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CHEMISTRY FINALTEST PART -A Q.1 If % ionic character of O–H bond is 10% & O–H bond distance is 1 4.8 Å, then calculate dipole moment of H2O molecule (in Debye) given H–O–H bond angle is 105°. [Use sin 52.5° = 0.8] [5] [Ans. 0.12 D] [Sol.  q × d = 4.8 × 10–10 × 0.1 × 1 × 10–8 4.8 = 0.1 × 10–18 ese cm = 0.1 D  {2}  net = 2 cos 52.5 = 2 × 0.1 × (1 – 0.64) D = 2 × 0.1 × 0.6  net = 0.12 D  {3} 2 Marks if only bond dipole is calculated ] Q.2 Arrange in ascending order (reasons not required) (a) | Hhydrogenation| of : 1, 2 Pentadiene, 1, 3 Pentadiene & 1, 4 Pentadiene (b) Ionic conductance of : F–, Na+ & Cs+. (c) Dipole Moment of : , , , (d) Boiling point : n-pentane, isopentane, neo-pentane [1.5+1+1.5+1] [Sol.(a) Reasons not required for marks. No marks for partially correct order. (1) (2) (3) C = C = C – C – C, C = C – C = C – C, C = C – C – C = C Most stable (2) > (3) > (1) |HHE|  (2) Tcritical, T = Tcritical, T

PART -A Q.1 If % ionic character of O–H bond is 10% & O–H bond distance is 1 4.8 Å, then calculate dipole moment of H2O molecule (in Debye) given H–O–H bond angle is 105°. [Use sin 52.5° = 0.8] [5] [Ans. 0.12 D] [Sol.  q × d = 4.8 × 10–10 × 0.1 × 1 × 10–8 4.8 = 0.1 × 10–18 ese cm = 0.1 D  net = 2 cos 52.5 = 2 × 0.1 × (1 – 0.64) D = 2 × 0.1 × 0.6  net = 0.12 D ] Q.2 Arrange in ascending order (reasons not required) (a) | Hhydrogenation| of : 1, 2 Pentadiene, 1, 3 Pentadiene & 1, 4 Pentadiene (b) Ionic conductance of : F–, Na+ & Cs+. (c) Dipole Moment of : , , , (d) Boiling point : n-pentane, isopentane, neo-pentane [1.5+1+1.5+1] [Sol.(a) (1) (2) (3) C = C = C – C – C, C = C – C = C – C, C = C – C – C = C Most stable (2) > (3) > (1) |HHE|  (2) Tcritical, T = Tcritical, T

SYLLABUS XI (PQRS) REVIEW TEST-6 PHYSICS Whatever has been taught upto Rotational Mechanics & Elasticity, Expansion & Calorimetery. MATHEMATICS 1. Compound angles 2. Logarithm 3. Quadratic Equation 4. Sequence and progression 5. Trigonomertry equation and inequation 6. Solution of triangle 7. Straight Line 8. Circle and whatever has been taught in class till date. CHEMISTRY 1. Mole concept 2. Gaseous state 3. Atomic structure 4. Periodic preoperties 5. Chemical Bonding (whatever has been taught in class till date) SYLLABUS XI (PQRS) REVIEW TEST-6 PHYSICS Whatever has been taught upto Rotational Mechanics & Elasticity, Expansion & Calorimetery. MATHEMATICS 1. Compound angles 2. Logarithm 3. Quadratic Equation 4. Sequence and progression 5. Trigonomertry equation and inequation 6. Solution of triangle 7. Straight Line 8. Circle and whatever has been taught in class till date. CHEMISTRY 1. Mole concept 2. Gaseous state 3. Atomic structure 4. Periodic preope...

speed = | v | = = 13 v ] 2 0 Q.4 A force of (3 ˆi 1.5ˆj) N acts on a 5 kg body. The body is at a position of (2 ˆi  3ˆj) m and is travelling at 4 ms–1. The force acts on the body until it is at the position (ˆi + 5ˆj) m. Assuming no other force does work on the body, find the final speed of the body. [Ans. [Sol. Given ms–1] Mass of the body = 5 kg Force F = 3ˆi 1.5ˆj Work energy theorem 1 × 5 × v2 – 1 × 5 × 16 2 2 = (3 ˆi 1.5ˆj) [ (ˆi + 5ˆj) – (2 ˆi  3ˆj) ] = – 3 – 12  v2 = (40 – 15) × 2 5  v = ms–1 ] Q.5 A particle of mass m is moving in a straight line under the influence of conservative force such that its velocity v varies with displacement x from a fixed point as v2 =  – x2 where  and  are constants. Considering that fixed point as reference for zero potential energy, find the total mechanical energyof the 1 particle at any instant t. [Ans. 2 m] [Sol. E = U + K = const. = U0 + K0  1 = 0 + 2 mv2 At x = 0 v2 =  –  × 02  v2 =  1 E = 2 m ] Q.6 A car begi...

REVIEW TEST-6 / 5 Q.1QE Consider the quadratic polynomial f (x) = x2 – 4ax + 5a2 – 6a. (a) Find the smallest positive integral value of 'a' for which f (x) is positive for every real x. (b) Find the largest distance between the roots of the equation f (x) = 0. [2.5 + 2.5] [Ans. (a) 7, (b) a = 6] [Sol. (a) D 0 a(a – 6) > 0  a > 6 or a 0 logx3 (x2  x) 1  x > – 2  x  (– 2, 0)  (1, ) Case-II: 0 x + 3 x2 – x 0 x2 – 2x – 3 0 (x – 3)(x + 1)

Q.1 (a) The first ionisation energy of Mg is higher than that of Na, but on the other hand the second ionisation energy of Na is very much higher than that of Mg. Explain [2] (b) Select the one having lower first ionisation energy in each given pair with explanation (i) I and I– (ii) Ba and Sr (iii) Be and B [3] [Sol:(a) Mg  1s2, 2s2 2p6, 3s2 , Na  1s2, 2s22p6, 3s1 Mg+  1s2, 2s22p6, 3s1 , Na+  1s2, 2s22p6 Electronic configuration of Mg (3s2) is more stable than that of Na(3s1) is less stable than Na+ (2, 8) so IE1 of Mg is higher than IE1 of Na but IE2 if Na is very much higher than IE2 of Mg. (b) (i) I–, since Zeff decreases due to increase in size. (ii) Ba, since in the same group IE decreases down the group. (iii) B, since B has EC 1s2, 2s22p1 & B has EC 1s2, 2s2 which is more stable . ] Q.2 Select those species from the following having X–X linkage (X-represent central atom) and draw their Lewis structure. [5] (i) H N O (ii) N O (iii) H P O (iv) S O2 2 2 2...

REVIEW TEST-6 Class : XI (PQRS) Time : 100 min Max. Marks : 100 General Remarks: INSTRUCTIONS 1. The question paper contain 20 questions and 24 pages. All questions are compulsory. Please ensure that the Question Paper you have received contains all the QUESTIONS and Pages. If you found some mistake like missing questions or pages then contact immediately to the Invigilator. 2. Each question should be done only in the space provided for it, otherwise the solution will not be checked. 3. Use of Calculator, Log table and Mobile is not permitted. 4. Legibility and clarity in answering the question will be appreciated. 5. Put a cross ( × ) on the rough work done by you. 6. Last three pages are Extra pages. You may use them for any unfinished question(s) mentioning the page number with remark "continued on page " Name Father's Name Class : Batch : B.C. Roll No. Invigilator's Full Name For Office Use ……………………………. Total Marks Obta...