DPP-25 to 26 With Answer

FACULTY COPY DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 25 Class : XIII Course : DPP No.1 Total Marks : 54 Max. Time : 54 min. Single choice Objective ('–1' negative marking) Q.1 to Q.3 (3 marks 3 min.) [9, 9] Multiple choice objective ('–1' negative marking) Q.4 to Q.5 (4 marks 4 min.) [8, 8] Subjective Questions ('–1' negative marking) Q.6 (a,b,c) (4 marks 4 min.) [12, 12] Comprehension ('–1' negative marking) Q.7 to Q.10 (3 marks 3 min.) [12, 12] Match the Following (no negative marking) Q.11 (4 marks 4 min.) [4, 4] BooSt YoUr PreViouS ConCept Comprehension ('–1' negative marking) Q.12 to Q.14 (3 marks 3 min.) [9, 9] 1. The dipole moments of the given molecules are such that : (1) BF3 > NF3 > NH3 (2) NF3 > BF3 > NH3 (3*) NH3 > NF3 > BF3 (4) NH3 > BF3 > NF3 . Sol. (3) The dipole moment of BF3 , NF3 and NH3 respectively is zero, 0.24D and 1.46D. 2. The molecule having least dipole moment (Assume benzene molecule be a regular hexagon) (1) (2) (3) (4*) 3. A polar molecule AB have dipole moment 3.2 D (Debye) while the bond length is 1.6 Å. Find the percentage ionic character in the molecule. (1) 31% (2*) 41.6% (3) 39.6% (4) None observed dipole moment Sol. % Ionic character = calculated dipole moment 3.2 10 18 esu cm  100 × 100 = 4.8  1010  1.6 10 8 esu cm = 41.66% 4. Which of the following molecule is/are non polar (1) XeF2 (2) PCl3F2 (3) XeF4 (4*) All 5. Which of the following is correct about the bond angle. (1*) OF < H O < Cl O < ClO (2*) PF < PCl < PBr < P 2 2 2 2 3 3 3 3 (3*) NH3 > NF3 (4*) KrF4 < SF2 < N2H2 Sol. All given order are correct OF2 < H2O < Cl2O < ClO2 sp3 sp3 sp3 sp3 low bp-bp larger repulsion size of Cl PF3 < PCl < PBr3 < PI3 97.7º 100.3º 101.0º 102º NH3 > NF3 107º 104º KrF4 > SF2 < N2H2 sp3d2 Sp3 sp2 (< 120º) (90ºC) (< 109º28’) sp2(< 120º) All gives order are order are correct 6.(a). Arrange in order of increasing dipole moment : BF3, H2S, H2O. Sol. BF3 < H2S < H2O BF3 has a zero dipole moment because of its symmetry. H2S has a lower dipole moment than H2O because of the much lower bond polarity of H–S compared to H–O. 6.(b). Which out of NH3 and NF3 has higher dipole moment and why? Ans. NH3 has higher dipole moment than NF3 . 6.(c) The dipole moment of HBr is 2.60 × 10–30 C.m, and the interatomic spacing is 1.41Å. What is the percent ionic character of HBr? Sol. The dipole moment of a 100% ionic “molecule” at the given internuclear distance would be (1.60 × 10–19C) (1.41 × 10–10m) = 2.26 × 10–29 C.m The actual dipole is less; the percent ionic character is given by 2.60  1030 C.m 2.26 1029 C.m × 100% = 11.5% Passage : (Read the following passage and answer the questions number 7 to 10) It was noted previously that a trigonal bipyramid is not a regular shape since the bond angles are not all the same. It therefore follows that the corners are not equivalent. Lone pairs occupy two of the corners, and F atoms occupy the other three corners. There different arrangements are theoretically possible, as shown in figure. The most stable structure will be the one of lowest energy, that is the one with the minimum repulsion between the five orbitals. The greatest repulsion occurs between two lone pairs. Lone pair bond pair repulsions are next strongest, and bond pair- bond pair repulsions the weakest. Groups at 90° to each other repel each other strongly, while groups 120° apart repel each other much less. Structure 1 is the most symmetrical, but has six 90° repulsions between lone pairs, and atoms. Structure 2 has one 90° repulsion between two lone pairs, plus three 90° repulsions between lone pairs and atoms. Structure 3 has four 90° repulsions between lone pairs and atoms. These factors indicate that structure 3 is the most probable. The observed bond angles are 87°40', which is close to the theoretical 90°. This confirms that the correct structure is (3), and the slight distortion from 90°. This confirms that the correct structure is (3), and the slight distortion from 90° is caused by the presence of the two lone pairs. As a general rule, if lone pairs occur in a trigonal bipyramid they will be located in the equatorial positions (round the middle) rather than the apical positions (top and bottom), since this arrangement minimizes repulsive forces. A rule of thumb can therefore be theorised, that the position having maximum repulsion amongst them are occupied at equitorial points. 7. If the repulsion produced by double bonds is more than single bonds, triple bonds more than double bond, then on the basis of the passage the geometry of XeOF2 would be : F O (1*) (2) (3) (4) F 8. If the bonds having maximum repulsion amongst them are found at the equitorial positions then the structure of PCl3F2 would be (1*) F C𝑙 F C𝑙 F (2) C𝑙 P C𝑙 C𝑙 F F (3) C𝑙 F C𝑙 (4) None of these 9. The structure of C𝑙 + would be (1) (2) (3) (4*) None of these 10. For the tri-iodide ion, on the basis of the above pasage the structure would be (1) (2) (3) (4*) 11. Match the following Column (Ι) Column (ΙΙ) (1) P4 (p) 7  - bond (2) SO 2– (q) central atom is in sp3 hybridisation (3) C2 H6 (r) No, ‘P–P’ bond (4) P4O10 (s) No, ‘O–O’ bond Ans. [1 – q, s] ; [2 – r, s, q] ; [3 – p, q, r, s] ; [4 – q, r, s]. BooSt YoUr PreViouS ConCept Comperhension : (Q.12 to Q.14) NH3 is formed in the following steps : I : Ca + C  CaC 2 50% yield II : CaC2 + N2  CaCN2 + C 100% yield III : CaCN + H O  NH + CaCO 50% yield 2 2 3 3 12. To obtain 2 mol NH3, calcium required is : (1) 1 mol (2) 2 mol (3) 3 mol (4*) 4 mol 13. If 50 g of CaCO3 is recovered in III step then what wt. of Ca will be required in 1st step. (1) 20 g (2) 40 g (3*) 80 g (4) 100 g 50 1 Sol. Moles of CaCO3 produced = 100 = 2  moles of CaCN = 1 x 2 = 1 2  moles of CaCl = 1  moles of Ca = 1 x 2 = 2  mass of Ca = 2 x 40 = 80 g 14. If above reaction are carried out by taking. 80g of Ca and 36 g of C then NH3(g) produced was completely dissolved in 500 ml of water then the molar concentration of NH4OH will be : (1) 6 M (2) 3 M (3*) 1.5 M (4) 1 M 80 Sol. Moles of Ca = 40 = 2 Moles of C = 36 = 3 12  'C' is the limiting reagent 3  Moles of CaC formed in 1st reaction = 2 x 0.5 = 0.75  moles of NH formed = 0.75 x 2 x 1 0.75 = 0.75  molarity of NH OH = x 1000 = 1.5 M 3 2 4 500 FACULTY COPY DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 26 Class : XIII Course : DPP No.2 Total Marks : 48 Max. Time : 48 min. Single choice Objective ('–1' negative marking) Q.1 to Q.2 (3 marks 3 min.) [6, 6] Multiple choice objective ('–1' negative marking) Q.3 to Q.4 (4 marks 4 min.) [8, 8] True or False (no negative marking) Q.5 to Q.6 (2 marks 2 min.) [4, 4] Subjective Questions ('–1' negative marking) Q.7 (4 marks 4 min.) [4, 4] Comprehension ('–1' negative marking) Q.8 to Q.10 (3 marks 3 min.) [9, 9] Match the Following (no negative marking) Q.11 (4 marks 4 min.) [4, 4] BooSt YoUr PreViouS ConCept Comprehension ('–1' negative marking) Q.12 to Q.14 (3 marks 3 min.) [9, 9] Subjective Questions ('–1' negative marking) Q.15 (4 marks 4 min.) [4, 4] 1. According to molecular orbital theory which of the following statement about the magnetic character and bond order is correct regarding O +. (1) Paramagnetic and bond order < O2 (2*) Paramagnetic and bond order > O2 . (3) Dimagnetic and bond order < O2 (4) Dimagnetic and bond order > O2 . 2. The bond order in NO is 2.5 while that in NO+ is 3. Which of the following statements is true for these two species? (1) Bond length is unpredictable (2*) Bond length in NO is greater than in NO+ (3) Bond length in NO+ is equal to that in NO (4) Bond length in NO+ is greater than in NO 3. Which of the following is/are correct : (1) Carbon-carbon bond length in CaC2 will be more than in CH2CCH2 (2) O–O bond length in KO2 will be more than in Na2 O2 . (3*) O–O bond length in O2 [PtF6] will be less than that in KO2 (4*) N–O bond length in NO gaseous molecule will be smaller than in NOCl gaseous molecule. Sol. In CaC there is C  C , whide in CH CCH , there is only C = C. KO2 2 = K+ + O – 2 2 Bond order = 1.5 Na2O2 = 2Na+ + O 2– Bond order = 1.0 O (Pt F ] = O + + [Pt F ]– 2 6 2 6 Bond order = 2.5 N Bond order = 2.5 while in NOCl, bond order = 2. 4. Which is incorrect order for net dipole moment - (1) HF > HCl > HBr > HI (2*) CH3– F > CD3 – F (3*) SO3 > SO2 (4*) CH3 – CH = CHCl (cis) > CH3 – CH = CHCl (trans) Sol. Correct orders of dipole moment are HF > HCl > HBr > HI (decreasing bond polarity) CD3F > CH3F (D is more electro +ve than hydrogen) SO2 > SO3 (SO3 is symmetircal so dipole moment - 0) CH3 H C = C (cis) Cl CH3 H H H C = C Cl (trans) 5. S1 : The HOMO in F – is *2px = *2py molecular orbitals. S2 : Bond order of O2– is more then O2+ . S : NO+ is more stable than N + S : C is more stable than C + State, in order, whether S1, S2, S3, S4 are true or false (1) FFFT (2) FTTT (3) FTFT (4*) FFTT 6. True or False (a) The dipole moment of HCl is 1.07 D and its internuclear separation is 1.2738 Å. The charge effectively 7 held by the chlorine atom is 40 times the electronic charge? (e = 4.8 × 10–10 esu) (b) All the N–N bond lengths are same in azide ion and hydroazoic acid. Ans. (a) True (b) False Sol. (a) 1.07 × 10–8 esu –cm =  × 1.2738 × 10–8  8.4 × 10–11 esu 8.4 1011 4.8 10 10 7 = 40 th 0.175. (b) azide ion hydroazoic acid. 7. The gaseous potassium chloride molecule has a measured dipole moment of 10.0 D, which indicates that it is a very polar molecule. The separation between the nuclei in this molecule is 2.67 × 10–8 cm. Calculate the percentage ionic character in KCl molecule. Ans. 78% Sol. Dipole moment of compound would have been completely ionic = (4.8 × 10–10 esu) (2.67 × 10–8 cm) = 12.8 D 10.0 so % ionic character = Comprehension # (8 to 10) 12.8 × 100% = 78.125 % ~ 78% Molecular orbital theory was put forward by Hund and Mulliken. According to this theory, all the atomic orbitals of the atoms participating in molecule formation get disturbed when the concerned nuclei approach nearer. They all get mixed up to give rise to an equivalent number of new orbitals that belong to the molecule . Now they are called molecular orbitals. The molecular orbitals involved in the formation of most of the bimoleculer molecules are given below Bonding orbitals  Antibonding orbitals * 2s 2Px * * 2s 2Px 2Py * 2Py 2Pz * 2Pz The energy of these orbitals has been calculated on the basis of ultravoilet molecular spectra and the sequence of increasing energy is as follows  < * < * <  2py <  *2py  < * (For more than or equal to 15 electrons) 1s 1s 2s 2Px 2pz  *2p z 2Px For diatomic molecules obtained from atoms having more than one electronic shell, in inner shell do not appreciably affect the bonding and may be omitted. 8. The molecular orbital electronic configuration of NO is similar to : (1) KK (2s)2 *(2s)2 (2p )2 (2p )2 = (2p )2 *(2p )1 = *(2p )1 z x y x y (2*) KK (2s)2 *(2s)2 (2p )2 (2p )2 = (2p )2 *(2p )1 = *(2p )0 z x y x y (3) KK (2s)2 *(2s)2 (2p )2 (2p )2 = (2p )2 *(2p )0 = *(2p )0 z x y x y (4) KK (2s)2 *(2s)2 (2p )2 = (2p )2 (2p )2 *(2p )1 = *(2p )0 x y z x y 9. Which of the following statement is not true (1) Stability of N2– is less than that of N2+ (2) Bond order of CO is 3 (3) NO+ is more stable than NO (4*) Bond dissociation energy of O – should be smaller than that of O 2– 10. In the following compounds (i) O2 (ii) O2[BF4] (iii) KO2 Identify the incorrect statement (1) O–O bond length order is ii < i < iii (2) O–O bond order is iii < i < ii (3*) O–O bond energy order is ii < iii < i (4) Unpaired electrons in anti bonding molecular orbital of species of O2 involved in i, ii and iii are 2 ,1, 1 respectively 11. Match the following : Column -I Column-II (1) N + is stable than N – (p) due to one have higher electrons in antibonding than 2 2 other (2) NO can easily loss its electron then N2 (q) one have B.O. 3 and other have 2.5 (3) NO have large bond length than NO+ (r) both are paramagnetic with same bond order (4) He + exist but less stable than H + (s) one paramagnetic and other diamagnetic Ans. [1 – p, r] ; [2 – p, q, s] ; [3 – q, s] ; [4 – p, r]. BooSt YoUr PreViouS ConCept 12. The rate of effusion of helium gas at a pressure of 1000 torr is 10 torr min–1. What will be the rate of effusion of hydrogen gas at a pressure of 2000 torr at the same temperature ? (1) 20 torr min–1 (2) 40 torr min–1 (3*) 20 torr min–1 (4) 10 torr min–1 13. If the number of molecules of SO2 (atomic weight = 64) effusing through an orifice of unit area of cross- section in unit time at 0°C and 1 atm pressure is n, the number of He molecules (atomic weight = 4) effusing under similar conditions at 273°C and 0.25 atm is : n (1*) 2 (2) n (3) 2n (4) n 2 14. The product of PV is plotted against P at two temperatures T1 and T2 and the result is shown in figure. What is correct about T1 and T2? (1) T1 > T2 (2*) T2 > T1 (3) T1 = T2 (4) T1 + T2 = 1 15. A compound of vanadium has a magentic moment 1.73 BM. Work out the electronic configuration of the vanadium ion in the compound. Sol. Number of unpiared electron are given by Magnetic moment = B.M. or 1.73 = where n is number of unpaired electrons or 1.73 × 1.73 = n2 + 2n  n = 1 Now Vanadium atom must have one unpaired electron and thus its configuration is 23V4+ : 1s2 , 2s2 2p6, 3s2 3p6 3d1