

InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
251. |
The correct order of radii isA. `N lt Be lt B`B. `F^(-) lt O^(2) lt N^(3-)`C. `Na lt Li lt K`D. `Fe^(3+) lt Fe^(2+) lt Fe^(4+)` |
Answer» Correct Answer - B Ionic radii decrese significantly from left to right in a period among representative elements. |
|
252. |
which of the following ions has the smallest radius?A. `Be^(2+)`B. `Li^(+)`C. `O^(2-)`D. `F^(-)` |
Answer» Correct Answer - A Size `prop 1/(E.N.C)` |
|
253. |
Which is helpful in the formation of ionic bondA. Only small cationB. Only small anionC. Small cation and small anion bothD. Low positive charge, large cation and small anion |
Answer» Correct Answer - D For ionic bond formation low I.E. High electron affinity and high lattice energy is needed. |
|
254. |
Which of the following has the highest ionic radius?A. `F^(-)`B. `B^(3+)`C. `O^(2-)`D. `Li^(+)` |
Answer» Correct Answer - C `O^(2-)` ion has the smallest effective nuclear charge. |
|
255. |
Mercury is the only metal which is liquid at `0^(@)C`.this is due toA. very high ionisation energy and weak metallic bondB. low ionisation energyC. high atomic weightD. hish vapour pressure |
Answer» Correct Answer - A factual |
|
256. |
There are 10 neutrons in the nucleus of the element `_(Z)M^(19)` . It belong toA. `f-block`B. `s-block`C. `d-block`D. None of these |
Answer» Correct Answer - D `._(z)M^(19) =19-10` `Z=9` it is p-block elements. |
|
257. |
The element with atomic number `9,17,35,53,85` are allA. noble gasesB. halogensC. heavy mentalsD. light mentals |
Answer» Correct Answer - B These atomic no. Gives the confinguration `ns^(2)np^(5)`which are of halogen group or seventeeth group. |
|
258. |
Assertion: The first ionisation energy of `Be` is greater than that of `B`. Reason: 2p-orbital is lower in energy than 2s-orbital.A. Statement-I is true, Statement-II is true ,Statement -II is the correct explanation for Statement-IB. Statement-I is true ,Statement -II is true, Statement -II is not the correct explanation for statement -IIC. Statement-I is true. Statement -II is false.D. Statement-I is false. Statement -II is true. |
Answer» Correct Answer - C Statement II is incorrect. The first ionisation energy of Be is greater than that of B because of its relatively stable electronic configuration. `Be=2s^(2), B=2s^(2) 2p^(2)` It is easier to remove electron from `2p` orbitals than `2s` orbitals since `2s` is much more penetrated towards nucleus. Hence `IE_(1)` of `Be gt IE_(1)` of B. Therefore, `2p`-orbital has rather higher energy than `2s`. |
|
259. |
Energy is released when electron is added to an isolated gases anion. |
Answer» Correct Answer - F False. For example, `O_((g))^(ɵ)+e^(-) rarr O_((g))^(2-)` `EA_(2)` is always positive, i.e. energy is absorbed. |
|
260. |
The energy released when an electron is added to a neutral gaseous atom is called`"………."`of atom |
Answer» Correct Answer - Electron gain enthalpy The energy released when an electron is added to a neutral gaseous atom is called electron gain enthalpy of the atom. |
|
261. |
Which one pair of atoms or ions will have same configuration ?A. `F^(o+)` and `Ne`B. `Li^(o+)` and `He^(ɵ)`C. `Na` and `K`D. `Cl^(ɵ)` and `Ar` |
Answer» Correct Answer - D `Cl = 3s^(2)3p^(5), Cl^(ɵ) = 3s^(2)3p^(6), Ar = 3s^(2)3p^(6)` |
|
262. |
Transition metals are characterised by which of the following properties ?A. Variable valencyB. Coloured compoundsC. High melting and boiling pointsD. Tendency to form complexes |
Answer» Correct Answer - A::B::C::D All are properties of transition elements. |
|
263. |
Which is true about the electronegative order of the following elements ?A. `P gt Si`B. `C gt N`C. `Br gt Cl`D. `Sr gt Ga` |
Answer» Correct Answer - A `P` and `Si` belong to group 4 and group 5 respectively and both of them are in the 3 rd period. `EN` of `P(2.1) gt Si (1.8)` `:. EN` of `P(2.1) gt Si(1.8)` |
|
264. |
The elements which are characterised by the outer shell configuration `ns^(1)` to `n p^(6)` are colectively calledA. Transition elementsB. Representative elementsC. LanthanidesD. Inner transition elements |
Answer» Correct Answer - B `ns^(1)` to `ns^(2)` (`s`-block) and `np^(6)` (`p`-block) are called representative elements. |
|
265. |
The inner electrons are shielded to a `"______"` extent than the outer electrons. |
Answer» Correct Answer - Lesser Lesser |
|
266. |
A cell wall material present only in blue green algea and bacteria isA. PectinB. CelluloseC. ChitinD. Muramic acid |
Answer» Correct Answer - D The cell wall material present only in bacteria and blue green algae is Muramic acid. |
|
267. |
Number of species that are isoelectronics with `Sr^(2+)` ion are four. |
Answer» Correct Answer - F False. `Sr(Z=38), Sr^(2+) = 38-2=36 e^(-1)s`. So, the number of species with `36 e^(-1)s` are only three. 1. `Br^(ɵ)(35+1=36)` , 2. `Kr(Z=36)` 3. `Rb^(1+)(37-1=36)` |
|
268. |
Bacterial viruses usually haveA. Single stranded RNAB. Double stranded RNAC. Single stranded DNAD. Double stranded DNA |
Answer» Correct Answer - D Bacterial viruses usually have Double Standard DNA. |
|
269. |
Second element of group `1` shows diagonal relationship with the first element of group `"_____"` . |
Answer» Correct Answer - Group 2 (`Li` shows diagonal relationship with `Mg`) Group 2 (`Li` shows diagonal relationship with `Mg`) |
|
270. |
Write short notes on restriction enzymes. |
Answer» Two enzymes responsible for restricting the growth of Bacteriophage in Escherichia coli were isolated in the year 1963. One of these added methyl groups of DNA and the other cut DNA. The latter was called restriction endonuclease. The first restriction endonuclease - The first restriction endonuclease - Hind II which cut DNA molecules at a particular point by recognising a specific sequence of six base pairs, called recognition sequence for Hind II. Today, more than 900 restriction enzymes were isolated from over 200 strains of Bacteria, each of which recognises a different recognition sequence. E CORI is a restriction enzyme in which, the first letter comes from the Genus (Escherichia). and the second two letters from the species of the Prokaryotic cell [coli], the letter 'R' is derived from the name of strain. Roman number indicate the order in which the enzymes were isolated from that strain of Bacteria. Restriction enzymes belong to a larger class of enzymes called nucileases. They are of two types. (a) Endonucleases which make cuts at specific location with in the DNA. (b) Exonucleases which remove nucleotides from the ends of the DNA. |
|
271. |
Why is the respiratory pathway referred to as amphibolic pathway ? Explain. |
Answer» Respiration involves the breakdown of substrates so traditionally called catabolic process and the respiratory pathway as a catabolic pathway. In respiration, different substrates enter into respiratory pathway at different points. If fatty acids were respired, they would be degraded to acetyl CoA and enter the pathway. Glycerol would enter the pathway after converted to PGAL. The proteins and the aminoacids would enter the pathway at pyruvate. Fatty acids would be brokendown to acetyl CoA before entering into respiratory pathway. But when the organism needs to synthesis fatty acids, Acetyl CoA would be withdrawn from the respiratory pathway for it. Hence the respiratory pathway comes into the picture both during the breakdown and the synthesis of fatty acids. In this issue, respiratory pathway is involved in both anabolism and catabolism, it would be better to consider it as amphibolic pathway rather than as a catabolic one. |
|
272. |
Write briefly about enzyme inhibitors. |
Answer» The chemical that can shut off enzyme activity are called inhibitors. They are of 3 types. (1) Competitive inhibitors : Substance which closely resembles the substrate molecules and inhibits the activity of the enzymes are called competitive inhibitors. Eg : Inhibition of succinic dehydrogenase by malonate which closely resembles the substrate succinate. (2) Non-competitive inhibitors : The inhibitor which have no structural similarity with the substrate and bind to an enzyme at locations other than the active site, so that the globular structure of the enzyme is changed are called non-competitive inhibitors. Eg : Metal ions of Copper, Mercury. (3) Feedback inhibitors : The end product of a chain of enzyme catalysed reactions inhibit the enzyme of the first reaction as a part of homoeostatic control of metabolism are called feed back inhibitors Eg : During respiration (Glycolysis) accumalation of Glucose-6 Phosphate occurs, it inhibits the Hexokinase. |
|
273. |
What will happen if a healthy plant is supplied with excess essential elements ? Explain. |
Answer» Any material ion concentration in tissues that reduces the dry weight of tissues by about 10 percent is considered toxic. Many a time, excess of an element may inhibit the uptake of another element. For example the Mn toxicity is the appearence of brown spots surrounded by chlorotic veins. Mn in excess, competes with iron and Mg for uptake and with Magnesium for binding with enzymes. Manganese also inhibits calcium translocation in the shoot apex. Therefore excess of manganese may infact, induce deficiencies of iron, magnesium and calcium. |
|
274. |
Under what conditions are C4 plants superior to C3? |
|||||||||||||||||||||||||||||||||
Answer»
|
||||||||||||||||||||||||||||||||||
275. |
Transpiration and photosynthesis - a compromise. Explain. |
Answer» Transpiration has more than one purpose; it (1) Creates transpiration pull for absorption and transportation in plants. (2) Supplies water for photosynthesis. (3) Transports minerals from the soil to all parts of the plant. (4) Cools leaf surface. (5) Maintains the shape and structure of the plants by keeping the cells turgid. An actively photosynthesising plant has an insatiable need for water. Photosynthesis is limited by available water which can be swiftly depleted by transpiration. C4 photosynthetic system is one of the strategies for maximising the availability of CO2 and minimizing water loss. C4 plants are twice efficient than C3 plants in fixing carbon and also water loss. |
|
276. |
Mention 3 traits in which Fungi resemble Animalia. |
Answer» Heterotrophy and glycogen as reserve food. | |
277. |
Which of the following statements is false ?A. TMV has a double stranded RNA moleculeB. Most plant viruses are RNA virusesC. The bacteriophage has a double stranded DNA moleculeD. Most animals viruses are DNA viruses. |
Answer» Correct Answer - A | |
278. |
Define the following terms :(a) Enthalpy(b) Entropy |
Answer» (a) Enthalpy : The amount of heat exchanged by a system with it's surroundings at constant pressure and temperature is called Enthalpy. Enthalpy (H) = U + PV Change in Enthalpy ∆H = ∆U + P∆U → It is a state function. ∆H = ∆HProducts – ∆Hreactants (b) Entropy : Entropy is taken as a Measure of disorder of molecules (or) randomness of the system. Greater the disorder of molecules in a system the higher is the Entropy. → It is a state function. → It depends on the temperature, pressure of the state. Entropy changed ∆S = qrev/T For a spontaneous process ∆S = +ve Solids (Low Entropy) liquids (Moderate Entropy) Gases (High Entropy) |
|
279. |
What is PH? Find the PH of a 0.1M HCl solution. |
Answer» PH : The negative logarithm of H+ ion concentration base ten expressed in Moles/Lit. PH = –log10[H+] Problem : Given 0.1M HCl solution. [H+] = 0.1M PH = –log[H+] = –log 0.1 = –log 10–1 = 1 PH = 1 |
|
280. |
Write about the biological importance of Calcium. |
Answer» Biological importance of Calcium : → Calcuim plays important role in neuromuscular function, inter neuromal transmission, cell membrane integrity and blood co-agulation. → Ca+2 ions are necessary for muscle contraction. → About 99% of body, calcuim is present in bones and teeth. |
|
281. |
An atom with high electronegativity hasA. Large sizeB. low electron affinityC. High ionisation enthalpyD. Low chemical reactivity. |
Answer» Correct Answer - C It is difficult to remove an electron from a highly electronegative element. |
|
282. |
Assertion:Electronegativity is not a measurable quantity.Reason:The electronegativity of any given element is not constant,itvaries depending on the element to which it bound.A. If both assertion and reason is the correct explanation of assertion.B. If both assertion and reason is the correct explanation of assertion.C. If both assertion and reason is the correct explanation of assertion.D. If both assertion and reason is the correct explanation of assertion. |
Answer» Correct Answer - A | |
283. |
Colletotrichum is an example ofA. BasidiomycetesB. DeuteromycetesC. AscomycetesD. Phycomycetes |
Answer» Correct Answer - B Colletotrichum is an example of Deuteromycetes. |
|
284. |
In the long or modern form of the periodic table, the element in the periodic table have been divided into four blocks, `s-p-d`-and `f`-. Each period begins with the filling of new energy shell. Two series of `f`-block elements are placed at the bottom of the periodic table. The element with `Z=113` has been discovered. Its block, group number, period and ourershell electronic configuration areA. `s`-block, group `2`, period `7,7s^(2)`B. `p`-block, group `13`, period `7,7s^(2)7p^(1)`C. `p`-block, group, 13`, period 6,6s^(2)6p^(1)`D. `d`-block, group `12`,period `6,5d^(10), 6s^(2)` |
Answer» Correct Answer - B `Z =113` means group `13`, and it is `p`-block, and `7thy` period. Therefore, electric configuration is `7s^(2)7p^(1)`. |
|
285. |
In the long or modern form of the periodic table, the element in the periodic table have been divided into four blocks, `s-p-d`-and `f`-. Each period begins with the filling of new energy shell. Two series of `f`-block elements are placed at the bottom of the periodic table. The last element of the `p`-block in the present periodic table is represented by the configuration, where `[X]` represents inert gas isA. `[X]7s^(2)p^(6)`B. `[X]5f^(14),6d^(10),7s^(2)7p^(5)`C. `[X]4f^(14),5d^(10),6s^(2)6p^(6)`D. `[X]` None of the above |
Answer» Correct Answer - C Last element of the `p`-block is `Rn` with `Z = 86`, group `18`, period `6`. Therefore, its valence electronic configuration is `6s^(2) 6p^(6)`. |
|
286. |
Which of the following is incorrect?A. Metals are electropositive elements.B. Metals usually have low ionzation enthalpies.C. As we move down a group, the metallic character the element decrease.D. Nonmetals are electronegative elements. |
Answer» Correct Answer - C Down the group, the matellic character of the elements increases on account of decrease of ionization enthalpy. |
|
287. |
Describe the ways in which Eddy currents are used to advantage. |
Answer» Eddy currents are used to advantage in (i) Magnetic braking in trains : A strong magnetic field is applied across the metallic drum rotating with the axle of the electric train. Thus large eddy currents are produced in the metallic drum. These currents oppose the motion of the drum and hence the axle of the train which ultimately makes the train come to rest. (ii) Induction Motor : Eddy currents are used to rotate the short circuited rotor of an induction motor. Ceiling fans are also induction motors which run on single phase alternating current. (iii) Electromagnetic damping : Certain galvanometers have a fixed core made of non magnetic metallic material. When the coil oscillates, the eddy currents generated in the core oppose the motion and bring the coil to rest quickly. (iv) Induction furnace : Induction furnace can be used to produce high temperatures and can be utilised to prepare alloys, by melting the constituent metals. A high frequency alternating current is passed through a coil. The eddy currents generated in the metals produce high temperatures sufficient to melt it. (v) Analogue energy meters : Concept of eddy currents is used in energy meters to record the consumption of electricity. Aluminium disc used in these meters get induced due to varying magnetic field. It rotates due to eddy currents produced in it. |
|
288. |
Identify the given figure of a protozoan protist and select the correct option. A. entamoeba histolyticaB. Plasmodium vivaxC. Giardia intestinalisD. Trypanosoma gambiense |
Answer» Correct Answer - D The given figure is of Trypanosoma gambiense that belongs to flagellated protozoans of Kingdom Protista. |
|
289. |
Which of the following is the atomic number of a metal?A. `32`B. `34`C. `36`D. `38` |
Answer» Correct Answer - D 38 is the atomic no. Of strontinum (sr) which is s-block element and all the elements of s-block are metals. |
|
290. |
Define Carbohydrates. |
Answer» The compounds which are primarily produced by plants and form a very large group of naturally occuring organic compounds are called Carbohydrates. Eg : Glucose, Fructose, Starch. → Carbohydrates are the polyhydroxy aldehydes (or) ketones. |
|
291. |
What are fibrous and globular proteins? |
Answer» Fibrous proteins : When the poly peptide chains run parallel and are held together by hydrogen and disulphide bonds then fibre - like structure is formed. These are called fibrous proteins. These are insoluble in water. Eg : keratin, myosin. Globular proteins : When the chains of polypeptides coil around to give a spherical shape then globular proteins are formed. These are usually soluble in water. Eg : insulin and albumins. |
|
292. |
What are barbiturates? |
Answer» Barbiturates : Derivatives of barbituric acid which functions as important class of tranquilizers are called barbiturates. E.g. : Veronal, Amytatt etc., |
|
293. |
What is tincture of iodine? What is its use? |
Answer» Tincture of iodine (antiseptic) is a mixture of 2 – 3% Iodine solution in alcohol-water. |
|
294. |
Classify each of the following as either a p – type or a n – type semiconductor.(1) Ge doped with In(2) Si doped with B. |
Answer» Both (1) and (2) come under "p – type semiconductors". Reason : In both the cases dopants (i.e.,) Indium in case – (1) and Boron in case – (2) belong to III (or) 13th group. Si (or) Ge doped with III group element is known as p-type semi conductor. Explanation : Doping the silicon or germanium element with III or 13th group element like "B", Al, Ga or "In" results in the substitution of some silicon atoms in its structure by the dopant. The dopant has only three valency electrons. The fourth valency electron is required. It is left as a vacant place on the atom. It is known as an 'electron vacancy' (or) a 'hole'. The electron vacancy on an atom in the structure migrates from one atom to another. Hence it facilitates the electrical conductivity. Si (or) Ge, doped with elements that create a hole in the structure, is known as p-type semi-conductor. |
|
295. |
What is the difference between a colloidal sol, gel, emulsion and a foam? |
|||||||||||||||||||||
Answer»
|
||||||||||||||||||||||
296. |
Explain the reaction of the following with water.(a) XeF2(b) XeF4(c) XeF6 |
Answer» (a) XeF2 is hydrolysed to form Xe, HF and O2 2XeF2 + 2H2O → 2Xe + 4HF + O2 (b) XeF4 is hydrolysed to give XeO3 6XeF4 + 12H2O → 4Xe + 2XeO3 + 24HF + 3O2 (c) XeF6 undergo hydrolysis to form XeO3 XeF6 + 3H2O → XeO3 + 6HF XeF6 undergo partial hydrolysis to form XeOF4 + XeO2F2 XeF6 + H2O → XeOF4 + 2HF XeF6 + 2H2O → XeO2F2 + 4HF |
|
297. |
Statement-1: correct IUPAC name of this compound is 1-chloro-6-fluoro-3,4-epoxyhexane. Statement-2: CI and F both are groups without suffix, CI comes before F in alphabetical order so numbering is done from this end. |
Answer» Correct Answer - A |
|
298. |
Total number of functional groups present in following compound: |
Answer» Correct Answer - 4 |
|
299. |
Which of the following is/are a bicyclo compound (s)? A. R onlyB. R and S onlyC. Q,R and SD. P,Q,R and S |
Answer» Correct Answer - B |
|
300. |
Suffix of this compound in its correct IUPAC name is:A. enoic acidB. enecarboxylic acidC. enyloic acidD. ene |
Answer» Correct Answer - B |
|