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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.
| 901. |
Suggest a technique to a researcher who needs to separate fragments of DNA. |
| Answer» fragments of DNA can be separated using the technique called Gel Electrophrisis here ,DNA molecules are separated using electric filed DNA beng nagatively charged moves towards anode through a matrix. | |
| 902. |
Write the name of the organism that is referred to as the "Terror of Bengal". |
| Answer» Water hyacinth (Eichhornia Crassipes) is reffered to as the "Terror of Bengal". | |
| 903. |
Mention how does DNA polymorphism arise in a population. |
| Answer» DNA polymorphism is a genetic variant that appears in at least `1%` of a population. It can be introduced in a population by mutation and genetic drift. | |
| 904. |
Which of the following pairs , will have greater conduction ? (i) `0.1`M acetic acid solution or 1 M acetic acid solution . (ii) `0.1`M NaCl solution at `25^(@)C` and `0.1` M NaCl solution at `50^(@)C`. |
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Answer» (i) `0.1` M acetic acid because with dilution dissociation increases and therefore , conductivity increases. (ii) `0.1` M NaCl solution at `50^(@)`C because with increase in temperature , ionic mobilities of strong electrolytes increase. |
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| 905. |
A reaction is of second order with respect to its reactant. How will its reaction rate be affected if the concentration of the ractant is (i) doubled (ii) reduced to half ? |
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Answer» Since Rate = K `[A]^(2)` Let [A] = a `" " therefore " " Rate = Ka^(2)…. (1)` (i) `" "` If [A] = 2a `" " therefore " "` Rate = K `(2a^(2)) = 4 Ka^(2) = 4 ` times (ii) `" "` If [A] = `(a)/(2) " " therefore " " Rate = K ((a)/(2))^(2) = (1)/(4)` `Ka^(2) = (1)/(4)` th |
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| 906. |
A reactions is of second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is reduced to half? |
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Answer» According to given information Rate = K `["Reactant"]^(2)` If concentration of reactant becomes twice . (i) `(R_("new"))/(R_("old")) = ([2R]^(2))/([R]^(2)) = 4` times increases . If concentration of reactant becomes half . (ii) `(R_("new"))/(R_("old")) = [(R)/(2R)]^(2) = (1)/(4)` times increases. |
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| 907. |
Define ionisation energy. What is the value for a hydrogen atom? |
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Answer» Ionisation energy : The minimum energy required to free the electron from the ground state of the hydrogen atom is called ionisation energy of the hydrogen atom . It is 13.6 eV. |
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| 908. |
Two different wires X and Y of same diameter but different materials are joined in series across a battery. If the number density of electrons in X is twice that in Y, find the ratio of drift velocity of electrons in the two wires . |
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Answer» The current I through the wire is related with drift velocity vd is given by I = n A `e v_(d)`. Since two wires X and Y of same diameter but different materials are joined in series . `therefore " " I_(x) = I_(y)` `therefore " " n_(x) A ev_(dx) = n_y Ae v_(dy)` `2 n_(y) A e v_(dx) = n_(y) Ae v_(dy) implies (v_(dx))/(v_(dy)) = (1)/(2) v_(dx) : v_(dy) = 1 : 2` |
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| 909. |
Complete the following chemical equations : (i) `Cr_2O_6^(2-)+6Fe^(2+)+14H^+to` (ii) `2CrO_4^(2-)+2H^+to` (iii) `2MnO_4^(-)+5C_2O_4^(2-)+16H^+to` |
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Answer» (i) `6e^(2+)+Cr_2O_7^(2-)+14H^+to2Cr^(3+)+6Fe^(3+)+7H_2O` (ii) `2CrO_4^(2-)+2H^+toCr_2O_7^(2-)+H_2O` (iii) `2MnO_4^(-)+5C_2O_4^(2-)+16H^(+)to2Mn^(2+)+10CO_2+8H_2O` |
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| 910. |
(i) Complete the following equations : (a) `2MnO_(4)^(-)+ 5SO_(3)^(2-)+6H^(+)to` (b) `Cr_2O_(7)^(2-)+6Fe^(2+)+14H^(+)to` (ii) Based on the data, arrange `Fe^(2+), Mn^(2+) and Cr^(2+)` in the increasing order of stability of `+2` oxdiation state . `E^(@) Cr^(3+)//Cr^(2+)=-0.4V` `E^(@) Mn^(3+)//Mn^(2+)=01.5V` `E^(@) Fe^(3+)//Fe^(2+)=+08V` |
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Answer» (i)(a) `2MnO_4^(-)+5SO_(3)^(2-)+6H^(+)to 2Mn^(2+)+3H_2O+5SO_(4)^(2-)` (b) `Cr_2O_7^(2-)to 6Fe^(2+)+14H^(+)to 2Cr^(3+)+6Fe^(3+)+7H_2O` (ii) `E^(@) Cr^(3+)//Cr^(2+)=-0.4V` `E^(@) Mn^(3+)//Mn^(2+)=01.5V` `E^(@) Fe^(3+)//Fe^(2+)=+08V` Increasing order of stability of `+2` oxidation state is : |
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| 911. |
Explain aminoacylation of t-RNA. |
| Answer» Amino acid are activated in the presence of ATP and linked to their congnate t-RNA, a process called charging of t-RNA or aminoacylation of t-RNA. | |
| 912. |
List the different parts of the human oviduct through which the ovum travels till it meets the sperm for fertilization. |
| Answer» Ovum passes through infundibulum and crosses ampulla to reach the ampullary-isthmic junction of the fallopian tube where it meets the sperm for fertilisation. | |
| 913. |
Differentiate between the genetic codes given below : (a) Unambiguous and Universal (b) Degenerate and Initiator : |
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Answer» (a) One codon codes for only one amino acid , hence , it is unambiguous and specific. The code is nearly universal : For example , fron bacteria to human UUU would code for Phenylalanine (Phe). Some exception to this rule have been found in mitochondrial codons and in some protozoans. Some amino acids are coded by more than one codon hence , the code is degenerate .AUG has dual functions It codes for Methionie (met) and it also act as initiator codon. |
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| 914. |
List any four techniques where the principle of ex-situ conversion of biodiversity has been employed. |
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Answer» Four techiques are : (i) Sperm Bank /Seed bank (ii) Cryopreservation techniques (iii) In-Vitro fertilization , (iv) Tissue Culture . |
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| 915. |
List six advantages of "ex-situ" approach to conservation of biodiversity. |
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Answer» Ex-situ conservation - It is a conservation of selected threatened plant and animal species in places outside their natural habital. Population is conserved under simulated conditions that closelly resemble their natural habit. The gametes of endangered species can be preserved by methods such as cryo-preservation. Whenever required, gametes can be fertilised invitro and transfer to the suruogate to produce endangered animals. Cryo-preserved plant material whenever required can be propagated through tissue culture methods. Seeds of rare varieties can be cryo-preserved. The commonly used ex-situ conservation methods are sperm bank/seed bank, cryo-preservation, invitro fertilization, tissue culture, botanical gardens, zoological parks, wild life safari. |
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| 916. |
Name two commonly used bioreactors. State the importance of using a bioreactor. |
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Answer» The two most commonly used bioreactors are simple stirred-tank bioreactor and sparged stirred -tank bioreactors. The importance of using bioreactors is as follows : (i) It provides large volume for cultures. Thus, products are obtained in high quantity. (ii) It provides optimal conditions like temperatures and pH for growth of desired product. |
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| 917. |
How do mycorrhizae act as biofertilizers? Explain. Name a genus of fungi that forms a mycorrhizal association with plants. |
| Answer» Mycorrhizae are associations between fungi and the roots of higher plants. The fungi help the plant in the absorption of essential nutrients from the soil while the plant in turn provides the fungi with energy yielding carbohydrates. Boletus is the soil fungus that forms a mycorrhizal association with plants. | |
| 918. |
Name the source of streptokinase. How does this bioreactor molecule function in our body? |
| Answer» Streptokinase produced by the bacterium streptococcus and modified by genetic engineering is used as a clot buster for removing clots from the blood vessels of patients who have undergone myocardial infection leading to heart attack. | |
| 919. |
Name the bacterium that causes typhoid. Mention two diagnostic symptoms. How is this disease transmitted to others? |
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Answer» Salmonella Tpr is a pathogenic bacterium which causes typhoid in human beings. Its common symptoms are (i) Constipation, loss of appetite. (ii) High fever upto 39° to 40° It is spread by contaminated food and water, it generally enters in our small intesine and migrates to other organs through blood. |
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| 920. |
How are productivity, gross productivity, net primary productivity, and secondary productivity interrelated? |
| Answer» Productivity: It is the rate of production of biomass per unit area over a period of time by plants during photosynthesis. Gross Productivity: It is the rate of production of organic matter by green plants per unit time per unit area. Net Primary Productivity: Net primary productivity is the available biomass for the consumption of heterotrophs. Net primary Productivity: Gross Primary Productivity-Respiratory losses, `N_(PP) = G_(PP) - R.` Secondary Productivity: It is the rate of formation of new organic matter by consumers. | |
| 921. |
Write the equation that helps in deriving the net primary productivity of an ecosystem. |
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Answer» Net Primary Productivity (NPP) of an ecosystem can be derived using the following equation : NPP= GPP-R = (Gross Primary Productivity)-(Respiratory Losses) |
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| 922. |
(i) `[-CH_(2)-overset(CH_(3))overset(|)(C)H-]_(n)` a homopolymer or copolymer ? Give reason. (ii) which the monomers of the following polymer. (iii) What is the role of Sulphur in vulcanization of rubber? |
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Answer» `[-CH_(2)-overset(CH_3)overset(|)(CH)-]_(n)` is a homopolymer because it is polymerised by single monomer. (iii) See Q.20 (i) Paper 2016, Outside Delhi, Set-I, [Page 359]. |
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| 923. |
Write the name of linkage joining two amino acids. |
| Answer» Amino acid can combined to form amides called peptide. The - CO-NH-bond or linkage is called the peptide bond. Formation of a peptide bond by a loss of water from 2-amino acid. | |
| 924. |
Why does a doctor administer tetanus antitoxin and not a tetanus vaccine to a child injured in a roadside accident with a bleeding wound? Explain. |
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Answer» If a person is infected with some deadly microbes to which quick immune response is required as is tetanus, we need to directly inject the preformed antibodies or antitoxin. Even in cases of snake bites, the.injection which is given to the patients contain preformed antibodies against the snake venom. This type of immunisation is called passive immunisation. |
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| 925. |
What is the pathogenic property of baculovirus, used as a biological agents ? Name the genus of these organisms. |
| Answer» Baculoviruses are pathogenus that attack insects and other arthopods. The majority of baculoviruses used as biological control agents are in the genes "Nucleopolyhedrovirus". | |
| 926. |
Name the chief ores of aluminium and zinc. |
| Answer» Aluminum =bauxite Zinc=sphalerite. | |
| 927. |
Write the IUPAC name of the complex `[Cr(NH_(3))_(4)Cl_(2)]^(+)` . What type of isomerism does it exhibit ? |
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Answer» The IUPAC name of the complex `[Cr(NH_(3))_(4)Cl_(2)]^(+) `is Tetraamminedichloridochromium (III) ion. This complex exhibits geometrical isomerism. `[Cr(NH_(3))_(4)Cl_(2)]^(+) is a [MA_(4)B_(2)]` type of complex, in which the two chloride ligands may be oridented cis ans trans to each other. See also Q. 23, Set - III, Outside Delhi - 2012. |
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| 928. |
(i) What type of isomerism is shown by the complex `[Cr(H_(2)O)_(6)]Cl_(3)`? (ii) On the basis of crystal field theory, write the electronic configuration for `d^(4)` ion if `Delta_(0)gtP.` (iii) Write the hybridization and shape of `[CoF_(6)]^(3-)`. (Atomic number of Co=27) |
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Answer» (i) Solvate Isomerism. (ii) `[Cr(H_(2)O)_(6)]Cl_(3)` Violet `[Cr(H_(2)O)_(5)Cl]Cl_(2)-H_(2)O` (Blue green) `[Cr(H_(2)O)_(4)Cl_(2)]Cl. 2H_(2)O` Dark green (iii) `[CoF_(6)]^(3-)` is an outer orbital or high spin complex involving `Sp^(3)d^(2)` hybridisation and octahedrol shape. |
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| 929. |
(a) Account for the following : (i) Electrophilic substitution reactions in haloarenes occur slowly. (ii) Haloalkanes, through polar, are insoluble in water. (b) Arrange the following compounds in increasing order of reactivity towards `S_(N)2` displacement: 2-Bromo-2-Methylbutane, 1-Bromopentane, 2-Bromopentane |
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Answer» (a) (i) Halogen atoms of haloarenes has -I effect which has some tendency to withdraw electrons from benzene ring. As a result, the ring gets somewhat deactivated as compared to benzene and hence electrophilic substituation reactions in haloarenes occur slowly. (ii) Haloalkanes, though polar, are insoluble oon water because less energy is released when new attractins are set up between the haloalkanes and water molecules because these are not as strong as the hydrogen bonds already existing between `H_(2)O` molecules. As a result, holoalkanes are insoluble in water. (b) `{:(" "CH_(3)),(" "|),(H_(3)C-HC-C-CH_(3)H_(3)CH_(2)CH_(2)C-Br","H_(3)CH_(2)CH_(2)CH_(2)C-CHCH_(3)),(""" |" " "(B)(1^(@))" "|" "(C)(2^(@))),(" "Br " "Br),(" "(A)(3^(@))):}` For `SN_(2)` mechanism, order of reactivity is `1^(@)gt2^(@)gt3^(@)` i.e ` BgtCgtA` i.e 1-Bromopentane `gt2-` Bromopentane `gt2-`Bromo-2-Methylbutane. |
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| 930. |
Define modulation index. Why is its value kept in practies, less than one ? A carrier wave of frequency 1.5 MHz and amplitude 50 V is moduclated by a sinusoidal wave of frequency 10KHz producing 50% ampitude moducation. Calculate the amplitude of the AM wave and frequencies of the side bands produced. |
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Answer» Frequency of carrier, ` " " f_(c)= 1.5 MHz = 1500 KHz ` Frequnency of signals,` f_(s) = 10 KHz` Modulation factor M = 50% Amplitude of carrier, ` E_(c) = 50V` The side and ferquency are : ` f_(c) -f_(s) and f_(c) +f_(s)` (1500 -10) and (1500 +5) (1490 Hz) and ( 1505 KHz) Amplitude of each sideband term, ` = (mE_(c))/2= ( 0.5 xx 50)/2 = 1.25 ` V. |
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| 931. |
State clearly how an unpolarised light gets linearly polarised when passed through a polaroid. Unpolarised light intensity `I_(0)` is incident ` P_(1)` which is kept near another polaroid `P_(2)` whose pass axis is parallel to that of `P_(1)`. How will the intensities of light, `I_(1) and I_(2)` , transmitted by the polarids `P_(1) and P_(2)` respectively, change on rotating `P_(1)`[ without disturbing `P_(2)` ? |
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Answer» The light having vibrations of electric field vector in all possible directions perpendicular to the direction of wave progation is called ordinary light. The light having vibrations field vectors in only one direaction perpendicular to the direction of progation of light is called plane polarised light. (ii) The value of the sound intensity inversely squared with increasing distance from sound source, i.e with ` 1/r^(2)` ` I alpha 1/(r^(2)) = I_(2)/I_(1) = r_(1)^(2)/r_(1)^(2) Rightarrow I_(2) =I_(1) (r_(1)/r_(2))^(2)` |
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| 932. |
Name the type of waves which are used for line of sight (LOS) communication. What is the range of their frequencies? A transmitting antenna at the top of a tower has a height of 20m and the height of the receiving antenna is 45m. Calculate the maximum distance between them for satisfactory communication in LOS mode. (Radius of the Earth `= 6.4 xx 10^(6)` m) |
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Answer» Space waves are used for the line of sight (LOS) communication. The range of their frequencies is 40 MHz and above. We have, height of transmitting antenna, `" "h_(T) = 20m` and height of receiving antenna, `" "h_(R) = 45 m`. Then, maximum distance between the two antennas, `dm = sqrt(2Rh_(T))+sqrt(2Rh_(R))` `dm = sqrt(2 xx 6.4 xx 10^(6) xx 20)+sqrt(2 xx 6.4 xx 10^(6) xx 4.5)` `" "=2 xx 8 xx 10^(3) + 3 xx 8 xx 10^(3) = 40 km`. Thus, the maximum distance between the antennas is 40 km. |
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| 933. |
What does the term LOS communication mean? Name the types of waves that are used for this communication. Which of the two height of tansmitting antenna and height of receiving antenna can effect the range over which this mode of communication remains effective? |
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Answer» LOS means Line-of-sight communication space waves. These are used for line of sight (LOS) communication. If `h_(T) and h_(R)` be the heights of transmitting antenna and the receiving antenna respectively, then the maximum Line-of-sight distance `d_(M)` between the two antennas is given by `d_(M)=sqrt(2Rh_(T))+sqrt(2Rh_(R))` where `R rarr` radius of the earth. The height of transmitting antenna can effect the communication system but the height of receiving antenna cannot effect the communication system. |
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| 934. |
A parallel beam of light of wavelength 600 nm is incident normally on a slit of width d. If the distance between the slits and the screen is 0.8 m and the distance of `2^(nd)` order maximum from the centre of the screen is 15 mm. The width of the slit is |
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Answer» Difference between interference and diffraction : Interference is due to superposition of two distinct waves coming from two coherent sources and diffraction is produced as a result of superposition of the secondary wavelets coming from different parts of the same wavefront. Given : `lambda = 600 nm = 6xx10^(-7)m, D=0.8 m, y_(2)=15xx10^(-3)m` To calculation :` y_(2)=(5)/(2)xx (lambda D)/(d)` `rArr d= (5)/(2) xx (6xx10^(-7)xx0.8)/(15xx10^(-3)) :. ` Distance, `d=8xx10^(-5)=80 mu m`. |
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| 935. |
What chemcial principal is involved in choosing a reducing agent for getting the metal from its oxide ore? Consider the metal oxides `Al_(2)O_(3) and Fe_(2)O_(3)` and justify and choice of reducing agent in each case. |
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Answer» (i) Thermodynamci factor helps us in choosing a suitable reducing agent for the reducing agent for the reducation of a particular metal oxide to the metallic state (ii) Any metal will reduce the oxides of other metals which lie aboe it in the Ellingham diagram because the standard free energy change `(Delta_(r)G^(@))` of the combined redox reaction will be-ve by an amount equal to the differenc in `Delta_(f)G^(@)` fo two metal oxides. Thus, both Al and Zn can reduces `Fe_(2)O_(3)` to Fe but Fe but Fe cannot reduce `Al_(2)O_(3)` to Al. |
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| 936. |
Define the following terms: (i) Sorption (ii) Tyndall effect (iii) Electrophoresis |
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Answer» (i) Sorption is a physical and chemical process by which one substance becomes attached to naother specific cases of sorption are treated in the following way: Asborption the incorporation of a substance in one state into ajotehr of different state (e.g liquied being absorbed by a solid or gases being absorbed by liquid ) Adsorption the physical adherence or bonding of ions and molecules on to the surface of anothe phase (e.g reagents absorbed to solid catalyst surface) Ion exchange an exchange of ions between two electrolytes or between an electorlyte solution and a complex (ii) Tydall effect : The phenomeonon of scattering of light by colloidal particles as a result of which the path of beam becomes visibles is called tyndall effect (iii) Eelctrophoroesis : when electric potential is applied across two platinum electrodes dipping in a colloidal soluton the collidal particlus move towards one or the other electrode .The movement of colloidal particles under an appied electric potential is called electrophoresis |
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| 937. |
How many atoms constitute one unit cell of a face-centered cubic crystal ? |
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Answer» Number of atoms in one face centered cubic unit cell can be determined from the number of contributed from the face and the corners of the unit cell as : `(8" corners "xx1/8" atom per corner"=8xx1/8=1" atom")+6" faces"xx1/2" atom per unitface"=6xx1/2=3" atoms")` |
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| 938. |
What are following ? Give one example of each (i) Sweetening agents (ii) Food preservatives (iii) Antibiotics |
| Answer» These are the drugs used to kill or stop growth of micro organism .These drug are produced by micro organisms [expenicillin] | |
| 939. |
Write the name and structure of the monomer of each of the following polymers : (i)Neoprene (ii) Buna-S(iii) Teflon |
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Answer» (i)Neoprene : Monomer is Chorophene `CH_(2)=C(CI)-CH=CH_(2)` (ii) Buna-S Monomers are 1,3- butadiene and styrene `CH_(2)=CH-CH=CH_(2)and C_(6)H_(5)-CH=CH_(2)` (iii) Teflon : Monomer is Tetrafluoroethene `CF_(2)=CF_(2)` |
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| 940. |
Give names of the monomers of the following polymers: (i) Neoprene (ii) Polystyrene (iii) Polypropene |
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Answer» (i) Neoprene -Chloroprene-`H_(2)C=overset(CI)overset(|)C-HC=CH_(2)` (ii) Polystyrene: styrene-`CH_(2)=CH-C_(6)H_(5)` (iii) polypropene: propene-`H_(3)C-HC=CH_(2)` |
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| 941. |
Name any two vertabrate body parts that are homologous to humen forelimbs. |
| Answer» Forelimb of frog, Flipper of whale and Wing of birds are holologous to humen forelimbs. | |
| 942. |
(a) Which solution is used for the leaching of silver metal in the presence of air in the metallurgy of silver ? (b) Out of C and CO, which is a better reducing agent at the lower temperature rangein the blast furnace to extract iron form the oxide ore ? |
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Answer» (a) Dilute solution of NaCl and KCN is used for leaching of silver metal in the presence of air in the metallurgy of silver. (b) Out of C and CO, Cois a better reducing agent at the lower temperature range because `DeltaG_((CO,CO_2))lt DeltaG_((Fe,FeO))`. So, CO will reduce FeO, will its self be oxidised to `CO_2`. |
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| 943. |
(a) (i) Complete the following chemical equations : (i)`NaOH_((aq))+Cl_(2(g)) rightarrow` (Hot and conc.) (ii) `Xef_(6)(s)+H_(2)O(l) rightarrow` (b) How would you account for the following ? (i) The value of electron gain enthalpy with negative sign for sulphur is higher than that for oxygen. (ii) `NF_(3)` is an exothermic compound but `NCl_(3)` is endothermic compound. (iii)`ClF_(3)` molecule has a T-shaped structure and not a trigonal planar one. |
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Answer» (a)(i) `6NaOH +3Cl_(2) rightarrow 5 NaCl +NaCl O_(3)+ 3H_(2)O` (ii)` Xe F_(6)+ H_(2)O rightarrow Xe OF_(4)+2HF` (b) (i) Because of larger size of sulphur atom than oxygen atom. (ii) Because bond energy of `F_(2)` is lowerthan `Cl_(2)` and N-F bond is smaller and stronger than N-Cl bond. (iii) Because it has `dsp^(3)` hybridization. |
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| 944. |
Arrange the following in order of property indicated for each set: `(i) H_(2)O,H_(2)S,H_(2)Se,H_(2)Te` - Increasing acidic character (ii) `HF,HCl,HBr,HI` - decreasing bond enthalpy |
| Answer» (i) `H_(2)OltH_(2)SltH_(2)SeltH_(2)Te` (ii) `HFltHClltHBrltHI` | |
| 945. |
Write balanced chemical equations for the following processes: (i) `XeF_(2)`, undergoes hydrolysis. (ii) `MnO_(2)`, is heated with conc. HCI |
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Answer» (i) `XeF_(2)` is hydrolysed to give Xe, HF and `O_(2)` `2XeF_(2)(s)+2H_(2)O(l)to2Xe(g)+4HF(aq)+O_(2)(g)` (ii) When maganese dioxide is heated with conc. hydrochloric acid it gives `MnO_(2)+4HCltoMnCl_(2)+Cl_(2)+2H_(2)O` |
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| 946. |
Draw the structure of the following (i) `HClO_3` (ii) `H_2S_2O_8` ( b) Give reasons for the following (i) Above 1000 K sulphur shows paramagnetism. (ii)Although electron gain enthalpy of fluorine is less negative of chlorine ,yet flourine is a better oxidising agent than chlorine . (iii) In solid state `PCl_5` exists as an ionic compound |
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Answer» SeeQ.29 (or) (ii) Paper - 2009 ,Outside Delhi ,Set-I[Page 57] See Q.29 (i) Paper - 2009 Outside Delhi ,Set-I [Page 56] ( b) Refer Q.29(i) Paper - 2008 ,Outside Delhi Set-I[page 22] (ii) Refer Q.25 (i) Paper - 2009 ,Outside Delhi Set-I [page 90] (iii) Refer Q.24 (i) Paper - 2016 ,Outside Delhi Set-I[page 346] |
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| 947. |
How would you account for the following? (i)The atomic radii of the metals of the third (5d) series of trasition elements are virtually the same as those of the corresponding members of the second (4d) series. (ii)The `E^(@)` value for the `Mn^(3+)//Mn^(2+)` couple is much more positive than that for `Cr^(3+)//Cr^(2+)` couple or `Fe^(3+)//Fe^(2+)` couple. (iii)The highest oxidation state of a metal is exhibited in itsoxide or fluoride. |
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Answer» (i) Due to Lanthanoid Contraction. (ii) Due to stable-filled `3d^(5)` configuration of `Mn(2+)` or 3rd ionisation enthalpy of Mn. (iii) Because oxygen or Fluorine is highly electronegative element and small size. |
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| 948. |
Give reason: (i) Transition metals show variable oxidation states. (ii) Actinoids show wide range of oxidation states. |
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Answer» (i) Transition elements have their valence electrons in `(n-1)` d and ns orbitals. Since there is very little difference in the energies of these orbitals, both energy levels can be used for bond formation and thus, give variable oxidation states. (ii) The energy gap between 5f, 6d and 7s subshells is very small and hence all the electrons present in these subshells can participate in bonding. Thus, Actinoids show wide range of oxidation states. |
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| 949. |
Write the definitions of the following : (i) Racemic mixture (ii) Chiral compound Or What are ambident nucleophiles ? Give an example. |
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Answer» (i) Racemic mixture : A mixture of equal amounts of two enantiomers which are non-superimposable mirror images of each other is called racemic mixture. (ii) Chiral compound : Compounds which are non-superimposable on their mirror images are called chiral compound. Ambidentate ligand : Unidentate ligands containing more than one coordinating atoms are called ambidentate ligands. Example : `CN^(-)` having Carbon and Nitrogen coordinating sites. |
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| 950. |
Write the types of sex-determination mechanisms the following crosses show. Give an example of each type. i) Female XY with Male XO ii) Female ZW with male ZZ |
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Answer» i) The type of sex determination mechanism shown in female XX with male XO is male heterogamety. Example- Human ii) The type of sex determination mechanism shown in female ZW with male ZZ is female heterogamety. Example- Birds. |
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