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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.
601. |
Classification given by Bentham and Hooker isA. ArtificalB. NaturalC. PhylogeneticD. Numerical. |
Answer» Correct Answer - B | |
602. |
Entities excluded from five kingdom classification of Whittaker areA. VirusesB. MouldsC. Algal fungiD. Fungi Imperfecti. |
Answer» Correct Answer - A | |
603. |
In three-kingdom classification, fungi are included underA. Kingdom FungiB. Kingdom ProtistaC. Kingdom AnimaliaD. Kingdom Plantae. |
Answer» Correct Answer - D | |
604. |
In two kingdom classification, bacteria and fungi are included amongstA. Animal kingdomB. Plant kingdomC. Both Plant and Animal kingdomsD. None of the two kingdoms. |
Answer» Correct Answer - B | |
605. |
The sex orange are absent, but plasmogamy is brought about by fusion of two vegetative or somatic cells of different genotypes. It is the feature ofA. PhycomycetesB. BasidiomycetesC. AscomycetesD. All of these |
Answer» Correct Answer - B | |
606. |
The element or elements whose position is anomalous the periodic table isA. HalogensB. Fe, Co and NiC. Inert gasesD. Hydrogen |
Answer» Correct Answer - D It show similarities with both alkali metals as well as halogens. |
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607. |
In the periodic table, the element with atomic number 16 will be placed in the groupA. ThirdB. FourthC. FifthD. Sixth |
Answer» Correct Answer - D `16-1s^(2)2s^(2)2p^(6)3s^(2)3p^(4)` there are `6e^(-)` in outer most shell therefore its group is `VI^(th)` A. |
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608. |
In third row of periodic table the atomic radii from Na and ClA. Continuosly decreasesB. Continuosly increasesC. Remains constantD. Increases but not continuously |
Answer» Correct Answer - A Continuously decreases as the effective nuclear charge increases. |
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609. |
The group having isoelectronic species isA. `O^(-), F^(-), Na,Mg^(+)`B. `O^(2-), F^(-), Na,Mg^(2+)`C. `O^(-), F^(-), Na^(+),Mg^(2+)`D. `O^(2-), F^(-), Na^(+), Mg^(2+)` |
Answer» Correct Answer - D Isoelectronic species `O^(2-),F^(-),Na^(+),Mg^(2+)` (All contain 10 electrons) |
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610. |
The ionic radii `(Å)" of "C^(4+) and O^(2-)` respectively are 2.60 and 1.40. The ionic radius of the isoelectronic ion `N^(3-)` would beA. 2.6B. 1.71C. 1.4D. 0.95 |
Answer» Correct Answer - B The ionic radii of isoelectronic ions decrease with the increase in the magnitude of the nuclear charge. So, decreasing order of ionic radii is `C^(4-)gtN^(3-)gtO^(2-)`. |
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611. |
The first ionisation potential of `Na,Mg,Al` and `Si` are in the orderA. `NaltMg gtAlltSi`B. `NagtMggtAlgtSi`C. `NaltMgltAlltSi`D. `NagtMggtAlltSi` |
Answer» Correct Answer - A `IE_(1)` of Mg is higher than that of Na because of increased nuclear charge and also that of Al because in Mg a 3s-electron has to be removed while in Al it is the 3p-electron. The `IE_(1)` of Si is, however, higher than those of Mg and Al because of it increased nuclear charge. Thus, the overall order is `NaltMg gtAl ltsi`. |
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612. |
The ionic radii of `N^(3-), O^(2-)` and `F^(-)` are respectively given by:A. 1.36, 1.40 and 1.71B. 1.36, 1.71 and 1.40C. 1.71, 1.40 and 1.36D. 1.71, 1.36 and 1.40 |
Answer» Correct Answer - C These are isoelectronic species As negative charge increases, ionic radius increases |
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613. |
Ionic radii of :A. `Ti^(4+) lt Mn^(7+)`B. `.^(35)Cl^(-)lt.^(37)Cl^(-)`C. `K^(+)gtCl^(-)`D. `F^(3+)gtF^(5+)` |
Answer» Correct Answer - D Nuclear charge per electron is greater in `P^(5+)`. Therefore, its size is smaller. |
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614. |
The order of magnitude of ionic radii of ions `Na^(+),Mh^(2+),Al^(3+)` and `Si^(4+)` isA. `Na^(+)ltMg^(2+)ltAl^(3+)ltSi^(4+)`B. `Mg^(2+)gtNa^(+)gtAl^(3+)gtSi^(4+)`C. `Al^(3+)gtNa^(+)gtSi^(4+)gtMg^(2+)`D. `Na^(+)gtMg^(2+)gtAl^(3+)gtSi^(4+)` |
Answer» Correct Answer - D `Na^(+)gtMg^(2+)gtAl^(3+)gtSi^(4+)`. All are isoelectronic but nuclear charge per electron is greatest for `Si^(4+)`. So it has smallest size and nuclear charge per electron for `Na^(+)` is smallest. So it has largest size. |
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615. |
Which of the following pair of ions have the same number of unpaired electronsA. `Mn^(2+) and Fe^(2+)`B. `Mn^(2+) and Fe^(3+)`C. `Ti^(2+) and Ni^(2+)`D. `Co^(3+) and Cr^(3+)` |
Answer» Correct Answer - B::C `Mn^(+2)` and `Fe^(+3)` have 5 unpaired electrons each `: Ti^(+2)` and `Na^(+2)` have 2 unpaired electrons each. |
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616. |
Which of the following sets will have highest hydration energy and highest ionic radiiA. Na and LiB. Li and RbC. K and NaD. Cs and Na |
Answer» Correct Answer - B Smaller the size, greater is the hydration enthalpy. So, Li will have highest hydration enthalpy. Down the group ionic radius increases, so Rb will have highest ionic radius. |
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617. |
Statement 1 : If a metal forms a number of ions such as `M^(+), M^(2+), M^(3+)` etc. then the hydration energies associated with these ions increase in that order. Statement 2 : The energies required to produce `M^(+), M^(2+)` and `M^(3+)` ions in gaseous state from metal increase in the order `M^(+) lt M^(2+) lt M^(3+)`. |
Answer» Correct Answer - B Both Statement are correct but Statement-2 is not the correct explanation of Statement-1. The large solvation energies are associated with ions possessing high charge and small size. |
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618. |
The energy required to pull the most loosely bound electron from an atom is known as ionization potential. It is expressed in electron volts. The value of ionization potential depends on three factors: (i) the charge on the nucleus (ii) the atomic radius and (iii) the screening effect of inner electron shells. The ionization energies, electron affinities, electronegativities, atomic and ionic radii and other physical properties usually shown a regular pattern of change within a group or along period with some irregularities. Ionization energies of five elements in kcal/mol are given below `{:("Atom","I","II","III"),("P",300,549,920),("Q",99,734,1100),("R",118,1091,1652),("S",176,347,1848),("T",497,947,1500):}` Which element is a noble gasA. SB. TC. RD. P |
Answer» Correct Answer - B T has abnormally higher `I.E._(1)` value. |
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619. |
The energy required to pull the most loosely bound electron from an atom is known as ionization potential. It is expressed in electron volts. The value of ionization potential depends on three factors: (i) the charge on the nucleus (ii) the atomic radius and (iii) the screening effect of inner electron shells. The ionization energies, electron affinities, electronegativities, atomic and ionic radii and other physical properties usually shown a regular pattern of change within a group or along period with some irregularities. Ionization energies of five elements in kcal/mol are given below `{:("Atom","I","II","III"),("P",300,549,920),("Q",99,734,1100),("R",118,1091,1652),("S",176,347,1848),("T",497,947,1500):}` Which element form stable unipositive ionA. RB. QC. PD. T |
Answer» Correct Answer - B There is sudden jump in `I.E._(2)` of Q. i.e., unipositive has noble gas configuration. |
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620. |
The energy required to pull the most loosely bound electron from an atom is known as ionization potential. It is expressed in electron volts. The value of ionization potential depends on three factors: (i) the charge on the nucleus (ii) the atomic radius and (iii) the screening effect of inner electron shells. The ionization energies, electron affinities, electronegativities, atomic and ionic radii and other physical properties usually shown a regular pattern of change within a group or along period with some irregularities. Ionization energies of five elements in kcal/mol are given below `{:("Atom","I","II","III"),("P",300,549,920),("Q",99,734,1100),("R",118,1091,1652),("S",176,347,1848),("T",497,947,1500):}` Which of the following pair represents elements of same groupA. Q, RB. P, SC. P, QD. Q, S |
Answer» Correct Answer - A Both Q and R shows group in `I.E._(2)` values and belong to alkali metals. |
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621. |
The energy required to pull the most loosely bound electron from an atom is known as ionization potential. It is expressed in electron volts. The value of ionization potential depends on three factors: (i) the charge on the nucleus (ii) the atomic radius and (iii) the screening effect of inner electron shells. The ionization energies, electron affinities, electronegativities, atomic and ionic radii and other physical properties usually shown a regular pattern of change within a group or along period with some irregularities. Ionization energies of five elements in kcal/mol are given below `{:("Atom","I","II","III"),("P",300,549,920),("Q",99,734,1100),("R",118,1091,1652),("S",176,347,1848),("T",497,947,1500):}` If Q reacts with fluorine and oxygen, the molecular formula of fluoride and oxide will be respectivelyA. `QF_(2), QO`B. `QF, Q_(2)O`C. `QF_(3), Q_(2)O_(3)`D. None of these |
Answer» Correct Answer - B Q is alkali metal as it shows group in `I.E._(2)` value. |
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622. |
The energy required to pull the most loosely bound electron from an atom is known as ionization potential. It is expressed in electron volts. The value of ionization potential depends on three factors: (i) the charge on the nucleus (ii) the atomic radius and (iii) the screening effect of inner electron shells. The ionization energies, electron affinities, electronegativities, atomic and ionic radii and other physical properties usually shown a regular pattern of change within a group or along period with some irregularities. Ionization energies of five elements in kcal/mol are given below `{:("Atom","I","II","III"),("P",300,549,920),("Q",99,734,1100),("R",118,1091,1652),("S",176,347,1848),("T",497,947,1500):}` The element having most stable oxidation state +2 isA. TB. QC. SD. R |
Answer» Correct Answer - C `I.E._(3)` of S is abnormally higher. |
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623. |
Assertion : Noble gases have maximum electron affinity. Reason : High electron affinity shows that the electron is loosely bonded to the atom.A. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not a correct explanation of the assertionC. If assertion is true but reason is falseD. If the assertion and reason both are false |
Answer» Correct Answer - D All noble gases have stable configuration. Therefore, they can not take any electron means that they have no affinity for electrons. High electron affinity shows that electron is strongly bonded to the atom. Here both assertion and reason are false. |
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624. |
The increasing order of ionic mobilities is/are given byA. `K^(+)ltNa^(+)ltLi^(+)`B. `Li^(+)ltNa^(+)ltK^(+)`C. `Na^(+)ltMg^(2+)ltAl^(3+)`D. `Al^(3+)ltMg^(2+)ltNa^(+)` |
Answer» Correct Answer - B::C The extent of hydration increases with decreases in ionic size. Thus extent of hydration varies in the order : `Li^(+)gtNa^(+)gtK^(+)` and so also size of ions. |
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625. |
The correct order of the increasing ionic character isA. `BeCl_(2)ltMgCl_(2)ltCaCl_(2)ltBaCl_(2)`B. `BeCl_(2)ltMgCl_(2)ltBaCl_(2)gtCaCl_(2)`C. `BeCl_(2)ltBaCl_(2)ltMgCl_(2)ltCaCl_(2)`D. `BaCl_(2)ltCaCl_(2)ltMgCl_(2)ltBeCl_(2)` |
Answer» Correct Answer - A Alkali metals have the configuration `(n-1)s^(2)p^(6),ns^(1)`. |
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626. |
In the periodic table, the metallic character of elementsA. Decreases from left to right across a period and on descending a groupB. Decreases from left to right across a period and increases on desecending a groupC. Increases from left to right across a period and on descending a groupD. Increases from left to right across a period and decreases on descending a group |
Answer» Correct Answer - B Fe belongs to first transition series. |
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627. |
Who is considered to be the father of virologyA. lvanowskiB. StanleyC. BeijerinckD. Pasteur |
Answer» Correct Answer - B | |
628. |
Two kingdom classification was given byA. LinnaeusB. HaeckelC. CopelandD. Whittaker |
Answer» Correct Answer - A | |
629. |
Three kingdom classification was proposed byA. LinnaeusB. HaeckelC. WhittakerD. Lamarck |
Answer» Correct Answer - B | |
630. |
Mitochondria probably originated from bacteria because both haveA. Similar type of circular DNAB. Similar pigmentsC. Same functionD. All the above. |
Answer» Correct Answer - A | |
631. |
Heackel kept Bacteria, Algae, Fungi and Protozoans in kingdomA. PlantaeB. ProtistaC. Protista and plantaeD. Animalia |
Answer» Correct Answer - B | |
632. |
Four kingdom classification was proposed byA. WhittakerB. CopelandC. HaeckelD. Linnaeus. |
Answer» Correct Answer - B | |
633. |
Three kingdom system of classification was proposed byA. HaeckelB. LinnaeusC. StanierD. Copeland |
Answer» Correct Answer - A | |
634. |
Decomposers belong to kingdomA. Monera and ProtistaB. Protista and Fungi (Mycota)C. Monera, Protista and FungiD. Protista, Fungi and Animalia |
Answer» Correct Answer - C | |
635. |
The protists haveA. Only free nucleic acid aggregatesB. Membrane bound nucleoproteins lying embedded in the cytoplasmC. Gene containing nucleoproteins condensed together in loose massD. Nucleoprotein in direct contact with the rest of the cell substance |
Answer» Correct Answer - B | |
636. |
Which one single organism or the pair of organisms is correctly assigned to its or their named taxonomic group?A. Yeast used in making bread and beer is a fungusB. Nostoc and Anabaena are examples of protistaC. Paramecium and Plasmodium belong to the same kingdom as that of PenicilliumD. Lichen is a composite organism formed from the symbiotic association of an alage and a protozoan |
Answer» Correct Answer - A | |
637. |
What is correct about classification ?A. It allows to confirm the already existing identificationsB. It helps in new identificationsC. It is a systematic arrangement of taxaD. All of the above |
Answer» Correct Answer - D | |
638. |
The cyanobacteria are also referred to as:-A. Slime mouldsB. Blue green algaeC. ProtistsD. Golden algae |
Answer» Correct Answer - B | |
639. |
Bacteria are grouped in which one of the following kingdom ?A. ProtistaB. PlantaeC. AnimaliaD. Monera |
Answer» Correct Answer - D | |
640. |
Autotrophic nutrition includesA. ChemosynthesisB. PhotosynthesisC. Both A and BD. Ingestive nutrition. |
Answer» Correct Answer - C | |
641. |
`(9+2)` arrangement of microtubules is present inA. Kingdom -PlantaeB. kingdom - AnimaliaC. kingdom- ProtistaD. All of these |
Answer» Correct Answer - D | |
642. |
Which of the following have cellulosic cell wall ?A. MoneransB. PlantsC. FungiD. Animals |
Answer» Correct Answer - B | |
643. |
Which of the following species will have the largest and the smallest size `Mg, Mg^(2+), Al, Al^(3+)`? |
Answer» Atomic radius decreases across the period. Here Mg and Al belong to the same period, so out of these two, Mg will have the larger size. Cation are smaller than their parent atoms due to higher effective nuclear charge. Among the isoelectronic species, the one with a greater positive charge will have smaller radius . Hence, we can say out of `Mg, Mg^(2+), Al " and " Al^(3+)` the largest size will be of Mg and smallest size is of `Al^(3+)`. |
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644. |
Mark the mis-matched pairA. Pyrocystis - BioluminescenceB. Whirling whips - Soap box like body structureC. Diatomite - KieselguhrD. Gymnodinium - Zygotic meiosis |
Answer» Correct Answer - D | |
645. |
Which taxon has suffix -ales ?A. FamilyB. ClassC. GenusD. Order |
Answer» Correct Answer - D | |
646. |
Pigment-containing membranous extensions in some cyanobacteria areA. Basal bodiesB. PneumatophoresC. ChromatophoresD. Heterocysts |
Answer» Correct Answer - C | |
647. |
The correct order in which the first ionisation potential increases isA. K, Be, NaB. Be, Na, KC. Na, K, BeD. K, Na, Be |
Answer» Correct Answer - D The electronic configuration of the elements are `._(4)Be-1s^(2)2s^(2)` `._(11)Na-1s^(2)2s^(2)2p^(6)3s^(1)` `._(19)K-1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)4s^(1)` The first ionisation energy of Be is maximum because electron is to be drawn from stable (fully filled) orbital. The `I^(st)` ionisation energy of Na is greater than K because size of the K is bigger than Na which facilitates easy removal of electron from its outermost shell. So, the sequence is `KltNaltBe`. |
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648. |
Generally, the first ionisation energy increases along a period. But there are some exceptions one which is not an exception isA. N and OB. Na and MgC. Mg and AlD. Be and B |
Answer» Correct Answer - B Be, N and Mg have higher values than expected, this is due to stable configurations of these elements. Be and Mg have fully-filled orbitals, in nitrogen 2p-orbitals are half-filled. |
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649. |
Statement I `LiCl` is predomionantly a covalent compound. Statement II Electronegatvity difference between `Li` and `Cl` is too smallA. Statement 1 is true, statement 2 is true , statement 2 is a correct explanation for statement 1B. Statement 1 is true, statement 2 is true , statement 2 is not a correct explanation for statement 1C. Statement 1 is true, statement 2 is falseD. Statement 1 is false, statement 2 is true |
Answer» Correct Answer - C Cl is more electronegative than Li. Although the difference is not much. Therefore the electron pair moves equally to both and thus forming a covalent compound. |
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650. |
The correct decreasing order of first ionisation enthalpies of five elements of second period isA. `BegtBgtCgtNgtF`B. `NgtFgtCgtBgtBe`C. `FgtNgtCgtBegtB`D. `NgtFgtBgtCgtBe` |
Answer» Correct Answer - C As we move along the period, the nuclear charge increases and size of atom decreases. Therefore, it is more difficult to remove electron from an atom. Hence the sequence of first ionisation enthalpy in decreasing order is `FgtNgtC`. But ionisation enthalpy of boron is less as compared to beryllium because first electron in boron is to be removed from p-orbital while in beryllium, is to be removed from s- orbital. More energy is required to remove electron from s- orbital as compared to p-orbital. |
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