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Here, you can find comprehensive NCERT solutions for Class 12 Science Chemistry Chapter 9 on Coordination Compounds. These solutions include easy-to-follow, step-by-step explanations. Widely appreciated by class 12 Science students, these Chemistry Coordination Compounds Solutions are invaluable for efficiently completing homework assignments and preparing for exams. All the questions and answers from the NCERT Book of Class 12 Science Chemistry Chapter 9 are available here at no cost. Page No 244: Question 9.1: Write the formulas for the following coordination compounds: (i) Tetraamminediaquacobalt(III) chloride (ii) Potassium tetracyanonickelate(II) (iii) Tris(ethane−1,2−diamine) chromium(III) chloride (iv) Amminebromidochloridonitrito-N-platinate(II) (v) Dichloridobis(ethane−1,2−diamine)platinum(IV) nitrate (vi) Iron(III) hexacyanoferrate(II) ANSWER: (i)  (ii)  (iii)  (vi)  (v)  (vi)  Page No 244: Question 9.2: Write the IUPAC names of the following coordination compounds: (i) [Co(NH3)6]Cl3 (ii) [Co(NH3)5Cl]Cl2 (iii) K3[Fe(CN)6] (iv) K3[Fe(C2O4)3] (v) K2[PdCl4] (vi) [Pt(NH3)2Cl(NH2CH3)]Cl ANSWER: (i) Hexaamminecobalt(III) chloride (ii) Pentaamminechloridocobalt(III) chloride (iii) Potassium hexacyanoferrate(III) (iv) Potassium trioxalatoferrate(III) (v) Potassium tetrachloridopalladate(II) (vi) Diamminechlorido(methylamine)platinum(II) chloride Page No 247: Question 9.3: Indicate the types of isomerism exhibited by the following complexes and draw the structures for these isomers: ANSWER: Trans-isomer is optically inactive. On the other hand, cis-isomer is optically active. (ii) Two optical isomers for  exist. Two optical isomers are possible for this structure. (iii)  A pair of optical isomers: It can also show linkage isomerism. and It can also show ionization isomerism. (iv) Geometrical (cis-, trans-) isomers of can exist. Page No 247: Question 9.4: Give evidence that [Co(NH3)5Cl]SO4 and [Co(NH3)5SO4]Cl are ionization isomers. ANSWER: When ionization isomers are dissolved in water, they ionize to give different ions. These ions then react differently with different reagents to give different products. Page No 254: Question 9.5: Explain on the basis of valence bond theory that [Ni(CN)4]2− ion with square planar structure is diamagnetic and the [NiCl4]2− ion with tetrahedral geometry is paramagnetic. ANSWER: Ni is in the +2 oxidation state i.e., in d8 configuration. There are 4 CN− ions. Thus, it can either have a tetrahedral geometry or square planar geometry. Since CN− ion is a strong field ligand, it causes the pairing of unpaired 3d electrons. It now undergoes dsp2 hybridization. Since all electrons are paired, it is diamagnetic. In case of [NiCl4]2−, Cl− ion is a weak field ligand. Therefore, it does not lead to the pairing of unpaired 3d electrons. Therefore, it undergoes sp3 hybridization. Since there are 2 unpaired electrons in this case, it is paramagnetic in nature. Page No 254: Question 9.6: [NiCl4]2− is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. Why? ANSWER: Though both [NiCl4]2− and [Ni(CO)4] are tetrahedral, their magnetic characters are different. This is due to a difference in the nature of ligands. Cl− is a weak field ligand and it does not cause the pairing of unpaired 3d electrons. Hence, [NiCl4]2− is paramagnetic. In Ni(CO)4, Ni is in the zero oxidation state i.e., it has a configuration of 3d8 4s2. But CO is a strong field ligand. Therefore, it causes the pairing of unpaired 3d electrons. Also, it causes the 4s electrons to shift to the 3d orbital, thereby giving rise to sp3 hybridization. Since no unpaired electrons are present in this case, [Ni(CO)4] is diamagnetic. Page No 254: Question 9.7: [Fe(H2O)6]3+ is strongly paramagnetic whereas [Fe(CN)6]3− is weakly paramagnetic. Explain. ANSWER: In both and , Fe exists in the +3 oxidation state i.e., in d5 configuration. Since CN− is a strong field ligand, it causes the pairing of unpaired electrons. Therefore, there is only one unpaired electron left in the d-orbital. Therefore, On the other hand, H2O is a weak field ligand. Therefore, it cannot cause the pairing of electrons. This means that the number of unpaired electrons is 5. Therefore, Thus, it is evident that is strongly paramagnetic, while is weakly paramagnetic. Page No 254: Question 9.8: Explain [Co(NH3)6]3+ is an inner orbital complex whereas [Ni(NH3)6]2+ is an outer orbital complex. ANSWER: Oxidation state of cobalt = +3 Oxidation state of Ni = +2 Electronic configuration of cobalt = d6 Electronic configuration of nickel = d8 NH3 being a strong field ligand causes the pairing. Therefore, Cobalt can undergo d2sp3 hybridization.Hence, it is an inner orbital complex. If NH3 causes the pairing, then only one 3d orbital is empty. Thus, it cannot undergo d2sp3hybridization. Therefore, it undergoes sp3d2 hybridization.Hence, it forms an outer orbital complex. Page No 254: Question 9.9: Predict the number of unpaired electrons in the square planar [Pt(CN)4]2− ion. ANSWER: In this complex, Pt is in the +2 state. It forms a square planar structure. This means that it undergoes dsp2 hybridization. Now, the electronic configuration of Pd(+2) is 5d8. CN− being a strong field ligand causes the pairing of unpaired electrons. Hence, there are no unpaired electrons in Page No 254: Question 9.10: The hexaquo manganese(II) ion contains five unpaired electrons, while the hexacyanoion contains only one unpaired electron. Explain using Crystal Field Theory. ANSWER: Mn is in the +2 oxidation state. Mn is in the +2 oxidation state. The electronic configuration is d5. The electronic configuration is d5. The crystal field is octahedral. Water is a weak field ligand. Therefore, the arrangement of the electrons in is t2g3eg2. The crystal field is octahedral. Cyanide is a strong field ligand. Therefore, the arrangement of the electrons in  isT2g5eg0.     Hence, hexaaquo manganese (II) ion has five unpaired electrons, while hexacyano ion has only one unpaired electron. Page No 256: Question 9.11: Calculate the overall complex dissociation equilibrium constant for the Cu(NH3)42+ ion, given that β4 for this complex is 2.1 × 1013. ANSWER: β4 = 2.1 × 1013 The overall complex dissociation equilibrium constant is the reciprocal of the overall stability constant, β4. Page No 258: Question 9.1: Explain the bonding in coordination compounds in terms of Werner’s postulates. ANSWER: Werner’s postulates explain the bonding in coordination compounds as follows: (i) A metal exhibits two types of valencies namely, primary and secondary valencies. Primary valencies are satisfied by negative ions while secondary valencies are satisfied by both negative and neutral ions. (In modern terminology, the primary valency corresponds to the oxidation number of the metal ion, whereas the secondary valency refers to the coordination number of the metal ion. (ii) A metal ion has a definite number of secondary valencies around the central atom. Also, these valencies project in a specific direction in the space assigned to the definite geometry of the coordination compound. (iii) Primary valencies are usually ionizable, while secondary valencies are non-ionizable. Page No 258: Question 9.2: FeSO4 solution mixed with …

NCERT Solutions for Class 12 Science Chemistry Chapter 9 – Coordination Compounds Read More »

Explore the NCERT Solutions for Class 12 Science Chemistry Chapter 8: “The D And F Block Elements,” featuring clear and concise step-by-step explanations. Widely embraced by class 12 Science students, these solutions prove invaluable for completing homework efficiently and preparing for exams. Free access to all questions and answers from the NCERT Book of class 12 Science Chemistry Chapter 8 is available here, making it a convenient resource for students. Page No 212: Question 8.1: Silver atom has completely filled d orbitals (4d10) in its ground state. How can you say that it is a transition element? ANSWER: Ag has a completely filled 4d orbital (4d10 5s1) in its ground state. Now, silver displays two oxidation states (+1 and +2). In the +1 oxidation state, an electron is removed from the s-orboital. However, in the +2 oxidation state, an electron is removed from the d-orbital. Thus, the d-orbital now becomes incomplete (4d9). Hence, it is a transition element. Page No 215: Question 8.2: In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomization of zinc is the lowest, i.e., 126 kJ mol−1. Why? ANSWER: The extent of metallic bonding an element undergoes decides the enthalpy of atomization. The more extensive the metallic bonding of an element, the more will be its enthalpy of atomization. In all transition metals (except Zn, electronic configuration: 3d10 4s2), there are some unpaired electrons that account for their stronger metallic bonding. Due to the absence of these unpaired electrons, the inter-atomic electronic bonding is the weakest in Zn and as a result, it has the least enthalpy of atomization. Page No 217: Question 8.3: Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why? ANSWER: Mn (Z = 25) = 3d5 4s2 Mn has the maximum number of unpaired electrons present in the d-subshell (5 electrons). Hence, Mn exhibits the largest number of oxidation states, ranging from +2 to +7. Page No 217: Question 8.4: The Eθ(M2+/M) value for copper is positive (+0.34V). What is possibly the reason for this? (Hint: consider its high ΔaHθ and low ΔhydHθ) ANSWER: The Eθ(M2+/M) value of a metal depends on the energy changes involved in the following: 1. Sublimation: The energy required for converting one mole of an atom from the solid state to the gaseous state. 2. Ionization: The energy required to take out electrons from one mole of atoms in the gaseous state. 3. Hydration: The energy released when one mole of ions are hydrated. Now, copper has a high energy of atomization and low hydration energy. Hence, the Eθ(M2+/M) value for copper is positive. Page No 219: Question 8.5: How would you account for the irregular variation of ionization enthalpies (first and second) in the first series of the transition elements? ANSWER: Ionization enthalpies are found to increase in the given series due to a continuous filling of the inner d-orbitals. The irregular variations of ionization enthalpies can be attributed to the extra stability of configurations such as d0, d5, d10. Since these states are exceptionally stable, their ionization enthalpies are very high. In case of first ionization energy, Cr has low ionization energy. This is because after losing one electron, it attains the stable configuration (3d5). On the other hand, Zn has exceptionally high first ionization energy as an electron has to be removed from stable and fully-filled orbitals (3d10 4s2). Second ionization energies are higher than the first since it becomes difficult to remove an electron when an electron has already been taken out. Also, elements like Cr and Cu have exceptionally high second ionization energies as after losing the first electron, they have attained the stable configuration (Cr+: 3d5 and Cu+: 3d10). Hence, taking out one electron more from this stable configuration will require a lot of energy. Page No 220: Question 8.6: Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only? ANSWER: Both oxide and fluoride ions are highly electronegative and have a very small size. Due to these properties, they are able to oxidize the metal to its highest oxidation state. Page No 220: Question 8.7: Which is a stronger reducing agent Cr2+ or Fe2+ and why? ANSWER: The following reactions are involved when Cr2+ and Fe2+ act as reducing agents. Cr2+ Cr3+ Fe2+ Fe3+ The value is −0.41 V and  is +0.77 V. This means that Cr2+ can be easily oxidized to Cr3+, but Fe2+ does not get oxidized to Fe3+ easily. Therefore, Cr2+ is a better reducing agent that Fe3+. Page No 222: Question 8.8: Calculate the ‘spin only’ magnetic moment of M2+(aq) ion (Z = 27). ANSWER: Z = 27  [Ar] 3d7 4s2  M2+ = [Ar] 3d7 3d7 =  i.e., 3 unpaired electrons n = 3 μ ≈ 4 BM Page No 224: Question 8.9: Explain why Cu+ ion is not stable in aqueous solutions? ANSWER: In an aqueous medium, Cu2+ is more stable than Cu+. This is because although energy is required to remove one electron from Cu+ to Cu2+, high hydration energy of Cu2+ compensates for it. Therefore, Cu+ ion in an aqueous solution is unstable. It disproportionates to give Cu2+ and Cu. Page No 232: Question 8.10: Actinoid contraction is greater from element to element than lanthanoid contraction. Why? ANSWER: In actinoids, 5f orbitals are filled. These 5f orbitals have a poorer shielding effect than 4f orbitals (in lanthanoids). Thus, the effective nuclear charge experienced by electrons in valence shells in case of actinoids is much more that that experienced by lanthanoids. Hence, the size contraction in actinoids is greater as compared to that in lanthanoids. Page No 234: Question 8.1: Write down the electronic configuration of: (i) Cr3++ (iii) Cu+(v) Co2+ (vii) Mn2+ (ii) Pm3+(iv) Ce4+ (vi) Lu2+(viii) Th4+ ANSWER: (i) Cr3+: 1s2 2s2 2p6 3s2 3p6 3d3 Or, [Ar]18 3d3 (ii) Pm3+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 4f4 Or, [Xe]54 3d3 (iii) Cu+: 1s2 2s2 2p6 3s2 3p6 3d10 Or, [Ar]18 3d10 (iv) Ce4+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 Or, [Xe]54 (v) Co2+: 1s2 2s2 2p6 3s2 3p6 3d7 Or, [Ar]18 3d7 (vi) Lu2+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 4f14 5d1 Or, [Xe]54 2f14 3d3 (vii) Mn2+: 1s2 2s2 2p6 3s2 3p6 3d5 Or, [Ar]18 3d5 (viii) Th4+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 4f14 5s2 5p6 5d10 6s2 6s6 Or, [Rn]86 Page No 234: Question 8.2: Why are Mn2+compounds more stable than Fe2+towards oxidation to their +3 state? ANSWER: Electronic configuration of Mn2+ is [Ar]18 3d5. Electronic configuration of Fe2+ is [Ar]18 3d6. It is known that half-filled and fully-filled orbitals are more stable. Therefore, Mn in (+2) state has a stable d5 configuration. This is the reason Mn2+ shows resistance to oxidation to Mn3+. Also, Fe2+ has 3d6 configuration and by losing one …

NCERT Solutions for Class 12 Science Chemistry Chapter 8 – The D And F Block Elements Read More »

Find comprehensive solutions for Class 12 Science Chemistry Chapter 7 – “The P Block Elements” in the NCERT book. These step-by-step explanations are highly sought after by Chemistry students, aiding in homework completion and exam preparation. The popularity of these solutions among class 12 Science students is attributed to their effectiveness. Access free answers to all questions from the NCERT Book of Class 12 Science Chemistry Chapter 7, ensuring a valuable resource for your studies. Page No 169: Question 7.1: Why are pentahalides more covalent than trihalides? ANSWER: In pentahalides, the oxidation state is +5 and in trihalides, the oxidation state is +3. Since the metal ion with a high charge has more polarizing power, pentahalides are more covalent than trihalides. Page No 169: Question 7.2: Why is BiH3 the strongest reducing agent amongst all the hydrides of Group 15 elements? ANSWER: As we move down a group, the atomic size increases and the stability of the hydrides of group 15 elements decreases. Since the stability of hydrides decreases on moving from NH3 to BiH3, the reducing character of the hydrides increases on moving from NH3 to BiH3. Page No 170: Question 7.3: Why is N2 less reactive at room temperature? ANSWER: The two N atoms in N2 are bonded to each other by very strong triple covalent bonds. The bond dissociation energy of this bond is very high. As a result, N2 is less reactive at room temperature. Page No 172: Question 7.4: Mention the conditions required to maximise the yield of ammonia. ANSWER: Ammonia is prepared using the Haber’s process. The yield of ammonia can be maximized under the following conditions: (i) High pressure (∼ 200 atm) (ii) A temperature of ∼700 K (iii) Use of a catalyst such as iron oxide mixed with small amounts of K2O and Al2O3 Page No 172: Question 7.5: How does ammonia react with a solution of Cu2+? ANSWER: NH3 acts as a Lewis base. It donates its electron pair and forms a linkage with metal ion. Page No 173: Question 7.6: What is the covalence of nitrogen in N2O5? ANSWER: From the structure of N2O5, it is evident that the covalence of nitrogen is 4. Page No 177: Question 7.7: Bond angle in is higher than that in PH3. Why? ANSWER: In PH3, P is sp3 hybridized. Three orbitals are involved in bonding with three hydrogen atoms and the fourth one contains a lone pair. As lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion, the tetrahedral shape associated with sp3 bonding is changed to pyramidal. PH3 combines with a proton to form  in which the lone pair is absent. Due to the absence of lone pair in, there is no lone pair-bond pair repulsion. Hence, the bond angle in is higher than the bond angle in PH3. Page No 177: Question 7.8: What happens when white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO2? ANSWER: White phosphorous dissolves in boiling NaOH solution (in a CO2 atmosphere) to give phosphine, PH3. Page No 178: Question 7.9: What happens when PCl5 is heated? ANSWER: All the bonds that are present in PCl5 are not similar. It has three equatorial and two axial bonds. The equatorial bonds are stronger than the axial ones. Therefore, when PCl5 is heated strongly, it decomposes to form PCl3. Page No 178: Question 7.10: Write a balanced equation for the hydrolytic reaction of PCl5 in heavy water. ANSWER: Page No 180: Question 7.11: What is the basicity of H3PO4? ANSWER: H3PO4 Since there are three OH groups present in H3PO4, its basicity is three i.e., it is a tribasic acid. Page No 180: Question 7.12: What happens when H3PO3 is heated? ANSWER: H3PO3,on heating, undergoes disproportionation reaction to form PH3 and H3PO4. The oxidation numbers of P in H3PO3,PH3, and H3PO4 are +3, −3, and +5 respectively. As the oxidation number of the same element is decreasing and increasing during a particular reaction, the reaction is a disproportionation reaction. Page No 183: Question 7.13: List the important sources of sulphur. ANSWER: Sulphur mainly exists in combined form in the earth’s crust primarily as sulphates [gypsum (CaSO4.2H2O), Epsom salt (MgSO4.7H2O), baryte (BaSO4)] and sulphides [(galena (PbS), zinc blends (ZnS), copper pyrites (CuFeS2)]. Page No 183: Question 7.14: Write the order of thermal stability of the hydrides of Group 16 elements. ANSWER: The thermal stability of hydrides decreases on moving down the group. This is due to a decrease in the bond dissociation enthalpy (H−E) of hydrides on moving down the group. Therefore, Page No 183: Question 7.15: Why is H2O a liquid and H2S a gas? ANSWER: H2O has oxygen as the central atom. Oxygen has smaller size and higher electronegativity as compared to sulphur. Therefore, there is extensive hydrogen bonding in H2O, which is absent in H2S. Molecules of H2S are held together only by weak van der Waal’s forces of attraction. Hence, H2O exists as a liquid while H2S as a gas. Page No 185: Question 7.16: Which of the following does not react with oxygen directly? Zn, Ti, Pt, Fe ANSWER: Pt is a noble metal and does not react very easily. All other elements, Zn, Ti, Fe, are quite reactive. Hence, oxygen does not react with platinum (Pt) directly. Page No 185: Question 7.17: Complete the following reactions: (i) C2H4 + O2→ (ii) 4Al + 3O2→ ANSWER: (i)  (ii)  Page No 187: Question 7.18: Why does O3 act as a powerful oxidising agent? ANSWER: Ozone is not a very stable compound under normal conditions and decomposes readily on heating to give a molecule of oxygen and nascent oxygen. Nascent oxygen, being a free radical, is very reactive. Therefore, ozone acts as a powerful oxidising agent. Page No 187: Question 7.19: How is O3 estimated quantitatively? ANSWER: Quantitatively, ozone can be estimated with the help of potassium iodide. When ozone is made to react with potassium iodide solution buffered with a borate buffer (pH 9.2), iodine is liberated. This liberated iodine can be titrated against a standard solution of sodium thiosulphate using starch as an indicator. The reactions involved in the process are given below. Page No 189: Question 7.20: What happens when sulphur dioxide is …

NCERT Solutions for Class 12 Science Chemistry Chapter 7 – The P Block Elements Read More »

Here, you will find NCERT solutions for Class 12 Science Chemistry Chapter 6 – “General Principles And Processes Of Isolation Of Elements” with clear and straightforward explanations. These solutions are widely favored by Class 12 Science students studying Chemistry. They serve as a valuable resource for completing homework efficiently and preparing for exams. All the questions and answers from Chapter 6 of the NCERT Book for Class 12 Science Chemistry are available here at no cost. Page No 150: Question 6.1: Which of the ores mentioned in Table 6.1 can be concentrated by magnetic separation method? ANSWER: If the ore or the gangue can be attracted by the magnetic field, then the ore can be concentrated by the process of magnetic separation. Among the ores mentioned in table 6.1, the ores of iron such as haematite (Fe2O3), magnetite (Fe3O4), siderite (FeCO3), and iron pyrites (FeS2) can be separated by the process of magnetic separation. Page No 150: Question 6.2: What is the significance of leaching in the extraction of aluminium? ANSWER: In the extraction of aluminium, the significance of leaching is to concentrate pure alumina (Al2O3) from bauxite ore. Bauxite usually contains silica, iron oxide, and titanium oxide as impurities. In the process of leaching, alumina is concentrated by digesting the powdered ore with a concentrated solution of NaOH at 473-523 K and 35-36 bar. Under these conditions, alumina (Al2O3) dissolves as sodium meta-aluminate and silica (SiO2) dissolves as sodium silicate leaving the impurities behind. The impurities are then filtered and the solution is neutralized by passing CO2 gas. In this process, hydrated Al2O3 gets precipitated and sodium silicate remains in the solution. Precipitation is induced by seeding the solution with freshly prepared samples of hydrated Al2O3. Hydrated alumina thus obtained is filtered, dried, and heated to give back pure alumina (Al2O3). Page No 157: Question 6.3: The reaction, is thermodynamically feasible as is apparent from the Gibbs energy value. Why does it not take place at room temperature? ANSWER: The change in Gibbs energy is related to the equilibrium constant, K as . At room temperature, all reactants and products of the given reaction are in the solid state. As a result, equilibrium does not exist between the reactants and the products. Hence, the reaction does not take place at room temperature. However, at a higher temperature, chromium melts and the reaction takes place. We also know that according to the equation Increasing the temperature increases the value of making the value of  more and more negative. Therefore, the reaction becomes more and more feasible as the temperature is increased. Page No 157: Question 6.4: Is it true that under certain conditions, Mg can reduce SiO2 and Si can reduce MgO? What are those conditions? ANSWER: The temperature range in which  is lesser than, Mg can reduce SiO2 to Si. On the other hand, the temperatures range in which  is less than, Si can reduce MgO to Mg. The temperature at which ΔfG curves of these two substances intersect is 1966 K. Thus, at temperatures less than 1966 K, Mg can reduce SiO2 and above 1966 K, Si can reduce MgO. Page No 163: Question 6.1: Copper can be extracted by hydrometallurgy but not zinc. Explain. ANSWER: The reduction potentials of zinc and iron are lower than that of copper. In hydrometallurgy, zinc and iron can be used to displace copper from their solution. But to displace zinc, more reactive metals i.e., metals having lower reduction potentials than zinc such as Mg, Ca, K, etc. are required. But all these metals react with water with the evolution of H2 gas. As a result, these metals cannot be used in hydrometallurgy to extract zinc. Hence, copper can be extracted by hydrometallurgy but not zinc. Page No 163: Question 6.2: What is the role of depressant in froth floatation process? ANSWER: In the froth floatation process, the role of the depressants is to separate two sulphide ores by selectively preventing one ore from forming froth. For example, to separate two sulphide ores (ZnS and Pbs), NaCN is used as a depressant which selectively allows PbS to come with froth, but prevents ZnS from coming to froth. This happens because NaCN reacts with ZnS to form Na2[Zn(CN)4]. Page No 163: Question 6.3: Why is the extraction of copper from pyrites more difficult than that from its oxide ore through reduction? ANSWER: The Gibbs free energy of formation (ΔfG) of Cu2S is less than that of and. Therefore, H2 and C cannot reduce Cu2S to Cu. On the other hand, the Gibbs free energy of formation of  is greater than that of. Hence, C can reduce Cu2O to Cu. Hence, the extraction of copper from its pyrite ore is difficult than from its oxide ore through reduction. Page No 163: Question 6.4: Explain: (i) Zone refining (ii) Column chromatography. ANSWER: (i) Zone refining: This method is based on the principle that impurities are more soluble in the molten state of metal (the melt) than in the solid state. In the process of zone refining, a circular mobile heater is fixed at one end of a rod of impure metal. As the heater moves, the molten zone of the rod also moves with it. As a result, pure metal crystallizes out of the melt and the impurities pass onto the adjacent molten zone. This process is repeated several times, which leads to the segregation of impurities at one end of the rod. Then, the end with the impurities is cut off. Silicon, boron, gallium, indium etc. can be purified by this process. (ii) Column chromatography: Column chromatography is a technique used to separate different components of a mixture. It is a very useful technique used for the purification of elements available in minute quantities. It is also used to remove the impurities that are not very different in chemical properties from the element to be purified. Chromatography is based on the principle that different components of a mixture are differently adsorbed on an adsorbent. In chromatography, there are two phases: mobile phase and stationary phase. The stationary phase is immobile and immiscible. Al2O3 column …

NCERT Solutions for Class 12 Science Chemistry Chapter 6 – General Principles And Processes Of Isolation Of Elements Read More »

Find comprehensive NCERT Solutions for Class 12 Science Chemistry Chapter 5 on Surface Chemistry, featuring clear, step-by-step explanations. These solutions are widely favored by Class 12 Science students as they offer a convenient way to complete homework assignments and prepare for exams. All questions and answers from Chapter 5 of the NCERT Book for Class 12 Science Chemistry are available here at no cost, providing valuable assistance to students in their studies. Page No 127: Question 5.1: Write any two characteristics of Chemisorption. ANSWER: 1. Chemisorption is highly specific in nature. It occurs only if there is a possibility of chemical bonding between the adsorbent and the adsorbate. 2. Like physisorption, chemisorption also increases with an increase in the surface area of the adsorbent. Page No 127: Question 5.2: Why does physisorption decrease with the increase of temperature? ANSWER: Physisorption is exothermic in nature. Therefore, in accordance with Le-Chateliere’s principle, it decreases with an increase in temperature. This means that physisorption occurs more readily at a lower temperature. Page No 127: Question 5.3: Why are powdered substances more effective adsorbents than their crystalline forms? ANSWER: Powdered substances are more effective adsorbents than their crystalline forms because when a substance is powdered, its surface area increases and physisorption is directly proportional to the surface area of the adsorbent. Page No 133: Question 5.4: Why is it necessary to remove CO when ammonia is obtained by Haber’s process? ANSWER: It is important to remove CO in the synthesis of ammonia as CO adversely affects the activity of the iron catalyst, used in Haber’s process. Page No 133: Question 5.5: Why is the ester hydrolysis slow in the beginning and becomes faster after sometime? ANSWER: Ester hydrolysis can be represented as: The acid produced in the reaction acts as a catalyst and makes the reaction faster. Substances that act as catalysts in the same reaction in which they are obtained as products are known as autocatalysts. Page No 133: Question 5.6: What is the role of desorption in the process of catalysis? ANSWER: The role of desorption in the process of catalysis is to make the surface of the solid catalyst free for the fresh adsorption of the reactants on the surface. Page No 145: Question 5.7: What modification can you suggest in the Hardy-Schulze law? ANSWER: Hardy-Schulze law states that ‘the greater the valence of the flocculating ion added, the greater is its power to cause precipitation.’ This law takes into consideration only the charge carried by an ion, not its size. The smaller the size of an ion, the more will be its polarising power. Thus, Hardy-Schulze law can be modified in terms of the polarising power of the flocculating ion. Thus, the modified Hardy-Schulze law can be stated as ‘the greater the polarising power of the flocculating ion added, the greater is its power to cause precipitation.’ Page No 145: Question 5.8: Why is it essential to wash the precipitate with water before estimating it quantitatively? ANSWER: When a substance gets precipitated, some ions that combine to form the precipitate get adsorbed on the surface of the precipitate. Therefore, it becomes important to wash the precipitate before estimating it quantitatively in order to remove these adsorbed ions or other such impurities. Page No 146: Question 5.1: Distinguish between the meaning of the terms adsorption and absorption. Give one example of each. ANSWER: Adsorption is a surface phenomenon of accumulation of molecules of a substance at the surface rather than in the bulk of a solid or liquid. The substance that gets adsorbed is called the ‘adsorbate’ and the substance on whose surface the adsorption takes place is called the ‘adsorbent’. Here, the concentration of the adsorbate on the surface of the adsorbent increases. In adsorption, the substance gets concentrated at the surface only. It does not penetrate through the surface to the bulk of the solid or liquid. For example, when we dip a chalk stick into an ink solution, only its surface becomes coloured. If we break the chalk stick, it will be found to be white from inside. On the other hand, the process of absorption is a bulk phenomenon. In absorption, the substance gets uniformly distributed throughout the bulk of the solid or liquid. Page No 146: Question 5.2: What is the difference between physisorption and chemisorption? ANSWER: Physisorption Chemisorption 1. In this type of adsorption, the adsorbate is attached to the surface of the adsorbent with weak van der Waal’s forces of attraction. In this type of adsorption, strong chemical bonds are formed between the adsorbate and the surface of the adsorbent. 2. No new compound is formed in the process. New compounds are formed at the surface of the adsorbent. 3. It is generally found to be reversible in nature. It is usually irreversible in nature. 4. Enthalpy of adsorption is low as weak van der Waal’s forces of attraction are involved. The values lie in the range of 20-40 kJ mol−1. Enthalpy of adsorption is high as chemical bonds are formed. The values lie in the range of 40-400 kJ mol−1. 5. It is favoured by low temperature conditions. It is favoured by high temperature conditions. 6. It is an example of multi-layer adsorption It is an example of mono-layer adsorption. Page No 146: Question 5.3: Give reason why a finely divided substance is more effective as an adsorbent. ANSWER: Adsorption is a surface phenomenon. Therefore, adsorption is directly proportional to the surface area. A finely divided substance has a large surface area. Both physisorption and chemisorption increase with an increase in the surface area. Hence, a finely divided substance behaves as a good adsorbent. Page No 146: Question 5.4: What are the factors which influence the adsorption of a gas on a solid? ANSWER: There are various factors that affect the rate of adsorption of a gas on a solid surface. (1) Nature of the gas: Easily liquefiable gases such as NH3, HCl etc. are adsorbed to a great extent in comparison to gases such as H2, …

NCERT Solutions for Class 12 Science Chemistry Chapter 5 – Surface Chemistry Read More »

Find comprehensive NCERT Solutions for Class 12 Science Chemistry Chapter 4 on Chemical Kinetics, featuring clear step-by-step explanations. These solutions have gained immense popularity among Chemistry students in Class 12 Science, serving as valuable resources for homework completion and exam preparation. Free access to all questions and answers from Chapter 4 of the NCERT Book for Class 12 Science Chemistry is provided here, ensuring a convenient and effective study aid for students. Page No 98: Question 4.1: For the reaction R → P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds. ANSWER: Average rate of reaction  = 6.67 × 10−6 M s−1 Page No 98: Question 4.2: In a reaction, 2A → Products, the concentration of A decreases from 0.5 mol L−1 to 0.4 mol L−1 in 10 minutes. Calculate the rate during this interval? ANSWER: Average rate  = 0.005 mol L−1 min−1 = 5 × 10−3 M min−1 Page No 103: Question 4.3: For a reaction, A + B → Product; the rate law is given by,. What is the order of the reaction? ANSWER: The order of the reaction = 2.5 Page No 103: Question 4.4: The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y? ANSWER: The reaction X → Y follows second order kinetics. Therefore, the rate equation for this reaction will be: Rate = k[X]2 (1) Let [X] = a mol L−1, then equation (1) can be written as: Rate1 = k .(a)2 = ka2 If the concentration of X is increased to three times, then [X] = 3a mol L−1 Now, the rate equation will be: Rate = k (3a)2 = 9(ka2) Hence, the rate of formation will increase by 9 times. Page No 111: Question 4.5: A first order reaction has a rate constant 1.15 10−3 s−1. How long will 5 g of this reactant take to reduce to 3 g? ANSWER: From the question, we can write down the following information: Initial amount = 5 g Final concentration = 3 g Rate constant = 1.15 10−3 s−1 We know that for a 1st order reaction, = 444.38 s = 444 s (approx) Page No 111: Question 4.6: Time required to decompose SO2Cl2 to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction. ANSWER: We know that for a 1st order reaction, It is given that t1/2 = 60 min Page No 116: Question 4.7: What will be the effect of temperature on rate constant? ANSWER: The rate constant of a reaction is nearly doubled with a 10° rise in temperature. However, the exact dependence of the rate of a chemical reaction on temperature is given by Arrhenius equation, Where, A is the Arrhenius factor or the frequency factor T is the temperature R is the gas constant Ea is the activation energy Page No 116: Question 4.8: The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate Ea. ANSWER: It is given that T1 = 298 K ∴T2 = (298 + 10) K = 308 K We also know that the rate of the reaction doubles when temperature is increased by 10°. Therefore, let us take the value of k1 = k and that of k2 = 2k Also, R = 8.314 J K−1 mol−1 Now, substituting these values in the equation: We get: = 52897.78 J mol−1 = 52.9 kJ mol−1 Note: There is a slight variation in this answer and the one given in the NCERT textbook. Page No 116: Question 4.9: The activation energy for the reaction 2HI(g)→ H2 + I2(g) is 209.5 kJ mol−1 at 581K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy? ANSWER: In the given case: Ea = 209.5 kJ mol−1 = 209500 J mol−1 T = 581 K R = 8.314 JK−1 mol−1 Now, the fraction of molecules of reactants having energy equal to or greater than activation energy is given as: Page No 117: Question 4.1: From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants. (i) 3 NO(g) → N2O(g) Rate = k[NO]2 (ii) H2O2 (aq) + 3 I− (aq) + 2 H+ → 2 H2O (l) +  Rate = k[H2O2][I−] (iii) CH3CHO(g) → CH4(g) + CO(g) Rate = k [CH3CHO]3/2 (iv) C2H5Cl(g) → C2H4(g) + HCl(g) Rate = k [C2H5Cl] ANSWER: (i) Given rate = k [NO]2 Therefore, order of the reaction = 2 Dimension of  (ii) Given rate = k [H2O2] [I−] Therefore, order of the reaction = 2 Dimension of  (iii) Given rate = k [CH3CHO]3/2 Therefore, order of reaction =  Dimension of  (iv) Given rate = k [C2H5Cl] Therefore, order of the reaction = 1 Dimension of  Page No 117: Question 4.2: For the reaction: 2A + B → A2B the rate = k[A][B]2 with k = 2.0 × 10−6 mol−2 L2 s−1. Calculate the initial rate of the reaction when [A] = 0.1 mol L−1, [B] = 0.2 mol L−1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L−1. ANSWER: The initial rate of the reaction is Rate = k [A][B]2 = (2.0 × 10−6 mol−2 L2 s−1) (0.1 mol L−1) (0.2 mol L−1)2 = 8.0 × 10−9 mol−2 L2 s−1 When [A] is reduced from 0.1 mol L−1 to 0.06 mol−1, the concentration of A reacted = (0.1 − 0.06) mol L−1 = 0.04 mol L−1 Therefore, concentration of B reacted = 0.02 mol L−1 Then, concentration of B available, [B] = (0.2 − 0.02) mol L−1 = 0.18 mol L−1 After [A] is reduced to 0.06 mol L−1, the rate of the reaction is given by, Rate = k [A][B]2 = (2.0 × 10−6 mol−2 L2 s−1) (0.06 mol L−1) (0.18 mol L−1)2 = 3.89 mol L−1 s−1 Page No 117: Question 4.3: The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2 and H2 if k = 2.5 × 10−4 mol−1 L s−1? ANSWER: The decomposition of NH3 on platinum surface is represented by the following equation. Therefore, However, it is given that the reaction is of zero order. Therefore, Therefore, the rate of production of N2 is And, the rate of production of H2 is = 7.5 × 10−4 mol L−1 s−1 Page No 117: Question 4.4: The decomposition of dimethyl ether leads to the formation of CH4, H2 and CO and the reaction rate is given by Rate = k [CH3OCH3]3/2 …

NCERT Solutions for Class 12 Science Chemistry Chapter 4 – Chemical Kinetics Read More »

Explore comprehensive NCERT solutions for Class 12 Science Chemistry Chapter 3 on Electrochemistry. These step-by-step explanations are highly sought after by Chemistry students for quick completion of homework and effective exam preparation. The solutions, derived from the NCERT book, cater to the needs of class 12 Science students, offering valuable assistance in understanding Electrochemistry concepts. Access all questions and answers from Chapter 3 for free, making it a convenient resource for academic support and exam readiness. Page No 68: Question 3.1: How would you determine the standard electrode potential of the systemMg2+ | Mg? ANSWER: The standard electrode potential of Mg2+ | Mg can be measured with respect to the standard hydrogen electrode, represented by Pt(s), H2(g) (1 atm) | H+(aq)(1 M). A cell, consisting of Mg | MgSO4 (aq 1 M) as the anode and the standard hydrogen electrode as the cathode, is set up. Then, the emf of the cell is measured and this measured emf is the standard electrode potential of the magnesium electrode. Here, for the standard hydrogen electrode is zero. ∴ Page No 68: Question 3.2: Can you store copper sulphate solutions in a zinc pot? ANSWER: Zinc is more reactive than copper. Therefore, zinc can displace copper from its salt solution. If copper sulphate solution is stored in a zinc pot, then zinc will displace copper from the copper sulphate solution. Hence, copper sulphate solution cannot be stored in a zinc pot. Page No 68: Question 3.3: Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions. ANSWER: Substances that are stronger oxidising agents than ferrous ions can oxidise ferrous ions. ;  = −0.77 V This implies that the substances having higher reduction potentials than+0.77 V can oxidise ferrous ions to ferric ions. Three substances that can do so are F2, Cl2, and O2. Page No 73: Question 3.4: Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10. ANSWER: For hydrogen electrode, , it is given that pH = 10 ∴[H+] = 10−10 M Now, using Nernst equation: =  = −0.0591 log 1010 = −0.591 V Page No 73: Question 3.5: Calculate the emf of the cell in which the following reaction takes place: Given that = 1.05 V ANSWER: Applying Nernst equation we have: = 1.05 − 0.02955 log 4 × 104 = 1.05 − 0.02955 (log 10000 + log 4) = 1.05 − 0.02955 (4 + 0.6021) = 0.914 V Page No 73: Question 3.6: The cell in which the following reactions occurs: has = 0.236 V at 298 K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction. ANSWER: Here, n = 2,  T = 298 K We know that: = −2 × 96487 × 0.236 = −45541.864 J mol−1 = −45.54 kJ mol−1 Again, −2.303RT log Kc = 7.981 ∴Kc = Antilog (7.981) = 9.57 × 107 Page No 84: Question 3.7: Why does the conductivity of a solution decrease with dilution? ANSWER: The conductivity of a solution is the conductance of ions present in a unit volume of the solution. The number of ions (responsible for carrying current) decreases when the solution is diluted. As a result, the conductivity of a solution decreases with dilution. Page No 84: Question 3.8: Suggest a way to determine the value of water. ANSWER: Applying Kohlrausch’s law of independent migration of ions, the value of water can be determined as follows: Hence, by knowing the values of HCl, NaOH, and NaCl, the value of water can be determined. Page No 84: Question 3.9: The molar conductivity of 0.025 mol L−1 methanoic acid is 46.1 S cm2 mol−1. Calculate its degree of dissociation and dissociation constant. Given λ °(H+) = 349.6 S cm2 mol−1 and λ °(HCOO−) = 54.6 S cm2 mol ANSWER: C = 0.025 mol L−1 Now, degree of dissociation: Thus, dissociation constant: Page No 87: Question 3.10: If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire? ANSWER: I = 0.5 A t = 2 hours = 2 × 60 × 60 s = 7200 s Thus, Q = It = 0.5 A × 7200 s = 3600 C We know that  number of electrons. Then, Hence, number of electrons will flow through the wire. Page No 87: Question 3.11: Suggest a list of metals that are extracted electrolytically. ANSWER: Metals that are on the top of the reactivity series such as sodium, potassium, calcium, lithium, magnesium, aluminium are extracted electrolytically. Page No 87: Question 3.12: What is the quantity of electricity in coulombs needed to reduce 1 mol of ? Consider the reaction: ANSWER: The given reaction is as follows: Therefore, to reduce 1 mole of , the required quantity of electricity will be: =6 F = 6 × 96487 C = 578922 C Page No 91: Question 3.13: Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging. ANSWER: A lead storage battery consists of a lead anode, a grid of lead packed with lead oxide (PbO2) as the cathode, and a 38% solution of sulphuric acid (H2SO4) as an electrolyte. When the battery is in use, the following cell reactions take place: At anode:  At cathode:  The overall cell reaction is given by, When a battery is charged, the reverse of all these reactions takes place. Hence, on charging,  present at the anode and cathode is converted into and respectively. Page No 91: Question 3.14: Suggest two materials other than hydrogen that can be used as fuels in fuel cells. ANSWER: Methane and methanol can be used as fuels in fuel cells. Page No 91: Question 3.15: Explain how rusting of iron is envisaged as setting up of an electrochemical cell. ANSWER: In the process of corrosion, due to the presence of air and moisture, oxidation takes place at a particular spot of an object made of iron. That spot behaves as the anode. The reaction at the anode is given by, Electrons released at the anodic spot move through the metallic object and go to another spot of the object. There, in the presence of H+ ions, the electrons …

NCERT Solutions for Class 12 Science Chemistry Chapter 3 – Electrochemistry Read More »

Explore the comprehensive NCERT Solutions for Class 12 Science Chemistry Chapter 2 – Solutions, featuring easy-to-follow step-by-step explanations. These solutions have gained immense popularity among Class 12 Science students, serving as a valuable resource for completing homework efficiently and preparing for exams. All the questions and answers from Chapter 2 of the NCERT Book for Class 12 Science Chemistry are available here at no cost, ensuring convenient access for your academic needs. Page No 37: Question 2.1: Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride (CCl4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride. ANSWER: Mass percentage of C6H6  Mass percentage of CCl4 Alternatively, Mass percentage of CCl4 = (100 − 15.28)% = 84.72% Page No 37: Question 2.2: Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride. ANSWER: Let the total mass of the solution be 100 g and the mass of benzene be 30 g. ∴Mass of carbon tetrachloride = (100 − 30)g = 70 g Molar mass of benzene (C6H6) = (6 × 12 + 6 × 1) g mol−1 = 78 g mol−1 ∴Number of moles of  = 0.3846 mol Molar mass of carbon tetrachloride (CCl4) = 1 × 12 + 4 × 35.5 = 154 g mol−1 ∴Number of moles of CCl4 = 0.4545 mol Thus, the mole fraction of C6H6 is given as: = 0.458 Page No 37: Question 2.3: Calculate the molarity of each of the following solutions: (a) 30 g of Co(NO3)2. 6H2O in 4.3 L of solution (b) 30 mL of 0.5 M H2SO4 diluted to 500 mL. ANSWER: Molarity is given by: (a) Molar mass of Co (NO3)2.6H2O = 59 + 2 (14 + 3 × 16) + 6 × 18 = 291 g mol−1 ∴Moles of Co (NO3)2.6H2O = 0.103 mol Therefore, molarity  = 0.023 M (b) Number of moles present in 1000 mL of 0.5 M H2SO4 = 0.5 mol ∴Number of moles present in 30 mL of 0.5 M H2SO4 = 0.015 mol Therefore, molarity = 0.03 M Page No 37: Question 2.4: Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal aqueous solution. ANSWER: Molar mass of urea (NH2CONH2) = 2(1 × 14 + 2 × 1) + 1 × 12 + 1 × 16 = 60 g mol−1 0.25 molar aqueous solution of urea means: 1000 g of water contains 0.25 mol = (0.25 × 60)g of urea = 15 g of urea That is, (1000 + 15) g of solution contains 15 g of urea Therefore, 2.5 kg (2500 g) of solution contains  = 36.95 g = 37 g of urea (approximately) Hence, mass of urea required = 37 g Note: There is a slight variation in this answer and the one given in the NCERT textbook. Page No 37: Question 2.5: Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL-1. ANSWER: (a) Molar mass of KI = 39 + 127 = 166 g mol−1 20% (mass/mass) aqueous solution of KI means 20 g of KI is present in 100 g of solution. That is, 20 g of KI is present in (100 − 20) g of water = 80 g of water Therefore, molality of the solution  = 1.506 m = 1.51 m (approximately) (b) It is given that the density of the solution = 1.202 g mL−1 ∴Volume of 100 g solution  = 83.19 mL = 83.19 × 10−3 L Therefore, molarity of the solution  = 1.45 M (c) Moles of KI  Moles of water  Therefore, mole fraction of KI  = 0.0263 Page No 41: Question 2.6: H2S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H2S in water at STP is 0.195 m, calculate Henry’s law constant. ANSWER: It is given that the solubility of H2S in water at STP is 0.195 m, i.e., 0.195 mol of H2S is dissolved in 1000 g of water. Moles of water  = 55.56 mol ∴Mole fraction of H2S, x = 0.0035 At STP, pressure (p) = 0.987 bar According to Henry’s law: p = KHx = 282 bar Page No 41: Question 2.7: Henry’s law constant for CO2 in water is 1.67 × 108 Pa at 298 K. Calculate the quantity of CO2 in 500 mL of soda water when packed under 2.5 atm CO2 pressure at 298 K. ANSWER: It is given that: KH = 1.67 × 108 Pa = 2.5 atm = 2.5 × 1.01325 × 105 Pa = 2.533125 × 105 Pa According to Henry’s law: = 0.00152 We can write,  [Since, is negligible as compared to] In 500 mL of soda water, the volume of water = 500 mL [Neglecting the amount of soda present] We can write: 500 mL of water = 500 g of water = 27.78 mol of water Now,  Hence, quantity of CO2 in 500 mL of soda water = (0.042 × 44)g = 1.848 g Page No 47: Question 2.8: The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase. ANSWER: It is given that: = 450 mm of Hg = 700 mm of Hg ptotal = 600 mm of Hg From Raoult’s law, we have: Therefore, total pressure,  Therefore,  = 1 − 0.4 = 0.6 Now,  = 450 × 0.4 = 180 mm of Hg = 700 × 0.6 = 420 mm of Hg Now, in the vapour phase: Mole fraction of liquid A = 0.30 And, mole fraction of liquid B = 1 − 0.30 = 0.70 Page No 55: Question 2.9: Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH2CONH2) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering. ANSWER: It is given that vapour pressure of water, = 23.8 mm of Hg Weight of water taken, w1 = 850 g Weight of urea taken, w2 = 50 g Molecular weight of …

NCERT Solutions for Class 12 Science Chemistry Chapter 2 – Solutions Read More »

Explore comprehensive NCERT solutions for Class 12 Science Chemistry Chapter 1 – The Solid State, complete with clear step-by-step explanations. Widely favored among Chemistry students in Class 12 Science, these solutions for The Solid State serve as valuable resources for efficiently completing homework assignments and preparing for exams. Free access to all questions and answers from Chapter 1 of the NCERT Book for Class 12 Science Chemistry is provided here, ensuring a convenient and effective study experience. Page No 4: Question 1.1: Why are solids rigid? ANSWER: The intermolecular forces of attraction that are present in solids are very strong. The constituent particles of solids cannot move from their positions i.e., they have fixed positions. However, they can oscillate about their mean positions. This is the reason solids are rigid. Page No 4: Question 1.2: Why do solids have a definite volume? ANSWER: The intermolecular forces of attraction that are present in solids are very strong. The constituent particles of solids have fixed positions i.e., they are rigid. Hence, solids have a definite volume. Page No 4: Question 1.3: Classify the following as amorphous or crystalline solids: Polyurethane, naphthalene, benzoic acid, teflon, potassium nitrate, cellophane, polyvinyl chloride, fibre glass, copper. ANSWER: Amorphous solids Polyurethane, teflon, cellophane, polyvinyl chloride, fibre glass Crystalline solids Naphthalene, benzoic acid, potassium nitrate, copper Page No 4: Question 1.4: Why is glass considered a super cooled liquid? ANSWER: Similar to liquids, glass has a tendency to flow, though very slowly. Therefore, glass is considered as a super cooled liquid. This is the reason that glass windows and doors are slightly thicker at the bottom than at the top. Page No 4: Question 1.5: Refractive index of a solid is observed to have the same value along all directions. Comment on the nature of this solid. Would it show cleavage property? ANSWER: An isotropic solid has the same value of physical properties when measured along different directions. Therefore, the given solid, having the same value of refractive index along all directions, is isotropic in nature. Hence, the solid is an amorphous solid. When an amorphous solid is cut with a sharp edged tool, it cuts into two pieces with irregular surfaces. Page No 6: Question 1.6: Classify the following solids in different categories based on the nature of intermolecular forces operating in them: Potassium sulphate, tin, benzene, urea, ammonia, water, zinc sulphide, graphite, rubidium, argon, silicon carbide. ANSWER: Potassium sulphate → Ionic solid Tin → Metallic solid Benzene → Molecular (non-polar) solid Urea → Polar molecular solid Ammonia → Polar molecular solid Water → Hydrogen bonded molecular solid Zinc sulphide → Ionic solid Graphite → Covalent or network solid Rubidium → Metallic solid Argon → Non-polar molecular solid Silicon carbide → Covalent or network solid Page No 6: Question 1.7: Solid A is a very hard electrical insulator in solid as well as in molten state and melts at extremely high temperature. What type of solid is it? ANSWER: The given properties are the properties of a covalent or network solid. Therefore, the given solid is a covalent or network solid. Examples of such solids include diamond (C) and quartz (SiO2). Page No 6: Question 1.8: Ionic solids conduct electricity in molten state but not in solid state. Explain. ANSWER: In ionic compounds, electricity is conducted by ions. In solid state, ions are held together by strong electrostatic forces and are not free to move about within the solid. Hence, ionic solids do not conduct electricity in solid state. However, in molten state or in solution form, the ions are free to move and can conduct electricity. Page No 6: Question 1.9: What type of solids are electrical conductors, malleable and ductile? ANSWER: Metallic solids are electrical conductors, malleable, and ductile. Page No 12: Question 1.10: Give the significance of a ‘lattice point’. ANSWER: The significance of a lattice point is that each lattice point represents one constituent particle of a solid which may be an atom, a molecule (group of atom), or an ion. Page No 12: Question 1.11: Name the parameters that characterize a unit cell. ANSWER: The six parameters that characterise a unit cell are as follows. (i) Its dimensions along the three edges, a, b, and c These edges may or may not be equal. (ii) Angles between the edges These are the angle ∝ (between edges b and c), β (between edges a and c), and γ (between edges a and b). Page No 12: Question 1.12: Distinguish between (i)Hexagonal and monoclinic unit cells (ii) Face-centred and end-centred unit cells. ANSWER: (i) Hexagonal unit cell For a hexagonal unit cell, Monoclinic unit cell For a monoclinic cell, (ii) Face-centred unit cell In a face-centred unit cell, the constituent particles are present at the corners and one at the centre of each face. End-centred unit cell An end-centred unit cell contains particles at the corners and one at the centre of any two opposite faces. Page No 12: Question 1.13: Explain how much portion of an atom located at (i) corner and (ii) body-centre of a cubic unit cell is part of its neighbouring unit cell. ANSWER: (i)An atom located at the corner of a cubic unit cell is shared by eight adjacent unit cells. Therefore, portion of the atom is shared by one unit cell. (ii)An atom located at the body centre of a cubic unit cell is not shared by its neighbouring unit cell. Therefore, the atom belongs only to the unit cell in which it is present i.e., its contribution to the unit cell is 1. Page No 21: Question 1.14: What is the two dimensional coordination number of a molecule in square close packed layer? ANSWER: In square close-packed layer, a molecule is in contact with four of its neighbours. Therefore, the two-dimensional coordination number of a molecule in square close-packed layer is 4. Page No 21: Question 1.15: A compound forms hexagonal close-packed structure. What is the total number of voids in 0.5 mol of it? How many of these are tetrahedral voids? ANSWER: Number of close-packed particles = 0.5 × 6.022 × 1023 = 3.011 × 1023 Therefore, number of octahedral voids = 3.011 × 1023 And, number of tetrahedral voids = 2 × 3.011 × 1023 = 6.022 ×1023 Therefore, total number of voids = 3.011 × 1023 + 6.022 × 1023 = 9.033 × 1023 Page No 22: Question 1.16: A compound is formed by two elements M and N. The element N forms ccp and atoms of M occupy 1/3rd of tetrahedral voids. What is the …

NCERT Solutions for Class 12 Science Chemistry Chapter 1 – The Solid State Read More »