RE-NEET Chemistry UNIT 2 Quick revision MCQs | Dr Vishal Gundekar (NET, K-SET, TET) | SARVAJNA

Added: 2026-05-20

[00:00:00]Hello dear students, welcome to Sarvagna and uh Justice Shivraj Patil PU College of Science Kalaburagi and Raichur.

[00:00:07]>> [snorts] >> As you know, we are discussing about the NEET uh examination fast track preparation we are going.

[00:00:13]So, here in the today's class we are going to discuss about the united test syllabus. United two test syllabus.

[00:00:21]I have chosen two topics in the united two, that is classification of elements and periodicity in the properties. Second one is chemical bonding.

[00:00:30]Let us go with very important questions which has been asked in the previous examination, NEET examination.

[00:00:36]Let us go one by one. Question number one.

[00:00:39]You You can see here. As the S character of hybridized orbital increases, the bond angle See here.

[00:00:47]S character See. S character increases, bond angle also increases.

[00:01:00]See here, you can see in SP hybridized orbital, the bond angle is 120. In SP2 hybridized molecule, bond angle is approximately 120. And in SP3 hybridized orbital, the bond angle is approximately 109.5°.

[00:01:17]So, this is for AB2 type of molecule. This is for AB3 type of molecule. This is for AB4 type of molecule. Here S character is of about 50%. Here S character is about 33.33%.

[00:01:33]Here S character is of about 25%. See here. As the S character decreases, the bond angle also decreases.

[00:01:41]In the sense, S character increases, the bond length bond angle also increases.

[00:01:47]So, option A is the answer.

[00:01:50]Next.

[00:01:52]Second question.

[00:01:56]See.

[00:01:56]Second question is which of the following molecules does not have the linear arrangement of atom?

[00:02:03]Linear arrangement means AB2 type of the molecule, where the bond angle is 180°.

[00:02:08]So, see here, hydrogen sulfide, ethyne, beryllium hydride, and carbon dioxide.

[00:02:15]You can go with the answer what? Option A. In case of hydrogen sulfide, the sulfur is the central atom.

[00:02:25]Sulfur is the what? Central atom.

[00:02:28]Its atomic number is a 16, and its electronic configuration is neon, 3s2 3p4.

[00:02:36]So, it has a six valence electron.

[00:02:39]Out of six, it is a forming only two bonds, two bond pairs. So, therefore, two bond pair means two lone pair. So, it is having bond angle approximately 117°.

[00:02:53]So, it is not linear. It is having a which shape? Bent shape.

[00:02:59]So, therefore, here C2H2, ethyne is ethyne is linear with 180° bond angle.

[00:03:08]Here, carbon dioxide is linear, 180° bond angle.

[00:03:13]Here, beryllium hydride is linear with 180° bond angle, but hydrogen sulfide is non-linear. It is having a bent shape.

[00:03:23]It is a type of AB2 E2 type of the molecule.

[00:03:28]So, therefore, it is not a linear molecule.

[00:03:32]So, this is from the chemical bonding topic. The topic is here VSEPR theory, valence shell electron pair repulsion theory, the shape of the molecule.

[00:03:44]Next, let us go to the what? Next question.

[00:03:48]Consider the ions or the molecules like O2+, O2 O2- and O2-2 for increasing order of their bond order.

[00:03:59]And the correct option is increasing order.

[00:04:04]Increasing bond order means lower bond order to the higher bond order.

[00:04:10]So this is related to the topic molecular orbital theory.

[00:04:16]So you generally molecular orbital theory is useful for calculating the bond order. Here we are not going to follow the lengthy method to calculate the bond order.

[00:04:29]We will go with the what?

[00:04:30]A shortcut method where every student must have to know this one.

[00:04:36]To calculate the bond order in the diatomic molecule of a homonuclear or heteronuclear molecule or ion so we can go for calculation of the bond order the number of by using this number of electrons in molecule or ions and this side we have to take what? Bond order.

[00:05:04]See if the number of molecules are elec- number of mol- sorry number of electrons in the diatomic molecule is 10 then the bond order is one, 11 then the bond order is 1.5.

[00:05:16]Next if it is 12 bond order is two.

[00:05:24]If 13 bond order is 2.5 14 bond order is three.

[00:05:30]15 bond order is 2.5 16 bond order is two, 17 bond order is 1.5 and 18 bond order is one.

[00:05:41]See they have given here O2 See here. In the question itself they have given O2, O2 minus, O2 minus two and O2 plus.

[00:05:53]See here.

[00:05:54]O2 See.

[00:05:56]Total 8 plus 8 8 electron Sorry, 8 plus 8 number of electrons are here.

[00:06:02]16 16 means bond order is 1.5.

[00:06:07]O2 minus 8 plus 8 plus one extra electron 17 17 means the bond order is bond order for 17 17 is 1.5.

[00:06:26]For 16 bond order is two.

[00:06:30]For 10 one, for 11 1.5, 12 two, 13 2.5, 14 three, 15 2.5, 16 two, 17 1.5, 18 one.

[00:06:44]So see here. For 16 number of electron in the bond order is two. For 17 number of electrons bond order is 1.5.

[00:06:52]For 8 plus 8 plus two. So that is nothing but what? 18 the bond order is one.

[00:06:59]O2 plus 8 plus 8 minus one. So 15 the bond order is 2.5. So increasing order lower bond order to higher bond order.

[00:07:12]So lowest bond order is O2 minus two.

[00:07:15]After that O2 minus, after that O2 and highest bond order is O2 plus. So as per this the answer in the given question is Answer in the given question we will go with the option A is the answer.

[00:07:44]Option A will be the what? Answer here.

[00:07:49]Okay.

[00:07:50]So, you can check out in the question option A E E is having the same bond order.

[00:08:00]Let us go to the what? Next question.

[00:08:04]The next question is also from the chemical bonding. The topic is molecular orbital theory.

[00:08:09]Bonding in which of the following diatomic molecule becomes stronger on the basis of molecular orbital theory by the removal of an electron?

[00:08:18]See, by the removal of an electron from the diatomic molecule, which molecule bond angle Sorry, which uh which which molecule will become more stable?

[00:08:30]So, here stable or he's asking stronger.

[00:08:37]See here, bond order is directly proportional to bond strength.

[00:08:47]That is nothing but bond bond enthalpy.

[00:08:55]See, for example, NO.

[00:08:58]NO means 7 + 8.

[00:09:01]7 + 8 7 + 8 is nothing but total 15 electron. 15 electron means the bond order is 2.5.

[00:09:10]So, by removing the electron, one electron, 15 minus one, so we will be getting 14. So, the bond order will become three.

[00:09:19]See here, in which of the following diatomic molecules become stronger on the basis of molecular orbital theory by the removal of an electron? See, you can check here from the nitric oxide, removal of one electron from the nitric oxide, the bond order changes from 15 Sorry, number of electrons changes from 15 to 14. That is nothing but bond order changes to 2.5 to 3. So, here bond order has been increased. So, therefore, the bond strength also increases.

[00:09:49]You can check here N2. N2 is nothing but 7 + 7 14 14 minus 1 13. So, 13 bond order is 2.5.

[00:10:02]Earlier, it was 3. Now, the bond order has been decreased. So, bond order decreases, bond strength also increases.

[00:10:09]This is not an answer.

[00:10:11]O2 See here.

[00:10:13]In case of the what?

[00:10:16]O2 16 means bond order is 2.

[00:10:20]16 minus 1 15 16 minus 1 15. 15 15 means the bond order is 2.5. So, here in case of O2 also, the bond strength increases. Why? Because the bond order changes from 2 to 2.5 by the removal of one electron from the oxygen.

[00:10:41]So, here in case of NO and as well as in case of O2, the bond order increases, the bond strength increases by the removal of electron.

[00:10:52]In case of C2 C2 number of electrons are 6 + 6 12. 12 means 12 means the bond order is of about two.

[00:11:04]So, when you remove the one electron, 12 minus 1 11. So, 2 to 1.5. The bond order has been decreased. Earlier, it was two.

[00:11:11]After the removal of one electron, it has changed 2.5. So, for C2, the bond order is not going to be increased. So, therefore, it is not becoming In case of B2 B2 B2 means 10. Bond order is 1. By the removal of one electron, the bond order here decreases. Nine. The bond order here decreases. Bond order decreases.

[00:11:36]And it is not becoming stronger. So, therefore, A and B Sorry, A and C are the two molecules after removal of one electron, their bond order increases and they become stronger and stable. So, therefore, the option C is the answer.

[00:11:57]Let us go for the next question.

[00:12:00]See.

[00:12:01]The next question.

[00:12:03]Here.

[00:12:04]Here they have given two structures here.

[00:12:07]You can check out.

[00:12:09]The structure number one.

[00:12:13]This is ortho nitrophenol.

[00:12:18]Next.

[00:12:21]This is para nitrophenol.

[00:12:26]See here.

[00:12:27]The vapor pressure of ortho nitrophenol The vapor pressure of what? Ortho nitrophenol.

[00:12:36]See here.

[00:12:38]Higher than para nitrophenol.

[00:12:46]See here. Here the ortho nitrophenol molecules are held together with intramolecular hydrogen bond.

[00:12:56]And the para nitro phenol molecules are held together with intermolecular hydrogen bond.

[00:13:10]Ortho nitrophenol molecules are held together with intramolecular hydrogen bond.

[00:13:16]Intramolecular hydrogen bond means within a molecule.

[00:13:19]So, therefore, the escaping tendency in case of the ortho nitrophenol molecule is more because they are not going to be held together with each other through the any kind of intermolecular forces of attraction. So, therefore, the escaping tendency will become more. The more vapors of the ortho nitrophenols are formed because of the intramolecular forces of attraction, that is intramolecular hydrogen bond.

[00:13:45]Which makes to have a more vapors. More vapor means more vapor pressure. So, the vapor pressure of ortho nitrophenol is higher than the para nitrophenol because of intramolecular hydrogen bonding which is formed in between the ortho nitrophenol molecules. But, in the case of the para nitrophenol, we can see the intermolecular hydrogen bonds, which makes to have the higher and stronger forces of attraction. Escaping tendency becomes less. So, therefore, less escaping tendency of the para nitrophenol molecules, less vapors. Less vapor means less vapor pressure compared to the what? Ortho nitrophenol. So, basically, uh vapor pressure is depends on the forces of attraction. Here, the ortho nitrophenol is having a more volatility because of the weak intramolecular hydrogen bonding, which is not present in the what? Para nitrophenol. Phenol.

[00:14:46]Next, question number six. Sulfur tetrafluoride can be classified as See here.

[00:14:53]Sulfur tetrafluoride. Here, sulfur is central atom.

[00:15:03]Its atomic number is 16.

[00:15:05]Neon.

[00:15:07]3s2. 3p4. Valence electrons are six. Out of six, four electrons are in the bonding.

[00:15:16]So, see here.

[00:15:18]Out of six, A is the central atom, that is nothing but sulfur. It has four bonding pair of electrons means SF4. And it has four 6 - 4 2 2 means one lone pair of electron.

[00:15:34]So, therefore SF4 is of the type AB4E type means sulfur has four bond pair out of six electron four electron has been used for bond pair and one lone pair of electron is there. So, because of that it is of a AB4 type of the molecule AB4E type of the molecule which exhibit seesaw shape.

[00:16:06]Next, this is also from the chemical bonding from the VSEPR theory.

[00:16:14]Next topic.

[00:16:15]Seventh question.

[00:16:16]Pentagonal bipyramidal structure contains bond angle of approximately Pentagonal bipyramidal means it has AB7 type of the molecule. Where 1 2 3 4 5 6 and 7.

[00:16:33]So, the bond angle is 90 90° 72° and as well as 180° we can have. So, pentagonal bipyramidal AB7 type of the molecules are possessing 72° 90° and 180° bond angle between the BAB orbitals. Bond pair electrons with the central atom orbitals.

[00:17:04]Next.

[00:17:06]Next is which of the following represents the zero overlapping? We have three type of overlapping positive overlapping negative overlapping and as well as zero overlapping.

[00:17:17]Whenever the two atomic orbitals are what? Merging on one over another, they should have certain condition for merging. After merging also, they should have more electron density between the two orbital of the valence shell, then only they will be favored for the formation of a covalent bond.

[00:17:36]If the two atomic orbitals, s atomic orbital with s atomic orbital, px atomic orbital with the px atomic orbital, py atomic orbital with the py atomic orbital, means the atomic orbital must be available in the same energy level.

[00:17:55]Overlapping condition, same energy level.

[00:17:58]Number two, similar orientation they must have.

[00:18:06]Then only they will be favored for the formation of a covalent bond.

[00:18:09]So, see here. Here the s orbital and the pz orbital.

[00:18:13]s orbital and pz orbitals are like a railway track. They have a fixed what?

[00:18:19]Axis. That is nothing but internuclear axis. z orbital is available in the internuclear axis. s orbital is also available in the internuclear axis. They all They both can easily overlap with each other and they can form a what?

[00:18:32]Sigma bond. And this type of the overlapping involved with the same sign.

[00:18:36]So, therefore, it is said to be positive overlapping.

[00:18:40]Here, px-px overlapping.

[00:18:43]Here the orientation of the two atomic orbitals are same. The signs are also same during the overlapping. So, therefore, this is also considered as what? Positive overlapping, which will be favorable for the formation of a pi bond.

[00:18:58]Next, px-px orbital overlapping. The orientations are same. Their faces may may not be same. Positive and negative.

[00:19:05]This type of overlapping we call it as a negative overlapping. And this overlapping also favors for the formation of a covalent bond.

[00:19:14]Last one, D. It is a zero overlapping because the PX orbital is not available in the internuclear axis. It is lying above and below to the internuclear axis, but S orbital is available in the internuclear axis. Whenever they come together and overlapping with each other, it is highly impossible to overlap because they do not have a similar orientation. They have a opposite orientation. So, here the overlapping is unsuccessful. It is not possible, and it does not favor for the formation of a covalent bond, and it is said to be what? Zero overlapping.

[00:19:53]Next, ninth question.

[00:19:57]The incorrect statement among the following is We have to find out which is incorrect here.

[00:20:04]Option A, bonding orbital orbit molecular orbital possesses the less energy than combining atomic orbital.

[00:20:11]Yes, it is true. Whenever the two atomic orbitals are linearly combining according to the molecular orbital theory. So, here the bonding molecular orbital favors for the formation of a covalent bond. They loses their energy because they should have to come out of their repulsive forces. So, therefore, bonding molecular orbitals are possesses the less energy.

[00:20:36]Anti-bonding molecular orbitals are possessing the more energy because here the electron density is less. Here we find the nodal plane where the electron density is very, very poor, and here it favors for the repulsion of the electrons, and here attraction of the electron loss of energy and stability is more.

[00:20:55]Bonding molecular orbitals are denoted by the sigma star and pi star. Wrong.

[00:21:00]Sigma star and pi star are denoted by the anti-bonding molecular orbital, and sigma pi is denoted by the what bonding molecular orbital so therefore this is the answer which is incorrect among the four option.

[00:21:17]Third option every electron in the bonding molecular orbital contributes the attraction between the two atoms.

[00:21:23]Yes, in the bonding molecular orbital electrons are attracting because they loses their repulsive energy during the repulsive forces so therefore they are get attracted with each other and they get paired and confined with the more electron density.

[00:21:37]D. The bonding molecular orbital contributes towards the stability of the molecule. Yes, because of the loss of energy the stability will be increases repulsive force will be minimized here.

[00:21:49]So stability is observed in the bonding molecular orbital. So statement A, C and D are correct but the statement B is wrong.

[00:22:00]Let us go with the what ninth question.

[00:22:04]See here, which of the following have the identical bond order? CN- O2- NO- See here, CN- the carbon number of electrons are six, nitrogen number of electrons are seven, minus one means plus one extra so therefore 14 so the bond order is three.

[00:22:24]Next, O2- O2- uh 8 plus 8 plus one so that is 17. 17 is nothing but bond order is 1.5.

[00:22:39]Next, NO+ nitrogen atomic number is seven, oxygen atomic number is eight, plus one is nothing but minus one removal of one electron so therefore this is also 14 then the bond order is three.

[00:22:54]Next, CN+ carbon number of valence electron number of electrons are six, nitrogen electrons are seven, plus one means minus, then six plus plus six 12. So, therefore the bond order is 1.5. So, here 10 means one, 11 means 1.5, 12 means two. So, here the bond order is not 1.5.

[00:23:17]Here the bond order is what? Two.

[00:23:20]So, see here, the bond order similar for the two pairs, three and three. That is CN- and as well as NO+. CN- number is one, NO+ number is three. So, one and three will be the answer. Option A is the answer. Where cyanide and nitronium ion, nitrosonium ion is having the similar bond order. That is nothing but what? Three.

[00:23:46]So, this is also a topic of molecular orbital theory, calculation of the bond order on the basis of the table.

[00:23:57]See, 11th question is also belong to the chemical bonding. Among the sulfur tetrafluoride, boron tetrafluoride, xenon tetrafluoride, and ICl4-.

[00:24:12]The number of species which have the two lone pair of electrons in the central atom.

[00:24:18]See, we can calculate the number of lone pair of electrons by the formula half into number of valence electrons minus number of sigma bonds.

[00:24:48]See here, in case of sulfur tetrafluoride, valence electrons are six, number of sigma bonds are four divided by two. So, 6 - 4 is nothing but two. 2 divided by 2 is equal to 1. So, it has one lone pair of electron.

[00:25:07]In case of a boron tetrafluoride, in case of boron tetrafluoride, you can check out boron has five uh sorry, three valence electron.

[00:25:19]3 + 1 four valence electron. 4 - 4 0. It has a zero valence electron.

[00:25:26]Xenon Xenon has See, half into valence electrons are eight. Number of sigma bonds are four. So, 4 by 2 two. It has two lone pair of electron.

[00:25:41]Next, iodine. I Iodine is a halogen, which has seven valence electron plus one extra because minus one charge. So, 7 plus one eight.

[00:25:53]8 - 4, that is nothing but 4 by 2, that is 2. Two lone pair of electron. So, XeF4 and ICl4 minus are the two which has two compounds which has the what? Two lone pair of electron. Sulfur tetrafluoride has a one lone pair of electron. BF4 minus has the what? Zero lone pair of electron. But, only these two compounds are having the what? Two lone pair of electron. So, then answer is what? Option A. So, here this is also VSEPR theory chemical bonding chapter.

[00:26:27]Next question.

[00:26:29]Question number 12, you can check out.

[00:26:31]Which of the following pairs have the different hybridization and the same shape?

[00:26:36]Hybridizations are different, but the shapes are what? Same.

[00:26:41]You can check out here NO3 minus.

[00:26:46]NO3 minus.

[00:26:49]See, NO3 minus and CO3 minus two.

[00:26:57]See.

[00:26:59]Next SO2 and NH2- See here.

[00:27:06]Sulfur dioxide and NH2- See, sulfur has a six valence electron.

[00:27:20]Out of six, four it has used for bonding and one lone pair of electron. This is belonging AB2E type Nitrogen has a five valence electron.

[00:27:31]Five minus one means NH2- means one extra electron will be there. Six valence electrons. Six Out of six, two valence electrons are used for bonding means four valence electrons are there. So, therefore, it is AB2 E2 type of molecule.

[00:27:52]You can check out here. This is AB2E. It is having a bent shape.

[00:28:00]And this is also having the what? Bent shape.

[00:28:05]Yes? Here, these two are possessing the different hybridization of the central atom.

[00:28:11]See. Hybridization can be calculated by this formula.

[00:28:18]Hybrid orbitals is equal to number of valence electrons minus number of sigma bonds Sorry, plus number of number of lone pair of electrons plus number of sigma bonds.

[00:28:40]See here. This is sigma, this is pi, this is sigma, this is pi. It has a two sig It has a one lone pair of electron and two sigma bond then three, so therefore it the hybridization is sp2.

[00:28:56]Here, two sigma bond, two lone pair, 2 + 2 four, then it is sp3.

[00:29:04]See here, they have both same shape, bent shape and bent shape, but their hybridizations are not same.

[00:29:11]See, different hybridization, same shape. Both are AB2E and AB2E2, but the shapes are same.

[00:29:19]But the hybridizations are different, so this is the one of the answer.

[00:29:24]Then XeF2 and CO2.

[00:29:27]See, you can check out XeF2 and CO2.

[00:29:35]Xenon has eight valence electron, among eight, only two are involved in the bonding.

[00:29:41]You can check out here, three lone pair, two bond pair. It is linear shape.

[00:29:50]CO2 is also linear in shape.

[00:29:57]But, see here, it has a three lone pair of electrons, two sigma bonds, totally five means sp3d2 hybridization.

[00:30:09]sp3d hybridization pi orbitals xenon tetra difluoride In case of CO2, this is sigma sigma pi pi, it has zero lone pair and two sigma bonds, so therefore carbon dioxide has a two value, that is nothing but sp.

[00:30:30]So, you can check out here, carbon dioxide and a xenon difluoride. Both are having the same shape, linear shape, linear shape, but it is having sp hybridization and XeF2 is having sp3d hybridization.

[00:30:46]How to calculate hybridization? Number of lone pairs plus number of sigma bond.

[00:30:51]Whatever the value will be getting, that will be giving a hybridization. If you get two value, sp hybridization. If you get three value, sp2 hybridization. If it is getting four value, sp3 hybridization.

[00:31:04]Let us go to the what? Next question.

[00:31:07]13th question.

[00:31:09]The electronic configuration of the four elements are given below.

[00:31:13]Arrange these elements in the correct order of their magnitude of their electron affinity.

[00:31:19]See, magnitude of their electron affinity.

[00:31:22]So, this is from classification of elements and periodicity.

[00:31:29]The topic is classification of elements and periodicity.

[00:31:35]See, how to solve this question?

[00:31:38]Electron affinity means the amount of energy released whenever the electron is added to the outermost orbital.

[00:31:50]Whenever the electron is added to the outermost orbital, whatever the energy is released, that energy is said to be electron gain enthalpy or electron affinity.

[00:31:59]2s2 2p4 is nothing but chlorine. 3s2 3p 2s2 2p5 is nothing but fluorine. 3s2 3p5 is chlorine. 2s2 2p4 is sulfur. 3s2 3p 2s2 2p4 is oxygen.

[00:32:15]And 3s2 3p4 is sulfur.

[00:32:18]So, there are some exceptional case.

[00:32:20]Chlorine is having smaller size compared to the chlorine.

[00:32:26]Oxygen is having a smaller size compared to the what? Sulfur.

[00:32:30]Though fluorine is having the what?

[00:32:32]Smaller atomic size than compared to the what? Chlorine.

[00:32:35]It is possessing the less negative value of the electron gain enthalpy.

[00:32:40]Reason is fluorine is too small and it cannot attract the what electron to the outermost orbital.

[00:32:48]So, therefore, it releases less energy.

[00:32:51]Due to the smaller atomic size and higher effective nuclear charge, the electron of the valence shell present in the fluorine experiences the repulsive force from the upcoming electron because the outermost orbital electrons of the fluorine are too attracted towards the what nucleus due to the smaller atomic size. Because of the smaller atomic size of the fluorine, it cannot able to attract the upcoming electron to add in the valence shell.

[00:33:20]So, due to which it is released less energy not only in the fluorine, as well as same reason is available in the case of sulfur. So, then fluorine is having the less negative value of the electron gain enthalpy or electron affinity than chlorine in its own group. And oxygen is also having the same reason due to smaller atomic size and due to electron-electron repulsion, the oxygen valence shell electron could not able to attract the upcoming electron in the valence shell. So, therefore, it is also releases the less energy and less electron affinity value possesses compared to the what sulfur in its 16 group.

[00:34:01]So, then we will go chlorine is having the highest one. Why?

[00:34:07]Because of its smaller atomic size.

[00:34:09]Later on, we have to move towards the what sulf- fluorine, later on sulfur, later on oxygen.

[00:34:17]See, fluorine, chlorine belong to the 17 group.

[00:34:25]17 group. Oxygen, sulfur belongs to 16 group. First, chlorine, later on fluorine, later on sulfur, later on what? Oxygen. So, highest lowest value is observed in case of the what? Oxygen. Oxygen means three.

[00:34:46]So, this is the answer. After that, four, sulfur. After that, one, fluorine.

[00:34:53]And the highest negative value is observed in the where? Chlorine. So, this is the answer. The topic is electron gain enthalpy.

[00:35:02]Next question.

[00:35:04]You can see which among the following statements are correct?

[00:35:08]Carbon monoxide is neutral for, whereas SO3 is acidic.

[00:35:13]See here.

[00:35:14]They're asking correct, but this is incorrect.

[00:35:20]See, generally nonmetal oxides nonmetal oxides are acidic.

[00:35:32]Metal oxides are basic.

[00:35:39]And metalloid oxides are amphoteric.

[00:35:48]So, you can some you can see some exceptional cases where N2O, NO, carbon monoxide are neutral even though they are nonmetal oxides.

[00:36:03]They are said to be neutral oxide.

[00:36:05]And zinc oxide, Al2O3, Cr2O3, and as well as vanadium pentoxide are amphoteric. So, you have to remember this some most important example. Carbon monoxide is neutral, whereas SO3 is acidic.

[00:36:28]See here.

[00:36:29]This is correct only.

[00:36:31]This is correct statement only.

[00:36:34]Carbon monoxide is neutral. SO3 is acidic because it is nonmetal oxide.

[00:36:40]Potassium oxide is basic whereas the nitrous oxide is acidic. This is wrong.

[00:36:45]NO is also neutral.

[00:36:49]So, this is incorrect statement.

[00:36:52]Aluminum and zinc oxides are amphoteric.

[00:36:54]Yes. Zinc oxide aluminums are amphoteric. That is correct.

[00:36:58]Sulfur dioxide is acidic where phosphorus pentoxide is basic. This is wrong.

[00:37:03]This is wrong. See, phosphorus pentoxide is also acidic.

[00:37:09]Next, carbon dioxide is a neutral. This is wrong. See, the only correct statements are found in one and two. One and two are the only correct answers. See, sorry, not one and two. One and three are only the correct answers. Carbon monoxide is neutral whereas SO3 is acidic. Aluminum and zinc oxides are amphoteric. So, therefore, the option is nothing but what?

[00:37:37]Let us go to the what? Question number 15. This is also from the what?

[00:37:41]Classification of elements and periodicity. Which of the following is the correct order of the radius?

[00:37:47]H- is greater than H and greater than H+. So, this is the parent atom.

[00:37:54]Atom, it has one electron and one proton.

[00:37:59]So, see here.

[00:38:00]This is having one proton, zero electrons.

[00:38:06]See, this is having one proton and two electrons.

[00:38:12]More the electrons compared to the proton, repulsive force will be more.

[00:38:18]Attractive effective nuclear charge will be less.

[00:38:21]More the number of protons compared to the what? Electrons.

[00:38:26]Protons increases the attractive force, effective nuclear charge increases, the size decreases. So, therefore, here the protons are more compared to the electrons, so therefore the effective nuclear charge increases, the size decreases compared to the parent atom and the hydrogen ion.

[00:38:43]Here the electrons are more, the effective nuclear charge is less, the repulsive force is more, so therefore the size is bigger here.

[00:38:52]Due to the presence of the any extra one electron, the size increases here. So, therefore the answer is which one? A.

[00:39:00]So, generally anions are bigger in size compared to the parent atom.

[00:39:08]And cations are smaller compared to their what?

[00:39:13]Parent atoms because of the more number of protons.

[00:39:19]16th question.

[00:39:21]Outermost electronic configuration of four elements A, B, C, and D are given.

[00:39:26]This is 3s2.

[00:39:28]3s2 generally we find in the where?

[00:39:31]Magnesium.

[00:39:32]3s2 3p1, aluminum. 3s2 3p3, phosphorus.

[00:39:37]3s2 3p4 is nothing but sulfur.

[00:39:41]The correct order of first ionization enthalpy.

[00:39:44]See here.

[00:39:45]In case of 3s2 3p3, the 3p3 p orbitals are three, it has a one one electron in the each orbital. It is a possessing half filled electronic configuration.

[00:39:58]So, due to the presence of the half filled electronic configuration, the phosphorus become extra stable, so it possesses the highest ionization enthalpy among all. Which one? First one is what? C.

[00:40:12]Okay.

[00:40:13]So, after that, it is we will go for what?

[00:40:18]A that is 3s2 where 3s2 is having S last electron is available in the S orbital. The penetration power is more in case of magnesium.

[00:40:34]And it is moreover fulfilled electronic configuration. The energy is more required because it is fulfilled extra stable penetration power is more compared to the aluminum.

[00:40:51]S orbital electrons are more than P orbital than D orbital than F orbital.

[00:40:55]So therefore the S orbital electron is difficult to break compared to the what?

[00:41:00]P orbital electron. So therefore C First one is phosphorus, second one is magnesium, third one is aluminum, and fourth one is what? Sulfur. So therefore We will go with the what? B. Last one is aluminum because it is bigger in size compared to the what? Sulfur.

[00:41:27]So here ionization enthalpy depends on the what? Size. Lower the size, higher the ionization enthalpy, higher the penetration power, higher the ionization enthalpy. Half-filled or full-filled electronic configuration containing molecule also possess higher ionization enthalpy. Why here the phosphorus is having the highest ionization enthalpy?

[00:41:49]Because it is not only smaller in atomic size compared to all, it is also possessing the what? Half-filled electronic configuration.

[00:41:58]Later on, magnesium. Why magnesium?

[00:42:02]Because magnesium because magnesium is having the high penetration power. Its electron has to be removed from the S orbital and moreover it is fulfilled electronic configuration. It require more energy.

[00:42:16]Why later on D, that is nothing but for sulfur? Because sulfur is a smaller in atomic size compared to the what aluminum. Smaller the atomic size, greater is ionization enthalpy compared to the what aluminum. So therefore we will go with the option C, not with the option B. So aluminum least, later on sulfur, later on magnesium and the highest ionization enthalpy is found in the where phosphorus.

[00:42:46]Let us go to the what question number 17.

[00:42:49]The total number of acidic oxides. See here, NO is a neutral.

[00:42:56]N2O laughing gas is also neutral.

[00:43:00]B2O3 is a purely acidic.

[00:43:03]N2O5 is a purely acidic.

[00:43:07]Carbon monoxide is a neutral.

[00:43:10]SO3 is acidic.

[00:43:13]And the P4O10 is also acidic. So therefore the answer is total number of acidic oxide 1 2 3 4. So option B is the answer.

[00:43:26]Next question number 18. The correct pair of electron affinity order.

[00:43:31]See as you know that the sulfur is possessing the more electron affinity than oxygen. And chlorine is possessing the more electron affinity than what fluorine. This is due to due to smaller atomic size of fluorine and oxygen than chlorine and sulfur respectively.

[00:44:08]As well as electron electron repulsion. This is also one of the reason to have the what?

[00:44:20]Lower electron affinity than what?

[00:44:23]Fluorine is having the lower electron affinity than chlorine.

[00:44:27]And it's own group.

[00:44:30]Oxygen is having a lower electron affinity compared to the what? Sulfur.

[00:44:34]In its own group. Reason, smaller atomic size and electron electron repulsion.

[00:44:40]Next one. 19. Which of the following represents the correct order of increasing first ionization enthalpy?

[00:44:48]So, as you know the ionization enthalpy is inversely proportional to size.

[00:44:53]Ionization enthalpy is directly proportional to penetration power.

[00:45:01]And ionization enthalpy increases either because of half filled or full [snorts] filled atomic orbitals.

[00:45:11]So, these are the reasons.

[00:45:13]See here, calcium calcium, barium, sulfur, selenium, argon. Argon is a noble gas. It has the highest Sorry, highest uh first ionization enthalpy because of full filled electronic configuration.

[00:45:28]See, argon comes in 18 group. Selenium comes in 17 group. Sulfur comes in 16 group.

[00:45:37]Sulfur, selenium.

[00:45:39]Both comes in 16 group.

[00:45:42]And uh second group.

[00:45:47]Calcium and barium comes in the where?

[00:45:51]Second group. So, along the period, the ionization enthalpy increases. So, therefore, the 18th group element is possessing the highest first ionization enthalpy. So, first one is argon.

[00:46:05]After that, sulfur, selenium, calcium, and as well as barium. So, then answer is what?

[00:46:13]D.

[00:46:14]Along the period and along the group. Along the group, first electron ionization enthalpy decreases here.

[00:46:23]Along the period, increases.

[00:46:28]Last question. Which of the following statements are not correct?

[00:46:32]The electron gain enthalpy is of fluorine is more negative than chlorine.

[00:46:37]Yes.

[00:46:38]Sorry, it is wrong. Electron gain enthalpy of chlorine is more negative than fluorine because now only we discussed that smaller atomic size due to electron due to which the electron of the valence shell of the fluorine experiences a more repulsion whenever the electron is added to it.

[00:46:54]So, this is incorrect.

[00:46:57]Ionization enthalpy decreases in a group along the period. Yes, down the group ionization enthalpy decreases with increase in the size. As the size increases, the electrons are very easy to remove. So, therefore the ionization enthalpy decreases down the group.

[00:47:11]The electronegativity of an atom depends on the atoms bonded to it. Yes.

[00:47:17]Two atoms in a covalent bond will decide the electronegativity.

[00:47:22]Al2O3 and NO are the examples of amphoteric oxide. No, only Al2O3 is amphoteric but not NO. NO is a neutral oxide. So, we have to choose incorrect.

[00:47:33]So, this is A is incorrect.

[00:47:36]And C D is incorrect.

[00:47:42]A is incorrect and as well as D is incorrect and even C is also incorrect.

[00:47:50]Not only the what bonded atoms, it also depend on the what size.

[00:47:55]Electronegativity depends on the size, not on the bonded atoms. So the Therefore, so option A, C, D is the answer. A, C, D means option C.

[00:48:05]are incorrect.

[00:48:08]So, these all are the questions which has been previous which has been asked in the previous year NEET examination.

[00:48:15]So, go through with these topics, specially periodicity in the properties in the classification of elements and periodicity and IUPAC nomenclature of the newly discovered elements from atomic number 100 to the what? Atomic number 100 and 18. Their original established names and the IUPAC names.

[00:48:36]Both you have to remember. And in the chemical bonding more focus you have to do on the what?

[00:48:42]Uh major topics VBT, VSEPR theory, molecular orbital theory and calculation of the formal charge.

[00:48:50]Thank you.