my dear students this particular video is going to give you for sure some 40 marks in just 10 to 15 minutes so I would want you guys to watch this particular video till the end in this video rules spectral lines quantum numbers calculation of valence electrons identification of period and group pin magnetic moment boiling point from vanto factor order of the reaction from half-life all these short short topics I shall be discussing with you in very small interval of time so without wasting any sort of time let's get going let's get started my dear students there is a short short question asked everywhere every year from your fajan's rules as per your fajan's rules are concerned coalent character that is directly proportional to polarizing power of kine and this coalent character is also directly proportional to polarizability of anion polarizing power of kine depends on two factors charge on kine size of kine polarizability of nine depends on charge on nine and also is directly proportional to size of nine these are just the factors which you need to know and you can easily solve the questions based on fajan's rules have a look exactly what I'm saying my dear students if you look at all these questions over here this is al positive f negative al try positive cl negative al positive Br negative al positive I negative so kine everywhere is same I'm not going to decide on the bas of kine Talk in terms of anion charge on every anion is same so decide on the bas of size of anion more the size of anion more the polarizability of anion and more the polarizability means more the coalent character that's it similarly here you have got la positive na positive k positive rb positive cs positive in every kion here charge is same so talk in terms of size lesser the size of kine more the polarizing power of kion hence more the coalent character so this is going to be the order of your coalent character similarly this is B F negative CL negative Br negative I negative again charge on every anion is same so talk in terms of size more the size of anion more the polarizability hence more the coent character and this particular question right you are going to tell me its answer in the comment section of this particular video and over here let me tell you let me give you a hint over here you get pseudo inert gas configuration right based on pseudo inert gas configuration let me know the answer of this particular question moving on to the concept number two as you all must be knowing this is the formula to calculate total number of spectral lines when the electron makes a transition now now my dear students out of the total spectral lines how many spectral lines will fall into LMAN series how many spectral lines will fall into Balmer series how many spectral lines will fall into series that's given by this particular formula which is N_sub_2 high energy state minus ground state of that particular series which we are talking about for example what does it mean let's say the electron is moving from fifth shell to first shell and when the electron moves from fifth to first how many spectral lines in total we are going to get we are going to get some 10 spectral lines out of these 10 spectral lines how many will fall into Lman series it is going to be equal talk about the formula N_sub_2 n_sub_2 is higher energy state minus ground state of Lman is 1 so 5 - 1 is four n_sub_2 here is five minus ground state of Balmer is two so 5 - 2 is three higher energy state N2 that's five minus ground state of position is three the value comes out be two higher energy state is five minus ground state of bracket is your four that is one and n_sub_2 is five and ground state of fun series is 5 so 5 - 5 is 0 if you take the sum of these 4 + 3 is 7 7 + 2 is 9 9 + 1 makes it 10 so overall 10 spectral lines are obtained and out of these 10 spectral lines four spectral lines will fall into lyman three into balmer two into passion one into bracket and zero in fund so this is how you are going to solve this sort of a short short question as well my dear students there is one more thing one more frequently asked question if you all must be knowing first of all you must know quantum number L its value is from 0 to n minus one magnetic quantum number m its values from minus L to plus L spin quantum number can have two values only one is + 1x2 and another one is minus 1 by 2 now I have given a set of quantum numbers you have to tell me which of the following is correct and which of the following is incorrect my dear students look at the first option n values N value is four l value is three that's correct because L is from 0 to N minus 1 0 to 4 - 1 that means 0 to 3 either 0 or 1 or 2 or 3 correct m is from minus L to plus L so m will be from minus3 to + 3 yes -3 is given and spin quantum number is + 1x2 so this is the correct one n value is four look at this m l value here l value is four can l have same value as that of n no l value is from 0 to n minus 1 0 to 4 - 1 that means 0 to 3 so 0 1 2 3 are the permissible values of l over here so this set of quantum numbers is not possible see n value is four correct l value is three correct but m is from minus l to + l so minus3 to + 3 -4 does not in come into that category so this is not the correct one last one spin quantum number value is coming out to be zero right but we know spin always has two values only either plus 1x2 or minus 1x2 so there is only one option that is going to be correct among all so this is how you are going to tackle this sort of equation based on these three things if you remember and if you remember these three things you're sorted over here now my dear students let's say you have got an element whose first ionation energy is 5 electron volt second is 10 third is 20 and fourth ionation energy is 2,000 electron volt you have to tell me how many valence electrons are there in the atom my dear students it's pretty much simple let's say this is the nucleus and for example this is the outermost shell right to take the first electron out from outermost shell to take the first electron out from the outermost shell how much energy is required 5 electron volts to take the second electron out how much energy is required 10 electron volt to take the third third electron out how much energy is required 20 electron volt to take the fourth electron out how much energy is required there's a huge jump from here to here there's a huge jump from here to here that means the fourth electron will be taken out from the inner subshell that right that means the fourth electron has to be taken out from the penultimate shell basically right not from the valence shell from the penultimate shell that's why more energy is required to take the fourth electron so how many electrons are there in the valence shell there are three electrons in the valence shell so the element will have the three valence electrons that's something which you are supposed to check calculate the number of valence electrons here and the number of valence electrons here will be three because the fourth electron its ionation energy is 2,000 electron volts what does that mean that means lot of energy is required so this energy this electron won't be present in the last shell it'll be present in for one shell before that that's why more energy is required to take that fourth electron out so only there are three electrons in the outermost shell so three valence electrons in total this is again a short short question similarly how do we calculate spin magnetic moment my dear students spin magnetic moment is given by the result root of n into n +2 where n represents number of unpaired electrons n represents number of unpaired electrons for example you have got k4 fee cn6 iron here is in dos state and fee dipositive configuration is this you know it here sinite is the strong field liant if sinet is a strong field it is going to cause a pairing of electrons and when it causes a pairing of electrons there won't be any unpaired electron that means n value is zero if n value is zero my dear students spin magnetic moment here is going to be zero if n was one if there was one unpaired electron then put the value of n as one here if there were two unpad electrons put the value of ns2 here so this is your homework question do let me know its spin magnetic moment in the comment section of this particular video moving on my dear students to one more thing remember when the solute under goes dissociation when the solute under goes dissociation boiling point at that time is directly proportional to vanto factor and vanto factor is directly proportional to the number of ions produced by the solute for example you have got nacl2 al3 na3 p4 so when na goes into the solution how many ions it's going to give two ions na positive and cl negative ba2 will give one ba positive and two cl negatives so in total three ions this will give three cl negatives one trip positive four ions and this will give how many exactly this will give how many let me just erase this option let me keep only three options over here so that it'll be easier for you so basically this is giving two ions this giving three ions this is giving four ions more the ions produced more is going to be the vanto factor more is going to be the boiling point so boiling point of the third one is going to be maximum followed by second one followed by first one right if this was first if this was second if this was third I would say third one will have maximum boiling point followed by your second one followed by your first one your freezing point of the solution that is inversely proportional to factor so my dear students if I ask you what would be the freezing point order freezing point order is going to be reverse of it freezing point order is going to be reverse of it so more the ions produce lesser is going to be the freezing point so remember this particular question as well similarly my dear students there is one more frequently asked question that is calculation of order from half life calculation of order from half life i want you guys to remember this particular result okay halflife 1 divide by half life 2 is equal initial concentration 1 initial concentration 2 whole raised^ 1 - n what does it mean exactly for example you have got a question here you have got the reaction A gives B when the initial concentration of reactant was kept as 50 molar at that time halfife of of the reaction was 4 seconds when the initial concentration is 100 molar half life is 2 seconds when the initial concentration is 200 molar half life is 1 second so we have to check basically what is going to be the order of the reaction my dear students let me just do one thing let me call this as a1 so this will be this will be t1 let me call this as a2 this will be t2 so what I'll be doing exactly I'll be putting these values into this expression so t1 is 4 t2 is 2 is equal to a1 by a2 a1 is 50 a2 is 100 whole raised^ what 1 - n where n represents order so this value comes out be 2 this value comes out be 1x 2^ 1 - n so I would say 2 is equal 2 is equal take the reciprocal it becomes 2^ n minus one so my dear students from here you can say n minus 1 value is equal to 1 so n value comes out be two so I would say order of the reaction whose data is given to me that is two so remember this particular result this is a very important question my dear students which is frequently asked nowadays Now in order to calculate the period number how do you calculate period number of the element highest value of principal quantum number that gives me the period number of an element let's say you have got an element whose configuration is given to me right so what is the highest value of quantum number here highest value of principal quantum number is three so period number of this element is three highest value of principal quantum number is four so period number of this element is four right similarly how do we check the group number for SB block elements for SB block elements group number is number of electrons in ns shell let's say this is the electronic configuration of some element right and the element belongs to S block perfect so how many electrons are there in the last S only one electron so the element belongs to group number one how many electrons are there in Ns one electron so this element belongs to group number one perfect for S block for P block it's going to be 10 plus sum of NS and NP electrons so NS is equal to two electrons np is equal 6 so 6 + 2 makes it 8 so its group number is going to be 10 + 8 that makes it 18 similarly it's going to be 10 + 7 that makes it 17 so these are the group numbers of the respective elements whose configurations are given to us when it comes to DB block elements it's going to be simply sum of n minus 1d and ns electrons so this is your n minus 1d and this is ns n minus 1d carries six electrons ns carries six two electrons so 6 + 2 makes it 8 so the element belongs to group number eight similarly all your eB block elements my dear students they belong to the third group all your EB block elements they belong to the third group so this was a short short session my dear students so please and please do watch this session on priority make its notes i'll be providing the session PDF of this particular session on my telegram channel which is Vasim Chemistry Official w A S I M B H A T Chemistry Official is the Telegram channel name wherein I keep on sharing all the session PDFs of these sessions in that particular Telegram group so do join that telegram group and I'll keep on coming with these sort of sessions again and my dear students Sunday uh 5:00 p.m i would want you guys to join us live because we are going to conduct the most predicted need 2025 paper right okay so do join me tomorrow exactly at 5:00 p.m all the educators of this particular channel will be present and showing you all the most predicted need 2025 questions uh when it comes to your preparation so with this take a leave take care god bless you all and love you all guys bye-bye