Let the longest wavelength of Balmer series is

1: If the radius of first Bohr orbit of H-atom is x, then de-broglie wavelength of electron in 3 rd orbit is nearly: (A) 2 x (B) 6 x (C) 9x x (D) 3 3h Solution: mvr 3 = 2 9 x 3h mv = mv = 1 2 h = = 6 x mv h 6 x (B) 2: If the shortest wavelength of H-atom in Lyman series is x, then the longest wavelength in Balmer Series of He + is 9x 36x (A) (B) 5 5 x 5x (C) (D) 4 9 Solution: For the shortest wave length of Lymann series in H-atom 1 1 1 1 RH 2 2 = RH x 1 x Let the longest wavelength of Balmer series is 1 1 1 RH 4 max 4 9 9x max = 5 (A) 3. The nucleus of an atom is located at x = y = z = 0. If the probability of finding an s-orbital electron in a tiny volume assumed x = a, y = z = 0 is 1 10 5, what is the probability of finding of the electron in the same sized volume around x = z = 0, y = a? (A) 1 10 5 (B) 1 10 5 a (C) 4 10 5 a 2 (D) 1 10 5 a 1 Solution: Since s-orbital is spherically symmetrical and has equi-distance from the nucleus. Hence the probability is identical (A) 4. To which of the following is Bohr s model not applicable: (a)he + (b)li +2 (c) Deuterium (d) Be +2

Solution: Be +2 contains two electrons, rest all contain one electron each. 5. Which of the following curves may represent the speed of the electron in a hydrogen atom as a function of principal quantum no. n. (A) Straight line passing through origin (C) Rectangular hyperbola (B) Circle (D) Parabola Ans: (C) Speed of an electron is inversely proportional to n. So, v x n = constant, resulting in a rectangular hyperbola 6. Photoelectric emission is observed from a surface for frequencies v, and v 2 of the incident radiation (where v 1 v 2 ). If the maximum kinetic energy of the photoelectrons in two cases are in the ratio of 1:k, then the threshold frequency v 0 is given by v1 v2 Kv1 v2 (A) (B) K 1 K 1 Kv 2 v1 v2 (C) (D) v1 K 1 K Ans: (B) Use Einstein s P.E. Equation, K.E = hv - hv 0 for both the cases. Then solve. For Questions, 7 to 9, read the passage given below: In the ionic bond, the cation tends to polarise the electron cloud of the anion by pulling electron density towards itself. This causes development of covalent character in ionic bond because the electron density get s localised in between the nuclei. The tendency of cation to bring about the polarisation of anion is expressed as its polarising power. The ability of ion to undergo polarisation is called its polarisability. The polarising power of cation and polarisability of anion are decided on the basis of Fajan s rules as given below: (i) Smaller the cation higher is its polarising power. (ii) Cations with pseudo noble gas configuration (ns 2 np 6 nd 10 ) have relativity high polarising power than those with noble gas configuration (ns 2 np 6 ). (iii) Large the size of anion higher is its polarisability. 7. Arrange the following species Ag +, Cu +, K + in increasing order of polarising power? (A) K + > Cu + > Ag + (B) Ag + > K + > Cu + (C) Cu + > Ag + > K + (D) Ag + > Cu + = K + Ans: (C) While both Cu & Ag have a pseudo noble gas configuration (ns 2 np 6 nd 10 ), Cu has a smaller size than Ag hence, Cu + > Ag +

8. Among the following MgCl 2, NaCl, Na 2 S, MgS, the compounds having the least melting point and the highest melting point are respectively: (A) MgS & NaCl (B) NaCl & Na 2 S (C) MgCl 2 & NaCl (D) NaCl & MgS Ans: (A) Higher the ionic nature of the bond, higher will be the melting point. MgS has the highest polarization while NaCl has the least. 9. The correct order of polarisability I, Br, Cl, F is : (A) I > Br > Cl > F (B) I > Br = Cl > F (C) I = Br = Cl > F (D) I = Br < Cl = F Ans: (A) Larger the size of the anion, greater is its polarisability. For Questions, 10 to 12, read the passage given below: Valence shell electron-pair repulsion (VSEPR) theory predicts the shape of a molecule by considering the most stable configuration of the bond angles in the molecule. The main points of the theory are : (i) Electron pairs in the valence shell of the central atom of a molecule, whether bonding or lone pairs are regarded as occupying localised orbitals. These orbitals arrange themselves in such a manner so as to minimise the mutual electronic repulsions. (ii) The magnitude of the different types of electronic repulsions follow the order given below: lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding pair. These repulsive forces alter the bond angles of the molecule or ion. (iii) The electron repulsion between two pairs of electrons will be minimum if they are as far apart as possible. On this basis, the following geometrical arrangements are most suited. 10. The correct order of the bond angles is : (A) NH 3 > PH 3 > AsH 3 > SbH 3 (B) PH 3 > AsH 3 > SbH 3 > NH 3 (C) AsH 3 > SbH 3 > NH 3 > PH 3 (D) NH 3 > PH 3 = AsH 3 = SbH 3 Ans: (A) The bond pairs being closer to the central atom in N as compared to elements as we go down, the group, the bond angle is the highest for NH 3 and decreases as we go down the group. 11. Which of the following statement is correct? Compound Structure Compound Structure (A) H 2 O Tetrahedral (B) H 2 O Linear (C) NH 3 Tetrahedral (D) NH 3 Pyramidal Ans: (D) 12. Which of the following is correct for the bond angle. (A) ClO 2 > Cl 2 O (B) H 2 O > NH 3 (C) OF 2 > H 2 O (D) CH 4 < PH 3

Ans: A O has two lone pairs while N has one, hence B.A. of NH 3 is greater than H 2 O F is more electronegative than O, hence the bp-bp repulsion is less than in case of H 2 O. Hence, B.A. of H 2 O is greater. P has a lp, hence its B.A. is less than methane While O has 2 lps in Cl 2 O, Cl has three electrons in ClO 2 For Questions, 13 to 15, read the passage given below: The minimum amount of energy required to remove the most loosely b ound electron from an isolated atom in the gaseous state is known as ionisation energy or first ionisation energy or ionisation enthalpy (IE 1 ) of the element. The energy required to remove the second electron from the monovalent cation is called second ionisation enthalpy (IE 2 ). Similarly, we have third, fourth,... ionisation enthalpies. The values of ionisation energy depends on a number of factors such as (i) size of the atom (ii) screening effect (iii) nuclear charge (iv) half filled and fully filled orbitals (v) shape of orbital In a group, the ionisation energy decreases from top to bottom. In a period, the value of ionisation energy increases from left to right with breaks where atoms have somewhat stable configurations. 13. In a period, the ionisation energy is lowest for the: (A) noble gases (B) halogens (C) alkaline earth metals (D) alkali metals Ans: (D) Its easiest to take an electron out of an alkali metal 14. Which order for ionisation energy is correct? (A) Be > B > C > N > O (B) B < Be < C < O < N (C) B < Be < C < N < O (D) B < Be < N < C < O Ans: (B) Ionization energy generally increases as we go from left to right. Atoms that have half-filled and full-filled sub-shells though have higher IE than the next element with higher atomic number. 15. The first ionisation energy of Mg, Al, P and S follows the order : (A) Mg < Al < P < S (B) Al < Mg < P < S (C) Al < Mg < S < P (D) Mg < Al < S < P Ans: C Similar logic as that of Q. 14

For Questions, 16 to 18, read the passage given below: The valence bond theory was the first quantum mechanical theory of bonding to be developed, and can be regarded as a way of expressing Lewis concepts in terms of wave functions. In valence bond theory, the wave function of an electron pair is formed by superimposing the wave function for the separated fragments of the molecule. And, the shape of the molecule depends on the orientation of the hybridized orbitals. 16. The maximum number of combinations of atomic orbitals that are possible in the valence shell of C-atom is (A) 4 (B) 3 (C) 2 (D) 1 Ans: (A) Carbon by excitation of its electron can create 4 orbitals that contain 1 electron each leading to covalent bonding. 17. Trigonal pyramidal structure is associated with (A) 1 only (B) 1 & 3 only (C) 2 &3 only (D) 1 & 2 only Ans: (A) Sulphur has 1 lp and forms 3 sigma bonds. This leads to Trigonal pyramidal structure. 18. Which one of the following is correct? (A) are both linear with hybridization (B) are both linear with hybridization respectively (C) is angular and is linear with and hybridization respectively (D) are both linear with and hybridization respectively Ans: For Questions, 19 to 21, read the passage given below: Difference in polarities of bonds is expressed on a numerical scale. The polarity of a molecule is indicated in terms of dipole moment ( ). To measure dipole moment, a sample of the substance is placed between two electrically charged plates. Polar molecules orient themselves in the electric field causing the measured voltage between the plates to change. The dipole moment is defined as the product of the distance separating charges of equal magnitude and opposite sign, with the magnitude of the charge. The distance between the positive and negative centres is called the bond length thus, = electric charge bond length = q d 19. Which of the following molecule has zero dipole moment (A) PCl 5 (B) BF 3 (C) CCl 4 (D) All of these Ans: (D) All the molecules have symmetry leading to cancellation of dipole moments.

20. The correct order of dipolemoment is: (A) CH 3 Cl > CH 3 F (B) HCl > HF (C) NF 3 > NH 3 (D) NH 3 > H 2 O Ans: (A) In case of CH 3 Cl, the value of d overpowers the value of q. In case of HF and HCl this does not happen. Water is more polar than NH 3. In NH 3, the dipole moments are directed towards N atom while they are oppositely directed in NF 3 21. Which of the following option is correct? [State T for true and F for false ] (I) Dipole moment is a vector quantity, i.e. it has both magnitude as well as direction. (II) CO 2 is non polar because its dipole moment 0 (III) Generally in any molecule, if bond angle increases, then dipole moment will be also increase (IV) We can determine percentage ionic character with the help of dipole moment (A) TTFF (B) TFFT (C) FFTT (D) FTTF Ans: B For Questions, 22 to 24, read the passage given below: Glenn Theodore Seaborg was an American chemist whose involvement in thesynthesis, discovery and investigation of ten transuranium elements earned him a share of the 1951 Nobel Prize in Chemistry. His work in this area also led to his development of the actinide concept and the arrangement of the actinide series in the periodic table of the elements. He was also the only person after whom an element was named while the person was alive. Assuming that in the time to come, students reading this question paper now became scientists who would discover newer elements and come up with path-breaking theories, the periodic table would change as compared to what it looks today. So, how many elements would there be in the periodic table if: 22. There were 8 periods and the 8 th period was completely full (A) 150 (B) 156 (C) 168 (D) 180 Ans: (C) 2,8,8,18,18,32,32 these are the number of elements in the first seven periods. The next term would be 50. Hence the number of elements would be 118+50 = 168 23. Each orbital instead of accommodating 2 electrons could accommodate 3 electrons? (A) 59 (B) 108 (C) 177 (D) 200 Ans: (C) With two electrons/orbital, the number of elements is 118. With three electrons/orbital it would be 177 24. Negative values of m are not allowed (m = magnetic quantum number)? (A) 76 (B) 78 (C) 84 (D) 100 Ans: Total elements in s-block = 14, p-block = 36, d block = 40, f-block = 28

Now, the elements where m has non-negative values are all s-block (14), 2/3 of p-block (24), 3/5 of d-block (24) and 4/7 of f-block (16) = 78 In each of the above questions, assume you start with 118 element s (as it exists in the current version of the periodic table). 25. How many electrons in sodium (Na)will satisfy this equation: (where n,l,m,s are the four quantum numbers) (A) 10 (B) 0 (C) 11 (D) 6 Ans: (B) LHS is an integer, RHS a fraction (due to s). 26. Two ice cubes when pressed, unite to form one cube. Which force is responsible for holding them together : (A) van der Waal s forces (B) Covalent attraction (C) Hydrogen bond formation (D) Dipole-dipole attraction Ans: (C) 27. The nodal plane in the -bond of ethene is located in (A) the molecular plane (B) a plane parallel to the molecular plane (C) a plane perpendicular to the molecular plane which bisects, the carbon-carbon bond at right angle. (D) a plane perpendicular to the molecular plane which contains, the carbon-carbon bond. Ans: (A) -bonds are perpendicular to the molecular plane. Hence their nodal plane will be located in the molecular plane. 28. The decreasing size of K +, Ca 2+, Cl & S 2 follows the order: (A) K + > Ca +2 > S 2 > Cl (B) K + > Ca +2 > Cl > S 2 (C) Ca +2 >K + > Cl > S 2 (D) S 2 > Cl > K + > Ca +2 Ans: Each of these have the same number of electrons. The radius will be the smallest in the ion where the number of protons is the highest. Hence, D Directions: The questions below to consist of an assertion in column-1 and the reason in column-2. Against the specific question number, write in the appropriate space. (A) If both assertion and reason are CORRECT, and reason is the CORRECT explanation of the assertion. (B) If both assertion and reason are CORRECT, but reason is not the CORRECT explanation of the assertion. (C) If assertion if CORRECT but reason is INCORRECT (D) If assertion is INCORRECT reason in CORRECT. Q.29 Assertion: F atom has a less negative electron gain enthalpy than Cl atom.

Ans: Reason: Additional electron is repelled more efficiently by 3p electron in Cl atom than by 2p electron in F atom. A Q.30 Assertion: Al(OH) 3 is amphoteric in nature. Reason: Al O and O H bonds can be broken with equal ease in Al(OH) 3. Ans: A