CBSE Class 12 th Chemistry Solved Guess Paper

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2 Solved Guess Paper 216 Subject hemistry lass XII Set II Time allowed: 3 hours Maximum Marks: 7 General Instructions: (i) All questions are compulsory. (ii) Q. No. 1 to 5 are very short answer type, carrying 1 mark each. (iii) Q. No. 6 to 1 are short answer type, carrying 2 marks each. (iv) Q. No. 11 to 22 are short answer type, carrying 3 marks each. (v) Q. No. 23 is a value based question carrying 4 marks. (vi) Q. No. 24 to 26 are long answer type, carrying 5 marks each. (vii) Use log tables if necessary, use of calculators is not allowed. Q.1 Why is HF acid stored in wax coated glass bottles? HF reacts with glass. It dissolves SiO 2 present in glass and forms hydrofluorosilicic acid. SiO 2 + 6HF H2SiF 6 + 2H2O However, HF is absolutely inactive towards wax. That s why it is stored in wax coated glass bottles. Q.2 Write the IUPA name of the following compound: The IUPA name of the given compound is: 5-ethyl-3-methyl-4-propyloctane

3 Q.3 What is Steam Reforming? In Haber s process, hydrogen is obtained by reacting methane with steam in presence of NiO as catalyst. The process is known as steam reforming. Q.4 Write the basic difference between primitive unit cell and centered unit cell. Primitive Unit ell: When constituent particles are present only on the corner positions of a unit cell, it is called as primitive unit cell. entered Unit ell: When a unit cell contains one or more constituent particles present at positions other than corners in addition to those at corners, it is called a centered unit cell. Q.5 Write the structure of Ethyl 3-methylbut-3-en-oate Structure of Ethyl 3-methylbut-3-en-oate is: Q.6 (a) Why do the transition elements have higher enthalpies of atomisation? (b) In 3d series (Sc to Zn), which element has the lowest enthalpy of atomisation and why? (a) Transition elements have higher enthalpies of atomisation due to presence of strong metallic bond. Stronger the metallic bond in between the atoms of an element, higher the enthalpy of atomisation. (b) The metallic bond arises due to presence of unpaired electrons in the (n - 1) d subshell. In 3d series (Sc to Zn) all elements have some unpaired electrons except zinc. Hence Zn has weakest metallic bonding and lowest enthalpy of atomisation.

) 2 4 3 Trisoxalatochromium (III) anion Q.8 E for r 3+ + 3e - r and r 3+ + e - r 2+ are.74 V and.4 V respectively.

4 Q.7 Draw the structures of the following coordination compounds along with giving their names: (i) o( en) l en ethan,2 diamine (ii) 3 r( O ) (i) o( en) 2 l 2 is-dichlorobis (ethylenediamine) cobalt (III) chloride (ii) 3 r( O ) Trisoxalatochromium (III) anion Q.8 E for r e - r and r 3+ + e - r 2+ are.74 V and.4 V respectively. Find E for r e - r. We have, E for r e - r =.74 V (i) E for r 3+ + e - r 2+ =.4 V (ii) From (i) and (ii), we get

5 r 3 e r; G 3.74 F 3 1 r e r ; G 1.4 F r 2 e r; G 2 E F E F F F E 1.82 E E F F.91V V OR Q8. Determine the values of equilibrium constant (K c ) and ΔG for the following reaction: Ni s + 2Ag + s Ni 2+ aq + 2Ag s, E =1.5 V (1F=965 /mol) + 2+ Ni s + 2Ag s Ni aq + 2Ag s, E =1.5 V E.25V E Ni Ag /Ni /Ag 2+ + ell R L ell ell.8 V Galvanic cellof the given cell reaction is: Ni s Ni aq Ag aq Ag s Now, E = E E E =.8.25 E r =1.5V G = nfe ell r r G = G = Kj/mol Q.9 Name the reagents used in following reactions: (i)

6 H 3 H 3 O OH (ii) OOH lh 2 OOH (i) Reagent used: This can be achieved through Grignard reaction. 1. MgI 2. H 2 O/H + H 3 O 1. MgI 2. H 2 O/H + (ii) When carboxylic acids react with chlorine or bromine in the presence of red phosphorous, the alpha-hydrogen atoms are replaced by halogen atoms. It is called Hell-Volhard- Zelinsky(H.V.Z) reaction. Reagent used for given conversion is: P/ l 2 OOH lh 2 OOH P/l 2 H 3 Q.1 (a) State the positive and negative deviation from the Roult s Law. (b) When chloroform and acetone are mixed, the temperature of new solution formed is increased. Is it positive or negative deviation? (a)the deviation is positive when partial pressure of each component and the resultant total pressure are greater than pressure expected on the basis of Roult s Law. This happens because the intermolecular force between solute-solvent is weaker than solute-solute or solvent-solvent molecule. For example: solution formed by mixing ethyl alcohol and water. The deviation is negative when partial pressure of each component and the resultant total pressure are less than pressure expected on the basis of Roult s Law. This happens because the intermolecular force between solute-solvent is stronger than solute-solute or solventsolvent molecule. For example: solution formed by mixing chloroform and acetone. (b) In case of negative deviation H mix is negative that means heat is released during the formation of solution. Since the temperature of the solution formed is increased hence it shows the negative deviation from the Roult s Law. OH

7 Q.11 (a) How is wrought iron different from steel? (b) What is the role of O in extraction of Iron? (c) Reduction of metal oxide to metal becomes easier if the metal is in liquid state. Why? (a) Wrought Iron is the purest form of iron & is prepared from cast iron. On the other hand, Steel is an alloy of iron. Steel possesses greater tensile strength than wrought iron. (b) Iron is extracted in pure form from its oxides. The most commonly used iron ores are hematite, Fe 2 O 3, and magnetite, Fe 3 O 4. The chemistry involve behind the extraction of iron is reduction of iron oxide in blast furnace in presence of coke. But coke is not the actual reducing agent. oke is produced by heating coal in the absence of air. The coke burns in the presence of hot air to form carbon dioxide. At the high temperature, carbon dioxide reacts with carbon to produce carbon monoxide. It is the carbon monoxide which is the main reducing agent in the furnace. + O 2 2O Fe 2 O 3 +3O 2Fe+3O 2 (c) The entropy of metal in liquid state is higher than that of the same metal in solid state. For reduction of metal oxide the Gibb s energy should be negative. When the metal formed is in liquid state, and metal oxide being reduced is in solid state the value of ΔS will be more on positive side for the reduction. Metal oxide to Metal Solid to liquid ΔS solid < ΔS liquid As the value of TΔS increases it leads to the more ve value of Gibb s free energy and makes the reaction feasible. Q.12 Give reasons for the following statements: (a) opper (I) ion is not known in aqueous solution (b) Transition metals form complex compounds (c) Actinoids exhibit greater range of oxidation states than lanthanoids. (a) In aqueous solution, u + ion undergoes oxidation to u 2+ ion. u e u E V ( aq) ( s), red.52 u 2 e u, E.34V 2 ( aq) ( s) red

u + has more reduction electrode potential value. Thus, it undergoes oxidation reaction quite feasibly. Thus, u + ion is not known in aqueous solution.

-Availability of empty d-orbitals of suitable energy to accept the non-bonding electrons from ligand.

8 u + has more reduction electrode potential value. Thus, it undergoes oxidation reaction quite feasibly. Thus, u + ion is not known in aqueous solution. (b) Transition metals form complexes due to the following reasons: - Small size of the metal ion and high charge density is suitable for forming bonds with ligands. -Availability of empty d-orbitals of suitable energy to accept the non-bonding electrons from ligand. (c) The actinoids exhibits greater range of oxidation states because there is very small energy gap between the 5f, 6d and 7s sub shells. Thus, all their electrons can take part in bond formation. Q.13 (a) Write the hybridisation of Ni in case of [Ni(l) 4 ] 2- and [Ni(N) 4 ] 2- with explanation. (b) On the basis of crystal field theory, write the electronic configuration for d 4 ion Δ P (c) [Nil 4 ] 2- is paramagnetic while Ni(O) 4 is diamagnetic though both are tetrahedral. Why? (Atomic no. Ni =28) (a) Hybridisation of Ni in [Ni(l)4]2- is sp 3 Hybridisation of Ni in [Ni(N)4]2- is dsp 2 In both the cases oxidation state of Ni is +2 with d 8 configuration. Since N is strong field ligand it force the pairing of electrons in d-orbitals leads to dsp 2 hybridisation while l is weak field ligand and it does not force for pairing of electrons. So, d- orbital does not take part in hybridisation and outer s and p orbitals are used to form sp 3 hybridisation. (b) Given that FSE is greater than pairing energy then it will cause the pairing of electrons rather than transferring to next higher level. So all 4 electrons will occupy t 2g orbital and will form low spin octahedral complex.

(c) The valence shell electronic configuration of ground state Ni atom is 3d 8 4s 2. In [Nil 4 ] 2-, Ni is in +2 oxidation state and electronic configuration will be 3d 8. l - is a weak field ligand.

Ni 2+ undergoes sp 3 hybridization to make bonds with l - ligands in tetrahedral geometry. As there are unpaired electrons in the d-orbitals, [Nil 4 ] 2- is paramagnetic.

9 (c) The valence shell electronic configuration of ground state Ni atom is 3d 8 4s 2. In [Nil 4 ] 2-, Ni is in +2 oxidation state and electronic configuration will be 3d 8. l - is a weak field ligand. In presence of weak field ligand, NO pairing of d-electrons occurs. So, there is 2 unpaired electrons in d-orbital. Ni 2+ undergoes sp 3 hybridization to make bonds with l - ligands in tetrahedral geometry. As there are unpaired electrons in the d-orbitals, [Nil 4 ] 2- is paramagnetic. In Ni(O) 4, Ni is in oxidation state and electronic configuration will be 3d 8 4s 2. O is a strong field ligand. It pushes the 4s electrons to 3d orbital and also cause the paring of all 3d electrons. The empty 4s and three 4p orbitals undergo sp 3 hybridization and form bonds with O ligands to give Ni(O) 4 a tetrahedral geometry. Since there is no unpaired electrons present in Ni(O) 4 it is diamagnetic. Q.14 Predict the products of following reactions: (i) H 3 O HN (ii) 6 H 5 H 2 (a) KMnO 4 /KOH (b) H + (iii) OOH NH 3 /heat

10 (i) This is an example of Nucleophilic addition of N - to carbonyl carbon. H 3 O HN H 3 N OH (ii) When alkyl benzene is treated with alkaline KMnO 4, it forms potassium salt of benzoic acid that yield benzoic acid on treatment with dilute acid 6 H 5 H 2 (a) KMnO 4 /KOH (b) H + OOH (iii) arboxylic acid on treatment with ammonia form ammonium salt of acid which on heating gives amide. OOH NH 3 /heat ONH 2 + H 2 O Q.15 State and explain the difference between α-form of glucose and β-form of glucose. α -form of glucose and β -form of glucose can be distinguished by the position of hydroxyl group on the first carbon atom. In open chain β -glucose, the hydroxyl group on the first carbon atom is present towards the left side whereas in the closed ring β glucose, the hydroxyl group on the first carbon atom is above the plane of the ring. In open chain α glucose, the hydroxyl group on the first carbon atom is towards the right whereas, in the closed ring α -glucose, the hydroxyl group on the first carbon atom is below the plane of the ring.

11 BSE lass 12th hemistry Solved Guess Paper 216 (Set-II) 4% OFF Publisher : Jagran Josh Author : Jagran Josh Expert Type the URL : Get this ebook

SIR MICHELANGELO REFALO SIXTH FORM

SIR MICHELANGELO REFALO SIXTH FORM SIR MIELANGELO REFALO SIXT FORM alf-yearly Exam 2016 Subject: hemistry ADV 1 ST Time: 3 hours Answer 6 questions. All questions carry equal marks. You are reminded of the importance of clear presentation