Chemical Bonding part 37

Ajouté : 2026-05-15

[00:00:00]now coming to the different other type of hybridization is your nothing but we will see sp3d hybridge sp3d so sp3d what i said it will be nothing but that that is trigonal bipyramidal got it sp 3d hybridization i have said you already it is trigonal bipyramidal so which all molecules will be coming under it it will be pcl5 it will be sbcl5 pf5 all right it will be all right this all molecules will come under it now here it is not possible to draw the floral diagram because d subshell has got five orbitals because of that that is v x ah d x y d x z d x square y square d z square got it so it is not possible to ah draw what you can say all those on vitals and show it it becomes ah possible means it becomes very clumsy it will become got it it will become very complex you will not understand anything in the diagram so we will be doing drawing here equivalent diagram and we will show you got it now let us take the example of pcl5 molecule as i said you so pcl5 phosphorus has got atomic number 15 so it belongs to group 15 also phosphorus got it so if you write the electronic configuration what you find is that phosphorus 15 is 1 s 2 2 is 2 2 p 6 3 s 2 and 3 p all right three s two and three p three all right i said you that it is not necessary that all the uh what you can say uh it will be uh it should be uh unpaired paired electrons can also participate in hybridization how it is seated so what will happen is that one electron from this suction will go to the d substance it will be three s one and three d one it will be three s one one electron will jump from the 3s substitute with the 3d substrate so what you got it here so see here 3s1 3p you got it 3 unpaired alright and 3d you got this one arcade so one two three four five so in this five what against amped electron five chlorine atom will come here one chlorine near one chlorine near one chlorine here one chlorine here one chlorine so how many chlorine one two three four five so hence it is going to form what pcl5 why chlorine because 17 c atomic number 1 is 2 2 is 2 2 p 6 3 s 2 3 p 5 so chlorine is having deficiency how many electron one electron so you just get one electron from phosphorus here another chlorine will get one electron from phosphorus in this way five chlorine atom will undergo bond formation so here also same thing who is undergoing hybridization phosphorus molecule phosphorous atom so what are the subshells involved in habitation one s all right and three p and one d so hence it is what sp3 d for that reason it is sp3d hybridization because one as you can see very clearly one s and three p substitute is involved and one t substitution is involved in hybridization so it is a security hybridization all right now next is here we will see what you will see that what are the subshells i already said you that is 3px 3py 3pz all right 3s option 3px 3py 3pz you know that that you have seen it already in sp 2 sp3 sp you have seen it that how p sub shell we are drawing it now here this option which this option will undergo hybridization in pcl5 molecules vcl5 molecule or bf5 molecule that is dz square why because it has got similar energy almost comparable energy as that of 3px 3py and 3pz understood what i said which subshell that is nothing but three which orbital it is rather i should say which orbital it is nothing but three d sorry i am drawing this three d z square is involved in hybridization here three d z square so how do you represent you represent in this way you know that three d is x square all right it is involving hybridization so three d z square because a 3d subshell means 3d orbital means which particular orbital is participating in hybridization we have got 5 orbitals isn't it in 3d i said you already we are having d x y d y z d x z d x square y square d z square so which one this one y the reason is that this particular sub shell almost have got similar energies as that of 3px 3py 3p but you can say z as an s3s almost similar comparable energies because they only participate in hybridization because whenever the what you can say redistribution of energy takes place almost the difference in energy should be not be so large it should be little bit slightly the difference in energy is going to decrease all right so hence this particular subshell is involved got it now so hence what are those orbitals involved here or what are the sub shells here involved so sub shell is three a sub shell and three piece option three p is three uh under three p substitute we have got three orbitals three p x three p y three p zero all right and under three d sub shell one orbital then three d is so here what i said what are the what we can draw we cannot draw the floral diagram so we will be drawing the what you can say the equivalent diagram equivalent diagram will be drawing it let us say pcl5 molecule you can say that so this is the x this is your axial bond which is slightly bigger and this two this three are the equatorial parts these three are the equatorial bonds we can say that these three are the equatorial bonds you have read it already these three are the equatorial bonds and this is the axial bond so this angle is 90 degree and this angle is 120 degree isn't it this is 120 degree and this angle is 90 degree and axial bond you can see i have done it slightly bigger generally in case of pcl5 molecule the axial bond length in case of pcl5 molecule the axial bond length is 215 picometer and equatorial bond length is 204 picometer we can say slightly bigger 250 and this is 2 4 means slightly bigger we can say that why it is slightly bigger i have said you already in this video 3 once again i am saying you because this this particular atom will suffer repulsion from these two chlorinator so as to have minimum repulsion and maximum stability this bond will be going little bit outside so as to have minimum repulsion and maximum surgery this bond will be slightly outside will be stretched little bit outside all right so that the bond length is slightly bigger for that particular and this is called trigonal this is trigonal and these two hands you can see which is stretched outside they are called as bi pyramidals it is trigonal bipyramidal this is your pcl5 molecule same thing ah you will see in this is your sp3d hybridization we got it sp 3d i've already studied 1s 3p and 1d sub-shell is involved you can see that same thing uh we can draw it for sp3 d2 which will be octahedral in shape sp3 d2 sp3d is trigonal bipyramidal sp3 d2 is octahedral instead what are the examples you can give it we can give the example here is sf6 all right we can give it scl6 we can say that sf6 we can say that ah sc f6 also we can say that clear or you can say tf6 also we can say that or tcl6 you can say that clear so it will be octahedral in nature so here what are the substances involved here same thing here you will find it out that is nothing but three s subshell is involved then three p subshell is involved and three d subshell is important all right so let us write the electronic configuration it will be clear 16 so now 16 means it will be 1s2 2s2 2p6 3s2 and it will be 3 p 4 see 6 4 10 10 to 12 for 16 all right so here what you will find that one electron from the s option all right one electron from the substitution and one electron from p sub shell will go to the d subshell so hence it will be what 3 s 1 then 3 p 3 and 3 d 2 all right 3 s 1 3 p 3 and 3 d 2 so c l 3 s 3s1 this is unpaid one unpaid 3p 3 3 unpaired and 3d here 5 should be there 5 out of that 2 is unpaid so what you are getting it here 1 s option three p sub shell and two d substitute is participating one is three p and two d so hence it is what s p three d two so that is s p three d two is involved so in this way what we find is that why the electron will jump i've already said you from s sub shell to d sub shell from p sub shell to d subject what is going to jump because when the fluorine atom will approach chlorine is 9 1 is 2 2 is 2 2 p sorry 2 p 5 all right so it is when the fluorine atom is going to approach the sulfur atom so when the fluorine atom is approaching absolutely then the attractive forces is taking place due to which the energy is released and this energy helps to jump the what you can say this electrons from s suction to d suction and p substitution to d suction we can say vacant d because this option is vacant it will jump for that reason and hence you can say that redistribution of energy is going to take place and the hybrid orbitals will develop and by the hybrid orbitals this sort of hybridization is going to take place what is that sp3 d2 hybridization what is the same octahedraline shape got it clear in case of sf6 molecule i have taken the example or any other molecule you can take which is octahedral in shape is also sp3 d2 hybridization either off either the molecule is sp3 d2 and octahedral or we can say that if you know the shape of criterion then the hybridization will be sp 3 d 2 and the bond angle will be 90 degree we can say that all right so now what are the sub shell involved i already said you s subshell and three p sub shell so what are the orbitals in three p three p x three p y and three p zero and two d sub shell is involved we can say that clear so here what you find is that that is two d sub shell is involved 2d substitution now which are the subshells of 2d is involved already i said you dz square in the previous example i said dz square is having similar energies as that of what you can say 3p here also another is there that is dx square y square d x square y square all right that particular subshell is involved in hybridization we consider d x square y square the subshell is involved in hybridization all right how is d x square y square you know that this is your d x square y square d x square y square this one is this subshell and d z square already we have seen it so what are the ah sorry substitution d x square y square ah that is nothing but orbitals will be involved so in two d subshell d x square y square and d z square will be participating it and in three p sub shell all the orbitals three p x three by three preset will be important and s options will be involved in hybridization why the reason is same again because they have got similar energies all right they have got similar energies we can say that clear so you can see it also that the energies are similar how it is now here it is 3s isn't it clear it is 3p okay so three p means l value is how much one so n plus l will be three plus one four again it is two d so s p all right 0 1 2 so l values here are 2 so 2 plus 2 it is 4 3 plus 1 it is 4 here 3 s 3 it is 1 that is 3 plus 0 it is 3 so almost similar energies are there for that particular purpose they are involved in averages got it understood now you will see that that that is your diagram equivalent diagram here also we cannot do the flora diagram you will see the equivalent diagram so sf6 how it will be see s f 6 6 fluorine atoms i will make this a little bit bigger so that it will be clear to you so this is your sf6 molecule we can say that which is octahedraline shape we can say clear up fine this is sp3d2 hybridization similarly we have also got ah that is d3 hybridization sp3 d3 so sp3 d3 we can have if7 if7 so uh what we find is that i that is atomic number will be quite bigger one so that is ah here what you will find is that in case of if7 now we need to understand that iodine atom has got seven electrons in the outer motion all right now fluorine atom also has also got seven electrons in the outer motion both they require one one electrons got it now in iodine what are the uh what you can say suctions involved in hybridization you will find it out that is i means it belongs to which period let us count it and say fluorine is second period chlorine is third period bromine is fourth period iodine is fifth period that means 5s suction 5p subshell and 4d subshell is involved in hydration 5s 5p and 4d subshell is involved in hydration so which five is subshell 1 5 is 5s is only one isn't it 5p subshell how many orbitals three orbitals five p x five p y five p z and five these sub shell which are the which are the three because what i said s p three d three isn't it clear in case of iodine i am drawing only the outer most i am writing the electronic configuration that is 5 s 1 5 p x 1 sorry 5 p y 1 5 p z 1 then 5 d which one x y all right then 5 d x z and 5 d y z all right you know there are two in case of d there are five orbitals out of that three are planar and two are axial already i have talked about the two all right that is d x square y square and d z square they are axial over now those two are having same energies now this three are having the same energies because these three are planar we can say that these three are planar you can see nothing i am writing there is also one understood so in this way you can see that is 5s 1s subshell 3 p sub shell and one two three three p sub shell and three d subshell so as a result what is that sp3 d3 it is is involved in hydrogen correct this is sort of iodine the last outer most substance which is participating hybridization i am doing it i did not write the whole atomic number you know that i have counted it and i have showed it to you similarly for the fluorine atom also seven fluorine atoms will participate because one two three four five six seven each iodine will give one electron to one one fluorine and hence seven chlorine atoms will participate and it will undergo bond formation all right so what is the shape of sp3 d3 it is nothing but pentagonal bipedal hope you are following it i i f7 it is pentagonal by pyramidal we can say that clear bond again will be one it will be 90 degree another will be nothing but 72 degrees so to say i'm trying and show to you what are the sub shell involved i've studied what are the orbitals in all cells i have showed you why it is involved that also i said because they have got the similar energies the comparable energies we can say that and how is the sp3 d3 you counted in c you'll be getting it or you put it in the formula you'll be getting it if7 molecule is put in the formula you'll be getting it all right now let us see the diagram for if7 what i said it is nothing but ah here it is you can see i 1 2 3 4 5 6 and 7 these are longer these are little bit longer stretched this reason is the same as i have given in pcl5 because all these are the equatorial bonds and this is the axial bond these two are the axial bond which is slightly bigger than the equatorial bond the reason is the same because it will suffer more repulsion from these two chlorine atom so as to have less repulsion and maximum stability it is little bit outside so you can see this pentagonal this diagram you can see it is pentagonal i am just this is not required now this is not the actual bond so i am making a dotted line this one i am rubbing it means i am making making it as a dotted line dotted line means which is not required all right so you can see that so hence it is pentagonal bipyramidal we can say that the molecule is pentagonal bipedal in this way if7 clear so now the last one in hybridization we will come to is dsp2 hybridization so basically ah if you see the number the value of x for sp3 object is 4 and for dsp 2 is also 4 but dsp 2 is a special case exceptional case we can say that i have said you there are many exceptional chemistry is full of exception definitely you know that so in this way you will find uh those molecules which are chart species like nicn four two minus generally you will be studying this one in class 12 also a chapter will come coordination compounds so the example which i am going to give it to you now they are the coordination compounds all right so what is that ni cn whole 4 2 minus we can say that or cu cn whole 4 so basically that is the dsp to hybridization listen carefully ish is is being shown by those species which are charged you can see is that not neutral species neutral species will show sp3 type of vibration already we have discussed with you neutral species will generally show and chart species will generally show that is 2 kind of hybridization so to say we can say that but the condition is that in those atom will ah what you can say show dsp hybridization which is having d sub shell present in it suppose a chart species is there but it does not have a deception means the atom the atomic number is not such that it is having a deception then how it will show the dsp to hybridization are you getting me point my point yes nickel the atomic number is 28 copper the atomic number is 29 so when you write the atomic number 28 and 29 you will see d subshell is coming here in it suppose you will take a molecule all right like oxygen is having this option no then how it will so that is dsp hybridization so use the common sense use the common sense you will come to know the answer all right so hence those molecule only will show which contains at least a deception or at least it should have a vacant d sub shell then also it is being present that means elements belonging from third period onwards can show the what you can say dsp2 kind of hybridization but it should be charged again i am saying it should be chart species because you can see i have given the two examples there you can have any other examples many other examples you can have it so but it should be chart species now how will you draw it now before ah what you can say going here i am i i have to calculate the oxidation number of this central atom this is on a central metal atom i said you know this particular examples again i will teach you in class 12 the chapter will come coordination compound there i'm going to teach you again i'm going to tell you again about it and this is known as ligands this is called as cn is a ligand what i said ligand l i g and d what are elegance ligands are the donor atoms it will donate to whom it will donate to the central metallic why because central metal atom is right the electronic configuration and see it is having deficiency of electrons all right for that reason it donates electrons to the central metal atom which is central metal atom this one is a central metal atom metal now this is metal atom central metal atom it is going to donate electrons to what is that nothing but to the nickel delay copper we can say that who is going to donate cn is going to donate so hence cn is called as ligands ligands means donor atoms which donates electrons to the central metal atom so what i said donates electrons so how many electrons it is donating at least one pair that is it donates lone pair of electrons hence what i told the name of the chapter to you coordination compounds means it forms coordinate bonds with a central metal atom cn forms coordinate bond with central metal atoms c in forms coding bond with nickel atom all right coordinate bond which is noted by an arrow you already already alright so hence it is said to be s all right now let us see the formation so i have to first find the oxidation number so you know that how to find the observation number is x c n the valence is minus 1 how many c is there 4 cm so it will be minus 4 and the total charge of the species is minus 2 c is equal to minus 2 total charge is minus 2 so how much you got x is equal to nothing but if this side will go it is plus 4 minus 2 it is nothing but plus 2. so oxidation number of nickel is plus two you find the oxidation of one in this way only the one which you do not know your degree is x five then after that this is one c n the valency is minus one our oxidation number is minus one we can say that so four c n so x minus 4 is equal to minus 2 x equal to plus 2. so you got it the oxidation number of nickel is plus 2 that means in plus 2 oxidation state nickel reacts with cyanide ion but it undergoes hybridization or bond formation and c and iodine to form n i c n for two minus species we can say that clear now nickel atomic number i said you there is 28 1 is 2 2 is 2 2 p 6 3 is 2 3 p 6 4 is 2 3 d 8 6 8 but i have nothing to do with nickel i have to do with ni 2 plus so n i 2 plus how it will be formed m 1 s 2 2 is 2 2 p 6 3 s 2 3 p 6 outermost shell is this c highest principle number 4 this is outermost this is not the outermost even if it comes last but it is not the outermost i've taught you already so it is from here two electrons will be lost from here four s because highest principle quantum number determines outer muscle so this is outer motion so two electrons will be lost from here but let's see carefully understand it carefully we calculate which i have taught you in atomic structure so it will be 38 all right so hence what we find is that what you are getting it you draw uh the spin quantum number you will find it out that vacant orbital sum is there one two three four one more i want your five all right it's a little bit bigger let me make it this one done so and four s also this weekend four is a second because two electrons is lost it is vacant now three d eight this is three eight one two three four five six seven eight all right so now not only it contains that is uh this unpaired electrons and here is vacant or vitals also definitely you know that along with forest four p subshell is also there which is also make it four p is there now of course four p is not involved here in uh bond formation but that does not mean it is not there in the fourth cell 4s is also there 4p is also there 4d is also there but i don't want 4d here i want up to 4p only got it why why i wanted to 4p because i want how many 4 vacant orbitals i want why four because how many ligands are there four ligands each ligand will give how many pair of electrons two because what i said ligands are donor atoms what it donates it donates lone pair of electrons to the central metal atom lone pair lone pair means two electrons it will donate so hence how many unpaired uh how many and what you can say vacant orbitals i want vacant orbitals how many i got four white is four see here four ligands are there four ligands are there each will give two two electrons to the vacant orbital isn't it now you cannot have here one electron so two more electron will come here you know that is not possible all right each orbital can accommodate maximum two electrons of opposite sphere so hence what will happen now you will study there in classroom that cn is a ligand and it is a very strong ligand so as it is strong ligand as i said this is an exceptional compound as a strong ligand it will go against the hunt's rule it will go against the hand soon and the pairing is going to take place see this electron will come here generally electron jumps from lower to higher but it here it will go against the hand sole because the strong ligand strongly again what it will do pairing it will do the pairing automatically it will do the pairing the pairing is going to take place and the four vacant orbitals are created one two three and four got it now what i said this is a strong ligand cn how will you know strongly again and we click and you will study everything there all right but here just i am saying you for your knowledge it is a strong ligand so strongly and what it will do it will go against the hands rune and it will pair up the unpaired electrons listen carefully it is going to pair up the unpaired electrons all right so two unpaired electrons is there the pairing is going to take place why it is going to do so as to create a vacant orbital so in the vacant orbital two electron can be donated see here i have donated two two electrons to each so i have donated one to d sub shell another two s suction another two two p subshell so what are the orbitals involved here see here what are the orbitals involved here s then it is 2p and it is nothing but 2d all right that is sorry what i wrote i am very sorry this is your d d sub shell and here it is 1 s and 1 it is 2p very sorry this is your 3d substitution this is isn't it this is your 3d this is 4s and 4p so 1d one s subshell and two piece options it is nothing but dsp 2 for that the hydration is 2 all right that's one thing you need to remember that is strongly again pairing is going to take place correct it is go it is go against the hunt suit because already steady this is an exceptional compound all right so now if it is dsp2 what is the shape i said square planar so how that i will be drawing it ni then square planar the shape is this one cn here one cn here one cm here one cn here this is i will write it here two n's so n i c n four two minus this is shape it is square planar what i said square planar p l a n a r plane r square planar molecular first i wrote the nickel configuration then i wrote ni 2 plus y because observation number i got it s2 so hence i created the orbitals here all right i want four vacant orbitals because four ligands are there so it will going to donate six so four vacant orbitals is created now what are the orbitals d s p two one d one s and two p what it signs it is called d s b two hydration which is nature this is all about your we can say hybridization thank you