NCERT Solutions | Class 12 Chemistry Chapter 7 | The p Block Elements
CBSE Solutions | Chemistry Class 12
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NCERT | Class 12 Chemistry
| Book: | National Council of Educational Research and Training (NCERT) |
|---|---|
| Board: | Central Board of Secondary Education (CBSE) |
| Class: | 12th |
| Subject: | Chemistry |
| Chapter: | 7 |
| Chapters Name: | The p Block Elements |
| Medium: | English |
The p Block Elements | Class 12 Chemistry | NCERT Books Solutions
NCERT Exercises
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 1.
Solution.
Higher the positive oxidation state of central atom, more will be its polarising power which, in turn, increases the covalent character of bond formed between the central atom and the other atom.NCERT Solutions for Class 12 Chemistry Chapter 7, Question 2.
Solution.
Because BiH3 is the least stable among the hydrides of group 15.NCERT Solutions for Class 12 Chemistry Chapter 7, Question 3.
Solution.
Because of strong pπ – pπ overlap resulting into the triple bond. N = N due to which the bond dissociation energy of N2 is very high rendering it less reactive.NCERT Solutions for Class 12 Chemistry Chapter 7, Question 4.
Solution.
N2 + 3H2 \(\rightleftharpoons \) 2NH3; ∆fH = -46.1 kj mol-1In accordance with Le Chatelier’s principle high pressure and low temperature would favour formation of ammonia. The optimum conditions for the production of ammonia are a pressure of about 200 atm, a temperature of about 700K and use of a catalyst such as iron oxide with small amount of K2O and Al2,O3 as promoters.
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 5.
Solution.
NH3 in the form of solution reacts with Cu2+ to form a complex with deep blue colour.
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 6.
Solution.
From the structure of N2O5, it is evident that covalence of nitrogen is four.
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 7.
Solution.
Both are sp3 hybridised. In PH4+ all the four orbitals are bonded whereas in PH3 there is a lone pair of electrons on P, which is responsible for lone pair-bond pair repulsion in PH3 reducing the bond angle to less than 109° 28′.NCERT Solutions for Class 12 Chemistry Chapter 7, Question 8.
Solution.
When white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO2, phosphine gas is liberated.
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 9.
Solution.
On heating PCl5 sublimes and is converted to PCl3 on stronger heating.
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 10.
Solution.
PCl5 + D2O → POCl3, + 2DClNCERT Solutions for Class 12 Chemistry Chapter 7, Question 11.
Solution.
Three P – OH groups are present in the molecule of H3PO4. Therefore, its basicity is three.NCERT Solutions for Class 12 Chemistry Chapter 7, Question 12.
Solution.
On heating phosphorous acid dispropotionates to give orthophosphoric acid and phosphine.4H3PO3 → 3H3PO4 + PH3
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 13.
Solution.
Combined sulphur exists as sulphates, such as gypsum, epsom, baryte and sulphides such as galena, zinc blende, copper pyrites, etc. Traces of sulphur occur as hydrogen sulphide in volcanoes. Few organic materials like eggs, proteins, garlic, onion, mustard, hair and wool contain sulphur. 0.03 – 0.1% sulphur is present in the earth’s crust.NCERT Solutions for Class 12 Chemistry Chapter 7, Question 14.
Solution.
H2O > H2S > H2Se > H2TeNCERT Solutions for Class 12 Chemistry Chapter 7, Question 15.
Solution.
Because of small size and high electronegativity of oxygen, molecules of water are highly associated through hydrogen bonding resulting in its liquid state.NCERT Solutions for Class 12 Chemistry Chapter 7, Question 16.
Solution.
Platinum does not react with oxygen directlyNCERT Solutions for Class 12 Chemistry Chapter 7, Question 17.
- C2H4 + O2 →
- 4Al + 3 O2 →
Solution.
- C2H4 + 3O2 → 2CO2 + 2H2O
- 4Al + 3O2 → 2Al2O3
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 18.
Solution.
Due to the ease with which ozone liberates nascent oxygen atoms, it acts as a powerful oxidising agent.O3 → O2 + O
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 19.
Solution.
When ozone reacts with an excess of KI solution buffered with a borate buffer (pH = 9.2), iodine is liberated which can be titrated against standard solution of sodium thiosulphate. This is used as a method of estimation of ozone quantitatively.NCERT Solutions for Class 12 Chemistry Chapter 7, Question 20.
Solution.
When sulphur dioxide is passed through an aqueous solution of ferric ions, ferric ions are reduced to ferrous ions.
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 21.
Solution.
Both the S—O bonds are covalent and have equal strength due to resonating structures.
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 22.
Solution.
Presence of SO2 is detected by bringing a paper dipped in acidified potassium dichromate near the gas. If the paper turns green, it shows the presence of SO2 gas.
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 23.
Solution.
- In manufacture of fertilisers.
- In manufacture of pigments, paints and dyestuff intermediates.
- In detergent industry
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 24.
Solution.
The key step in the manufacture of sulphuric acid is oxidation of SO2 to SO3 in presence of V2O5 catalyst.
The reaction is exothermic and reversible. Hence, low temperature and high pressure are the favourable conditions for maximum yield of SO3. In practice a pressure of 2 bar and temperature of 720 K is maintained.
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 25.
Solution.
H2SO4 is a very strong acid in water largely because of its first ionisation to H3O+ and HSO4– The ionisation of HSO4– to H3O+ and SO42- is very very small. That is why, Ka2« Ka1.NCERT Solutions for Class 12 Chemistry Chapter 7, Question 26.
Solution.
Fluorine is a better oxidising agent than chlorine because E°F2/F- is higher than E°Cl2/Cl- It is mainly due to low bond dissociation energy, high hydration energy and lower electron gain enthalpy, non-availability of d-orbitals in valence shell, that results in higher reduction potential of F2 than chlorine.NCERT Solutions for Class 12 Chemistry Chapter 7, Question 27.
Solution.
- Ionisation enthalpy, electro-negativity and electrode potential are higher for fluorine than the expected trends of other halogen.
- Fluorine does not show any positive oxidation state except in HOF.
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 28.
Solution.
Sea water contains chlorides, bromides and iodides of sodium, potassium, magnesium and calcium but sodium chloride being the maximum makes sea water saline. Various sea weeds contain upto 0.5% iodine.NCERT Solutions for Class 12 Chemistry Chapter 7, Question 29.
Solution.
Chlorine bleaches by oxidation Cl2 + H2O → HCl + HOCl → HCl + [O]The nascent oxygen reacts with dye to make it colourless.
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 30.
Solution.
COCl2 (phosgene), CCl3NO2 (tear gas)NCERT Solutions for Class 12 Chemistry Chapter 7, Question 31.
Solution.
In general, interhalogen compounds are more reactive than halogens due to weaker X-X’ bonding than X-X bond. Thus, ICI is more reactive than I2.NCERT Solutions for Class 12 Chemistry Chapter 7, Question 32.
Solution.
A mixture of helium and oxygen does not cause pain due to very low solubility of helium in blood as compared to nitrogen.NCERT Solutions for Class 12 Chemistry Chapter 7, Question 33.
XeF6 + H2O → XeO2F2 + 4HF
Solution.
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 34.
Solution.
Radon is radioactive with very short half-life which makes the study of chemistry of radon difficult.NCERT Exercises
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 1.
Solution.
The valence shel 1 electronic configuration of group 15 elements is ns2np3. Due to half- filled p-orbitals, these elements have extra stability associated with them.The common oxidation states of these elements are – 3, +3 and +5. The tendency to exhibit -3 oxidation state decreases down the group. The stability of +5 state decreases and that of +3 state increases down the group due to inert pair effect.
The size of group 15 elements increases down the group. There is a considerable increase in covalent radius from N to P. However, from As to Bi only a small increase in covalent radius is observed. This is due to presence of completely filled d and or f orbitals in heavier members.
Down the group, ionisation enthalpy decreases due to increase in atomic size. Due to stable half-filled configuration, they have much greater value than that of group 14 elements.
The electronegativity value, in general, decreases down the group with increasing atomic size. However, amongst the heavier elements, the difference is not that much pronounced.
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 2.
Solution.
Nitrogen has a unique ability to form pπ – pπ multiple bonds with itself and with other elements having small size and high electronegativity. Consequently, its bond enthalpy is very high and reactivity is less. Another factor which affects the chemistry of N is the absence of d-orbitals in its valence shell. As a result, not only the covalency of N is restricted to four, it can also not form dn – pn bonds. P on the other hand, does not form pn – pn bonds and hence it only forms a P – P single bond, which can easily be broken. Also, phosphorus has vacant d-orbitals and can form dn – dn bond with transition metals to form compounds which can act as ligands.NCERT Solutions for Class 12 Chemistry Chapter 7, Question 3.
Solution.
Nitrogen has very low reactivity due to unavailability of vacant d-orbital and high bond dissociation energy of N ≡ N bond.(a) Hydrides : General formula for hydrides is MH3, e.g., NH3, PH3, ASH3, SbH3, BiH3. All these hydrides are covalent in nature and have pyramidal structure (sp3 hybridized).
(b) Halides – Elements of group 15 form two types of halides viz. trihalides and pentahalides. The halides are predominantly basic (Lewis bases) in nature and have lone pair of electrons (central atom is sp3 hybridized). The pentahalides are thermally less stable than the trihalides.
(c) Oxides – All the elements of this group form two types of oxides ie., M2O3 and M2O5 and are called trioxides and pentoxides
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 4.
Solution.
Hydrogen bond is formed between electronegative atom and hydrogen atom. Nitrogen is an electronegative atom and electronegativity decreases down the group so PH3 cannot form H – bond.NCERT Solutions for Class 12 Chemistry Chapter 7, Question 5.
Solution.
In the laboratory, dinitrogen is prepared by treating an aqueous solution of ammonium chloride with sodium nitrite.
Small amounts of NO and HNO3 are also formed in this reaction. These impurities can be removed by passing the gas through aqueous sulphuric acid containing potassium dichromate. It can also be obtained by the thermal decomposition of ammonium dichromate.
Very pure nitrogen can be obtained by the thermal decomposition of sodium or barium azide.
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 6.
Solution.
Industrially ammonia is manufactured by Haber’s process.
In accordance with Le-Chatelier’s principle, high pressure and low temperature would favour the formation of ammonia. The optimum conditions for production of ammonia are a pressure of 200 × 105 Pa (about 200 atm), a temperature of -700 K and the use of a catalyst such as iron-oxide with small amounts of K2O and Al2O3 to increase the rate of attainment of equilibrium.
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 7.
Solution.
acid is a strong oxidising agent and attacks most metals. The products of oxidation depend upon the concentration of the acid, temperature and the nature of the material undergoing oxidation. Copper with . cold dil. HNO3 forms nitric oxide (NO).
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 8.
Solution.
Resonating structures
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 9.
[Hint : Can be explained on the basis of sp3 hybridisation in NH3 and only s-p bonding between hydrogen and other elements of the group].
Solution.
The actual bond angles are
The decreased bond angle in other hydrides can be explained by the fact that the sp3 hybridisation becomes less and less distinct with increasing size of the central atom i.e., pure p-orbitals are utilised in M-H bonding.
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 10.
Solution.
In R3 N = O, covalency required is five. The maximum covalency of nitrogen is four as it does not possess d-orbitals in the valence shell i.e., it cannot extend its valency beyond four. On the other hand, other members have d-orbitals and can utilise these orbitals to show covalency of five or six e.g., R3P = O, PCl5, [SbF6]– etc. These can form dπ – pπ bonds.NCERT Solutions for Class 12 Chemistry Chapter 7, Question 11.
Solution.
The basic character decreases from NH3 to BiH3. The basic nature is due to the presence of lone pair of electrons on the central atom. NH3 is the strongest electron pair donor due to its small size as the electron density of the electron pair is concentrated over a small region. As the size increases the electron density gets diffused over a large region and hence the ability to donate the electron pair (basic nature) decreases.NCERT Solutions for Class 12 Chemistry Chapter 7, Question 12.
Solution.
Nitrogen is diatomic gaseous molecule at ordinary temperature due to its ability to form pπ – pπ multiple bonds. The molecule has one σ and two π – bonds. Phosphorus exists as discrete tetratomic tetrahedral molecules as these are not capable of forming multiple bonds due to repulsion between non-bonded electrons of the inner core.NCERT Solutions for Class 12 Chemistry Chapter 7, Question 13.
Solution.
Structure of white and red phosphorus are given below :
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 14.
Solution.
The single N-N bond is weaker than the single P-P bond because of high interelectronic repulsion of the non-bonding electrons, owing to the small bond length. As a result the catenation tendency is weaker in nitrogen.NCERT Solutions for Class 12 Chemistry Chapter 7, Question 15.
Solution.
The acids in +3 oxidation state of phosphorus tend to disproportionate to higher and lower oxidation states, e.g., phosphorous acid on heating disproportionates to give orthophosphoric acid and phosphine.4H3PO3 → 3H3PO4 + PH3
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 16.
Solution.
The oxidation state of P in PCl5 is +5. As P has five electrons in its valence shell, it cannot increase its oxidation state beyond +5 by donating electrons, therefore, PCl5 cannot act as a reducing agent. However, it can decrease its oxidation number from +5 to +3 or some lower value, so, PCl5 acts as anoxidising agent. For example, it oxidises Ag to AgCl, Sn to SnCl4.
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 17.
Solution.
All these elements have same ns2np4 (n = 2 to 6) valence shell electronic configuration and hence are justified to be placed in group 16 of the periodic table.
(ii) Oxidation states : They need two more electrons to form dinegative ions by acquiring the nearest noble gas configuration. So, the minimum oxidation state of these elements should be -2. Oxygen predominantly and sulphur to some extent being electronegative show an oxidation state of -2. Since these elements have six electrons in the valence shell, therefore, the maximum oxidation state they can show is, + 6. Other positive oxidation states shown by these elements are +2 and +4. Although, oxygen due to the absence of d-orbitals does not show oxidation states of +4 and +6. Thus, on the basis of minimum and maximum oxidation states, these elements are justified to be placed in the same group i.c., group 16 of the periodic table.
(iii) Formation of hydrides : All the elements complete their respective octets by sharing two of their valence electrons with ls-orbilal of hydrogen to form hydrides of the general formula EH2 i.c., H2O, H2S, H2Se, H2Te and H2Po. Therefore, on the basis of formation of hydrides of the general formula, EH2, these elements are justified to be placed in group 16 of the periodic table.
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 18.
Solution.
Oxygen atom has the tendency to form multiple bonds (pπ – pπ interaction) with other oxygen atom on account of small size while this tendency is missing in sulphuratom. The bond energy of oxygen-oxygen double bond (0 = 0) is quite large (about three times that of oxygen-oxygen single bond, O – O = 34.9 kcal mol-1) while sulphur-sulphur double bond (S = S) is not so large (less than double of sulphur-sulphur single bond, S – S = 63.8 kcal mob1). As a resul —O—O—O— chains are less stable as compared to O = O molecule while —S—S—S— chains are more stable than S = S molecule. Therefore, at room temperature, while oxygen exists as a diatomic gas molecule, sulphur exists as S8 solid.
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 19.
Hint : Consider lattice energy factor in the formation of compound.
Solution.
This can be explained with the help of electronic configuration.
As O2- has most stable configuration amongst these. So, formation of O2- is much more easier. In solid state, large amount of energy (lattice enthalpy) is released to form divalent O2- ions. It is greater lattice enthalpy of O2- which compensates for the high energy required to remove the second electron.
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 20.
Solution.
CFC – Chlorofluorocarbon (freons) : These compounds commonly known as freons are introduced into the atmosphere from aerosol sprays and refrigerating equipments. They undergo photochemical decomposition and destroy ozone as shown by the following sequence of reactions.
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 21.
Solution.
Sulphuric acid is manufactured by the Contact process which involves three steps :- burning of sulphur or sulphide ores in air to generate SO2.
- conversion of SO2 to SO3 by the reaction with oxygen in the presr nee of a catalyst (V2O5), and
- absorption of SO3 in H2SO4 to give oleum (H2S2O7).
A flow diagram for the manufacture of sulphuric acid is shown in the figure. The SO2 produced is purified by removing dust and other impurities such as arsenic compounds. The key step in the manufacture of H2SO4 is the catalytic oxidation of SO2 with O2 to give SO3 in the presence of V2O5 (catalyst).
The reaction is exothermic, reversible and the forward reaction leads to a decrease in volume. Therefore, low temperature and high pressure are the favourable conditions for maximum yield. But the temperature should not be very low otherwise rate of reaction will become slow.
In practice, the plant is operated at a pressure of 2 bar and a temperature of 720 K. The SO3 gas from the catalytic converter is absorbed in concentrated H2SO4 to produce oleum. Dilution of oleum with water give H2SO4 of the desired concentration. In the industry, two steps are carried out simultaneously to make the process a continuous one and also to reduce the cost.
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 22.
Solution.
SO2 present in atmosphere combines with water to give sulphuric acid.2SO2 + O2 + 2H2O → 2H2SO4
This is called acid rain. Acid rain causes extensive damage to building and statues made of marble, limestone, etc. due to its reaction to give CaSO4.
CaCO3 + H2SO4 → CaSO4 + CO2 + H2O
This way SO2 acts as pollutant.
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 23.
Solution.
General electronic configuration of halogen is
It will easily accept one electron to fulfill its shell. This easily acceptance of electron makes halogens strong oxidising agents.
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 24.
Solution.
Due to high electronegativity and absence of d orbitals, F does not form oxoacids such as HOFO, HOFO2 and HOFO3 in which the oxidation state of F is +3, +5 and +7. It just forms one oxoacid, i.e., HOF in which the oxidation state of F is + 1.NCERT Solutions for Class 12 Chemistry Chapter 7, Question 25.
Solution.
Although O and Cl have about the same electronegativity, yet their atomic size (covalent radii) are much different: O = 66 pm and Cl = 99 pm. Thus, electron density per unit volume of oxygen atom is much higher than that of chlorine atom. Hence, oxygen forms hydrogen bonds while chlorine does not though both have approx, the same electronegativity.NCERT Solutions for Class 12 Chemistry Chapter 7, Question 26.
Solution.
ClO2 is a powerful oxidising agent. ClO2 is used as a bleaching agent for paper pulp and textiles and in water treatment.NCERT Solutions for Class 12 Chemistry Chapter 7, Question 27.
Solution.
All halogens are coloured because of absorption of radiations in visible region which results in the excitation of outer electrons to higher level. By absorbing different quanta of radiation, they display different colours asF2 → yellow
Cl2 → greenish yellow
Br2 → red
I2 → violet
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 28.
Solution.
Fluorine reacts vigorously with water and oxidises water to oxygen.
Chlorine dissolves in water to form chlorine water. It slowly reacts with the water to form a mixture of hydrochloric acid and hypochlorous acid.
Hypochlorous acid is very unstable. In presence of sunlight, it decomposes to give HCl and nascent oxygen.
This oxygen is responsible for oxidising and bleaching properties of chlorine.
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 29.
Solution.
Preparation of chlorine by Deacon’s process : By oxidation of hydrogen chloride gas by atmospheric oxygen in the presence of CuCl2 at 723 K
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 30.
Solution.
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 31.
- H3PO3
- PCl3
- Ca3P2
- Na3PO4
- POF3?
Solution.
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 32.
- NaCI is heated with sulphuric acid in the presence of MnO2.
- Chlorine gas is passed into a solution of Nal in water.
Solution.
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 33.
Solution.
Xenon forms three binary fluorides, XeF2, XeF4 and XeF6 by the direct reaction of elements under appropriate experimental conditions.
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 34.
Solution.
Replace O– (9 electrons) in ClO– by F (9 electrons). The resulting neutral molecule is ClF. Since ClF can combine further with F to form ClF3, so, ClF is a Lewis base.NCERT Solutions for Class 12 Chemistry Chapter 7, Question 35.
Solution.
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 36.
- F2, Cl2, Br2, l2 – increasing bond dissociation enthalpy.
- HF, HCI, HBr, HI – increasing acid strength.
- NH3, PH3, ASH3, SbH3, BiH3 – increasing base strength.
Solution.
- l2 < Br2 < F2 < Cl2
- HF < HCI < HBr < HI
- NH3 > PH3 > AsH3 > SbH, > BiH3
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 37.
(i) XeOF4
(ii) NeF2
(iii) XeF2
(iv) XeF6
Solution.
(ii) Amongst all noble gases, only Xe (except KrF2) forms compounds.NCERT Solutions for Class 12 Chemistry Chapter 7, Question 38.
- lCl–4
- IBr–2
- Br0–3
Solution.
(iii) Structure of BrO–3 : The central atom Br has seven electrons. Four of these electrons form two double bonds or coordinate bonds with two oxygen atoms while the fifth electron forms a single bond with O– ion. The remaining two electrons form one lone pair. Hence, in all there are three bond pairs and one lone pair around Br atom in BrO–3 Therefore, according to VSEPR theory, BrO–3 s should be pyramidal.
Here, BrO–3 has 26(7 + 3 × 6 + 1 = 26) valence electrons. A noble gas species having 26 valence electrons is XeO3 (8 + 3 × 6 = 26). Thus, like BrO–3, XeO3 is also pyramidal.
NCERT Solutions for Class 12 Chemistry Chapter 7, Question 39.
Solution.
The atomic radii of noble gases are by far the largest in their respective periods. This is due to the reason that noble gases have only van der Waals radii while others have cova lent radii, van der Waals radii, by definition are larger than covalent radii.NCERT Solutions for Class 12 Chemistry Chapter 7, Question 40.
Solution.
Uses of neon :- Neon is used in discharge tubes and fluorescent bulbs for advertisement display purposes.
- Glow of different colours ‘neon signs’ can be produced by mixing neon with other gases.
- Neon bulbs and used in botanical gardens and in green houses.
Uses of argon:
- Argon is used mainly to provide an inert atmosphere in high temperature metallurgical processes such as arc welding of metals and alloys. In the laboratory, it is used for handling substance which are air sensitive.
- It is used in filling incandescent and fluorescent lamps where its presence retards the sublimation of the filament and thus increases the life of the lamp.
- It is also used in “neon signs” for obtaining lights of different colours.
NCERT Class 12 Chemistry
Class 12 Chemistry Chapters | Chemistry Class 12 Chapter 7
Chapterwise NCERT Solutions for Class 12 Chemistry
-
NCERT Solutions For Class 12 Chemistry Chapter 1 The Solid State
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions
NCERT Solutions For Class 12 Chemistry Chapter 3 Electro chemistry
NCERT Solutions For Class 12 Chemistry Chapter 4 Chemical Kinetics
NCERT Solutions For Class 12 Chemistry Chapter 5 Surface Chemistry
NCERT Solutions For Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements
NCERT Solutions For Class 12 Chemistry Chapter 7 The p Block Elements
NCERT Solutions For Class 12 Chemistry Chapter 8 The d and f Block Elements
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds
NCERT Solutions For Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes
NCERT Solutions For Class 12 Chemistry Chapter 11 Alcohols Phenols and Ethers
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines
NCERT Solutions For Class 12 Chemistry Chapter 14 Biomolecules
NCERT Solutions For Class 12 Chemistry Chapter 15 Polymers
NCERT Solutions For Class 12 Chemistry Chapter 16 Chemistry in Everyday Life
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