Chemistry Discussion Test-6 2026

[00:00:00]Welcome to NGcon online Academy. In this lecture, we are going to discuss the Sunday test number six.

[00:00:07]So, the first MCQ says, which transition element does not belong to 4D series?

[00:00:15]So, the correct answer is tungsten because tungsten belongs to chromium family and in chromium family, chromium belongs to 3D series, while platinum belongs to 4D series and tungsten belongs to 5D series, not 4D series.

[00:00:32]Which of the following is classified as inorganic compound even though it contains carbon?

[00:00:37]So, methane is organic compound, ethanol is also organic, acetic acid is organic.

[00:00:43]Carbon dioxide contains carbon, but it is inorganic compound.

[00:00:50]Difference between first and second ionization energy is greatest for So, in the sodium metal, we have three shells.

[00:00:59]Now, in the sodium metal, the first electron is moved from valence shell.

[00:01:04]But, the second electron is removed from second last or penultimate shell. So, first ionization energy of sodium is not that much high, but removing second electron is very difficult in sodium.

[00:01:17]Therefore, the difference between first and second ionization energy is highest in sodium because its second electron is being removed from penultimate shell, which requires a lot of energy.

[00:01:30]Electronic configuration of transition metal with atomic number 29 is So, 29 is atomic number of copper. Its expected configuration is But, instead, it shows a configuration of 4s1 3d10 because it wants to stabilize its D orbital by making it complete. Therefore, this is the correct configuration of copper.

[00:01:57]Which group 2A element shows abnormal trend in melting and boiling point? So, if we see the book table, magnesium is showing abnormal trend in both melting and boiling point due to changes in its structure.

[00:02:12]Carbon forms enormous number of inorganic compounds mainly because of its So, carbon forms a large number of compounds because of its ability to link with other carbons and form long chains and big huge rings. Carbon can form large chains and rings containing millions of carbons because carbon can bind with carbon through the process of catenation or tetravalency.

[00:02:41]First ionization energy of beryllium is greater than boron because So, beryllium is S block member, its configuration is S2.

[00:02:50]Boron is P block member, its configuration is S2P1.

[00:02:55]Although boron has smaller size than beryllium and its ionization energy should be higher.

[00:03:01]But, in beryllium electron is removed from S orbital and in boron electron is removed from P orbital. So, removing electron from P orbital is easy, therefore beryllium has greater ionization energy.

[00:03:15]So, electron is removed from less penetrating orbital in boron.

[00:03:21]Which pair of ions are diamagnetic? So, ions having no unpaired electrons are diamagnetic. So, it's very easy to spot.

[00:03:31]First of all, we have Uh we can see here scandium and zinc.

[00:03:36]So, scandium is 4S2 3D1. But, scandium +3 is 4S0 3D0.

[00:03:45]And zinc is normally 4S2 3D10.

[00:03:49]And zinc +2 is 4S0 3D10. So, both scan and zinc plus two have no unpaired electrons, so both are diamagnetic.

[00:04:01]The general formula CnH2nO is followed by So when we have general formula of alkanes CnH2n+2O.

[00:04:10]This is the formula of alcohols and ethers.

[00:04:15]When we have general formula of alkenes CnH2n and along with that we have O.

[00:04:20]This is the formula of aldehydes and ketones. And whenever we have CnH2n and double O. This is the formula of carboxylic acids and esters. So this formula is followed by aldehydes and ketones.

[00:04:36]Which of the following oxides are amphoteric in nature? So oxides of lead, oxides of aluminum, oxide of tin, oxide of beryllium.

[00:04:48]So keep these oxides in mind. These oxides are amphoteric. So here we have been given uh beryllium and aluminum as the option, so this is correct.

[00:05:00]Incorrect statement regarding transition metals. They are used as catalyst, it is correct. Their ions are colored, it is also correct. They do not show variable oxidation states, this is wrong. They show variable oxidation states.

[00:05:14]Which of the group four chlorides is least stable? So the order of group four stability is that it decreases down the group.

[00:05:24]So down the group the lowest chloride is PbCl4, so it is least stable.

[00:05:31]Amino carbonyl is functional group of So this is amino and this is carbonyl. So this is functional group of amides. D is the correct option.

[00:05:42]The best index of metallic character is Now this is a pretty famous question.

[00:05:47]Ionization energy is indexed to metallic character.

[00:05:53]Which of the following 4D elements follows the expected configuration without exception?

[00:05:58]So, yttrium is the element that follows its correct configuration. So, this is the position of yttrium. It is right below scandium.

[00:06:09]So, yttrium is expected to be 5s2 and 4d1.

[00:06:15]So, it follows this configuration and it has normal configuration.

[00:06:21]The heterocyclic compound pyrrole has molecular formula. So, this is the compound pyrrole.

[00:06:28]It contains nitrogen in the ring and this is the structure of pyrrole.

[00:06:35]So, pyrrole consists of four carbons, five hydrogens, and one nitrogen. So, D is the correct option.

[00:06:44]Correct order of basic strength of group 1 oxides is So, down the group, the basic strength of oxides increases.

[00:06:51]So, among lithium, sodium, and potassium, potassium oxide is most basic followed by sodium, followed by lithium. So, C is the correct option.

[00:07:02]M + 3 ion So, a transition metal has ion M + 3. It has configuration of 4s0 3d3.

[00:07:11]So, M in neutral state would be So, we put two electrons in S orbital and one electron in D orbital, but this configuration does not exist. Instead, the correct configuration is 4s1 3d5 and this element is chromium. So, C is the correct option.

[00:07:30]Which of the following can show cis-trans isomerism? Ethene does not show cis-trans isomerism. Chloroethene does not show cis-trans isomerism.

[00:07:38]2-Butene shows cis-trans isomerism. It can exist in two forms. So, this is the cis form of two butene.

[00:07:47]And again, looking at this one, this is the trans form of two butene.

[00:07:53]So, two butene can show cis-trans isomerism.

[00:07:58]Which of the following statements is not correct for alkali metals?

[00:08:02]Radius increases down the group, this is correct. Ionization potential decreases, this is correct. Density decreases, this is wrong. Down the group, the density of transition Sorry, group one metals increases. So, C is the wrong option.

[00:08:19]Which statement is false about transition elements? They often form colored compound, this is correct. They follow off for principle strictly, this is wrong. They can form complex ion, this is correct. They may exhibit variable oxidation state, this is correct. All of them do not follow off for principle, there are many exceptions to off for principle in transition metals.

[00:08:42]Which of the following is alicyclic compound? Alicyclic compound is a cyclic compound whose cycle contain all of the carbons.

[00:08:50]So, furan does not have all of the carbons. Pyrrole does not have all of the carbons. Cyclobutane has four carbons in the cycle, so it is alicyclic. Toluene is aromatic, it is not alicyclic.

[00:09:04]For the same atom, which relationship between the covalent radius and van der Waals radius is correct? For So, for the same atom, van der Waals radius is measured from from the atom in solid state.

[00:09:17]So, van der Waals radius is measured by dividing this distance by two, which is actually larger than the original atom.

[00:09:24]Whereas, covalent radius is measured when the elements bond with each other.

[00:09:28]So, this distance is smaller compared to van der Waals radius. So, van der Waals is greater than covalent.

[00:09:36]A tripositive ion has electronic configuration 4s03d3.

[00:09:41]Which is correct about this configuration? It is diamagnetic. No. It has three unpaired electrons in D orbitals because it is D3.

[00:09:50]So, it is paramagnetic. It is colorless.

[00:09:53]No. It is a D ion, so it is bound to be colored. Electronic configuration of neutral atom is 4s23d4. No. Neutral atom has configuration of 3d54s one.

[00:10:05]It contains three unpaired electrons.

[00:10:07]This is correct.

[00:10:10]Electron affinity is the energy change that occurs. So, electron affinity is either the energy released or absorbed when an electron adds to valence shell of isolated gaseous atom. So, A is the correct option to that when electron is added to isolated gaseous atom.

[00:10:28]Geometrical isomerism is not possible in in 2-butene. Geometrical isomerism is possible. It is the commonest question asked in geometrical isomerism. So, this has geometrical isomerism.

[00:10:41]1,2-dimethylcyclobutane.

[00:10:43]Cycloalkanes also have geometrical isomerism. And in consecutive carbons of cycloalkane, if we have alkyl groups, it also shows geometrical isomerism.

[00:10:54]1,2-dichloroethene.

[00:10:56]So, we have ethene.

[00:10:58]And this is 1,2-dichloro. So, it also shows geometrical isomerism.

[00:11:03]2-methylpropene.

[00:11:05]So, this is propene.

[00:11:09]And here we have methyl group. So, this compound has two hydrogens and two carbons, but the same groups are on same carbon. So, such compounds do not show geometrical isomerism.

[00:11:22]Which of the following give red color in flame test? So, this is lithium and rubidium. This has been taken from flame test in the book.

[00:11:32]Which of the following ions has completely empty 3D orbital? So, let's see. Titanium in neutral state is 4S0 Yeah, 4S2 3D2. So, titanium four in plus four state is 4S0 3D0. So, it has completely empty 3D orbital.

[00:11:52]SNO reacts with NaOH to produce. So, in the textbook, the reaction has been given SNO plus NaOH forms Na2SnO2.

[00:12:03]So, B is the correct option.

[00:12:06]A carbon atom that gives rise to optical isomerism is one that is bonded to. So, the carbon atom that gives rise to optical isomerism is asymmetric carbon.

[00:12:16]Asymmetric carbon is the carbon having four single bonds and bonded to four different groups.

[00:12:22]So, it is bonded to four different atoms or groups.

[00:12:27]Arrange the isoelectronic species in order of decreasing ionic radius.

[00:12:33]So, oxygen and fluorine belong to period number two, and sodium and magnesium belong to period number three. But, positive ions are smaller in size, and negative ions are larger in size. In negative ions, the larger the charge, the larger is the size. So, O2- is larger than F-. And in positive ions, the larger the charge, the smaller is the size. So, sodium is larger compared to magnesium.

[00:13:02]So, negative ions are larger than positive ion. In negative ions, larger larger charge, larger size. And in positive ions, larger charge, smaller size. So, this is the correct order. B.

[00:13:18]The general configuration of elements in first transition series is So, first transition series belongs to fourth period. So, in fourth period, we can have one or two electrons in S and in D, we can have 1 to 10 electrons. So, D is the correct option.

[00:13:36]Which group one metal forms super oxide when heated in excess oxygen? Lithium forms normal oxide, sodium forms per oxide, potassium forms super oxide.

[00:13:48]Two stereo isomers that are non-superimposable mirror image, so such isomers are called optical isomers or enantiomers.

[00:13:58]The most electronegative element in the periodic table is So, this is simply fluorine.

[00:14:04]Which of the following transition metal ions has no unpaired electrons? So, Fe + 2 is 4s0 3d6. It has four unpaired electrons. Mn + 2 is 4s0 3d5. It has five unpaired electrons. Chromium + 3 is 4s0 3d3. It has three unpaired electrons. Zinc + 2 is 4s0 3d10. So, it has no unpaired electrons.

[00:14:31]The reactivity of group one metals with water increases in which order? So, down the group, reactivity with water increases. So, lithium is least reactive followed by sodium, potassium, rubidium, and cesium. So, D is the correct option.

[00:14:45]Which is the reason behind irregular electronic configuration of chromium and copper? So, chromium should be 4s2 3d4, but it is 4s1 3d5 because half-filled D orbital is stable.

[00:14:57]Copper should be 4s2 3d9, but it is 4s1 3d10.

[00:15:02]So, this is because completely filled D orbital is again stable. So, it is due to stability of half-filled and fully filled D subshells.

[00:15:12]Which group two elements is not white?

[00:15:14]So, all of the elements are white except beryllium, which is grayish silver in color. This has previously been asked in entry test exam.

[00:15:24]Which of the following is the correct configuration for palladium? So, palladium belongs to the group of nickel. So, nickel is 4s2 3d8. Palladium should be 5s2 3d 4d8.

[00:15:40]But, instead of this configuration, palladium is 5s0 4d10.

[00:15:45]So, C is the correct option.

[00:15:49]Which of the following best explains why sodium gives off intense orange flame during its reaction with both oxygen and chlorine?

[00:15:57]So, it is because that sodium gets excited to higher shell, and when it excites down, it emits orange light. So, B is the correct option.

[00:16:09]On removing electron from transition element, electrons are removed from So, this is an important concept. According to Aufbau principle, 4s is filled first, followed by 3d due to Aufbau principle. So, 3d is filled after 4s. But, when electrons are being removed, they will be removed from 4s first because 4s is the outer shell. So, electrons are first removed from 4s.

[00:16:39]Magnesium reacts with steam and produces So, magnesium is reacted with steam.

[00:16:45]So, it produces magnesium oxide along with hydrogen gas. So, D is the correct option.

[00:16:53]The electronic configuration of Fe+3 is So, neutral iron is 4s2 3d6. Fe+3 is 4s0 3d5. So, C is the correct option.

[00:17:05]On heating in excess of oxygen, which group two metal most readily forms peroxide? So, in group two, we have beryllium, magnesium, calcium, strontium, and barium. So, these two are have ability to form peroxide. So, barium is the correct option here.

[00:17:23]Thank you.