Test Date: 6.10.2013 (Class +1)

Dr. Sangeeta Khanna Ph.D

Test Date: 6.10.2013 (Class +1)

Topic: Basic concept of Chemistry, Atomic Structure, Chemical Bonding, Periodic Table and Periodic Properties & Gas Law

Total Marks = 127

Section – A (Single Correct Choice Type)

This Section contains 17 multiple choice questions. Each question has four choices A), B), C) and D) out of which ONLY ONE is correct. (Mark only One choice) 17 × 3 = 51 Marks

1. The ratio of time periods (T1/T2) in second orbit of hydrogen to third orbit of He+ ion is:

a. 8/27 b. 32/27 c. 27/32 d. 24/36

Sol.

2. Which of the following pairs, both the species have the same hybridization?

(I) SF4, XeF4 (II) (III) (iv)

a. I, II b. II, III c. II, IV d. I, II, III

Sol. and XeF2 have sp3d hybridization

ClO and have sp3 hybridization

3. KMnO4 reacts with ferrous ammonium sulphate according to the following equation.

Here, 10 mL of 1 M KMnO4 is equivalent to

a. 20 mL of 0.1 M FeSO4 b. 30 mL of 0.1 FeSO4

c. 50 mL of 0.5 M H2C2O4 d. 20 mL of 0.1 M H2C2O4

4. Arrange the following ions in the order of decreasing X – O bond length where X is the central atom

a. b.

c. d.

Sol.

Greater the bond order, shorter is the bond length. Hence X – O bond lengths will be in the order:

5. Which of the following statement is true?

a. the cage-lattice structure of ice involves true covalent bond

b. the cage-lattice structure of ice is the result of dipole – dipole interaction

c. the cage-lattice structure of ice is the result of intra and inter-molecular hydrogen bond formation

d. the cage – lattice structure of ice is only due to inter –molecular hydrogen bonding.

6. KF combines with HF to form KHF2. The compound contains ………… and …………… species.

a. b. K+, [HF2]– c. d. KHF–, F+

7. Column I Column II

(a) pyramidal (p)

(b) Triangualr planar (q)

(d) See-saw (s)

a. (a – p), (b – q), (c – s), (d – r) b. (a – q), (b – p), (c – s), (d – r)

c. (a – p), (b – q), (c – r), (d – s ) d. (a – s), (b – r), (c – p), (d – q)

8. Column I Column II

(a) paramagnetic with bond order 2 (p)

(b) paramagnetic with bond order 2×5 (q) O2

(d) paramagnetic with bond order 0×5 (s)

a. (a – q), (b – p), (c – s), (d – r ) b. (a – s), (b – p), (c – r), (d – q)

c. (a – q), (b – p), (c – r), (d – s) d. (a – s), (b – r), (c – p), (d – q)

9. Match the following

(a) Energy of ground state of He+ (i) + 6.04 eV

(b) Potential energy of 1s orbital of H – atom (ii) – 27.2 eV

(d) Ionisation potential of He+ (iv) – 54.4 eV

a. (a) – (i), (b) – (ii), (c) – (iii), (d) – (iv) b. (a) – (iv), (b) – (iii), (c) – (iii), (d) - (i)

c. (a) – (iv), (b) – (ii), (c) – (i), (d) – (iii) d. (a) – (ii), (b) – (iii), (c) – (i), (d) – (iv)

10. Match the column – I and Column II

Column – I / Column – II
(a) / The radial node of 4s atomic orbital is / (p) / 0
(b) / The angular node of 3d atomic orbital is / (q) / 3
(c) / The sum of angular node radial node of 5d atomic orbital / (r) / 2
(d) / The angular node of 4s atomic (s) orbital is / (s) / 4

a. (a – q), (b – r), (c – s), (d – p) b. (a – r), (b – q), (c – s), (d – p)

c. (a – r), (b – s), (c – q), (d – p) d. (a – r), (b –s), (c – p), (d – q)

11. How many bonds are there in

a. 14s, 8p b. 18s, 8p c. 19s, 4p d. 14s. 2p

Sol.

12. On analysis, a certain compound was found to contain 254 g of X and 80 g of Y. If the atomic weight of X is 127 and that of Y is 16, then formula of the compound containing X and Y is:

a. XY b. X2Y c. X5Y2 d. X2Y5

13. Which of the following statements is incorrect about resonance?

a. Resonance structures must have nearly same energy

b. In resonance structures, the constituents atoms remain in the same position

c. Resonance structure differ only in the position of electrons around the constituent atoms

d. In resonance structures, a structure with more number of odd electrons will be more stable

14. The molecular weight of a gaseous substance is 80. The volume of the one gram of the gas at 0 C and 720 mm of mercury pressure will be approximately

a. 2950.7 mL b. 0.203 lit c. 3.1 lit d. 1.5 lit

Sol. Volume , V = ? P = 720 mm × ;

R = 0.0821 L atm K– mol–, T = 0 + 273 = 273 K.

We know that :

V » 3.1 lit

15. A 10 L cylinder contains 0.4 g of helium, 1.6 g of oxygen gas and 1.4 of nitrogen gas at 27°C. Calculate the partial pressure of helium gas in the cylinder. Assume ideal behaviour for gases. R = 0.082 L atm K– mol–.

a. 24.6 atm b. 0.0246 atm c. 0.492 atm d. 0.246 atm

Sol. PtV = ntRT

Partial pressure of He gas = pHe × total pressure

= 0.246 atm

16. At fixed temperature and 600 mm pressure, the density of gas is 42. At the same temperature and 700 mm pressure, what is the density of the gas?

a. 65 b. 60 c. 55 d. 49

Sol. According to Boyle’s law,

(at constant temperature)

Density (D) = or

\ P µ D or

\ D2 =

Given P1 = 600, D1 = 42, P2 = 700

\ D2 = . = 49

Hence, the density of the gas is 49.

17. Which of the following graphs is not consistent with ideal gas behaviour?

c. b.

c. d.

Sol. T1 > T2

SECTION – B (Paragraph Type)

This Section contains 3 comprehensions. Each of these questions has four choices A), B), C) and D) out of which ONLY ONE is correct. 3 × 9 = 27 Marks

Comprehension 1

Read the passage given below and answer the question that follow. There is no negative marking for wrong answer. Only one option is correct.

A single structure cannot explain all the properties of a particular compound. A collection of structure is thus proposed to explain all the properties. These collection of structures is known as canonical structures or resonating or resonance hybrid structures and the property is called resonance.

e.g.

If we assume a single structure of benzene then there would be two different bond lengths of C – C bond, but is has been observed that all the bond lengths are equal and intermediate between C – C single bond and double bond. This is due to delocalization of p - electrons due to which they a are equally distributed on all the carbon atoms.

The bond order in case of compounds exhibiting resonance can be calculated as :

Bond order

e.g., Assuming only the Kekule structure of benzene:

Bond order

1. The bond order of is equal to:

a. 1.5 b. 1.25 c. 1.75 d. 2

2. Which among the following pair is a resonating structure

3. Which of the following is correct order of stability of given resonating structure

a. I > II > III b. I > III > II c. III > II > I d. II > III > I

Sol. A structure with complete octet is more stable so structure III is more stable than II

Comprehension 2

Read the passage given below and answer the questions that follow. No negative marking for wrong answer. Only are option is correct.

The characteristics of ionic compounds can be illustrated as follows:

(i) The electrostatic force of attraction exists in all the directions due to which ionic compounds do not possess any directional property and hence do not exhibit any stereochemcial property.

(ii) The ionic compounds are soluble in solvents with high dielectric constant because force of attraction between oppositely charged ions , where q1 and a2 are opposite charges (in coulomb) r is distance between the opposite charges and D is the dielectric constant of the solvent. However for an ionic compound to be soluble in a solvent. Solvation energy is more than lattice energy.

The rule governing the transition from ionic to covalent bonding, one called Fajan’s rule. By this rule, the degree of covalency of a molecules are known, which one based on deformation of interacting ions in the bond. When an anion and cation each approach each other, then the electron cloud of anion is not only attracted by the nucleus, but also by the charge on cation.

At the same time, the cation also tends to repel, the positively charged nucleus of anions. The combined effect of these two forces is that the electron symmetrical, but is elongated towards the cation. This is called distortion of the anion by the cation. This ability of cation to polarize anion is known polarization of power of cation (f)

The ability of anion to get deformed is known as polarisability of anion.

4. Which of the following decreases the ionic character in an electrovalent bond?

a. low charge on anions only b. High charge on cations only

c. high charge on both cations and anions d. Low charge on both cations and anions

5. Which of the following is the increasing order of covalent character of the compound?

a. KCl < NaCl < MgO < AIN b. AIN < MgO < NaCl < KCl

c. MgO < AIN > KCl < NaCl d. KCl > NaCl > MgO > AIN

6. Which of the following anion has the highest polarisability?

a. I– b. Br – c. Cl – d. F Q

Comprehension – 3

Standard enthalpy of formation is used for stability of compound. More negative standard enthalpy of formation of a compound indicate the stability of compound. All alkali metal halides have negative enthalpy of formation therefore all are thermodynamically stable. Hydration energy of compound depend on attractive interaction between ion & H2O. It is inversely proportional to size of ion. Group II metal ion have higher hydration energy than gp I.

7. Standard enthalpies of formation of group I halides (all values, in kJ mol-1 are given as follows):

MF MCl MBr MI

Li – 612 – 398 – 350 – 271

Na – 569 – 400 – 360 – 288

K – 563 – 428 – 392 – 328

Rb – 549 – 423 – 389 – 329

Cs – 531 – 424 – 395 – 337

The second least stable halide is:

a. LiBr b. NaI c. LiI d. data insufficient

Sol. LiI is least stable & NaI & second least stable.

8. Cohesive energy of following alkali metal is given (kJ/mol)

A B C D E

161 108 82 90 78

Which of the above metal represent Cs.

a. A b. B c. C d. E

9. Hydration energies and ionic radius data of Group II A metal ions are given as:

Metal ion Ionic radius (Å) Hydration energy (kJ mol-1)

A 1.00 – 1577

B 0.72 – 1921

C 0.31 – 2494

D 1.35 – 1305

E 1.18 – 1443

Which of the above metal ions give scarlet Red colour to flame

a. A b. E c. D d. C

Sol. Increasing size & decreasing hydration energy; C; B; A; E; D [Be, Mg, Ca, Sr, Ba]

SECTION – C (More than One Answer Type)

This Section contains 6 multiple choice questions. Each question has four choices A), B), C) and D) out of which ONE OR MORE may be correct. 6 × 3 = 18 Marks

1. Which of the following molecules have axial and equatorial bonds of different length?

a. PCl5 b. SF6 c. IF7 d. BF3

A, C

2. Which of the following is the correct order as indicated?

a. Order of B.pt.

b. B.pt. CH3 – CH2 – CH2OH > CH3-CH2 – O – CH3 > CH3CH2 – CH2 – CH3

c. B.pt. I2 > Br2 > Cl2 > F2

d. M.pt. Li > Na > Rb > Cs

B,C,D

3. Which of the following is/are correct order of their dipole-moments (increasing)

b. PF2Cl3 < PF3Cl2 < BF3

c. BH3 < NH3 < H2O

A,C

4. Which of the following pair (S) is/are isostructral?

a. , b. SO3, c. d.