/ Preparation Chemistry NCERT Exemplar Solutions Class 11th Chapter Eleven
- The P-Block Elements Questions and Answers
- JEE Mains 2021
- NEET 2021
- 02 November 2025
The P-Block Elements Questions and Answers
| 1. Describe the general trends in the following properties of the elements in Groups 13 and 14. (i) Atomic size |
| Ans: Grp-13 The atomic size of the boron family follows the irregular trend. Generally, down the group the size increases but Gallium has a smaller atomic radius than Aluminium due to the poor shielding effect of 3d-orbitals. Order: B
Grp-14 The size of the carbon family is smaller than the modern family and as we move down the group the atomic size increases regularly. The increase in covalent radius from carbon to silicon is prominent while from Silicon to lead a small increase in covalent radius is observed; this is due to the presence of completely filled D and f-orbital in the heavier members. Order: C |
| 1. (ii) Ionisation enthalpy |
| Ionization enthalpy: In group 13, the standardised trend of decrease of Ionisation enthalpy is not monitored. As we move from Boron to Aluminium the atomic size increases and ionisation enthalpy decreases but when we move ahead from Aluminium to Gallium, the screening effect of 3d electrons comes into play. The poor shielding effect of the electrons lead to the increase in nuclear charge on the valence electrons and results in increase of ionisation enthalpy. Moving from gallium to Indium, due to The shielding effect of 4d electrons the ionisation enthalpy decreases. From Indium to thallium, 4f electrons come into action their poor shielding effect increases the effective nuclear charge on valence electrons hence the ionisation of Thallium energy further increases. The decreasing order of ionization enthalpy for group-13 elements will be: B>Ti>Ga>Al>In As we move down in group 14 the ionisation enthalpy decreases in the order: C>Si>Ge>Pb>Sn On moving down the group the size of the atom increases which results in the large decrease in ionisation energy from carbon to silicon but as we move from silicon to heavier metals small decrease in the ionisation energy is observed. This is mainly due to the less screening effect of d-electrons in germanium and tin, and due to fully filled d & f electrons in lead. |
| 1. (iii) Metallic character |
| Ans: Metallic character The metallic character in group 13 follows an unusual trend. Metallic character First increases from Boron to aluminium and then decreases from aluminium to thallium. This decrease is majorly due to the poor shielding effect of d and f electrons present in the heavier metals. In group-14, as we move down the group, the metallic character increases. |
| 1. (iv) Oxidation states |
| Ans: Possible oxidation states Group 13 elements are from +3 to -5. But the stability of +1 oxidation state increases down the group due to inert their effect than the stability of +3 oxidation state. Similarly, in group 14, The common oxidation states shown by the carbon family are +2 and +4. Due to the inert pair effect, stability of +2 oxidation state increases then the stability of +4. |
| 1. (v) Nature of halides |
| Ans: Except thallium, All the elements of the boron family combine with three atoms of halogens to form trihalides. All these halides exist as molecular species with hybridisation of sp2. All the trihalides are Lewis acid and their acidic strength of trihalides of boron is in order: Bf3
There are few exceptions also, SnF4 and Pf4 are ionic in nature, while the rest are covalent compounds. |
| 2. Account for the following observations: (i) AlCl3 is a Lewis acid |
| Ans: lnAlCl3, Aluminium is bonded to 3 chlorine atoms through three covalent bonds and six electrons are shared in the structure. To have complete Octet, AlCl3 lacks two electrons & hence act as an electron acceptor substance or Lewis acid. |
| 2. (ii) Though fluorine is more electronegative than chlorine yet BF3 is a weaker Lewis acid than BCl3 |
| Ans: Undoubtedly, fluorine has more electronegativity than chlorine, BF3 is stronger Lewis acid than BCl3 because of n−p π back bonding in BF3, both the constituting atoms boron and fluorine are involve p-orbital in back bonding. On moving down the group, the size of halogen atoms increases, back bonding decreases and Lewis acid character also decreases. |
| 2. (iii) PbO2 is a stronger oxidising agent than SnO2 |
| Ans: PbO2, lead is in +2 oxidation state and can be reduced to its most stable oxidation state of +2. Hence PbO2 can be further exidised or act as a strong oxidising agent. and in SnO, Sn is and +2 oxidation state and can extend its oxidation state up to +4. Thus, SnO can be used as a reducing agent. |
| 2. (iv) The +1 oxidation state of thallium is more stable than its +3 state. |
| Ans: As we move down the group in group 13 the participation of s-electrons in bond formation decreases the primary reason behind this is the inert pair effect. In this the p-electrons take part in bond formation and more energy is required to unpair the valence electrons to make them participate in bonding. Due to this the lower oxidation state of elements becomes stable than the higher oxidation state. As for thallium, +1 oxidation state is more stable than +3. |
| 3. When aqueous solution of borax is acidified with hydrochloric acid, a white crystalline solid is formed which is soapy to touch. Is this solid acidic or basic in nature? Explain. |
| Ans: When aqueous solution of borax is acidified with hydrochloric acid, boric acid is produced. As the name says, boric acid is acidic in nature but weak acid. Unlike protonic acid, boric acid is monobasic acid. It accepts electrons from the hydroxyl group of water and forms [B(OH)4]-. |
| 4. Three pairs of compounds are given below. Identify that compound in each of the pairs which has group 13 elements in a more stable oxidation state. Give reason for your choice. State the nature of bonding also. (i) TlCl3 ,TlCl |
| Ans: TlCl is more stable than TlCl3, due to inert pair effect, +1 oxile. oxidation state. |
| 4. (ii) AlCl3, AlCl |
| Ans: AlCl3, Al3+ is more stable than aluminum ions in +1 state. |
| 4. (iii) InCl3, InCl |
| Ans: Due to the inert pair effect, +1 oxidation state is more stable than the +3 oxidation state. So,InCl is more stable than InCl3. |
| (ii) What are boranes? Give chemical equation for the preparation of diborane. |
| Boranes correspond to alkane-like compounds of boron. They consist of boron and hydrogen. Most common borane existing is dibecane. 4BF3 + 3LiAlH4⟶2B2H6 + 3LiF + 3AIF3 |
Commonly asked questions
Explain the following :
(i) Gallium has higher ionisation enthalpy than aluminium (ii) Boron does not exist as B3+ ion (iii) Aluminium forms [AlF6] 3– ion but boron does not form [BF6] 3– ion (iv) PbX2 is more stable than PbX4. (v) Pb4+ acts as an oxidising agent but Sn2+ acts as a reducing agent. (vi) Electron gain enthalpy of chlorine is more negative as compared to fluorine (vii) Tl (NO3)3 acts as an oxidising agent. (viii) Carbon shows catenation property but lead does not. (ix) BF3 does not hydrolyse.
(x) Why does the element silicon, not form a graphite like structure whereas carbon does.
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(i) When moving down the periodic table within a group, ionization enthalpy generally decreases due to increased atomic radius. However, in the case of gallium (Ga) and aluminum (Al), gallium experiences a higher effective nuclear charge due to less shielding from its inner electrons, resulting in a higher ionization enthalpy compared to aluminum.
(ii) As boron is smaller in size and the sum of its first three ionization enthalpies i.e. ΔH1+ΔH2+ΔH3 is very large so boron does not allow to lose its all three valence electrons and exist as +3 ion rather it shares electrons to make covalent bonds.
(iii) As compared to boron, aluminium is larger in size and has vacant d-orbital through which it can expand its coordination number from +4 to +6. Can easily acquire the sp3d2 hybridisation State. On the other hand, boron is smaller in size with no d-orbitals and can expand its coordination number maximum up to +4 so boron can maximum the form BF4− but not BF62−.
(iv) Lead is extra stable in its +2 oxidation state than in +4 oxidation state due to inert pair effect.
(v) Due to the inert Pair effect, lead is more stable in its +2 oxidation state than in +4. So if it is present in +4 oxidation state in any compound it can readily reduce itself by gaining two electrons. Thus acting as an oxidising agent.
Reaction: Pb4++2e−→Pb2+
While for Sn2+, +4 oxidation state is extra stable than +2 oxidation state so in any compound where acid is present in +2 oxidation state it loses two more electron and oxidises itself to +4 oxidation state. Thus acting as a reducing agent.
Reaction: Sn2+→Sn4++2e−
(vi) Electronic configuration of F & Clare:
In fluorine the new electron added will be accommodated in 2p sub-shell while in chlorine the new electron goes to 3p subshell. The size of the 2p subshell is smaller than the size of 3p sub shell. So, the added electron in 2p sub shell experiences strong inter-electronic repulsion which is comparatively less in the case of 3p sub shell of chlorine. The added electron doesn't feel much nuclear attraction in fluorine, that's why fluorine doesn't easily gain new electrons. Therefore, fluorine has less negative electron gain enthalpy than chlorine.
(vii) When we move down the group in group 13 due to the inert pair effect for Thallium, +3 oxidation state is less stable than its +1 oxidation state. In Tl (NO3)3 the central atom Tl exist in +3 oxidation state which is not much stable so it readily gains two electrons to form compounds in which Thallium is in +1 oxidation state that is TINO3.
(viii) As we move down the group in group 14 size of element increases, electronegativity decreases which results in the decrease of M-M born strength hence the catenation decreases
(ix) Complete hydrolysis of BF3 does not occur, rather it gets partially hydrolysed. BF3 reacts with water to form boric acid and HF the HF further reacts with H3BO3 and prevents Hydrolysation of BF3.
(x) In group 14, as we move down the size of elements increases and electronegativity decreases. Silicon does not have sp2 state and does not form sp2 double bond like carbon does in graphite. Rather silicon exists in sp3 hybridisation state which forms 3-D structure more like diamond.
When aqueous solution of borax is acidified with hydrochloric acid, a white crystalline solid is formed which is soapy to touch. Is this solid acidic or basic in nature? Explain.
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When aqueous solution of borax is acidified with hydrochloric acid, boric acid is produced. As the name says, boric acid is acidic in nature but weak acid. Unlike protonic acid, boric acid is monobasic acid. It accepts electrons from the hydroxyl group of water and forms [B (OH)4]-.
Aluminium dissolves in mineral acids and aqueous alkalies and thus shows amphoteric character. A piece of aluminium foil is treated with dilute hydrochloric acid or dilute sodium hydroxide solution in a test tube and on bringing a burning matchstick near the mouth of the test tube, a pop sound indicates the evolution of hydrogen gas. The same activity when performed with concentrated nitric acid, reaction does not proceed. Explain the reason.
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The following reaction takes place in the test tube
- i) When Aluminium is treated with dilute hydrochloric acid.
Al + HCl -à AlCl3
ii) When Aluminium is treated with dilute sodium hydroxide solution.
Al + NaOH -àNaAlO2 + H2
In both the cases, the Hydrogen atoms are evolved. So, when a burning match stick is brought near the test tube we hear a pop sound.
Aluminium does not react with concentrated nitric acid. A "layer" of "Aluminium Oxide" is formed, when aluminium reacts with the nitric acid because "nitric acid" is an oxidising agent. Due to the "layer" of “Aluminium oxide”, nitric acid cannot react with the “inner aluminium metal.
Describe the general trends in the following properties of the elements in Groups 13 and 14.
(i) Atomic size (ii) Ionisation enthalpy (iii) Metallic character (iv) Oxidation states (v) Nature of halides
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(i) Grp-13 The atomic size of the boron family follows the irregular trend. Generally, down the group the size increases but Gallium has a smaller atomic radius than Aluminium due to the poor shielding effect of 3d-orbitals.
Order: B
Grp-14 The size of the carbon family is smaller than the modern family and as we move down the group the atomic size increases regularly. The increase in covalent radius from carbon to silicon is prominent while from Silicon to lead a small increase in covalent radius is observed; this is due to the presence of completely filled D and f-orbital in the heavier members.
Order: C
(ii) Ionization enthalpy: In group 13, the standardised trend of decrease of Ionisation enthalpy is not monitored.
As we move from Boron to Aluminium the atomic size increases and ionisation enthalpy decreases but when we move ahead from Aluminium to Gallium, the screening effect of 3d electrons comes into play. The poor shielding effect of the electrons lead to the increase in nuclear charge on the valence electrons and results in increase of ionisation enthalpy.
Moving from gallium to Indium, due to The shielding effect of 4d electrons the ionisation enthalpy decreases.
From Indium to thallium, 4f electrons come into action their poor shielding effect increases the effective nuclear charge on valence electrons hence the ionisation of Thallium energy further increases.
The decreasing order of ionization enthalpy for group-13 elements will be: B>Ti>Ga>Al>In
As we move down in group 14 the ionisation enthalpy decreases in the order: C>Si>Ge>Pb>Sn
On moving down the group the size of the atom increases which results in the large decrease in ionisation energy from carbon to silicon but as we move from silicon to heavier metals small decrease in the ionisation energy is observed. This is mainly due to the less screening effect of d-electrons in germanium and tin, and due to fully filled d & f electrons in lead.
(iii) Metallic character
The metallic character in group 13 follows an unusual trend. Metallic character First increases from Boron to aluminium and then decreases from aluminium to thallium. This decrease is majorly due to the poor shielding effect of d and f electrons present in the heavier metals.
In group-14, as we move down the group, the metallic character increases.
(iv) Possible oxidation states Group 13 elements are from +3 to -5. But the stability of +1 oxidation state increases down the group due to inert their effect than the stability of +3 oxidation state.
Similarly, in group 14, The common oxidation states shown by the carbon family are +2 and +4. Due to the inert pair effect, stability of +2 oxidation state increases then the stability of +4.
(v) Except thallium, All the elements of the boron family combine with three atoms of halogens to form trihalides. All these halides exist as molecular species with hybridisation of sp2. All the trihalides are Lewis acid and their acidic strength of trihalides of boron is in order: Bf3
There are few exceptions also, SnF4 and Pf4 are ionic in nature, while the rest are covalent compounds.
Account for the following observations:
(i) AlCl3 is a Lewis acid (ii) Though fluorine is more electronegative than chlorine yet BF3 is a weaker Lewis acid than BCl3
(iii) PbO2 is a stronger oxidising agent than SnO2 (iv) The +1 oxidation state of thallium is more stable than its +3 state
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(i) ln? AlCl3, Aluminium is bonded to 3 chlorine atoms through three covalent bonds and six electrons are shared in the structure. To have complete Octet, AlCl3 lacks two electrons & hence act as an electron acceptor substance or Lewis acid.
(ii) Undoubtedly, fluorine has more electronegativity than chlorine, BF3 is stronger Lewis acid than BCl3 because of n−p π back bonding in BF3, both the constituting atoms boron and fluorine are involve p-orbital in back bonding. On moving down the group, the size of halogen atoms increases, back bonding decreases and Lewis acid character also decreases.
(iii) PbO2, lead is in +2 oxidation state and can be reduced to its most stable oxidation state of +2. Hence PbO2 can be further exidised or act as a strong oxidising agent. and in SnO, Sn is and +2 oxidation state and can extend its oxidation state up to +4. Thus, SnO can be used as a reducing agent.
(iv) As we move down the group in group 13 the participation of s-electrons in bond formation decreases the primary reason behind this is the inert pair effect. In this the p-electrons take part in bond formation and more energy is required to unpair the valence electrons to make them participate in bonding. Due to this the lower oxidation state of elements becomes stable than the higher oxidation state. As for thallium, +1 oxidation state is more stable than +3.
Three pairs of compounds are given below. Identify that compound in each of the pairs which has group 13 elements in a more stable oxidation state. Give reason for your choice. State the nature of bonding also.
(i) TlCl3 ,TlCl (ii) AlCl3, AlCl (iii) InCl3, InCl (iv) What are boranes? Give chemical equation for the preparation of diborane.
(i) TlCl is more stable than TlCl3, due to inert pair effect, +1 oxile. oxidation state.
(ii) AlCl3, Al3+ is more stable than aluminum ions in +1 state.
(iii) Due to the inert pair effect, +1 oxidation state is more stable than the +3 oxidation state. So, InCl is more stable than InCl3.
(iv) Boranes correspond to alkane-like compounds of boron. They consist of boron and hydrogen. Most common borane existing is dibecane.
4BF3 + 3LiAlH4?2B2H6 + 3LiF + 3AIF3
BCl3 exists as monomer whereas AlCl3 is gimerised through halogen bridging. Give reason. Explain the structure of the dimer of AlCl3 also.
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Both the compounds, BCl3 and AlCl3 are electron deficient compounds. In BCl3, boron is smaller in size and cannot assemble four big chlorine atoms near it causing steric hindrance and making it unstable.
Hence, BCl3 exists as a monomer only.
In AlCl3, aluminum has 3p-orbitals through which chlorine atoms can be accommodated easily to complete its octet and dimer is formed.
Boron fluoride exists as BF3 but boron hydride doesn't exist as BH3. Give reason. In which form does it exist? Explain its structure.
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In BF3, due to n−pπ back bonding between the vacant p-orbital of boron and filled p-orbital of fluorine. This Π− pπ back bonding is absent in case of hydrogen as it is a single electron element.
Two BH3 molecules dimerise to form diborane.
In B2H6 There are two types of hydrogens present.
(I) Four hydrogens that are terminally bonded to each of two boron atoms.
(II) Two hydrogens that are bonded to both boron atoms forming a bridge in between.
The four terminal hydrogen atoms and two boron atoms lie in the same plane while bridging hydrogen lies in a plane perpendicular to them.
Two hydrogens forming a bridge in B2H6 are peculiar in bonding and can be termed as 3 -centered-2-electron bond or banana bond. 1sorbital of each hydrogen overlaps with the hybrid orbital of one of the boron then delocalising the 2e−over three atoms making 3 -centered- 2 -electron bond.
(i) What are silicones? State the uses of silicones. (ii) What are boranes? Give chemical equation for the preparation of diborane.
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(i) Organosilicon polymers having (R2SiO2) as monomer units are called silicones. Silicones contain organic side groups which surround it giving alkane like nature and makes it hydrophobic. Silicones are applicative in electrical insulators, water proofing, sealant. They have been utilized in the biological field in cosmetic implants and other surgeries.
(ii) Boranes correspond to alkane-like compounds of boron. They consist of boron and hydrogen. Most common borane existing is dibecane.
4BF3 + 3LiAlH4?2B2H6 + 3LiF + 3AIF3
A compound (A) of boron reacts with NMe3 to give an adduct (B) which on hydrolysis gives a compound (C) and hydrogen gas. Compound (C) is an acid. Identify the compounds A, B and C. Give the reactions involved.
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B2H6 + 2NMe3→ 2BH3NMe3 BH3NMe3+H2O → H3BO3 + NMe3 + 6H2
(A) (B) (C)
(A): B2H6
(B): 2BH3NMe3
(C): H3BO3
A non-metallic element of group 13, used in making bullet proof vests is an extremely hard solid of black color, It can exist in many allotropic forms and has unusually high melting points. Its trifluoride acts as Lewis acid towards ammonia. The element exhibits a maximum covalency of four. Identify the element and write the reaction of its trifluoride with ammonia. Explain why the trifluoride acts as a Lewis acid
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Boron is the only non-metallic and extremely hard element in group 13, and it is also used to make bulletproof vests. Boron exists in a variety of allotropic forms. It usually has a high melting point and no d orbital. Using 2s and 2p orbitals, it can achieve a maximum covalency of 4 . Because the octet of boron is not completed in trivalent halides of boron, it acts as Lewis acid. It forms an adduct when it reacts with Lewis base.
BF3+NH3→H3N−B−F3
A tetravalent element forms monoxide and dioxide with oxygen. When air is passed over a heated element (1273 K), producer gas is obtained. Monoxide of the element is a powerful reducing agent and reduces ferric oxide to iron. Identify the element and write formulas of its monoxide and dioxide. Write chemical equations for the formation of producer gas and reduction of ferric oxide with the monoxide
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2C (s) + O2 + 4N2 (g) 
2CO (g) + 4N2 ( g)
Fe2O3 (S)+3CO (g) 
2Fe (S)+3CO2
C= tetravalent carbon
CO= carbon monoxide
Fe2O3= ferric oxide
CO2= carbon dioxide
Tetravalent elements i.e. carbon combines with oxygen to produce carbon monoxide. The reaction occurs at high temperatures and in the presence of nitrogen gas, which acts as a producer gas only. It is not consumed in the reaction. The carbon monoxide formed acts as a reducing agent for ferric oxide, reduces the oxidation state of iron from +3 to zero, and oxidizes itself to form carbon dioxide.
Draw the structures of BCl3.NH3 and AlCl3 (dimer).
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The central B atom in BCl3 has six electrons in its valence shell. As a result, it is an electron-deficient molecule in need of two more electrons to complete its octet. To put it another way, BCl3 acts as a Lewis acid. NH3 on the other hand, has a lone pair of electrons that it can easily donate. As a result, NH3 serves as a Lewis base. As shown below, the Lewis acid (BCl3) and Lewis base (NH3) combine to form an adduct:
The valence shell of AlCl3 contains six electrons. As a result, it is an electron-deficient molecule that requires two additional electrons to complete its octet.
Chlorine, on the other hand, has three lone electron pairs. As a result, to complete its octet, the central Al atom of one molecule accepts a lone pair of electrons from the Cl atom of the other molecule, resulting in the formation of a dimeric structure, as shown below.
Explain the nature of boric acid as a Lewis acid in water
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H3BO3 or B (OH)3 is an electron deficient compound or Lewis acid which easily accepts electron from OH of water and releases its proton hence it is a Monobasic acid.
Reaction:
B (OH)3 + 2H2O → [B (OH)4]− + H3O+
Draw the structure of boric acid showing hydrogen bonding. Which species is present in water? What is the hybridisation of boron in this species?
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H3BO3
Boric acid forms a hexagon and rings through hydrogen bonding and has a layer-like structure.
Boric acid is present in water as [B (OH)4]−.H3BO3 electron from the OH of water and forms the complex BOH for negative for sp3 and is present in sp3 hybridisation.
Reaction:
B (OH)3+2H2O→ [B (OH)4]− + H3O+
Explain why the following compounds behave as Lewis acids?
(i) BCl3
(ii) AlCl3
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Both BCl3 and AlCl3 are electron deficient compounds that are central atom boron and aluminium have incomplete Octet. In each compound, a metal atom is surrounded by six electrons of three covalent bonds with 3 chlorine atoms.
Each chlorine atom has a complete Octet of eight electrons. The electron deficient compounds act as Lewis acid and readily accept two electrons to complete their octet.
Give reasons for the following:
(A) CCl4 is immiscible in water, whereas SiCl4 is easily hydrolysed.
(B) Carbon has a strong tendency for catenation compared to silicon.
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(A) sarbon in CCl4 does not have a vacant d-orbital to accommodate the electrons from OH of water molecules. Also CCl4 is nonpolar covalent compounds whereas H2O is polar. So, no strong interaction occurs between them. Hence CCl4 is miscible in water.
Whereas in SiCl4, silicon has bigger size than carbon and have d-orbitals for accommodation of electrons donated by OH of water in the process of hydroxylation. This leads to a strong interaction and silicon acid Is formed as a product. SiCl4 is completely miscible in water.
(B) As we move from carbon to silicon, the size increases and electronegativity decreases. This results in the decrease in M-M bond energy. Hence the catenation property decreases.
Bond energy is the amount of energy required to fragment the atoms combined in a molecule are born into separate atoms. Carbon has maximum bond energy as compared to silicon. Therefore, shows maximum catenation property.
Explain the following:
(A) CO2 is a gas whereas SiO2 is a solid.
(B) Silicon forms SiF62− ion whereas corresponding fluoro-compound of carbon is not known.
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(A) Carbon has small size and large electronegativity, it forms strong n−pπ bonding with two oxygen atoms forming a separate CO2 molecule.
While in SiO2 silicon is larger in size with comparatively less electronegativity than carbon it shows no tendency to form n−pπ bonding rather forms Single covalent bond with oxygen. Thus, SiO2 possess 3D network-like structure in which each Silicon is bonded to 4 oxygen atoms.
(B) Carbon is smaller in size and lacks d-orbitals hence can have a maximum coordination number of four and sp3 hybridisation only.
Whereas Silicon has d-orbital, it can amplify its hybridisation up to sp3d2 (CN=6) and can complete its Octet. Therefore, Silicon forms SiF62−.
whereas the corresponding flugre-compound of carbon is not known.
The +1 oxidation state in group 13 and +2 oxidation state in group 14 becomes more and more stable with increasing atomic number. Explain.
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As we move down the group in group 13 and 14 the participation of s-electrons in bond formation decreases the primary reason behind this is the inert pair effect.
In this the p-electrons take part in bond formation and more energy is required to unpack the valence electrons to make them participate in bonding. Due to this the lower oxidation state of elements becomes stable done the hire oxidation state. As for group 13, +1 oxidation state is more stable than +3 and for group 14, +2 oxidation state is more stable than +4.
Carbon and silicon both belong to the group 14, but in spite of the stoichiometric similarity, the dioxides (i.e., carbon dioxide and silicon dioxide) differ in their structures. Comment.
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As compared to carbon, silicon is bigger in size and is less electronegative. It shows resistance in forming p−p multiple bonding which is easily done by carbon. Thus, SiO2 is a 3−D network where each silicon is linked covalently to 4 oxygen atoms while in CO2, Carbon is linked with two oxygen atoms with double bond in a linear manner.
If a trivalent atom replaces a few silicon atoms in three dimensional network of silicon dioxide, what would be the type of charge on overall structure?
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When a trivalent atom is added to the crystal of SiO2, it substitutes silicon atoms which result in generation of holes. These holes make the crystal conductor of electricity. The overall Crystal is electrically neutral and is called the p-type conductor.
When BCl3 is treated with water, it hydrolyses and forms [B(OH)4]−only whereas AlCl3 in acidified aqueous solution forms [Al(H20)6]3+
Explain what is the hybridisation of boron and aluminium in these species?
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BCl3+3H2O→B (OH)3+3HCl
B (OH)3+2H2O→ [B (OH)4]−+H3O+
B (OH)3 due to its incomplete octet accepts an electron pair (OH)- to give has configucation [B (OH)4]−.
Boron in this ion involves one 2s orbital and three 2p orbitals. Thus, hybridization of B in [B (OH)4]− is sp3
AlCl3 + 6H2O [Al (H2O)6]3+ +3Cl-
Hence, hybridization of Al is sp3d2
Identify the compounds A, X and Z in the following reactions :
i) A + 2HCl + 5H2O → 2NaCl + X i )X 
HBO2 
Z
Kindly go through the resolution
i) 
ii) 
Complete the following chemical equations:
Z + 3LiAlH4 → X + 3 LiF 3AlF3
X + 6H2O → Y + 6H2
3X + 3O2
B2O3 + 3H2O
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4BF3 + 3LiAlH4? 2 B2H6 + 3LiF + 3AlF3
(Z) (X)
B2H6 + 6H2O?2H3BO3 + 6H2
(X) (Y)
B2H6 + 3O2? ΔB2O3 + 3H2O
(X)
The element which exists in liquid state for a wide range of temperature and can be used for measuring high temperature is
(A) B
(B) Al
(C) Ga
(D) In
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C
Gallium generally exists as solid at room temperature but melts on slight heating (melting point: 20? C ).
Whereas the boiling point of Gallium is very high around 2400? C. Gallium has large cohesive forces that hold its structure together and it is stable for a wide range of temperatures and can be used for measuring high temperatures.
Which of the following is a Lewis acid?
(A) AlCl3
(B) MgCl2
(C) CaCl2
(D) BaCl2
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A
Lewis acids are the species in which the state is not complete and ready to accept electrons. Because Al is surrounded by 6 electrons in AlCl3 and all three Cl atoms are surrounded by 8 electrons, AlCl3 is an electron acceptor. It is a covalent compound.
The geometry of a complex species can be understood from the knowledge of type of hybridisation
of orbitals of central atom. The hybridisation, of orbitals of central atom in [B(OH)4]− and the geometry of the complex are respectively
(A) sp3, tetrahedral
(B) sp3, square planar
(C) sp3 d2, octahedral
(D) dsp2, square planar
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A
Central atom in [B (OH)4 ]− no lone pair. Electronic conf In ground state: 1s2 2s2 2px1 2py0 2pz0
In excited state, One Electron from 2s shifts to 2py orbital and the configuration becomes:
1s2 2s2 2p1 x2 p1 y2 pz0
Now, one s and three p orbitals combined to give sp3 hybridisation and tetrahedral shape.
Which of the following oxides is acidic in nature?
(A) B2O3
(B) Al2O3
(C) Ga2O3
(D) In2O3
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A
The B2O3 reacts with water to form boric acid hence it is an acidic oxide. As we move down the group electronegativity decreases and the tendency to donate electrons increases therefore basic character increases or acidic character decreases. So, from the given options, the most acidic oxide is B2O3.
The exhibition of highest coordination number depends on the availability of vacant orbitals in the central atom. Which of the following elements is not likely to act as central atom in MF63− ?
(A) B
(B) AI
(C) Ga
(D) In
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A
Number of items/molecules bonded to the central atom is termed as coordination number. In the given MF63−
Coordination number of metal is six and boron can have topmost coordination number of four as it consists of s and p orbital only and lacks d-orbital. Therefore, boron cannot be compared in the form of MF63−.
Boric acid is an acid because its molecule
(A) Contains replaceable H+ ion
(B) Gives up a proton.
(C) Accepts OH−from water releasing proton.
(D) Combines with proton from water molecule.
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C
Boric acid is Lewis acid, having six electrons in its valence shell. It combines with water, accepts electrons
from OH−of water molecule and complete it octet to 8 and releases H+
Reaction: B (OH)3 + OH − H→ [B (OH)4]− + H+
Catenation, i.e. linking of similar atoms depends on size and electronic configuration of atoms. The tendency of catenation in Group 14 elements follows the order
(A) C>Si>Ge>Sn
(B) C>>Si>Ge≈Sn
(C) Si>C>Sn>Ge
(D) Ge>Sn>Si>C
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B
In group 14, as we move down from C to Sn, the size increases and electronegativity decreases. This results in the decrease in M-M bond energy. Hence the catenation property decreases.
Bond energy is the amount of energy required to fragment the atoms combined in a molecule are born into separate atoms. Carbon has maximum bond energy as compared to its other group elements. Therefore, shows maximum catenation property.
Silicon has a strong tendency to form polymers like silicones. The chain length of silicone polymer can be controlled by adding
(A) MeSiCl3
(B) Me2SiCl2
(C) Me3SiCl
(D) Me4Si
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C
Me3SiCl is a soleuless liquid That has a major application in the formation of silicones. On its addition in the process, it blocks the end and controls the chain length of the polymer.
Ionisation enthalpy (ΔtH1 kJ mol−1) for the elements of Group 13 follows the order.
(A) B>Al>Ga>In>Tl
(B) B
(C) BGa
(D) B>A1In
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D
In group 13, the standardised trend of decrease of ionization enthalpy is not monitored.
As we move from Boron to Aluminium the atomic size increases and jonjation enthalpy decreases but when we move ahead from Aluminum to Gallium, the screening effect of 3 d electrons comes into play. The poor shielding effect of the electrons lead to the increase in nuclear charge on the valence electrons and results in increase of jopisation enthalpy.
Moving from gallium to Indium, due to the shielding effect of 4 d electrons the ionization enthalpy decreases.
The decreasing order of ionization enthalpy for group-13 elements will be:
B>Tl>Ga>Al>In
From Indium to thallium, 4f electrons come into action their poor shielding effect increases the effective nuclear charge on valence electrons hence the jonjation of Thallium energy further increases.
In the structure of Diborane
(A) aAll hydrogen atoms lie in one plane and boron atoms lie in a plane perpendicular to this plane.
(B) 2 Boron atoms and 4 terminal hydrogen atoms lie in the same plane and 2 bridging hydrogen atoms lie in the perpendicular plane.
(C) 4 Bridging hydrogen atoms and boron atoms lie in one plane and two terminal hydrogen atoms lie in a plane perpendicular to this plane.
(D) All the atoms are in the same plane.
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B
The four terminal hydrogen atoms and two boron atoms are all in the same plane.
There are two bridging hydrogen atoms above and below this plane.
The four terminal B-H bonds are regular two-centre-two-electron bonds, whereas the two bridge (B-H-B) bonds are unique and can be described as three-centre-two-electron bonds, as shown in figure:
A compound X, of boron reacts with NH3 on heating to give another compound Y which is called inorganic benzene. The compound X can be prepared by treating BF3 with lithium aluminium hydride. The compounds X and Y are represented by the formulas.
(A) B2H6, B3 N3H6
(B) B203, B3 N3H6
(C) BF3, B3 N3H6
(D) B3 N3H6, B2H6
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A
A compound X, of boron, reacts with NH3 on heating to give another compound Y which is called inorganic benzene.
3H2H6 x + 6NH3 → 3 [BH2 (NH3)2]+ [BH4]− 2 B3N3H6 + 12H2
x ( (Diborane) Y (Borazole Inorganic Benzene)
4BF3 + 3LiAlH4 → 2 B2H6 + 3LiF + 3AlF3
Quartz is extensively used as a piezoelectric material, it contains
(A) Pb
(B) Si
(C) Ti
(D) Sn
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B
Quartz is a crystalline form of silica that can be converted into other crystalline forms at high temperatures. It's commonly used as a piezoelectric material.
The most commonly used reducing agent is
(A) AlCl3
(B) PbCl2
(C) SnCl4
(D) SnCl2
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D
Reducing agent are compound that donates electron to an electron recipient compound. Reducing agents oxidize themselves and reduce the other compounds. They easily lose electrons and increase their oxidation state; those electrons are accepted by electron needed compounds.
In SnCl2, Sn is in +2 oxidation state and can easily lose its two electrons and oxidize to +4 stable oxidation state.
Dry ice is
(A) Solid NH3
(B) Solid SO2
(C) Solid CO2
(D) Solid N 2
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C
Explanation:
Solid CO2 is referred to as dry ice because it is used in laboratories to create an ice bath for organic reactions. It is made by rapidly cooling high-pressure CO2 gas.
Cement, the important building material is a mixture of oxides of several elements. Besides calcium, iron and sulfur, oxides of elements of which of the group(s) are present in the mixture?
(A) Group 2
(B) Groups 2,13 and 14
(C) Groups 2 and 13
(D) Groups 2 and 14
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B
Cement is made by combining lime ( CaO ), clay with silica ( SiO), and oxides of aluminum, magnesium, and iron.
In the following questions two or more options may be correct.
16. The reason for small radius of Ga compared to Al is
(A) Poor screening effect of d and f orbitals
(B) Increase in nuclear charge
(C) Presence of higher orbitals
(D) Higher atomic numbeR
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(A) & (B)
The atomic radii decrease as one moves down the group from Al to Ga due to the shielding effect of electrons. As a result of this ineffective effect, the effective nuclear charge rises.
The linear shape of CO2 is due to
(A) sp3 hybridisation of carbon
(B) sp hybridisation of carbon
(C) n- p π bonding between carbon and oxygen
(D) sp2 hybridisation of carbon
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(B) & (C)
Hybridisation of any atom in a molecule corresponds to the number or σ bonds it is making covalently with other atoms.
In CO2, carbon is attached to two oxygen atoms through double bonds. Out of which two are σ bonds and the rest two are π bonds forming I−p π bonding between carbon and oxygen.
The two σ bonds correspond to the hybridization sp and give a linear shape to CO2.
Me3SiCl is used during polvmerisation of grgane silicones because
(A) The chain length of orange silicone polymers can be controlled by adding Me3
(B) Me3SiCl blocks, the end terminal of silicone polymer.
(C) Me3SiCl improves the quality and yield of the polymer.
(D) Me3SiCl acts as a catalyst during polymerization.
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(A) & (B)
Me3SiCl is a soleuless liquid and is stable in anhydrous condition. When added in the process of polymerization, it blocks the end terminal of the silicone polymer and determines its chain length.
Which of the following statements are correct?
(A) Fullerenes have dangling bonds.
(B) Fullerenes are cage-like molecules.
(C) Graphite is thermodynamically most stable allotrope of carbon.
(D) Graphite is slippery and hard and therefore used as a dry lubricant in machines.
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(B) & (C)
(B) Fullerene is an allotrope of carbon consisting of 60 carbon atoms bonded together through single and double bonds in such a manner that they form a hollow sphere like shape or a cage-like structure.
(C) In graphite, each carbon atom has one free/ delocalised electron that causes strong attraction between carbon atoms providing more stability to the overall structure of graphite
Which of the following statements are correct? Answer on the basis of figure.
(A) The two bridged hydrogen atoms and the two boron atoms lie in one plane.
(B) Out of six B - H, bonds two bonds can be described in terms of 3 centre 2-electron bonds.
(C) Out of six B- H bonds four B- H bonds can be described in terms of 3 centre 2 electron bonds.
(D) The four terminal B - H bonds are two centre-two electron regular bonds.This is a Multiple Choice Questions as classified in NCERT Exemplar
(A), (B) & (D)
Two hydrogens forming bridge in B2H6 are peculiar in bonding and can be termed as 3 -centered-2-electron bond or banana bond. 1 s orbital of each hydrogen overlaps with the hybrid orbital of one of the boron then delocalising the 2e−over three atoms making 3 -centered-2-electron bond.
Identify the correct resonance structures of carbon dioxide from the ones given below:
(A) O−C≡O
(B) O=C=O
(C) −O≡C−O+
(D) −O−C≡O+
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(B) & (D)
The following is the resonance structure of CO2.
Match the species given in Column I with the properties mentioned in Column II.
|
Column I |
Column II |
|
(i) BF4− |
(a) Oxidation state of central atom is +4 |
|
(ii) AlCl3 |
(b) Strong oxidising agent |
|
(iii) SnO |
(c) Lewis acid |
|
(iv) PbO2 |
(d) Can be further oxidised |
|
(e) Tetrahedral shape |
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(i)- (e), (ii)- (c), (iii)- (d), (iv)- (a) & (b)
Match the species given in Column I with properties given in Column II.
|
Column I |
Column II |
|
(i) Diborane |
(a) Used as a flux for soldering metals |
|
(ii) Gallium |
(b) Crystalline form of silica |
|
(iii) Borax |
(c) Banana bonds |
|
(iv) Aluminosilicate |
(d) Low melting, high boiling, useful for measuring high temperatures |
|
(v) Quartz |
(e) Used as catalyst in petrochemical industries. |
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(i)→ (c) (ii)→ (d) (iii)→ (a) (iv)→ (e) (v)→ (b)
Match the species given in Column I with the hybridisation given in Column II.
|
(i) Boron in [B(OH)4]− |
(a) sp2 |
|
(ii) Aluminium in [Al(H2O)6]3+ |
(b) sp3 |
|
(iii) Boron in B2H6 |
(c) sp3d2 |
|
(iv) Carbon in Buckminsterfullerene |
|
|
(v) Silicon in SiO44− |
|
|
(vi) Germanium in [GeCl6]2− |
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(i)→ (b), (ii) → (c), (iii) → (b), (iv) → (a) (v)→ (b) (vi)→ (c)
Assertion (A): If aluminium atoms replace a few silicon atoms in three dimensional network of silicon dioxide, the overall structure acquires a negative charge.
Reason (R): Aluminium is trivalent while silicon is tetravalent.
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is not the correct explanation of A.
(C) Both A and R are not correct.
(D) A is not correct but R is correct.
This is a Assertion and Reason Type Questions as classified in NCERT Exemplar
(A)
In aluminium silicate, silicon is doped with group 13 elements and as aluminium is trivalent and silicon is tetravalent so, on replacing aluminium by Silicon, a negatively charged structure will be obtained. Both the statements assertion and reason are correct and reason is the correct explanation of assertion.
Assertion (A): Silicones are water repelling in nature.
Reason (R): Silicones are organosilicon polymers, which have (-R2SiO-) as repeating unit.
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is not the correct explanation of A.
(C) Both A and R are not correct.
(D) A is not correct but R is correct.
This is a Assertion and Reason Type Questions as classified in NCERT Exemplar
(B)
Silicones are a class of silicon polymers that contain the repeating unit (-R 2 SiO-). Silicones are water repellent in nature because they are surrounded by non-polar alkyl groups, so A and R are both correct, but R is not the correct explanation of A.
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