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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Energy equal to the mass of one electron is |
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Answer» `8.2 XX 10^(-7)` erg `E = 9.1 xx 10^(-31) xx (3x 10^8 )^2 = 8.2 xx 10^(-7) ` erg |
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| 2. |
Energy associated with which of the following waves is not quantised? (i) Electromagnetic wave (ii) Matter wave. |
| Answer» SOLUTION :MATTER WAVES. | |
| 3. |
Energy barrier between staggered and eclipsed ethane is |
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Answer» 0.6 Kcal/mole |
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| 4. |
End products of the following sequence of reaction isunderset((ii)H^(+),Delta)overset((i)NaOI,Delta)toA+B, A& B are |
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Answer»
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| 5. |
End product of the following sequence of reaction isCH-=CHoverset(CH_(3)MgBr)tooverset(CO_(2)//H_(3)O^(+))tooverset(HgSO_(4)//H_(2)SO_(4))tooverset(Ag_(2)O)underset(Delta)to |
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Answer»
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| 6. |
Enamel on teeth is |
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Answer» `CaCO_(3)` |
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| 7. |
Formula of enamel on teeth is |
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Answer» `CaCO_(3)` |
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| 8. |
E_(n) = total energy , l_(n) angular momentum K_(n) = K.E., V_(n) = P.E., T_(n) = time period , r_(n)= radius of nth orbit |
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Answer» |
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| 9. |
{0} is a set contains the element…….. |
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Answer» `2E_1 - E_2` |
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| 10. |
EN of the element (A) is E_1 and IP is E_2. Hence EA will be |
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Answer» `2E_1 - E_2` |
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| 11. |
Emulsions of polyvinyl acetate are used in : |
| Answer» Answer :B | |
| 12. |
Empty space in ccp lattice is |
| Answer» Solution :In ccp lattice 74% of the space is occupied by spheres and 26% is empty. | |
| 13. |
Empty apace in ccp lattice is |
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Answer» `26%` |
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| 14. |
Empirical formula is the simplest formula of the compound which gives the atomic ratio of various elements present in one molecule of the compound. However the molecular formula ofthe compound gives the number of atoms of various elements present in one molecule of the compound. Molecular formula = (Empirical formula) xxn Which pair of species have same percentage composition of carbon? |
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Answer» HF `((10)/(M))` = MOLES = `(5.6)/(22.4)` `implies M=40=(HF)_(n)=nxx20impliesn=2` |
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| 15. |
Empirical Formulafor Molecular Formula. |
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Answer» If the mass per cent of various elements present in a compound is known, its empirical formula can be determined. Molecular formula can futher be OBTAINED if the molar mass is known. |
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| 16. |
Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as v = 3.29 xx 10^(15) (Hz) [1//3^(2) - 1//n^(2)] Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum. |
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Answer» Solution :`v = (c)/(lamda) = (3.0 xx 10^(8) ms^(-1))/(1285 xx 10^(-9) m) = 3.29 xx 10^(15) ((1)/(3^(2)) - (1)/(n^(2)))` or `(1)/(n^(2)) = (1)/(9) - (3.0 xx 10^(8) ms^(-1))/(1285 xx 10^(-9) m) xx (1)/(3.29 xx 10^(15)) = 0.111 - 0.071 = 0.04 = (1)/(25) or n^(2) = 25 or n = 5` The radiation corresponding to 1285 nm lies int eh infrared region. |
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| 17. |
Emissiontransitionin thepaschen seriesendat orbitn=3 andstartfromorbitn andcan berepresentedas v= 3.29 xx10^(15)(Hz)(1)/(3^(2)- (1)/(n^(2))) Calculatethe vallueof n ifthe transitionisobserved at 1285nm . findthe region of thespectrum |
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Answer» Solution :Equation forPaschen SERIES(n=30) `v=3.29 XX 10^(15) Hz (1)/(3^(2))- (1)/(n^(2))` Nowvalueof v PUTIN givenequation 2.3346 `xx 10^(14) Hz =3.29xx 10^(15)((1)/(3^(2) )-(1)/(n^(2)))` `(1)/(9)- (1)/(n^(2))=(2.3346 xx 10^(14) Hz)/(3.29 xx 10^(15 ) Hz)` `(1)/(9) -0.071 = (1)/( n^(2))` `(1)/(n^(2)) = 0.1111- 0.071 =0.040` sotransitionoccursn=-5 ton=3in PASCHEN seriesso infraredspectrum |
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| 18. |
Emission spectrum of gaseousatom is linearbut not cureve Why ? |
| Answer» SOLUTION :Electrongaseousatomemitsformdefinit energylevelas uncontinuousquantumand formspectra.So ITIS linearbut notcurve. | |
| 19. |
Emission of positron is found in |
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Answer» `""_1^2H + ""_1^2H to ""_2^4He` |
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| 20. |
Eludicate the differences in relative stability of conformations. |
Answer» SOLUTION :
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| 21. |
Elucidate the structure of benzene in detail. |
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Answer» Solution :`1.` Molecular formula : Elemental analysis and molecular weight determination have proved that the molecular formula of benzene is `(C_(6)H_(6)`. This indicates that benzene is a highly unsaturated compound. `2.` Straight chain structure not possible : Benzene could be contructed as a straight chain or ring compound but it not feasible since it does not show the properties of alkenes or alkynes. for example, it did not declourise bromine in carbon etrachloride or acidied `KMnO_(4)`. It did not react with water in the presence of acid. `3`. Evidence of cyclic structure: `I)` Substitution of benzene : Benzene reacts with bromine in the presence of `AlCl_(3)` to form mono BROMOBENZENE. `C_(6)H_(6)+3Br_(2)overset(AlCl_(3))(to)underset("Bromobenzene")(C_(6)H_(5)Br+HBr)` Formation of only one monobromo compound indicates that all the six hydrogen atoms in benzene were identical. This is possible only if it has a cyclic structure of six carbons each containing one hydrogen. `ii)` Addition of hydrogen : Benzene can add on to three moles of hydrogen in the presence of nickel catalyst to give cyclohexane. `C_(6)H_(6)+3H_(2)overset("Raney" Ni)(to)underset("Cyclohexane")(C_(6)H_(12)+HBr)` This confirms cyclic structure of benzene and the presence of three carbon-carbon double bond. `4.` Kekule's structure of benzene : In `1865`, August Kekule suggested that benzene consists of a cyclic planar structure of six carbon with alternate single and double bonds. `(i)` Benzene forms only one orthodisubtituted products whereas the Kekule's structure predicts two o-di substituted products as shown below.  `(ii)` Kekule's structure failed to explain why benzene with three double bonds did not give addition reactions like other alkenes. To overcome this OBJECTION, Kekule suggested that benzene was mixture of two forms (`1` and `2`) which are in rapid equilibrium.  `5.` Resonance description of benzene : The phenomenon in which two or more structures can be written for a substance which has identical position of atoms is called resonance. The actual structure of the molecule is said to be resonance hybrid of various possible alternative structures. In benzene, Kekule's structures `I` & `II` represented the resonance structure, and structure `III` is the resonance hybrid of structure `I` & `II`  The structures `1` and `2` exist only in theory. The actual structure of benzene is the hybrid of two hypothetical resonance structures. `6.` Spectrossopic measurments : Spectroscopic measurements show that benzene is planar and all of its carbon-carbon bonds are of equal length `1.40A^(@)`. This value lies between carbon-carbon single bond length `1.54A^(@)` and carbon-carbon double bond length `1.34A^(@)`. `7.` Molecular orbital structure : The structure of benzene is best described in terms of the molecular orbital theory. All the six carbond atoms of benzene are `SP^(2)` hybridized. Six `sp^(2)` hybrid orbitals of carbon linearly overlap with six one is orbitals of hydrogen atoms to form six `C-H` sigma bonds. Overlap between the remaining `sp^(2)` hybrid orbitals of carbon forms six `C-C` sigma bonds.  All the bonds in benzene lie in one plane with bond angle `120^(@)`. Each carbond atom in benzene POSSESS an un hybridized `p`-orbital containing one electron. The lateral overlap of their `p`-orbital produces `3 pi`-bond. The six electrons of the `p`-orbitals cover all the six carbon atoms and are said to be delocalised. Due to delocalization , strong -bond is formed which makes the molecule stable. Hence unlike alkenes and alkynes benzene undergoes substitution reactions rather addition reactions under normal conditions   `8`. Representation of benzene : Hence, there are three ways in which benzene can be represented.    Benzene and its homologeous series Benzene and its homologous series are colorless LIQUIDS with pleasant odour. They are lighter than water and insoluble in it. Their vapours are highly flammable, and volatile and toxic in nature. |
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| 22. |
Elemetal carbon appears in many structural forms or allotropes. Three of these forms are crystalline - diamond, graphite and the recently discovered fullerene (bucky ball) - while more than 40 others including coke and carbon black are amorphous. Now there seems to be a fourth crystalline allotrope of carbon, reported in 1995 by Rich and Lagow at the University of Texas Structures of different allotropes of carbon have been compared. Which represents incorrect comparison ? |
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Answer» ALLOTROPE DISCOVERD in 1995 sp-hybridised carbon |
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| 23. |
Elemetal carbon appears in many structural forms or allotropes. Three of these forms are crystalline - diamond, graphite and the recently discovered fullerene (bucky ball) - while more than 40 others including coke and carbon black are amorphous. Now there seems to be a fourth crystalline allotrope of carbon, reported in 1995 by Rich and Lagow at the University of Texas Newly discovered allotrope of carbon has the form |
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Answer» Polyene |
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| 24. |
Elements X, Y and Z have atomic numbers 7,15 and 33 respectively. Which of the following statements is correct? |
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Answer» ATOMIC size of X is greater than that of Y atom.  X, Y and Z are elements of same group `1S^(th)`: "Atomic size increase from top to bottom in a group". |
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| 25. |
Elements Zn, Cd and Hg with electronic configuration (n-1)d^(10) ns^(2) do not show most of transition elements properties. Give reason. |
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Answer» Solution :• ZN, Cd and Hg are having completely filled d-orbitals (`d^(10) electronic configuration). • They do not have PARTIALLY filled d-orbitals like other TRANSITION elements. So they do not show much of the transition elements PROPERTIES. |
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| 26. |
Elements X, Y and Z have atomic numbers 19, 37 and 55 respectively. Which of the following statements is true about them? |
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Answer» Their IONIZATION POTENTIAL would increase with increasing atomic number |
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| 27. |
Elements X, Y and Z have 4, 5 and 7 valence electrons respectively, (a) Write the molecular formula of the compounds formed by these elements individually with hydrogen. (b) Which of these compounds will have the highest dipole moment ? |
Answer» SOLUTION : Z has seven electrons in its VALENCE shell. It is the most electronegative element. THEREFORE, HZ will have the highest DIPOLE moment. |
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| 28. |
Elements X, Y and Z have 4,5 and 7 valence electrons repectively. (i) Write the molecular formula of the compounds formed by these elements indiviually with hydrogen. (ii) Which of these compounds will have the highest dipole moment ? |
Answer» SOLUTION :
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| 29. |
Elements with atomic numbers 7, 8, 9 and 10 are gases. Under similar conditions, which gas has highest rate of diffusion? Why? |
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Answer» Solution :Gas with atomic number 10 ( NEON) has highest rate of diffusion. This is because, gas with Z value 10, has LEAST molecular WEIGHT AMONG the four gases given. |
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| 30. |
Elements whose atoms have their s and p- sublevels complete are the_______ |
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Answer» NORMAL elements |
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| 31. |
Elements with atomic number 17 and 20 from compounds with hydrogen. Write the formulae of the two compounds and compare their chemical behaviour in water. |
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Answer» Solution :The elements with ATOMIC number 17 and 20 are CHLORINE and calcium. Their corresponding hydrids are HCl and `CaH_(2)`. While HCl dissociates in WATER to give acidic solution. `CaH_(2)` reacts with water to evolve hydrogen GAS and the resulting solution is basic in nature. `HCl overset(aq)to H^(+)(aq)+Cl^(-)(aq)` `CaH_(2)+2H_(2)O to Ca(OH)_(2)+H_(2)`. |
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| 32. |
Elements which generally exhibit multiple oxidation states and whose ions are usually coloured are |
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Answer» metalliods `=(7+39)/(2)=(46)/(2)=23` |
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| 33. |
Elements of which of the following group(s) of periodic table do not form hydrides. |
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Answer» Groups 7,8,9 |
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| 34. |
Elements of which of the following group(s) of periodic table do not form hydrides ? |
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Answer» Groups 7, 8, 9 |
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| 35. |
Elements of the same vertical group of the periodic table have |
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Answer» same ATOMIC SIZE |
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| 36. |
Elements of p-block are |
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Answer» Only non-metals |
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| 37. |
Elements of I B and II B are called |
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Answer» Normal elements |
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| 38. |
Elements of group - II are known as... |
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Answer» HALOGEN elements |
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| 39. |
Elements of group-I B and II B are known as… |
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Answer» ALKALI ELEMENTS |
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| 40. |
Elements of group - I-A are not found in free state, because... |
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Answer» EASILY lose electrone. |
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| 41. |
Elements of group - I-A is known as... |
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Answer» ALKALI elements |
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| 42. |
Elements of group- 15 form compounds in +5 oxidation state. However, bismuth forms only one wellcharacterised compound in +5 oxidation state. The compound is |
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Answer» `Bi_(2)O_(5)` |
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| 43. |
Elements of group 14 exhibit oxidation state of |
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Answer» `+4` only |
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| 44. |
Elements of group 14 (a) exhibit oxidation state of +4 only (b) exhibit oxidation state of +2and+4 (c) formM^(2–)" and "M^(4+) ions (d) formM^(2+)"and "M^(4+) ions |
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Answer» exhibit OXIDATION STATE of +4 only |
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| 45. |
Elements of group 14 …… |
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Answer» EXHIBIT OXIDATION state of +4 only |
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| 46. |
Elements of group 13 not reacting with water are |
| Answer» Solution :B, Ga and In do not REACT with WATER. | |
| 47. |
Elements of group 10 is called pseudo-inert group, Why? |
| Answer» Solution : PRESENCE of 18 electrons in the OUTER most SHELL is CALLED seudo-octet or pseudo-inert CONFIGURATION. Palladium, a member of group 10 has such configuration. | |
| 48. |
Elements of group 1 generally form chloride ores. The only other element which form chloride ores is |
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Answer» Cu |
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| 49. |
Elements of 6^(th) period are known as |
| Answer» Answer :a | |
| 50. |
Elements of a vertical group have |
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Answer» Same ATOMIC NUMBER |
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