1050 + Interview Questions in CLASS-11 IN GENERAL AWARENESS Page 1 InterviewSolution

1.	The trivalent ion having size in lanthanide series isA. `Ti`B. `Zr`C. `Hf`D. `La`
Answer» Correct Answer - D The trivalent ion having largest size in lanthanide series is lanthanum. This is due to lanthanide contraction .

Discussion

2.	The multinucleate slimy mass of protoplasm which forms the body of slime moulds is called asA. plasmodiumB. myxamoebaC. sprocytesqD. periplasmodium.
Answer» Correct Answer - A A free living thalloid body of the acellular slime moulds is called plasmodium. The plasmodium is wall-less mass of multinucleate protoplasm covered by sline.

Discussion

3.	During unfavourable conditions, slime moulds forms fruiting body after the differentation ofA. AscocarpB. BasidiocarpC. PlasmodiumD. Gemmule
Answer» Correct Answer - C During unfavourable condition, slime moulds forms fruiting body after the differentation of plasmodium.

Discussion

4.	(A) What are protozoans ? Give two examples each of free-living and parasitic forms. (B) What are slime-moulds ? How are they different from true fungi ? Give one example each of the cellular and ascellular slime and moulds. (C) Differentiate between ascus and basidium.
Answer» (A) Protozoans are unicellular protist animals. Each protozoans is microscopic organism in which a single cell performs all the vital activities. Examples of free living protozoans are Amoeba and Paramoecium. Example of parasitic forms are Trypanosoma and Plasmodium. (B) Slime moulds are consumer decomposer protists which occur in the form of slimy masses during vegetative phase. The vegetative parts do not possess cell walls and, therefore, they occur either as free-living multinucleate amoeboid mass of protoplasm (plasmodium) or aggregation of amoebae (pseudoplasmodium). Example of cellular slim mould is Dictyostrelium and that of acellular slime mould is Physarum. (C) The main differences between an ascus and a basidium are -(i) Ascus is formed in ascomycetes whereas basidium is formed in basidiomycetes, (ii) Ascus encloses 8 ascospores which are borne endogenously. The basidium produces 4 basidiospores which are borne exogenously, (iii) The basidiospres are attack basidia by means of sterigma whereas ascospores are filled inside the sac like ascus.

Discussion

5.	The IUPAC name of the compound `CH_(3)CH=CHCH=CHC=C CH_(3)` is:A. Octa-4,6-dine-2-yneB. Octa-2,4-diene-6-yneC. Oct-2-yne-4,6-dieneD. Oct-6-yne-2,4-diene
Answer» Correct Answer - B

Discussion

6.	Which one is the smallest organism capable of autonomous growth and reproduction Or Which among the following are the smallest living cells, known without a definite cell wall, pathogenic to plants as well as animals and can survive without oxygenA. PseudomonasB. MycoplasmaC. NostocD. Bacillus
Answer» Correct Answer - B Mycoplasma is the smallest living cells, known without a definite cell wall, pathogenic to plants as well as animals and can survive without oxygen.

Discussion

7.	The shape of the cocci bacteria isA. Rod shapedB. SphericalC. Comma shapedD. Spiral
Answer» Correct Answer - B The shape of the cocci bacteria is Spherical.

Discussion

8.	Cell walls of diatoms are indestructibel because of the presence ofA. `CaCO_(3)`B. SilicaC. ChitinD. All of the above
Answer» Correct Answer - B Diatoms are indestructible because of the presence of siliceous cell wall.

Discussion

9.	The wonder drug, penicillin is extracted from which of the following species of penicillium?A. Penicillium notatumB. P. chrysogenumC. Both a and bD. None of these
Answer» Correct Answer - C The beast known antibiotic, called penicillin is obtained from both Penicillium notatum and P. chrysogenum.

Discussion

10.	What are the physiological processes that are regulated by ethyline in plants ?
Answer» (1) Ethylene promotes the ripening of fruits. (2) Ethylene promotes the senescence and abscission of leaves and flowers. (3) Ethylene breaks seed and bud dormancy, initiates germination in peanut seeds and sprouting of potato tubers. (4) Ethylene promotes rapid internode/petiole elongation in deep water rice plants. (5) It also promotes root growth and root hair formation, thus helping plants to increase their absorption surface. (6) Ethylene is used to initiate flowering (mango) and for synchronising fruit set in pineapples. (7) It promotes female flowers in cucumbers, thereby increasing the yield.

10.

What are the physiological processes that are regulated by ethyline in plants ?

Answer»

(1) Ethylene promotes the ripening of fruits.

(2) Ethylene promotes the senescence and abscission of leaves and flowers.

(3) Ethylene breaks seed and bud dormancy, initiates germination in peanut seeds and sprouting of potato tubers.

(4) Ethylene promotes rapid internode/petiole elongation in deep water rice plants.

(5) It also promotes root growth and root hair formation, thus helping plants to increase their absorption surface.

(6) Ethylene is used to initiate flowering (mango) and for synchronising fruit set in pineapples.

(7) It promotes female flowers in cucumbers, thereby increasing the yield.

Discussion

11.	Give a brief account of Bt. Cotton.
Answer» Some strains of Bacillus thuringiensis produce proteins that kill certain insects such as lepidopterans (tabacco budworm, armyworm), coleopterans (beetles) and dipterans (files, mosquitoes). Bacillus thuringiensis forms protein. Crystals which contain a toxic insecticidal protein. The gene responsible for the production of this toxic protein is introduced genetically into the cotton seeds protects the plants from Bollworm, a Major pest of cotton. The worm feeding, on the leaves of Bt. Cotton plant becomes lethargic and sleepy thereby causing less damage to the plant. Use of Bt. Cotton has led to 3-27% increase in cotton yield in countries where it is grown. The toxin is coded by a gene named 'cry'. The proteins encoded by the genes cry IAc and cry IIAb control the cotton boll worms and cry IAb controls corn borer.

11.

Give a brief account of Bt. Cotton.

Answer»

Some strains of Bacillus thuringiensis produce proteins that kill certain insects such as lepidopterans (tabacco budworm, armyworm), coleopterans (beetles) and dipterans (files, mosquitoes). Bacillus thuringiensis forms protein. Crystals which contain a toxic insecticidal protein. The gene responsible for the production of this toxic protein is introduced genetically into the cotton seeds protects the plants from Bollworm, a Major pest of cotton. The worm feeding, on the leaves of Bt. Cotton plant becomes lethargic and sleepy thereby causing less damage to the plant. Use of Bt. Cotton has led to 3-27% increase in cotton yield in countries where it is grown.

The toxin is coded by a gene named 'cry'. The proteins encoded by the genes cry IAc and cry IIAb control the cotton boll worms and cry IAb controls corn borer.

Discussion

12.	Define transformation in Griffith's experiment. Discuss how it helps in the Identification of DNA as genetic material.
Answer» Frederick Griffith (1928) conducted experiments on streptococcus pneumoniae and observed a transformation in bacteria. When streptococcus were grown on a culture plate, some produced smooth shiny colonies (s) while others produced rough colonies (R). Mice injected with 's' shain (mucous coat) die from pneumonia infection but mice injected with R strain do not develop pneumonia. He injected heat killed 's' strain bacteria to mice, It is healthy. Finally he injected heat killed S and R strains, the mice died. He concluded that the R strain bacteria had been transformed by heat killed 's' strain bacteria. Some transforming principle transferred from heat killed strain to R strain to synthesize a mucous coat and become virulent. This is due to the transfer of genetic material.

12.

Define transformation in Griffith's experiment. Discuss how it helps in the Identification of DNA as genetic material.

Answer»

Frederick Griffith (1928) conducted experiments on streptococcus pneumoniae and observed a transformation in bacteria. When streptococcus were grown on a culture plate, some produced smooth shiny colonies (s) while others produced rough colonies (R). Mice injected with 's' shain (mucous coat) die from pneumonia infection but mice injected with R strain do not develop pneumonia.

He injected heat killed 's' strain bacteria to mice, It is healthy. Finally he injected heat killed S and R strains, the mice died. He concluded that the R strain bacteria had been transformed by heat killed 's' strain bacteria. Some transforming principle transferred from heat killed strain to R strain to synthesize a mucous coat and become virulent. This is due to the transfer of genetic material.

Discussion

13.	What is ICTV ? How are viruses named ?
Answer» International Committee on Taxonomy of Viruses [ICTV] regulates the norms of classification and nomenclature of viruses. The ICTV has only three hierarchial levels, The family, Genus, and Species. The family names end with the 'suffix viridae' while the Genus name ends with virus and the Species names are common English expressions . Viruses are named after the disease they cause. Eg : Polio virus.

13.

What is ICTV ? How are viruses named ?

Answer»

International Committee on Taxonomy of Viruses [ICTV] regulates the norms of classification and nomenclature of viruses. The ICTV has only three hierarchial levels, The family, Genus, and Species. The family names end with the 'suffix viridae' while the Genus name ends with virus and the Species names are common English expressions . Viruses are named after the disease they cause.

Eg : Polio virus.

Discussion

14.

With the help of an example, differentiate between incomplete dominance and Co-dominance.

Answer»

Incomplete Dominance	Co-Dominance
It is the condition when one allele of a gene is not completely dominant over the other allele and results in the hetero- zygotes having phenotype different from the dominant and recessive homozygotes.	The phenomenon where heterozygotes have features of both the homozygotes i.e., an alle is neither dominant nor recessive to the other.
Eg: Cross between red flowered (RR) and white flowered (rr) plants results in pink flowered plants (Rr).	Eg: Cross between spotted lentils and dotted lentils produce heterozygotes that are both spotted and dotted.

Discussion

15.	What is common about Trypanosoma, Noctiluca Monocystis and GiardiaA. These are all parasitesB. These are all uncellular protistsC. They have flagellaD. They produce spores
Answer» Correct Answer - B

Discussion

16.	Which one of the following is wrong for fungiA. They are heterotrophicB. They are both unicellular and multicellularC. They are eukaryoticD. All fungi possess a purely cellulosic cell wall
Answer» Correct Answer - D

Discussion

17.	(a) 3,6,-Bis-(1,1-dimetylethyl) octane (b) Octa-4,7-dien-2-amine (c) 3-Hydroxymethyl cyclohexanecarboxylicacid (d) 2-Nitropent-3-ene If the given IUPAC name is correct the write 1 and if it is wrong then write 2
Answer» Correct Answer - 2112

Discussion

18.	Write the correct IUPAC name of 2-amino-1, 3,7-trihydroxy hept-4-ene-1, 7-dione?A. 6-amino-5-hydroxy hept-3-ene-1,7-dioicacidB. 2-amino-3-hydroxy hept-4-ene-1,-7,dioicacidC. 6-amino-5-hydroxy hept-4-ene-1-,7-dicarboxylic acidD. 6-amino-5-hydroxy hept-3-ene-1,7-dicarboxylic acid
Answer» Correct Answer - A

Discussion

19.	Statement-1: IUPAC name is 1,4-epocy-1-ethyl-4-methylbutane. Statement-2: Substituents are written in alphabetic order.A. Both the statements are correct but Statement 2 is not the correct explanation of Statement 1B. Both the statements are correct but Statement 2 is the correct explanation of Statement 1C. Statement 1 is correct but Statement 2 is incorrectD. Both the statements are incorrect
Answer» Correct Answer - D

Discussion

20.	One among the following is the correct IUPAC name for N-ethyl-1-amino methanal.A. N-formyl aminoethaneB. N-ethyl formyl amineC. N-ethyl methanamideD. Ethylamino methanal
Answer» Correct Answer - C

Discussion

21.	The correct name of 1-methoxy-1-ethyl-2-methyl propane is:A. 2-methyl-3-methoxy pentaneB. 1-methoxy-1-iso-propyl methaneC. 3-methoxy-4-methylpentaneD. 3-methoxy-2-methyl pentane
Answer» Correct Answer - D

Discussion

22.	Statement-1: 4-ethyl-3-bromo pentanoic acid is not the correct IUPAC name. Statement-2: Selection of carbon chain in 3-bromo-4-ethyl pentanoic acid is not according to IUPAC rule.A. Both the statements are correct but Statement 2 is not the correct explanation of Statement 1B. Both the statements are correct but Statement 2 is the correct explanation of Statement 1C. Statement 1 is correct but Statement 2 is incorrectD. Both the statements are incorrect
Answer» Correct Answer - B

Discussion

23.	Observe the following compounds answer the following question Which one of these three compounds contains maximum number of side chains in its principal carbon chain ?A. IB. IIC. IIID. Equal in I and II
Answer» Correct Answer - A

Discussion

24.	Find out number of C-atoms present in principal chain according to IUPAC rules. [Write answer of part (a),(b),(c) and (d) in the same order and present the four digit number as answer in OMR sheet. For example. ]
Answer» Correct Answer - 7576

Discussion

25.	Write D.B.E (Double bond equivalent) for the following compound: [Write answer of part (a),(b),(c) and (d) in the same order and present the four digit number as answer in OMR sheet. For example.]
Answer» Correct Answer - 6959

Discussion

26.	Observe the following compounds answer the following question Which one of these three compounds contains maximum number of quarternary carbon atoms `(4^(@))` ?A. IB. IIC. IID. Equal in II and III
Answer» Correct Answer - B

Discussion

27.	Number of secondary carbon atoms present in benzene, o-xylene, touene and mesitylene are respectively:
Answer» Correct Answer - 6453

Discussion

28.	Viral capsid is made ofA. ProteinB. Protein + Nucleic acidC. Nucleic acidD. Either DNA or RNA
Answer» Correct Answer - A Viral capsid is made up of protein.

Discussion

29.	The capsid of viruses protects theA. NucleoproteinB. CapsomereC. Nucleic acidD. Protien
Answer» Correct Answer - C The capsomere of viruses protects the nucleic acid.

Discussion

30.	Diatomaceous earth is the deposition of cell wall in their habitat by Diatoms takes overA. Milions of yearsB. Billions of yearsC. Trillions of yearsD. Thousands of years
Answer» Correct Answer - B Diatomaceous earth is the deposition of cell wall in their habitat by diatoms and it takes over billions of year.

Discussion

31.	Consider the following four statement (1-4) and select the option which includes all the correct ones only (1) Single cell Spirulina can produce large quantities of food rich in protein, minerals, vitamins etc (2) Body weight-wise the microorganism Methylophilus methylotrophus may be able to produce several times more proteins then the cows pe day (3) Common button mushrooms are a very rich source of vitamin C (4) A rich variety has been developed which is very rich in calciumA. Statement (C) and (D)B. Statement (A), (C) and (D)C. Statement (B), (C) and (D)D. Statements (A) and (B)
Answer» Correct Answer - D

Discussion

32.	`IE_(1), IE_(2)` and `IE_(3)` values are `100, 150` and `1500 eV` respectively. The element can beA. `Na`B. `B`C. `Be`D. `F`
Answer» Correct Answer - C There is a high jump from `IE_(2)` to `IE_(3)`. Therefore, it is difficult to remove the 3rd valence electron. So, the element must be of group 2, e.g. `Be(2s^(2))`.

Discussion

33.	The value of `IE_(1), IE_(2), IE_(3)` and `IE_(4)` of an atom are respectively `7.5 eV, 25.6 eV, 48.6 eV` and `170.6 eV`. The electronic configuration of the atom will beA. `1s^(2), 2s^(2)2p^(6), 3s^(1)`B. `1s^(2)2s^(2)2p^(6)3s^(2) 3p^(1)`C. `1s^(2), 2s^(2)2p^(6)3s^(2)3p^(3)`D. `1s^(2), 2s^(2)2p^(6)3s^(2)`
Answer» Correct Answer - B There is a high jump from `IE_(3)` to `IE_(4)`. Therefore, it is difficult to remove the 4th valence electron. So, electronic configuration of element must have three valence electrons. Hence, the answer is (b).

Discussion

34.	Assertion (A): `IE_(1)` of `N` is lower than `O`. Reason (R ) : Across a period `Z_(eff)` decrease.A. If both Assertion (A) and (R ) are correct and Reason (R ) is the correct explanation of Assertion (A).B. If both (A) and (R ) are correct but (R ) is not the correct explanation for (A).C. If (A) is correct but (R ) is not correct.D. If both (A) and (R ) are correct
Answer» Correct Answer - C Correct assertion: `IE_(1)` of `IE_(1)` of `O`, due to stable half-filled configuration of `N`. Correct reason: `Z_(eff)` increases across a period.

Discussion

35.	Which of the following van der Waals radii is the largest ?A. `Ne`B. `Cl`C. `O`D. `F`
Answer» Correct Answer - B `Ne, O` and `F` are of `2nd` period elements whereas `Cl` is of `3rd` period. Hence have large van der Waals radii.

Discussion

36.	Assertion(A) : `IE_(2)` of `Li` is the highest in the second period. Reason (R ): `Li^(o+)` haqs nob,le gas i.e., `NE` gas configuration.A. If both Assertion (A) and (R ) are correct and Reason (R ) is the correct explanation of Assertion (A).B. If both (A) and (R ) are correct but (R ) is not the correct explanation for (A).C. If (A) is correct but (R ) is not correct.D. If (A) and (R ) are correct
Answer» Correct Answer - C Correct reason: `Li^(o+)` has noble gas (e.g. He gas configuration)

Discussion

37.	Compounds that formally contain `Pb^(4+)` are easily reduced to `Pb^(+)` .The stability of the lower oxidation state is due to `"…….."` .
Answer» Correct Answer - Inert pair effect Compounds that formally contain `Pb^(4+)` are easily reduced to `Pb^(2+)`. The stability of the lower oxidation state is due to inert pair effect. In the periodic table, inert pair effect increases down the group. So in group `14`, the stability of smaller oxidation state of `+2` goes on increasing down the group. `Pb^(4+)` has an oxidising nature and `Br^(ɵ)` and `I^(ɵ)` have reducing nature, so `Pb^(4+)` is reduced to `Pb^(2+)`. `Pb^(4+)+2Br^(ɵ) rarr Pb^(2+)+Br_(2)` `Pb^(4+)+2I^(ɵ) rarr Pb^(2+)+I_(2)`

Discussion

38.	What is the atomic number of the element present in the second period anf group 15.
Answer» `Z=7`, Element is nitrogen.

Discussion

39.	The angle between two covalent bonds is maximum for `(CH_(4), H_(2)O, CO_(2))` `"______"` .
Answer» Correct Answer - `CO_(2)` (because it is `sp` hybridised, bond angle `= 180^(@)`) `CO_(2)` (because it is `sp` hybridised, bond angle `= 180^(@)`)

Discussion

40.	Ionic bonds are non-directional while covalent bonds are directional.
Answer» Correct Answer - T True.

Discussion

41.	As the `s` character of the hybrid orbital decreases, the `EN` increases.
Answer» Correct Answer - F False. As the s-character of the hybrid orbital decreases, the EN decreases.

Discussion

42.	Along the period `(rarr)` atomic`//`ionic radii and metallic character decreases while `IE`, `EN` , non-metallic character and oxidising power increases. Down the group `(darr)`, atomic`//`ionic radii, metallic character and reducing character increase while `IE` and `EN` decrease. However, `Delta_(eg)H^(ɵ)` becomes less negative down a group but more negative along a period. If the ionic radii of `M^(o+)` and `X^(ɵ)` are about `135 p m`, then expected values of metallic radii of `M` and `X` should be respectively.A. `65` and `230 p m`B. `230` and `60 p m`C. `230` and `135 p m`D. `135` and `135 p m`
Answer» Correct Answer - B Since size of `M gt M^(o+) (135 pm)` and that of `X lt X^(ɵ)(135 p m)`, therefore answer (b) is correct.

Discussion

43.	Along the period `(rarr)` atomic`//`ionic radii and metallic character decreases while `IE`, `EN` , non-metallic character and oxidising power increases. Down the group `(darr)`, atomic`//`ionic radii, metallic character and reducing character increase while `IE` and `EN` decrease. However, `Delta_(eg)H^(ɵ)` becomes less negative down a group but more negative along a period. Which of the following isoelectronic species has lowest `IE_(1)` ?A. `K^(o+)`B. `Ca^(2+)`C. `S^(2-)`D. `Cl^(ɵ)`
Answer» Correct Answer - C `S^(2-)` have the largest size among the given isoelectronic ions, so it has the lowest `IE_(1)`.

Discussion

44.	The energy required to remove an electron from the outermost shell of an isolate gaseous atom is known as `IE_(1)` of that atom. Similarly, the enrgy required for the removal of the electron from the unipositive ion, diapositive ion and tripositive ion are known as `IE_(2),IE_(3)` and `IE_(4)` respectively, and are called successive ionisation energies. The magnitude of the charge depends on the size of the orbital of electron. Electrons in smaller orbitals are on average close with each other and have more repulsion. Thus for `Be(2s^(2))`, the `IE_(1)` and `IE_(2)` are `9.3` and `18.2 eV "atom"^(1)`, whereas for `Ca(4s^(2))`, the values are `6.1` and `11.9 eV`. Which of the following are isoelectronic species? `1 rarrCH_(3)^(o+),IIrarrNH_(2)^(ɵ),IIIrarrNH_(4)^(o+),IVrarrNH_(3)`A. `I, II` and `III`B. `II, III`and `IV`C. `I,II` and `IV`D. `II` and` I`
Answer» Correct Answer - B `II, III` and `IV` are isoelectronic species.

Discussion

45.	The screening effect of `d`-electron isA. Equal to `p`-electronB. Much more than `p`-electronC. Same as `f`-electronsD. Less than `p`-electrons
Answer» Correct Answer - D Order of screening effect is `s gt p gt d gt f`.

Discussion

46.	Ionic radii of :A. `.^(35)Cl^(ɵ) gt Cl(ɵ)`B. `Mn^(7+) gt Ti^(4+)`C. `K^(o+) gt Cl`D. `P^(3+) gt P^(5+)`
Answer» Correct Answer - D Size of less positive charge ion is greater than that of having more positive ion.

Discussion

47.	Calculate the screening constants of members of the `2nd` period for valency electrons.
Answer» Correct Answer - `Li = 1.7,Be 2.05, B = 2.40, C = 2.75, N = 3.10, O = 3.45, F = 3.80, Ne = 4.15` a. `Li(Z=3), 1s^(2) 2s^(1)` `(sigma)_(2s^(1))=[(0.85xx "No, of electrons in" (n-1)th "shell"]` `=0.85xx2=1.7` b. `Be(Z=4)=1s^(2) 2s^(2)` (one electron is left in `2s` shell) `(sigma)_(2s^(1))=[["("0.35 xx "No . of electrons left in the nth"],["shell i.e. in 2 s orbital"+"("0.85 xx"No . of"],["electron in (n-1)th shell"]]` `=(0.35xx1)+(2xx0.85)=2.05` c. `B(Z=5)=1s^(2) 2s^(2) 2p^(1)` `(sigma)_(2p^(1))=[["("0.35xx "No. of electrons left in"],["nth shell"+"("0.85 xx "No. of electrons"],["in (n-1)th shell"]]` `=(0.35xx2)+(0.85xx2)=2.40` d. `C(Z=6)=1s^(2) 2s^(2) 2p^(2)` (one electron in `2p` and `2` electrons in `2s` are left) `(sigma)_(2p^(1))=[["("0.35xx "No. of electrons left in nth shell +"],["("0.85 xx "No. of electrons in (n-1)th shell"]]` `=(0.35xx3)+(0.85xx2)=2.75` e. `N(Z=7)=1s^(2) 2s^(2) 2p^(3)` (two electrons in each `2p` and `2s` are left) e. `N(Z = 7)=1s^(2) 2s^(2) 2p^(3)` (two electrons in each `2p` and `2s` are left) `(sigma)_(2p^(2)) = [["("0.35 xx "No. of electrons left in nth shell+"],["(" 0.85 xx "No. of electrons in (n-1)th shell" ")"]]` `= (0.35 xx 4)+(0.85 xx 2) = 3.10` f. `O(Z=8) = 1s^(2) 2s^(2) 2p^(4)` (three electrons in `2p` and two electrons in `2s` are left). `(sigma)_(2p^(3)) = [["("0.35 xx "No. of electrons left in nth shell+"],["(" 0.85 xx "No. of electrons in (n-1)th shell" ")"]]` `= (0.35 xx 5)+(0.85 xx 2) = 3.45` g. `F(Z = 9) = 1s^(2) 2s^(2) 2p^(5)` (four electrons in `2p` and two electrons in `2s` are left). `(sigma)_(2p^(4)) = [["("0.35 xx "No. of electrons left in nth shell+"],["(" 0.85 xx "No. of electrons in (n-1)th shell" ")"]]` `=(0.35 xx 3)+(0.85 xx 2) = 2.75` h. `Ne(Z = 10) = 1s^(2) 2s^(2) 2p^(6)` (five electrons in `2p` and two electrons in `2s` are left). `(sigma)_(2p^(5)) = [["("0.35 xx "No. of electrons left in nth shell+"],["(" 0.85 xx "No. of electrons in (n-1)th shell" ")"]]` `= (0.35 xx 7) + (0.85 xx 2) = 4.15`

Discussion

48.	The group all members of which are in gaseous state under ordinary conditions is a `" ________"` group.
Answer» Correct Answer - Zero or `18` Zero or `18`

Discussion

49.	Among the radioactive elements, which is a liquid element?
Answer» Correct Answer - `Fr` Francium `(Fr)`

Discussion

50.	The elements which are radioactive and have been named after the name of planet areA. `Hg` (Mercury) (Hergentium)B. `Np` (Neptunium)C. `Pu` (Plutonium)D. `Ra`(Radium)
Answer» Correct Answer - B::C a. `Hg` (derived from the word Hergentum) is not radioactive. b. `Np` (Neptunium, `Z = 93`) is radioactive and named after the name of a planet. c. `Pu` (Plutonium, `Z = 94`) is also radioactive and named after the name of a planet. d. `Ra` (Radium, `Z = 88`) is also radioactive but not named after the name of any planet.

Discussion

Explore topic-wise InterviewSolutions in .

The trivalent ion having size in lanthanide series isA. `Ti`B. `Zr`C. `Hf`D. `La`

The multinucleate slimy mass of protoplasm which forms the body of slime moulds is called asA. plasmodiumB. myxamoebaC. sprocytesqD. periplasmodium.

During unfavourable conditions, slime moulds forms fruiting body after the differentation ofA. AscocarpB. BasidiocarpC. PlasmodiumD. Gemmule

(A) What are protozoans ? Give two examples each of free-living and parasitic forms. (B) What are slime-moulds ? How are they different from true fungi ? Give one example each of the cellular and ascellular slime and moulds. (C) Differentiate between ascus and basidium.

The IUPAC name of the compound `CH_(3)CH=CHCH=CHC=C CH_(3)` is:A. Octa-4,6-dine-2-yneB. Octa-2,4-diene-6-yneC. Oct-2-yne-4,6-dieneD. Oct-6-yne-2,4-diene

Which one is the smallest organism capable of autonomous growth and reproduction Or Which among the following are the smallest living cells, known without a definite cell wall, pathogenic to plants as well as animals and can survive without oxygenA. PseudomonasB. MycoplasmaC. NostocD. Bacillus

The shape of the cocci bacteria isA. Rod shapedB. SphericalC. Comma shapedD. Spiral

Cell walls of diatoms are indestructibel because of the presence ofA. `CaCO_(3)`B. SilicaC. ChitinD. All of the above

The wonder drug, penicillin is extracted from which of the following species of penicillium?A. Penicillium notatumB. P. chrysogenumC. Both a and bD. None of these

What are the physiological processes that are regulated by ethyline in plants ?

Give a brief account of Bt. Cotton.

Define transformation in Griffith's experiment. Discuss how it helps in the Identification of DNA as genetic material.

What is ICTV ? How are viruses named ?

With the help of an example, differentiate between incomplete dominance and Co-dominance.

What is common about Trypanosoma, Noctiluca Monocystis and GiardiaA. These are all parasitesB. These are all uncellular protistsC. They have flagellaD. They produce spores

Which one of the following is wrong for fungiA. They are heterotrophicB. They are both unicellular and multicellularC. They are eukaryoticD. All fungi possess a purely cellulosic cell wall

(a) 3,6,-Bis-(1,1-dimetylethyl) octane (b) Octa-4,7-dien-2-amine (c) 3-Hydroxymethyl cyclohexanecarboxylicacid (d) 2-Nitropent-3-ene If the given IUPAC name is correct the write 1 and if it is wrong then write 2

Write the correct IUPAC name of 2-amino-1, 3,7-trihydroxy hept-4-ene-1, 7-dione?A. 6-amino-5-hydroxy hept-3-ene-1,7-dioicacidB. 2-amino-3-hydroxy hept-4-ene-1,-7,dioicacidC. 6-amino-5-hydroxy hept-4-ene-1-,7-dicarboxylic acidD. 6-amino-5-hydroxy hept-3-ene-1,7-dicarboxylic acid

One among the following is the correct IUPAC name for N-ethyl-1-amino methanal.A. N-formyl aminoethaneB. N-ethyl formyl amineC. N-ethyl methanamideD. Ethylamino methanal

The correct name of 1-methoxy-1-ethyl-2-methyl propane is:A. 2-methyl-3-methoxy pentaneB. 1-methoxy-1-iso-propyl methaneC. 3-methoxy-4-methylpentaneD. 3-methoxy-2-methyl pentane

Observe the following compounds answer the following question Which one of these three compounds contains maximum number of side chains in its principal carbon chain ?A. IB. IIC. IIID. Equal in I and II

Find out number of C-atoms present in principal chain according to IUPAC rules. [Write answer of part (a),(b),(c) and (d) in the same order and present the four digit number as answer in OMR sheet. For example. ]

Write D.B.E (Double bond equivalent) for the following compound: [Write answer of part (a),(b),(c) and (d) in the same order and present the four digit number as answer in OMR sheet. For example.]

Observe the following compounds answer the following question Which one of these three compounds contains maximum number of quarternary carbon atoms `(4^(@))` ?A. IB. IIC. IID. Equal in II and III

Number of secondary carbon atoms present in benzene, o-xylene, touene and mesitylene are respectively:

Viral capsid is made ofA. ProteinB. Protein + Nucleic acidC. Nucleic acidD. Either DNA or RNA

The capsid of viruses protects theA. NucleoproteinB. CapsomereC. Nucleic acidD. Protien

Diatomaceous earth is the deposition of cell wall in their habitat by Diatoms takes overA. Milions of yearsB. Billions of yearsC. Trillions of yearsD. Thousands of years

`IE_(1), IE_(2)` and `IE_(3)` values are `100, 150` and `1500 eV` respectively. The element can beA. `Na`B. `B`C. `Be`D. `F`

Which of the following van der Waals radii is the largest ?A. `Ne`B. `Cl`C. `O`D. `F`

Compounds that formally contain `Pb^(4+)` are easily reduced to `Pb^(+)` .The stability of the lower oxidation state is due to `"…….."` .

What is the atomic number of the element present in the second period anf group 15.

The angle between two covalent bonds is maximum for `(CH_(4), H_(2)O, CO_(2))` `"______"` .

Ionic bonds are non-directional while covalent bonds are directional.

As the `s` character of the hybrid orbital decreases, the `EN` increases.

The screening effect of `d`-electron isA. Equal to `p`-electronB. Much more than `p`-electronC. Same as `f`-electronsD. Less than `p`-electrons

Ionic radii of :A. `.^(35)Cl^(ɵ) gt Cl(ɵ)`B. `Mn^(7+) gt Ti^(4+)`C. `K^(o+) gt Cl`D. `P^(3+) gt P^(5+)`

Calculate the screening constants of members of the `2nd` period for valency electrons.

The group all members of which are in gaseous state under ordinary conditions is a `" ________"` group.

Among the radioactive elements, which is a liquid element?

The elements which are radioactive and have been named after the name of planet areA. `Hg` (Mercury) (Hergentium)B. `Np` (Neptunium)C. `Pu` (Plutonium)D. `Ra`(Radium)