electron is filled.

Can You Recall? (Textbook Page No. 127)

Question 1. What is common between diamond and graphite?
Answer: Both diamond and graphite are made up of carbon atoms as they are two allotropes of carbon.
In simple words: The commonality between diamond and graphite is that they are both allotropes of carbon, meaning they are different structural forms of the same element, carbon.

🎯 Exam Tip: Allotropes are different physical forms of the same element; carbon's allotropes like diamond and graphite showcase distinct properties due to structural variations.

Can You Recall? (Textbook Page No. 129)

Question i. Which element from the following pairs has higher ionization enthalpy? B and Tl, N and Bi
Answer: Among B and Tl, boron has higher ionization enthalpy while, among N and Bi, nitrogen has higher ionization enthalpy.
In simple words: Generally, ionization enthalpy decreases down a group due to increased atomic size and shielding, so the lighter elements (Boron and Nitrogen) in these pairs have higher ionization enthalpies compared to their heavier counterparts (Thallium and Bismuth).

🎯 Exam Tip: Ionization enthalpy typically decreases down a group and increases across a period; exceptions often involve shielding effects or stable electronic configurations.

Question ii. Does boron form covalent compound or ionic?
Answer: Yes, boron forms covalent compound.
In simple words: Boron primarily forms covalent compounds because of its small size and high ionization energy, which makes it difficult to form ions.

🎯 Exam Tip: Elements with high ionization energies and small atomic radii tend to form covalent bonds rather than ionic bonds.

Try This. (Textbook Page No. 131)

Question 1. Find out the structural formulae of various oxyacids of phosphorus.
Answer:

No.	Oxyacids of Phosphorus	Structure
i.	Hypophosphorus (Phosphinic) acid (H\(_{3}\)PO\(_{2}\)): Phosphorus atom has +1 oxidation state. One P-OH bond, two P-H bonds, and one P=O bond are present.	ℹ️ चित्र व्याख्या (Diagram Explanation): यह हाइपोफॉस्फोरस एसिड (H\(_{3}\)PO\(_{2}\)) की संरचना है। इसमें एक केंद्रीय फॉस्फोरस परमाणु एक ऑक्सीजन परमाणु के साथ दोहरा बंधन (P=O), एक P-OH एकल बंधन, और दो P-H एकल बंधन बनाता है।
ii.	Orthophosphorus (Phosphonic) acid (H\(_{3}\)PO\(_{3}\)): Phosphorus atom has +3 oxidation state. Two P-OH bonds, one P-H bond and one P=O bond are present.	ℹ️ चित्र व्याख्या (Diagram Explanation): यह ऑर्थोफॉस्फोरस एसिड (H\(_{3}\)PO\(_{3}\)) की संरचना है। इसमें एक केंद्रीय फॉस्फोरस परमाणु एक ऑक्सीजन परमाणु के साथ दोहरा बंधन (P=O), दो P-OH एकल बंधन, और एक P-H एकल बंधन बनाता है।
iii.	Pyrophosphorus acid (H\(_{4}\)P\(_{2}\)O\(_{5}\)): Phosphorus atom has +3 oxidation state. Two P-OH bonds, two P-H and two P-O bonds are present. One P-O-P bond is also present.	ℹ️ चित्र व्याख्या (Diagram Explanation): यह पाइरोफॉस्फोरस एसिड (H\(_{4}\)P\(_{2}\)O\(_{5}\)) की संरचना है। इसमें दो फॉस्फोरस परमाणु एक P-O-P ब्रिज के माध्यम से जुड़े होते हैं। प्रत्येक फॉस्फोरस परमाणु एक P=O दोहरा बंधन, एक P-H एकल बंधन, और एक P-OH एकल बंधन बनाता है।
iv.	Hypophosphoric acid (H\(_{4}\)P\(_{2}\)O\(_{6}\)): Phosphorus atom has +4 oxidation state. Four P-OH bonds, two P=O bonds and one P-P bond are present.	ℹ️ चित्र व्याख्या (Diagram Explanation): यह हाइपोफॉस्फोरिक एसिड (H\(_{4}\)P\(_{2}\)O\(_{6}\)) की संरचना है। इसमें दो फॉस्फोरस परमाणु सीधे एक P-P बंधन से जुड़े होते हैं। प्रत्येक फॉस्फोरस परमाणु एक P=O दोहरा बंधन और दो P-OH एकल बंधन बनाता है।
v.	Orthophosphoric acid (H\(_{3}\)PO\(_{4}\)): Phosphorus atom has +5 oxidation state. Three P-OH bonds and one P=O bond are present.	ℹ️ चित्र व्याख्या (Diagram Explanation): यह ऑर्थोफॉस्फोरिक एसिड (H\(_{3}\)PO\(_{4}\)) की संरचना है। इसमें एक केंद्रीय फॉस्फोरस परमाणु एक ऑक्सीजन परमाणु के साथ दोहरा बंधन (P=O) और तीन P-OH एकल बंधन बनाता है।
vi.	Pyrophosphoric acid (H\(_{4}\)P\(_{2}\)O\(_{7}\)): Phosphorus atom has +5 oxidation state. Four P-OH bonds, two P=O bonds and one P-O-P bond are present.	ℹ️ चित्र व्याख्या (Diagram Explanation): यह पाइरोफॉस्फोरिक एसिड (H\(_{4}\)P\(_{2}\)O\(_{7}\)) की संरचना है। इसमें दो फॉस्फोरस परमाणु एक P-O-P ब्रिज के माध्यम से जुड़े होते हैं। प्रत्येक फॉस्फोरस परमाणु एक P=O दोहरा बंधन और दो P-OH एकल बंधन बनाता है।
vii.	Polymetaphosphoric acid (HPO\(_{3}\))\(_{n}\): Phosphorus atom has +5 oxidation state. For n = 3, it consists of three P-OH bonds, three P=O bonds and two P-O-P bonds.	ℹ️ चित्र व्याख्या (Diagram Explanation): यह पॉलीमेट्राफॉस्फोरिक एसिड (HPO\(_{3}\))\(_{n}\) की संरचना है, विशेष रूप से n=3 के लिए। यह एक चक्रीय संरचना दिखाता है जहाँ तीन फॉस्फोरस परमाणु ऑक्सीजन ब्रिज के माध्यम से जुड़े होते हैं। प्रत्येक फॉस्फोरस परमाणु एक P=O दोहरा बंधन और एक P-OH एकल बंधन बनाता है।

In simple words: Oxyacids of phosphorus vary in their structures and oxidation states, featuring P=O bonds, P-OH bonds, and sometimes P-H or P-O-P linkages, which determine their chemical properties.

🎯 Exam Tip: Familiarity with the structures, oxidation states, and common names of various oxyacids of phosphorus is crucial for understanding their reactivity and interconversions.

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