Chemical Bonding Solution.pdf | Chemical Polarity | Covalent Bond

Please download to get full document.

View again

of 9
All materials on our website are shared by users. If you have any questions about copyright issues, please report us to resolve them. We are always happy to assist you.
Information Report
Category:

Documents

Published:

Views: 6 | Pages: 9

Extension: PDF | Download: 0

Share
Related documents
Description
GET EQUIPPED FOR IIT – JEE Advanced Only one option correct 1) (B) 4 B.P and Zero L.P is SiF4. 2) (D) ICl2 has 2 B.P and 2 L.P. Sp 3 hybridization and bent shape. 3) (C) Sp 2 hybridization and trigonal planar geometry. 4) (D) BrF5 : 5 B.P + 1 L.P Sp 3d2 hybridization and square pyramidal geometry. 8 F – Por – F angles at 900. F F F Br F 5) (C) 6) (D) 7) (A) 8) F ClO4 : 4 B.P + Zero L.P Sp3 hybidiation. HS+ OS u C C S+ S u H O u Bond order of O22  1 H ; O2  2
Transcript
  FFFFFBr  uuu S+ H S O   CC H S O   S+ H C C     3 11 1 10 2 12 5 4 a   a 8 7 6  N 3 HC OH 3 HC 3 CH O OH 2  NH  OH OH  OHO OO GET EQUIPPED FOR IIT – JEE Advanced Only one option correct 1) (B) 4 B.P and Zero L.P is SiF 4 . 2) (D) ICl 2   has 2 B.P and 2 L.P. Sp 3  hybridization and bent shape. 3) (C) Sp 2  hybridization and trigonal planar geometry. 4) (D) BrF 5  : 5 B.P + 1 L.P Sp 3 d  2  hybridization and square pyramidal geometry. 8 F – Por – F angles at 90 0 . 5) (C) ClO 4   : 4 B.P + Zero L.P Sp 3  hybidiation. 6) (D) 7) (A) 8) Bond order of 222+22 1 ; O2O = 1.5 ; O2.5 O      Hence, Stability 22222 OOOO        ; Ans : (B) 17)  NO 2   Sp  ;  NO 3   Sp 2  ; 34  NHSp    (3) 18) (B) 33  NHSp   ;   224 PHldSp    (Square planar) 35 PClSpd    ; BCl 3    s  p 2 . 19) (A) All are 14 e   squares with B.O = 3. 20) (A) The e   density lies above and below the inter-nuclear / molecular axis. 21) (B) 22) (C) 2*22222*2 ://22222  y znnz Ossspzpzpp        21 22  y  zpyzpy  23) (B) More than number of H – atoms, more is the H – Bonding hence, Due to higher intermolecular forces of attraction, higher is the M. pt. 24) (A) Lead Oxide PbO 2 . 25) (D) 12  uuuu HC 3 CH 0 120 HCH26) (C) 27) (C) Dipole Moment of 31.5 D       2201212 2cos60  pppp     2.6 D.    12 PP.   28) (D) O – H bond is more polar than N – H bond. & SO + on H is more in O – H bond. Also N is  better donor than O 29) (C) Comprehension Type  Passage 1: 1)   (D) Both F – atoms are present in axial positions. 2)   (A) uF 3  has Sp 3 d hybridization ; 3 B. P + 2 L.P 3)   (D) PI 5  and PBr  6   e donot exist due to steric factors. In PH 5 , Because of its low electronegative H –atom cannot effect the contraction of dz 2  orbital to form Sp 3  hybridized P = atom. 4)   (B) PBr  6   e  doesnot exist since P cannot accommodate 6 Br – atoms around it due to steric hindrance. 5)   (D) 30 P 35 Cl   5  + 30 P 37 Cl   5  + 31 P 35 Cl   5  + 31 P 37 Cl   5  4 i) 31 P 35 Cl   137 Cl   4   31 P 35 Cl   ax 37 Cl   4 eq    = 2 forms 31 P 35 Cl 37 Cl 3   eq 37 Cl   ax  ii) 31 P 35 Cl   237 Cl   3   31 P ax35 Cl   2 eq    37 Cl   3  3 = 31 P 35 Cl 37 Cl   ax 35 Cl 37 Cl 2   eq  forms 31 P 37 Cl   ax 35 Cl 237 Cl   eq  iii) 31 P 35 Cl   337 Cl   2  3  forms ; same as (ii) iv) 31 P 35 Cl   437 Cl    2 forms ; same as (ii) Similarly with 30 P – atom as central Atom, 10 forms will be there. Total = 24.  OOS SOO  NC   NC  CN  CN  C  NC   NC  CN  CN  CCCHCHCHO = 32HHHH O H CCC H Propanal HHHHOHCCCH CHCOCH = 33 Propauone 6)   SF 4  has 4 G.P + 1 L. P ; Hence (C). 7)   (C)   32 IFSpdhas 2B.P + 3 L.P   8)   (C) In Sp 3 d hybridized, dz 2  orbital is utilized.  Passage 2: 1)   A) i) +33 I:2B.P2L.P;Sp   hybridization ii) SO 2  can be represented as and . It is considered that S being in 3 rd    period can utilized its d – orbitals to form pz – dx bonds with O atoms. iii) XeF 2  : 3 L.P and 2 B.P 3 Spd    hybridization. CO 2  : Sp hybridization 2 B.P   iv) SF 4  : Sp 3 d hybridization 4 B.P + 1 L.P   ICl 3  : Sp 3 d hybridization 3 B.P + 2 L.P   2)   3 I:L.P3; B.P2     Ratio 1.5   XeF 4  : L.P = 2 ; B.P = 4 Ratio 0.5   3)   (A) 02  NOlinear180        02  NOodd e systembent134        02  NO1 L.Pbent115    4)   (B) Bond angle PF 3 > PCl 3 < PBr  3 < PI 3 . Due to dx – px bonds in PF 3 , double bond character develops increase in B.P – B.P repulsions lead to increase is Bond Angle. 5)   (D) B.O of 2222222 O2.5;O1.5;O2,O3;1 O           Bond length 22222222 OOOOO          6)   (D) B.O of 2  N2.5    B.O of 2  N2.5;   es   occupy non – bonding Molecular orbital’s. +22  bond strength of N Bond strength of N     7)   (C) i)  bonds = 8      bonds = 9      bonds = 8  bonds = 9 ii) ;  uuuu 3 CH uu o-dichloco  benxene m-dichlone benzene toluene  p-dichloro  benzene  ii) XeF 4  is square planar in shape. SF 4  has See – saw Shape. iv) Due to smaller orbital size, effective overlapping take place in u – u bond.  Passage 3: 1)   (D) 2)   (A) In 2 HF H   bonding exists between HF and I.   3)   (B)  Passage 4: 1)   A) BeCl 2  has maximum covalent character because of smallest size of cation   2 Be   2)   D) I   is the largest Anion. It has maximum polar ability. 3)   C) Highest positive change and smallest size of cation 3 (Al),   hence maximum polarization and covalent character. 4)   B) Liu has maximum covalent character, hence it will be most soluble in non – polar solvent ether. 5)   D) CaI 2  has maximum covalent character hence least M.pt…  Passage 5: 1)   A) and B) These are planar molecules. XeF 4  has square planar shape. XeF 4  has trigonal planar shape. 2)    pql     188 1.210esu cm10cm  pql      10 1.210esu. q      Fraction of change 1010 1.21010.254.8104     . 3)   (C)
We Need Your Support
Thank you for visiting our website and your interest in our free products and services. We are nonprofit website to share and download documents. To the running of this website, we need your help to support us.

Thanks to everyone for your continued support.

No, Thanks