"H H + or H H N H N H H HO SO 3 O S OCl Cl Cl P Cl PCl 6 Cl ClF F F Sb F SbF 6 F FProperties of the coordinate compounds are intermediates of ionic and covalent compounds. Comparison of ionic, covalent & coordinate compoundsProperty Ionic Covalent Coordinate1. binding forceBetween ions Between molecules in between strong smaller (Vander Waal’s)(coulombic)2. mp/bpHigh less than ionicin between3. conditionconductor of bad conductor Greater thanelectricity in covalentfused state & in aqueoussolution 4. solubility in polarHig hLess in between solvent (H O) 2 5.Solubility in non Low High in betweenpolar solvent(ether)6.Physical state generally solidliquid & gaseous solid, liquidgas Example 2.Which of the following statement is/are not true for ? ?-bond. 1. It is formed by the overlapping of s ? s or s ? p orbitals 2. It is weaker than pi bond 3. It is formed when ? ?bond exists already. 4. It is resulted from partial overlapping of orbitals. (A) 1, 2, 3, 4 (B) 2, 3 and 4 (C) 2 and 4 (D) 1, 2 and 4 6 Solution: (B)Example 3.The types of bond present in ZnSO .7H O are only 4 2(A) Electrovalent and covalent (B) Electrovalent and co-ordinate (C) Electrovalent, Covalent and co- ordinate(D) None of these Solution:(C)Example 4. Classify the following bonds as ionic, polar covalent or covalent and give your answer(a) Si Si bond in Cl SiSiCl(b) SiCl bond in Cl SiSiCl3 3 3 3(c) CaF bond in CaF (d) NH bond in NH2 3 Solution: (a) Covalent due to identical electronegativity (b) One electron pair is shared between Si & Cl and thus, covalent bond isexpected but electron negativity of Cl is greater than that of Si & somepolarity develops giving polar – covalent nature(c) Ionic since Ca completes its octet by transfer of two outershell electrons thus,completing their octets2 2 5 Ca [Ar]4s , F[He)2s 2p (d) Polar covalent, explanation as in (b) Example 5. Arrange the bonds in order of increasing ionic character in the molecules: LiF, K O, N , SO and ClF2 2 2 3 Solution:N < ClF < SO < K O < LiF2 3 2 2 Example 6. Arrange the following in order of increasing ionic character: C ?H, F ?H, Br ?H, Na ?I, K ?F and Li ?ClSolution:C ?H < Br ?H < F ?H < Na ?I < Li ?Cl < K ?FHYBRIDIZATIONThe tetravalency shown by carbon is actually due to excited state of carbon which is responsiblefor carbon bonding capacity. C ?excited state ? 2s 1s 2P 2P 2P x y z If the bond formed is by overlapping then all the bonds will not be equivalent so a new conceptknown as hybridization is introduced which can explain the equivalent character of bonds. s and p orbital belonging to the same atom having slightly different energies mix together toproduce same number of new set of orbital called as hybrid orbital and the phenomenon is calledas hybridization. Important characteristics of hybridization (i) The number of hybridized orbital is equal to number of orbitals that get hybridized. (ii)The hybrid orbitals are always equivalent in energy and shape. (iii) The hybrid orbitals form more stable bond than the pure atom orbital. (iv) The hybrid orbitals are directed in space in same preferred direction to have some stablearrangement and giving suitable geometry to the molecule.7 Depending upon the different combination of s and p orbitals, these types of hybridization areknown. 3 3 (i) sp hybridization: In this case, one s and three p orbitals hybridise to form four sp hybrid3 orbitals. These four sp hybrid orbitals are oriented in a tetrahedral arrangement.For example in methane CH 4 0 109 28' 2s 2py 2pz 2px 2 2 (ii) sp hybridization: In this case one s and two p orbitals mix together to form three sp hybridorbitals and are oriented in a trigonal planar geometry. 0 120 2s 2py 2pxThe remaining p orbital if required form side ways overlapping with the other unhybridized porbital of other C atom and leads to formation of ? bond as in H C = CH 2 2 (iii) sp hybridization: In this case, one s and one p orbital mix together to form two sp hybridorbitals and are oriented in a linear shape. 0 180 2s 2psThe remaining two unhybridised p orbitals overlap with another unhybridised p orbital leadingto the formation of triple bond as in HC ? CH. ShapeHybridisationLinearsp2 Trigonal planarsp3 Tetrahedralsp3 Trigonal bipyramidal sp d3 2 Octahedralsp d3 3 Pentagonal bipyramidalsp dExample 7.Which of the following molecule has trigonal planer geometry?(A) CO (B) PCl2 5(C) BF (D) H O 3 22 Solution. BF has trigonal planer geometry (sp - hybridized Boron). 3 Hence (A) is correct. Rule for determination of total number of hybrid orbitals ? Detect the central atom along with the peripheral atoms.8 ?Count the valence electrons of the central atom and the peripheral atoms. ? Divide the above value by 8. Then the quotient gives the number of ? bonds and theremainder gives the non-bonded electrons. So number of lone pair non bondedelectrons= .2 ?The number of ? bonds and the lone pair gives the total number of hybrid orbitals. An example will make this method clearSF Central atom S, Peripheral atom F4? total number of valence electrons = 6+(4 ?7) = 34 2Now 34/8= 4 ? Number of hybrid orbitals = 4 ? bonds + 1 lone pair) 8 3 So, five hybrid orbitals are necessary and hybridization mode is sp d and it is trigonal bipyramidal(TBP).Note:Whenever there are lone pairs in TBP geometry they should be placed in equatorialposition so that repulsion is minimum.F F F S S F F F F F (A) (B) 1.NCl Total valence electrons = 26 3 NRequirement= 3 ? bonds + 1 lone pair 3Hybridization = spCl Cl ClShape= pyramidalBr 2.BBr Total valence electron= 243 Requirement= 3 ? bonds B 2 Hybridization = spBr BrShape= planar trigonal Cl 3. SiCl Total valence electrons= 324Requirement= 4 ? bondsSi 3Hybridization =spCl Cl ClShape= Tetrahedral l 4. CITotal valence electron = 324Requirements = 4 ? bonds C 3Hybridization = spl l lShape = Tetrahedral F 5. SFTotal valence electrons= 486 F FRequirement = 6 ? bonds S 3 2hybridization = sp dF Fshape= octahedral / square F bipyramidal6. BeF Total valence electrons : 162Requirement : 2 ? bondsF – Be – F Hybridization : sp 9Shape : Linear7. ClFTotal valence electrons : 283 FRequirement : 3 ? bonds + 2 lone pairs 3 ClHybridization : sp dFShape : T – shaped F We have already discussed that whenever there are lone pairs they should be placed inequatorial positions. Now a question that may come to your mind that though the hybridization is3 sp d, so the shape should be T.B.P. But when all the bonds are present the actual shape is TBP.But when instead of bond there are lone pairs in TBP the actual geometry is determined by thebonds not by the lone pairs. Here in ClF the bond present (2 in axial and 1 in equatorial) gives3 the impression of T shape.8.PF Total valence electrons : 405 F FRequirement : 5 ? bonds F 3 PHybridization : sp d Shape :Trigonal bipyramidal (TBP)F F 9.XeFTotal valence electrons : 364Requirement:4 ? bonds+ 2 lone pairs 3Hybridisation : sp d Shape :Square planar FFF FFFXeXe XeFFFFFF(A)(B)(C) Now three arrangements are possible out of which A and B are same. A and B can be interconverted by simple rotation of molecule. The basic difference of (B) and (C) is that in (B) thelone pair is present in the anti position which minimizes the repulsion which is not possible instructure (C) where the lone pairs are adjacent. So in a octahedral structure the lone pairs mustbe placed at the anti positions to minimize repulsion. So both structure (A) and (B) are correct.10.XeF Total valence electrons : 22F 2 Requirements: 2 ? bonds + 3 lone pairs 3 XeHybridisation: sp d Shape : Linear F [ ? l.p. are present in equatorial position and ultimate shapeis due to the bonds that are formed]Br F 11.PF Br Total valence electrons : 402 3Requirements : 5 ? bonds Br 3 PHybridisation: sp d Shape : trigonal bipyramidalF Br Here we see that fluorine is placed in axial position whereas bromine is placed in equatorialposition. It is the more electronegative element that is placed in axial position and lesselectronegative element is placed in equatorial position. Fluorine, being more electronegativepulls away bonded electron towards itself more than that is done by bromine atom which resultsin decrease in bp – bp repulsion and hence it is placed in axial position. 10 "