hello and welcome to lucid mind i'm Majid and in today's video i'll go through solving of multiple choice questions paper 1 variant 2 from march 2020. if you want to go to a particular question just follow timestamp in the video description let's begin question number one what number protons neutrons and electrons are present in the ion fe plus 3 now the atomic number is not given so we will be consulting periodic table so from the periodic table we come to know that the atomic number is 26 mass number is 54. so the number of protons is 26 number of electrons as the charge is plus 3 so it means 3 electrons are lost so it is 26 minus 3 equals 23 electrons and neutrons for neutrons we have to subtract mass number and atomic number 54 minus 26 the answer comes to be 28 so now it means that the answer is a second question for which hydrocarbon are the molecular and empirical formula the same so molecular formula is something that gives the actual ratio of atoms and empirical formula is the one which gives the simple ratio of atoms so first one is butane formula for butane is c4h10 this is the molecular formula so if you are going to simplify the number of atoms it comes to c2h5 for ethane there are two carbon atoms c2 h6 and the simplified formula is ch3 popentine is pent one e means the double bond is on first carbon atom so as it is an alkene there are five number of carbon atoms and 10 number of hydrogen atoms the empirical formula is ch2 for propane we know that there are three carbon atoms and eight hydrogen atoms now this ratio cannot be simplified further so the empirical and molecular formula are same so it means that the answer is d question number three which molecule does not have any 90 degrees or 180 degrees bond angles part a is c2h6 which is ethane in ethane carbon is making single covalent bonds each carbon is making four single covalent bonds so the bond angle is 109.8 and the structure is tetrahedral b is carbon dioxide now the bond angle in carbon dioxide is 180 degree because the structure is linear c is phosphorus penta fluoride now in phosphorus penta fluoride there are two different types of bond angles which are present three of the fluorine atoms are in the same plane while two are not so the bond angle here is 120 degrees and this one shows the bond angle of 90 degrees d is sf6 so in sfc bonds are like this two chlorine atoms four fluorine atoms are in the same plane so the bond angle is 90 degrees while the one above the plane and below the plane also has an angle of 90 degrees so the answer is a question number four the following data are needed for this question now two different types of enthalpies are given one is of sodium bicarbonate second is of sodium carbonate both are reacting with hcl on heating sodium hydrogen carbonate decomposed is as shown what is enthalpy change for this decomposition so we have to find the enthalpy change of decomposition as three different reactions are given so we can use hess's law so writing the main equation we have 2 n a h c o 3 solid gives sodium carbonate solid along with water liquid and carbon dioxide as gas now na8co3 also reacts with hcl it forms hcl and forms sodium chloride along with water and carbon dioxide similarly sodium carbonate also reacts with hcl and forms the same product just have to we just have to balance out here now it is balanced now let's see according to hess's law which is one step process and which is two-step process if this is the reactant and this is the final product so this is going to be a and two steps are involved here so this is going to be b and c so according to his law a is equal to b plus c now we have to find enthalpy changes in a b and c 2 nah co3 are giving 2 nacl and water and carbon dioxide so 2 moles of nhu 3 are involved so taking value from the first equation we have sodium carbonate with hcl and the enthalpy changes minus 38.97 so as two moles are involved it will be 2 into minus 38.97 it comes to be minus 77.94 b step is something that we have to find out c is also given in the question sodium carbonate plus 2 hcl gives minus 96.59 so c is minus nine six five nine just putting the values that we have minus seventy seven point nine four is equal to b plus minus 96.59 so b is equal to minus 77.94 plus 96.59 and therefore it is eight point six five kilojoules per mole the answer is positive so it is c question number five in the redox section shown how the oxidation states of vanadium and sulfur change so they have given us two elements and we have to find the change in oxidation states first finding the oxidation state of vanadium in vo2 plus so vanadium as oxygen is in group 6 so it has a charge of -2 and there are two oxygen atoms so two into minus two and the overall charge is plus one so v minus four is equal to plus one so v is equal to plus 5 now the answer could be c or d finding the oxidation state of sulfur in so2 again we have two oxygen atoms so sulfur minus four and the overall charge is zero sulfur is plus four now the oxidation state of vanadium in the product is plus three and oxidation state of sulfate sulfur and sulfate iron can be found like this sulfur there are four oxygen so it will be minus eight and the overall charge is minus two so sulfur is equal to plus eight minus two and plus six so it is obvious that the answer is d question number six the equation for the reaction between silver chloride and echos ammonia stone what the units of kc for this reaction so we have to find the units of kc as we know that kc is the concentration of product divided by concentration of reactants this is one of the product chloride is another product and the reactant is silver chloride solid and ammonia all have their powers equal to the number of moles as silver chloride is solid so it does not have any concentration unit the concentration is only possible for aqueous solution not for solids so as we know that the unit is mole by decimeter cube ammonia has got two number of moles so we squared this one now all of them will be cancelled out and there will be no unit left behind so the answer is part a no unit question number seven sodium azide and an3 decomposes as shown they have given the equation which volume of nitrogen measured at room temperature and pressure will be produced by the decomposition of 150 grams of sodium aside so sodium azide decomposes to give sodium atoms and nitrogen gas so first we are going to find the ratio between the given components so nan3 and nitrogen according to the balanced chemical equation 2 moles of nan and 3 produces 3 moles of nitrogen now they have given you the mass of sodium azide so first we have to find the moles of sodium azide as moles is equal to mass divided by molecular mass so mass is 150 grams molecular mass of nan3 is 23 plus 3 into 14. so moles is equal to 2.307 so if two moles of sodium azide produces three moles of nitrogen so then 2.307 moles of sodium enzyme will produce x moles of nitrogen cross multiply find out the value of x so it is 2x is equal to 3 into 2.307 divided by 2. so x is equal to three point four six zero now this is the number of moles of nitrogen produced from 150 grams of sodium azide but they have asked about the volume of nitrogen not the number of moles so we know the formula of volume is moles into 24 decimeter cube why have we chosen 24 decimeter cube because they told that this reaction was occurring at room temperature and pressure if it was standard temperature pressure we should have taken 22.414 decimeter cube so now 3.460 into 24 volume comes to be 83.05 decimeter cube so the answer is part b question number eight a stable iron n5 plus has been produced by research chemists which structure is most likely to show the electron arrangement of the eye so it says that the n5 is stable by stable we mean there the octet should be complete so there should be eight electrons around the nitrogen atom so in a part there are five nitrogen atoms and while counting the number of electrons we have one two three four five six electrons so this does not seems to be the stable arrangement coming to part b nitrogen has one two three four five six seven electrons so again this is not octet configuration part c we have one two three four five six seven eight electrons so yes c seems to have the complete octet so now finding out the next number of nitrogen atoms we have one two three four five six seven eight yes good the next one is one two three four five six seven eight again the octet is complete one two three four five six seven eight and the last one one two three four five six seven eight so yeah all atoms have complete octet so the answer is rc question number nine 1.8 grams of water 80 to 227 degree centigrade in a sealed container turns to steam with a pressure of 200 kilopascals what is the approximate volume of the container question number nine 1.8 grams of water heated to 227 degrees centigrade in a sealed container turns to steam with a pressure of 200 kilopascals what is the approximate volume of the container so now we'll see this is from gases so we are going to use the ideal gas equation pv is equal to nr t p is the pressure pressure should be in pascals v is the volume volume must be in meter cube n is the number of moles r is the ideal gas constant and t is the temperature temperature should be in kelvin so first finding the number of moles 1.8 grams of water means 1.8 divided by mr of water is 18 so it is 0.1 moles 227 degree centigrade is the temperature so converting this into kelvin we have 227 plus 273 and it is equal to 500 kelvins pressure is 200 kilopascals so we can also write as 200 exponential 3 pascals now we have to find the volume simply putting these values into the equation so we have volume is equal to n r d divided by p putting the values volume is to be found number of moles is 0.1 moles the value of r can be seen from the data booklet is it is 8.31 temperature is 500 kelvin and pressure is 200 exponential 3 pascals so calculating this the value of v comes to be 0.002 meter cube which can also be written as 2 exponential minus 3 meter cube so the answer is part b question number 10 when the equation is correctly balanced what is the value of c so they have given you a b c d e so many values are not given so what does the balanced chemical equation means it means that the number of atoms should be balanced on the rectangular product side and also the charges must be balanced charges means that the positive charge in the negative charge on both sides of the equation should be equal so as we know that there are two manganese atom on the rectangle side so we can balance the manganese ion by writing two in the place of a writing the chemical equation again we have c 2 h h4 plus h2o plus h plus plus 2 mno4 minus gibbs c2 h6 o2 and mn plus 2 which is 2. now finding the charges on each of these molecules or compounds we can see that c2h4 is neutral so the charge is zero overall charge is zero similarly h2 is neutral so the overall charge is zero h plus is positive and we have to find the number of moles so let's suppose suppose it to be x then we have two meganations mno4 minus we have already balanced this one now the negative charge is there are two mno4 minus ions so it means that the negative charge present on manganate is minus two for one magnetine it is minus one and as there are two moles so the charge is minus two now on the product side we have c2h6o2 which is neutral so the charge is 0 and mn plus 2 which are two number of moles so the charge is plus 4. so we have the equation x minus 2 is equal to 0 plus 4 or x minus 2 is equal to 4 or x is equal to plus 6 so this shows that the charge on hydrogen atom should be plus 6 so as we have one hydrogen atom there must be six hydrogen atoms so the charge should be plus six so now there are six h plus so c is six what is the value of c they were asking about the value of c so yes now the answer is six as the charges are balanced on both sides of the equation question number 11 the main stage in the contact process is an equilibrium reaction so2 reacts with oxygen and forms so3 which row describes the effect of the named condition on the equilibrium yield so now we have to find out whether the yield of this chem correction is increasing or decreasing or remaining the same first one is the presence of catalyst so as we know that catalyst only reduces time it does not have any effect on the yield of the product second is pressure as in this chemical equation all the reactants and product is in the gaseous form so it means that there will be an effect of pressure so finding the number of moles there are two number of moles of sulfur dioxide and one number of mole of oxygen on the rectangle and we have two moles of sulfur trioxide on the product side so overall there are three number of moles on the rectan side and two number of moles on the product side as we know that if the pressure is increased then the reaction moves in the forward direction because whenever the pressure is increased the equation moves from more number of moles to less number of moles as the number of moles of reactant is three and the number of moles of product is two so direction will move in the forward direction so it says high pressure so it should increase the yield third one is high temperature now the in contact process this chemical reaction is exothermic because the enthalpy change is negative so temperature you have to remember this point that whenever we increase the temperature for exothermic reaction direction goes backward while for the endothermic reaction the reaction goes forward or we can say that the exothermic reaction for the exothermic reaction the yield is decreased and for the endothermic reaction the yield is increased in this case we have exothermic year so the yield will decrease so the answer is part b question number 12 x and y are the oxides of period three elements if one mole of x is added to water the solution form is neutralized by exactly one mole of y so you have one mole of x direction with water and it is neutralized by one mole of y so we have to solve it for all of these so the first one is b4o10 access p4o10 so if we dissolve it in water it forms phosphoric acid h3po4 now balancing this chemical equation we have four phosphorous atoms on the right hand side so we can balance the number of phosphorus atoms we can also balance the number of hydrogen atoms 12 hydrogen atoms on the product side so we have 12 hydrogen atoms now we see whether it it it could be neutralized by one mole of al2o3 or not 4 h3po4 plus aluminium oxide it forms aluminium phosphate along with h2o now balancing this chemical equation we have four number of moles of h3po4 so it means we have 4 phosphate ions so just balancing the number of phosphate ions we can write 4 on the product side now aluminium is 4 on the product side so we can write two now there are two aluminium on the reactant side in the last balancing the number of oxygen and hydrogen atoms we have six h2o so now we can see that one mole of p4o10 produces four moles of h3po4 and 4 moles of h3po4 require 2 moles of aluminum oxide while in the question it says there should be one mole of y so it means it is not a coming to part b so3 reacts with water to form sulfuric acid h2so4 now sulphuric acid reacts with aluminium oxide and it forms aluminium sulfate now we can see the ratio balancing this chemical equation the ratio in the first one is 1 ratio 1 balancing the second chemical equation we have 3 sulfate ions in the product side so we can write 3 with h2so4 there are six hydrogen atoms on the left-hand side and it is balanced by writing three with h2 now we can see that three moles of sulfuric acid require one mole of aluminium oxide as one mole of sulfur trioxide produces one mole of sulfuric acid so it means three mole of sulfuric acid will be formed from three moles of sulfur dioxide so the ratio is not one ratio one so b is also wrong now coming to part c part c has p 4 o 10 so the equation is similar now four moles of h3po4 are formed it reacts with na2o and the product is sodium phosphate and h2o now balancing this chemical equation we have four phosphate on the rectangle so we can write 4 with the product in order to balance the phosphate ions now the number of sodium atoms on the product height is 4 into 3 12 so we can write 6 to make it 12. so again the ratio is 4 ratio 3. so the ratio is 4 ratio 6 or 2 ratio 3 again it is not 1 to 1 ratio so c is also wrong now coming to part d we have sulfur trioxide which reacts with water to give sulphuric acid and the ratio of rectangular product is one to one now h2so4 will react with sodium oxide and form sodium sulphate along with water now balancing this chemical equation we have one sulfate iron one sulfate iron two sodium atoms two sodium atoms two hydrogen one oxygen two hydrogen one oxygen so now the ratio comes to be one ratio one so it means that the answer is part d question number 13 compound z is insoluble in water but soluble at low ph what could be compounded number one is barium carbonate number two is barium chloride so first we see what are the properties of barium carbonate so barium carbonate is basically insoluble in water but it is soluble in assets so acids basically means low ph for example a barium carbonate is dissolved in hydrochloric acid it forms barium chloride plus water plus carbon dioxide and barium chloride is water soluble b part is barium chloride yes it is soluble in water barium hydroxide it is also soluble in water barium sulphate it is insoluble in water or assets so the answer is part a question number 14 hot aqueous sodium hydroxide rates with chlorine which statement is correct number eight oxidation numbers of chlorine and hydrogen both change in the reaction the oxidation numbers of chlorine in the products are minus 1 and plus 1. so let's see what are the oxidation states sodium is plus one oxygen is minus two hydrogen is plus one chlorine is zero because it is neutral again sodium is plus one chloride is minus one sodium is plus one there are three oxygen atoms so the overall charge is minus six so the positive charge should be plus five water oxygen is -2 so hydrogen is plus 1 again there are two hydrogen atoms so each hydrogen atom has plus one the oxidation numbers of chlorine and hydrogen both change we see chlorine changes from zero to minus one and zero to plus five while the oxidation state of hydrogen is plus 1 on the right hand side and plus 1 on the product side so both does not change only oxidation number of chlorine is changing so a is incorrect b is the oxidation numbers of chlorine in the products are minus one and plus one it is as we can see in sodium chloride it is minus one but in sodium chlorate it is plus five so it is not plus one it is plus five so this is also wrong c part if the aqua sodium hydroxide is gold the reaction produces ns yellow instead of an selo3 yes this is correct because chlorine when reacts with gold sodium hydroxide it produces sodium chloride along with naclo and h2o so c is correct d sodium undergoes disproportionation in this reaction disproportionation basically means self-oxidation reduction so in this reaction basically chlorine is getting oxidized and reduced so it is not sodium it is chlorine question 50 solid ammonium nitrate is brought into a test tube and solution x is added to it the resulting mixture is warmed and the gas given off is tested with damp red litmus paper the litmus paper changes from red to blue what could be the identity of x and its role in this reaction so let's see in order to solve this question you should know about one chemical reaction always remember that ammonium salt plus alkali it always produces ammonia gas any ammonium salt when reacts with alkali it produces ammonia in the form of gas second point is that ammonia is basic in nature so base turns red litmus to blue and third point is that according to bronsted-lowry concept of acid and base base is proton acceptor so considering these three points the answer is part b because sodium hydroxide it reacts with ammonium nitrate to form ammonium gas and ammonia gas turns red litmus to blue and sodium hydroxide is a base so yes it is a proton acceptor question 16 sodium magnesium and aluminium are three elements in the period three of the periodic table each elements forms an oxide which row is correct so first one is a sodium oxide it is basic yes metal oxides are basic in nature magnesium oxide it is also basic in nature so this is wrong aluminium oxide is amphoteric this is correct now the structure of sodium oxide is giant ionic yes it is an ionic compound so the structure is giant ionic similarly magnesium oxide is also an ionic compound so the structure is giant ionic aluminium oxide is also a giant ionic compound so this is wrong simple molecular is wrong now sodium oxide has a high melting point obviously when the structure is giant ionic so the melting point should also be high same for magnesium oxide the melting point is high but it says the low melting point so this is wrong because aluminum oxide is also giant ionic and the melting point is very high now the fourth point is sodium oxide readily reacts with water yes it reacts with water and produces sodium hydroxide magnesium oxide slight reaction with water yes it is correct and it forms magnesium hydroxide with water aluminium oxide has no reaction with water as it is stable to the dissolution or reaction with h2o so the answer is part d question 17 a student investigates calcium nitrate crystals by heating them in the apparatus so he is heating calcium nitrate a colorless gas leaves the apparatus at y a flame is held to this gas which observations would the student make so first of all we will see how the thermal decomposition of calcium nitrate goes so calcium nitrate is decomposed when heated to produce calcium oxide nitrogen dioxide and oxygen gas now about no2 we know that this is toxic gas and it is also brown in color while oxygen it helps the flame no2 also dissolves in water to form acid nitric acid no2 plus h2o it gives hno3 the property of acid is that acid turns blue litmus to red so from these points we can find out the answer first one is litmus solution so litmus solution changes to blue or changes to red so it should be changes to red and also oxygen gas is also formed as the product so the flame burns more brightly so the answer is part c question number 18 how does concentrated sulphuric acid behaves when it reacts with sodium chloride so let's write the chemical equation we have sulfuric acid and sodium chloride it gives sodium sulfate and hcl let's find out the oxidation states so the oxidation state of sodium is plus one chloride is minus one sulfur is plus six plus six plus one minus one sodium is also unchanged oxygen is also unchanged hydrogen is also plus one so it the concentrated sulfuric acid behaves as an acid only as it is true it acts as an acid and oxidizing agent so we can see that sodium is not oxidized the oxidation state of sodium is unchanged oxidation state of chloride and sulfur is not changed so it is wrong so it neither behaves as oxidizing agent nor reducing agent it only behaves as an acid so the answer is part a question number 19 which statement about nitrogen or its compounds is correct part a in the haber process the temperature is kept high to give a good equilibrium yield of ammonia now we know that the habers process is exothermic so the energy is negative the temperature is kept high to give a good equilibrium yield this is wrong the temperature is kept high too in order to keep the pace of the reaction at optimum speed because in exothermic reaction whenever you increase the temperature direction goes backward and the yield is actually decreased so part is wrong b nitrogen gas is unreactive yeah it is unreactive because of the strong nitrogen nitrogen double bond as nitrogen is in group 5 of the periodic table so it makes three bonds with itself so it is not a double bond it is a triple bond so b is also wrong part c says nitrogen monoxide will react with carbon monoxide under suitable conditions so if we have n o and c o this reaction takes place in the catalytic converter so it converts into nitrogen carbon dioxide so yes under suitable conditions given proper catalysts this reaction is possible d the formula of ammonium sulfate is nh4 so4 as we know that the charge on mine is plus one and on sulfate is minus two so the formula should be ammonium two sulfate so d is wrong so the correct answer is c question number 20 which pair of compounds are functional group isomers of each other now what is the functional group isomer functional group isomers are basically those compounds that have the same molecular formula but different functional groups involved let's see the answers part a says butane one old and butanol this is butane one hole butane consists of four carbon atoms and oh is attached to the first carbon butanol butanol also consists of four carbon atoms and the functional group is aldehyde now let's see the number of hydrogen atoms on each of them this one has three hydrogen two hydrogen atoms two hydrogens two hydrogens this carbon has three this has two this is two now the total number of hydrogens in butanol and butanol is three four five six seven eight nine ten ten with alcohol and eight with aldehyde so it means they are not isomers b is ethyl propanoate and bentanoic acid let's draw the structures of ethyl propanoate this is ethyl propanoid this one is pentanoic acid now let's draw the number of hydrogens with each three two two three three two two two so this one has ten hydrogen atoms this one has 10 hydrogen atoms both have five carbon atoms each two oxygen atoms so it means that they are functional group isomers because the functional group in both of them are different one is ester and the second one is carboxylic acid let's see how part c goes it says hex one in and hex to e so hex one e means that the double bond is with first and is in between first and second carbon atom this is hex one e and hex two in means that the double bond is in between carbon two and carbon 3. as the functional group in both of them are same so they are not functional group isomers they are actually same alkene compound d is propylamine and propane nitrile as you can see the number of hydrogens are not equal so therefore they are not the isomers of each other question number 21 which row identifies a suitable starting material and reagent that can be used to produce butane nitride so we have to take two things one is starting material and second is reagent let's see what is butane nitrile so butane means we have four carbon atoms and nitrile means that we have one nitrogen atom present so in order to make butane nitrile with a cyanide reagent we need starting material of three carbon atoms so the starting material must be of three carbon atoms because the fourth carbon is introduced by cyanide so it could be a or it could be b because a has three carbon atoms and b also has three carbonate then we have to find out the reagent whether it should be hcn or nacn so why do we take nacn and why shouldn't we take hcn actually for this reaction ethanolic nscn is used hcn is not used because hcn is poisonous and also its dissociation power is very low it is barely dissociated into ions while nsen dissociates into na plus and cn minus readily so therefore the answer should be nsen so the answer is part b question number 22 the diagram shows the structure of ethane and dioic acid means it has got two carbon atoms and two carboxylic acid groups are present ethane dioic acid reacts with ethanol ethanol is ch3 ch2 oh in the presence of few drops of concentrated sulfuric acid to form a diester so two esters are formed one in each carboxylic acid functional group the molecular formula of the diesel is c6h10o4 what is the structural formula of the diastere so let's see how this compound reacts first i'm going to draw ethane dioic acid oh on each side and we have carbon double bonded with oxygen atom now let's take ethanol on both sides ch3 ch2 oh and similarly we are going to draw an opposite one oh h you can say ch2 and ch3 now what happens when carboxylic acid and alcohol they react with each other a water molecule is eliminated so this is a water which is eliminated so there are two molecules of water that are eliminated and the rest of the compound is ch3 ch2 o then we have carbon then double bond o then carbon then again double bond o then oxygen then ch2 and then ch3 so what is the structural formula of the diasta part a says ch3 ch2 co2 this is not co2 this is oco part b it writes oco and then we have co2 oco and then we have co2 so yes part b is correct the part c says o2 which is wrong or d has co2 on the second place which is also incorrect question number 23 the infrared spectrum shown was obtained from a compound j so this is the infrared spectrum we can see that the peaks coming are in between 1600 and 1700 per centimeter and also between 20 800 to 3500 which statement about j is correct now we can see that in all four parts a b c and d only two different functional groups are given one is carbonyl and second is carbon triple bonded with nitrogen atom so let's see what is their values in the function in the data booklet so from the data booklet we know that carbonyl carbon comes between 1640 and 1750 with a strong peak while cn comes in between 2200 to 2250 per centimeter and the peak is white now this region is between 16 to 1700 degrees per centimeter so it means yes carbonyl is present because this shows the strong peak between 16 and 1700 carbon is present while see and there is no peak between 2200 and 2250 so we can say there is no peak here so it means carbonyl is present but cn is not present so the answer is part c question number 24 a section of a polymer chain is shown what is the correct monomer so from this chain we have to find out the monomer first let's find out the repeat unit as we know that the bond that is coming out of this square bracket is double bond in the monomer similarly this point was also the double bond when it was a monomer now let's find out the repeat unit so if this was a double bond then we have one carbon here second one which is not shown and again these two carbon must have double bond between them and then these two carbon atoms must have a double bond between them and then again this carbon as the and the carbon which is not shown had a double bond so let's find out the repeat unit we can take out the repeat unit so this is one unit this is another repeat unit let's draw it separately it is like this one bond is continuing like this and second bond is continuing like this so in order to make a more number we have to make a double bond now this is the repeat unit and this one is monument now in this monomer we can see that the one carbon which is doubly bonded has a methyl group ch3 group attached while the second carbon has ethyl group ch3 and also isopropyl group ch3 ch ch2 as three group now let's see which is the answer so in part a if this is the double bonded carbon atom we have one methyl group and with the other carbon atom we have propyl group and ethyl group so yes this one seems to be the answer let's see b par we have methyl group and there is also ethyl group with the same carbon atom which is wrong then on the c part we have ethyl groups on both sides of the double bond which is also wrong and d part we have methyl and propyl group on the same side which is wrong this should be ether in proper group on the same side so the only structure which is correct is a question number 25 structural isomerism and stereoisomerism should be considered when answering this question how many non-cyclic isomers so we all we have to draw are the non-cyclic isomers have the molecular formula c5h10 so let's find out c5 h10 means it is an alkene because the formula is cnh2n so this alkene has five carbon atoms let's draw the double bond on the first carbon atom as we know that there are two hydrogen atoms with the corner carbon so the name is pent one e and there is no sas trans isomer because with the double bonded carbon atom there are two hydrogen atoms the atoms with the double bonded carbon must be different for a geometric isomerism now let's draw another one let's take double bond to the carbon too this is bent two in and says entrance both are possible for this one because the doubly bonded carbon has two different atoms attached directly with it let's draw the third one again i am taking pent one in and i'm changing the long chain so this will be three methyl but one a this is the first carbon this is second this is the third one again stereoisomer is not possible because corner carbon has two same atoms attached with it so no cis trans let's move this substituent to another place so we have ch2 double bond c now the name of compound is two methyl but one in and again no cis trans is possible we can move double bond to the next carbon atom it will be something like this this is two methyl but two e again no c trans isomers because the double bonded carbon has same atoms attached now counting this is the first structure this is the second one which says and third one with trance this is the fourth one this is the fifth one and this is sixth so we have six non-cyclic isomers that are possible so it means that answer is d question number 26 an excess of dry hbr is warm with compound y what is the major product of this reaction now let's see what does dry hbr do first reaction is that it can replace oh of primary alcohol hpr can also be added to alkene now the addition to alkene should be according to marconi cough fruit marconi off rule says that hydrogen of hbr goes to the carbon atom having more number of hydrogen atoms so let's see how many hydrogen atoms are present on each carbon atom this has no hydrogen atom because all four bonds are made this one has one hydrogen atom this carbon does not have any hydrogen atoms this has two hydrogen atoms and this is oh of primary alcohol so hbr will replace this oh with br and according to marconi of addition we should have br over here and h on this carbon similarly according to marconi of addition vr must be here and hydrogen on this carbonate let's see which answer is correct we have oh unchanged so this is not the answer we have oh unchanged no this one is replaced by bromine so yes this is good this bromine is attached to this carbon atom it is also correct this bromine is attached to this carbon atoms it is also correct so the answer should be part c in part d we can see this one is correct on the position of bromine should not be here it should be on the other carbon atom according to markonikov fruit so therefore this is wrong question number 27 which reactions would produce propanoic acid as one of its products so there are four different types of reactions a b c and d let's solve each of them in detail first a heating with concentrated acidified kmno4 so we know that when concentrated acidified camino 4 is used it will break the double bond and oxidize each of the sides so one of the side will be converted into ketone while the other one is converted into aldehyde first one becomes ketone and second becomes aldehyde now as it is concentrated second one becomes aldehyde because of the concentrated camino 4 this aldehyde won't stop it will be further oxidized and converted into carboxylic acid now let's find out the names this one is propane to own this one is propanoic acid the question said that which reaction would produce propanoic acid as one of its products so one product is propanoic acid so part a is the possible answer now let's see part b part b says heating this is ester with aqua sodium hydroxide so it is hydrolysis of ester in the presence of equisodium hydroxide this is profile propanoid and in the presence of equal sodium hydroxide it will be hydrolyzed into sodium propanoid and propanol or you can say propane one all so neither of the product is propanoic acid so b is not the correct option now let's see part c it says heating ethanol with acidified potassium dichromate under reflux as this is ethanol it will be oxidized to aldehyde so ethanol to ethanol neither of the product is propanoic acid so c is also incorrect option part d says reacting ethanol with hcn and then heating the organic product with sulfuric acid so let's see what happens when aldehyde reacts with hcn one carbon atom will be increased now what will happen that it is converted into hydroxy nitride a double bonded oxygen hydrogen is going to be attached and with other carbon atoms cyanide this is ethanol this is hydrogen cyanide this one is two hydroxy propane nitride nitrile is because of the presence of carbon triple bond nitrogen now this is further reacted with sulfuric acid so in the presence of aqueous sulfuric acid following product is formed ch3 a which and cn is converted into carboxylic acid group what happens to the nitrogen atom is that it forms ammonium sulphate with sulfuric acid so none of the product is propanoic acid this the name of this compound is 2 hydroxy propanoic acid so therefore d is also incorrect so the only answer is part a question number 28 compound x is treated with an excess of lithium aluminium hydride the reaction is allowed to go to completion what is the structure of the organic product formed what is lithium aluminium hydride is it is lilh4 and it is a powerful reducing agent so what does lithium aluminum hydride does so what does lithium aluminum hydride do we have ketone in our given structure so ketone is reduced to secondary alcohol but on the other side of this compound we have carboxylic acid carboxylic acid will be reduced to primary alcohol so cwoh is reduced to primary alcohol let's see which of these four options are correct so in part a they haven't changed the ketone so this is wrong in part b ketone is not changed in part c ketone is converted into secondary alcohol this is correct but the carboxylic acid remains as it is so it is wrong now part d secondary alcohol and primary alcohol so the answer is part d question number 29 what is the skeletal formula of the compound formed when this compound is heated under reflux with potassium dichromate the presence of acid so let's see what have they given to us ch3 c double bond c c h 2 o h so we have primary alcohol and we also have a double bond alkene now potassium dichromate is oxidizing agent it oxidizes primary alcohol too aldehyde and aldehyde further to carboxylic acid while alkene is not affected by potassium dichromate so only this portion must be converted into carboxylic acid let's see which part is correct so in part a they have converted into aldehyde which is wrong because it won't stop at ldi it will convert into carboxylic acid part b they have converted into carboxylic acid and alkene is unaffected so it means it is the correct option in part c they have changed the alkene which is not possible in part d they also have changed the alkene which is not possible so the only possible answer is part b question number 30 the diagram shows the structure of compound q two of the rings x and y contain a carbon-carbon double bond which row is correct now the first row says first column says number of ester groups in one molecule of q how many acid groups are there let's see what is ester the functional group of ester is r c double bond o o r where r is the alkyl group and c double bond o o is the central functional group so let's see how many c double bond o o are present there is this one and then we have this one so one and two we have two number of faster molecules ester groups sorry so it could be c or d second part is description of rings x and y both are planar or neither is planner let's see what type of hybridization occurs in x and y there are two types of carbon atoms in x some having carbon-carbon single bonds and the others with carbon-carbon double bond and same is the case with wiring so the carbon atoms making single bonds are actually sp3 hybridized while the carbon atoms making double covalent bond are sp2 hybridized if in one ring sp2 and sp3 both type of hybridization are present then it means that it is not planar while if all the carbon atoms are of sp2 hybridization then the ring could be player for example in benzene we have a planar ring because all the carbon atoms are sp2 hybridized so we can say neither is planar so the answer is part d question number 31 carbon and nitrogen are adjacent in the periodic table which properties do they both have one there is an empty 2p orbital in each atom of the element so let's find out the electronic configuration of carbon and nitrogen the atomic number of carbon is 6 so it will be 1 is 2 2 s 2 2px1 2py1 2pz no electron now nitrogen is 7 1 is 2 2 s 2 2 p x 1 2 p y 1 2 p z 1 so there is an empty 2p orbital in each atom no the there is an empty orbital in carbon but not in nitrogen first is wrong the principal quantum number of the highest occupied orbital is two we can see that the highest occupied orbital is two in carbon case and it is also two in nitrogen case which is correct they can form compounds in which the red atoms form four bonds we do have an example for carbon atom we know that carbon can make ch4 which is methane and carbon atom is making four single covalent bonds while for nitrogen we know of a compound names ammonium ion in ammonia ions nitrogen is also making four bonds three of them are covalent and one is coordinate covalent bond this one is correct so we have options 2 and option 3 correct so the answer is c question number 32 the strong hydrogen bonding present in liquid water causes an increase in which properties number one viscosity viscosity is actually the resistance of a liquid to flow so higher the hydrogen bonding more will be the viscosity so yes stronger hydrogen bonding will increase the viscosity now boiling point is actually the the temperature at which intramolecular forces are broken which is actually hydrogen bond so higher the hydrogen bonding higher will be the boiling point third one is surface tension if the there is extensive hydrogen bonding present then the surface of the liquid will be more tense because of the attraction from the water molecules present in the lower layers so yes all of them are correct so we can say 1 2 3 and the answer is a question number 33 a reaction between carbon and oxygen is shown this is the chem correction carbon and oxygen they are forming carbon monoxide gas how can the standard enthalpy change of this reaction can be described correctly now one says standard enthalpy change of formation now what is enthalpy change of formation it is enthalpy change when one mole of a compound is formed from its element in standard state under standard conditions so yeah this is fulfilling the equation because one of one mole of carbon monoxide is being formed from carbon and oxygen and both are there both are in their elemental states so this could be the change of enthalpy change of formation second is the enthalpy change of combustion the enthalpy change of combustion is basically when one mole of a substance is burnt in excess oxygen now in this equation you can see that oxygen is not in excess but actually only 0.5 moles of oxygen gas are given that's why it is not properly combusted and carbon monoxide is formed instead of carbon dioxide so it is not enthalpy change of combustion third is the enthalpy change of atomization in enthalpy change of atomization one mole of gaseous atoms is formed from its elements under standard conditions so in this equation there are no atoms but actually carbon monoxide is formed which is a molecule so this is also wrong we just have one option which is one so the answer is d question number 34 hydrochloric acid reds with zinc so acid plus metal gives salt and hydrogen gas what will increase the rate of this reaction but will not change the boltzmann distribution of molecular energies so we know that the boltzmann distribution of energy is actually affected by temperature only it is not affected by catalyst part one addition of a suitable catalyst yes catalyst reduces the time taken in order to make the product so therefore it increases the rate of reaction second one is an increase in concentration of hydrochloric acid yes we know higher the concentration more will be the rate of reactions so it also increases the rate of this reaction an increase in temperature of hydrochloric acid so temperature also increases the rate but temperature also affects the old spin distribution of energies so therefore we will emit part 3 only 2 will be left behind 1 and 2 so the answer is part the answer is b actually question number 35 the element astatine is below iodine in group 17 of the periodic table which statement concerning astatine are likely to be correct number one it is a dark colored solid at room temperature yeah we know for the group 17 that the elements become darker down the group number two it is a more powerful oxidizing agent than iodine for halogens we know that the oxidizing power decreases down the group so this is not the case actually iodine is more powerful oxidizing agent than astatine three its hydride is more thermally stable than hbr now stronger the bond between hydrogen and halogen it will be more thermally stable so we know that this bond strength decreases down the group hf has the highest bond strength then hcl then hbr then hi and astatine is in the blast so it is not more thoroughly stable it is actually less thermally stable than hbr the only option which is correct is one so the answer is d question number 36 which statements are correct number one aluminum chloride dissolves in water to give an acidic solution so when aluminum chloride is dissolves in water it forms aluminium iron and chloride ion now aluminum iron form coordinate coordination compound with water molecules so it will be basically like this alh2o 6 and plus 3 charge for aluminum at so now this is also dissociated this compound forms al one of the water molecules is dissociated so five water molecules are left behind and the water molecule which is dissociated it forms o h negative and h positive now as we have minus one charge on the inside so the total charge is changed to plus two so because of the presence of this h plus iron the solution becomes acidic in nature so it says aluminum chloride dissolves in water to give an acidic solution so yes this is correct number two magnesium chloride salt dissolves in water to give a solution of ph close to seven as we know that magnesium chloride does not react with water it is just converted into ions magnesium ions and chloride ions so there is no effect on the ph so this is also correct now the third one says sodium chloride dissolves in water to give an alkaline solution now sodium chloride behaves exactly like magnesium chloride only the ions are formed and there is no change in ph so it does not give an alkaline solution it actually gives a neutral solution so only two options are correct one and two and the answer is b question number 37 ethanol and hydrogen cyanide react together the reaction mechanism involves cyanide ions this is the reaction mechanism the question says which statements about this mechanism are correct number one c and minus acts as a catalyst what is the definition of a catalyst catalyst is something that is reformed at the end of a reaction without being used up or it also increases the rate of chemical reaction we can see cyanide ion present in the reactant side and also formed in the product so yes it is behaving as a catalyst because it is not involved in the chemical reaction cyanide ion is a nucleophile now nucleophile is something that has negative charge or lone pair of electron present we can say we can see in cyanide ion there is a lone pair and also a negative charge so yes cyanide also acts as a nucleophile third one is an addition reaction it is an addition reaction let's see whether something is added or not so we had hcn and this is the reactant so hcn is added into the reactant and double bond is converted into single bond so yes it is the example of addition reaction so all three are correct which means that the answer is a question number 38 which compounds will produce a yellow precipitate with alkaline aqueous iod now this is a factual question you must remember that alkaline aqueous iodine gives yellow precipitate with three types of compounds so it gives positive result with number one if you have methyl ketone that is ch3 c double bond o r so on one side of the ketone there must be a methyl group present second is ethanol ch3 c double bond o h so this one gives positive result with aqueous iodine third one is with the secondary alcohol group we have ch3 coh and r if these three types of compounds are present then always the yellow precipitate is formed so let's see which one of these is present in the given compounds number one this is carbon atom so this oh is tertiary because this carbon atom is further combined with one two three other carbon atoms so tertiary does not gives positive result on the other side we have aldehyde which is which does not have any methyl group so this is unable to give the precipitate let's see the second one we have oh group attached to this carbon atom and this carbon atom is further attached to two other carbon atoms so it is secondary alcohol yeah it can give positive result now going to the third one we have secondary alcohol yeah this one gives positive and this one is tertiary alcohol it does not give positive result so it means compound 2 and compound 3 will give yellow precipitate with aqueous alkaline iodine so the answer is c question 39 which statements apply to tetrafluoromethane tetrafluoromethane is basically carbon bonded with four fluorine atoms let's see the options number one it is rapidly decomposed by ultraviolet radiations now cf bond is present in tetrafluoromethane which is a very strong bond so it is not easily decomposed by uv radiations number two it is less harmful to the ozone layer than dichlorodifluoromethane dichlorodiflora means there are two fluorine atoms and two chlorine atoms if the chlorine atoms are present then in the presence of uv light chlorine radical is formed easily so therefore cf4 is less harmful than cf2cl2 so part 2 is correct third one says it is a non-polar molecule so let's see fluorine is attached with carbon atom on all four sides as it is a tetrahedral structure third option says it is a non-polar molecule let's see carbon combines with four fluorine atoms and the structure is tetrahedral and bond angle is one zero nine point eight so now all four fluorine are electronegative and the dipole moment of this molecule is going to be zero because it is attracted from all four sides equally so dipole moment is zero so therefore this molecule is non-polar so it is correct so we have option number two and option number three it means the answer is c question number 40 which statements comparing 18 and ethane are correct number one the bond angles in ethene are larger than the bond angles in the ethane let's draw the structure of ethene molecule and the structure of ethane molecule now we can see that the bond angle here is 120 degrees because there are three bonds one double bond this one single sigma bond this one is also single sigma one so overall in 360 degree space the angle is 120 degrees while in ethane carbon atoms make four single covalent bonds so the structure is tetrahedral and the bond angle is 109.8 degrees so the bond angle in ethene is larger than the bond angle in ethane so this one is correct option second says 18 reacts much more quickly with bromine in the dark than ethane does so 18 when reacts with bromine what happens that the bromine actually converts the double bond of ethene into single bond and each double bonded carbon atom has now bromine atom now ethane does not react with bromine in the dark because ethane is a saturated compound and saturated compound does not add something else in it ethane only reacts when light is present because light makes the bromine radical which then substitutes the hydrogen atom present in the ethane so therefore b is second part is true now part three complete combustion of 0.01 mole of ethene or ethane produces the same volume of gas now this is the question of moles let's find out the number of moles of the gas formed or the volume of gas form direction of 18 with oxygen 18 is c2h4 it reacts with oxygen to give carbon dioxide and water molecules as in the question it says that measured at room temperature so when there is room temperature then water molecules will be converted into liquid form so the only gas that is left is carbon dioxide gas let's balance out this chemical equation we have two number of carbon atoms on the reactant side so two number of carbon atoms now on the product side and then we have four hydrogen atom on the right hand side let's balance the number of hydrogen now there are six oxygen atoms in the product so we can write three with oxygen at so if we have one mole of ethene then two moles of carbon dioxide are formed so if we have 0.01 mole of ethene then 0.02 moles of carbon dioxide formed we can also change this into volume just multiply 0.02 into 24 decimeter cube so it is zero point four eight decimeter cube gas now let's carry out the same reaction with ethane ethane c2h6 plus oxygen it will give carbon dioxide water molecule water molecule will be liquid only the gas is carbon dioxide less balance the chemical equation we again have two moles of carbon and three moles of hydrogen now four and three seven just you can write seven by two now we can see one mole of ethane produces two mole of carbon dioxide so 0.01 mole of ethane will produce 0.02 mole of carbon dioxide or when converted into volume it will form 0.48 decimeter cube which is the same as compared to 18 so third part is also correct so we have three options one two three the answer is a thank you for watching if this video was useful please drop a like and do subscribe to this channel for similar upcoming videos see you in the next one till then stay happy peace out