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. |
In order that a compound may show geometrical isomerism A Compound must have a chiral carbon (B) Each of the two doubly bonded carbon atoms must have different atoms or groups (C ) Compound must have a plane of symmetry |
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Answer» only A |
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| 2. |
In order of decreasing contribution to the green house effect, the three main gases are _____, __________ and ______ |
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Answer» |
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| 3. |
In open vessel sugar solution hArr "sugar"_((s)) which information is obtained in this reaction ? |
| Answer» SOLUTION :Here saturated solution of SUGAR is given. So at given equilibrium is ESTABLISHED between SOLID sugar and dissolve sugar. | |
| 4. |
In one organic compound the weight proportion of C, H and N is 9 : 1 :3.5respectivily. Its molecular mass is 108 mg/mole. Find the molecular formula. |
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Answer» |
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| 5. |
In one organic compound the proportion of C, H & Cl is 10%, 0.84% and 89.2 % respectively.Its empirical formula is..... |
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Answer» `C Cl_(4)` `=0.83:0.84:2.51` `= (0.83)/(0.83) : (0.84)/(0.83) : (2.51)/(0.83)` `= 1:1:3` `:.CHCl_(3)` |
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| 6. |
In one organic compound percentage of C& H are 54.55 and 9.06. What will be its epirical formula. |
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Answer» |
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| 7. |
In one of the following compounds, the oxidation number of sulphur is not a whole numbers |
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Answer» `Na_(2)S_(4)O_(6)` |
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| 8. |
In one experiment, certain amount of NH_(4)I(s) was heated rapidly in a closed container at 375^(@)C. The following equilibrium was established: NH_(4)I(s)hArrNH_(3)(g)+HI(g) Excess of NH_(4)I(s) remained unreacted in the flask and equilibrium pressure was 304mm of Hg. After some time, the pressure started increasing further owing to the dissociation of HI. 2HI(g)hArrH_(2)(g)+l_(2)(g) K_(C)=0.010 calculate final pressure. |
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Answer» `331mm "of" HG` |
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| 9. |
In one discharge tube, H_(2) gas is taken and in the other O_(2) gas is taken. Will the electrons an positive ions in the cathode rays and in the anode rays be same or different ? |
| Answer» SOLUTION :ELECTRONS will be same but POSITIVE IONS will not be same | |
| 10. |
In one compound 54.55 % C, 9.09 % H and 36.36 % O are present. Its empirical formula is..... |
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Answer» `C_(3)H_(5)O` |
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| 11. |
In oil fire extinguisher, the compound used pyrene is chemically ………………. . |
| Answer» SOLUTION :`C Cl_4` | |
| 12. |
In OF_(2) ,number of bond pairs and lone of electrons are respectively |
Answer» SOLUTION :
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| 13. |
In O_(2) molecule, the correct order of molecular orbitals is |
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Answer» `PI2 pygt pi2 pz` |
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| 14. |
In O_(2), H_(2)O_(2) and O_(3) the correct order of oxygen -oxygen bond length is |
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Answer» `O_(2)gtO_(3)gtH_(2)O_(2)` |
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| 15. |
In O_(2)^(-) and O_(2)^(2-) which has more bond order ? |
| Answer» SOLUTION :`O_(2)^(-) (1.5) and O_(2)^(2-) ` (1.0) THEREFORE BOND ORDER in `O_(2)^(-)` | |
| 16. |
In nuclear reactors, heavy water is used as a |
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Answer» Fuel |
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| 17. |
In nuclear transmutation tritium is obtained from |
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Answer» `""_7N^14` |
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| 18. |
In nomenclature of cyclic compound…… prefix is written before the name of simple chain containing alkane. |
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Answer» |
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| 19. |
In NO_(3)^(-) ion, the number of bond pairs and lonepairs of electrons on nitrogen atom are |
| Answer» Solution :`NO_(3)^(-) = [underset(. .) overset(. . ) O= underset(. .) underset(":O:")underset(|)N- underset(. .)overset(. . )O: ] ^(-)`, Bond PAIRS on N = 4, LONE pairs = 0 | |
| 20. |
In NO_(3)^(-)ion, the number of bond pairs and lone pa1rs of electrons on nitrogen atom are |
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Answer» 2,2 In N-atom, are = 5 valence electrons = 5 As the presence of one negative charge, NUMBER of valence electrons = 5 + 1 = 6 one O-atom forms TWO bond (= bond) and two O-atom shared with two electrons of N-atom So, 3 O-atoms shared with 8 electrons of N-atom. `therefore `Number of bond PAIRS (or shared pairs) = 4 Number of lone pairs = 0 |
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| 21. |
In {:(-NO_(2),-NH_(2),-SO_(3)H),(I,II,III):}, the decreasing order of -I effect is |
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Answer» `I GT II gt III` |
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| 22. |
In nitric oxide (NO), the oxidation state of nitrogen is |
| Answer» SOLUTION :`NO,x-2=0impliesx=+2` | |
| 23. |
In nitration of benzene ……is electrophile. |
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Answer» NITRATE ions |
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| 24. |
In neutralisation of KI by AgNO_(3) positive charge is due to adsorption of |
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Answer» `AG^(+)` ions |
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| 25. |
In nature, non metals are found in the |
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Answer» native STATE |
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| 26. |
In natue a decay chain sereis starts with ._(90)h^(232) and finally terminates at ._(82)Pb^(208). A throium ore sample was found to contain 8xx10^(-5) mL of He At STP and 5xx10^(7)g of Th^(232). Find the age of ore sample assuming that sourcesof He to be only due to decay of Th^(232) Also assumecomplete retention of He within the ore, t_(1//2) for Th^(232) = 1.39xx10^(10) year. |
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Answer» Solution :We KNOW, `._(90)Th^(232) rarr ``._(82)Pb^(208) + 6 ``._(2)He^(4) + 4 ``._(-1)e^(0)` `:' 6xx22400 mL` He is formed by `232g Th` decay `:. 8xx10^(-5) mL` He is formed by `= (232xx8xx10^(-5))/(6xx22400)g Th` decay `= 1.38xx10^(-7)g Th` decay At `t = t`, sample has `Th = 5xx10^(-7) g prop N` At `t = 0`, sample has `Th = 5xx10^(-7) + 1.385 xx 10^(-7) prop N_(0) = 6.38xx10^(-7)g` For `Th` decay `:' t = (2.303)/(LAMBDA) log_(10) (N_(0))/(N)` `= (2.303xx1.39xx10^(10))/(0.693) log_(10) (6.38xx10^(-7))/(5x10^(-7))` `=4.89xx10^(9)` year |
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| 27. |
In naturally available hydrogen, the percentage of tritium is |
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Answer» `99.986%` |
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| 28. |
In NaCl is doped with10^(-4) mol % ofScCl_(2) , the concentration of cation vacancies will be(N_(A) = 6.02 xx 10^(23) mol^(-1)) |
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Answer» ` 6.02 xx 10^(14) mol^(-1)` `SrCl_(2)` doped per mole of NaCl = `10^(-4) //100` ` 10^(-6)` mole =`10^(-6) xx ( 6.02 xx 10^(23)) Sr^(2+)` ions ` = 6.02 xx 10^(17) Sr^(2+)` ions Hence, concetration of cation vacancies ` 6.02 xx 10^(17) "mol"^(-1)` |
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| 29. |
In NaCl crystals , Cl^- ions are in FCC arrangement. Calculate the number of Cl^- ions in its unit cell |
| Answer» SOLUTION :`Cl^-` ion per UNIT CELL =`8xx1/8` (from corners ) + `6xx1/2` (from face CENTRES)=4 | |
| 30. |
In NaCl crystal, Ci^(-1) ions are I FCc arrangement, Calculate the number ofCl^(-)ions in its unit cell. |
| Answer» Solution :`CL^(-)` ION per unit CELL = ` 8 xx 1/8 "( from corners)" + 6 xx 1/2`( form face CENTRES ) = 4. | |
| 31. |
In NaCl, Cl^- ions are present in the _________ structure whereas Na^+ ions are present in the _______ voids. |
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Answer» |
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| 32. |
In Na and K metals, which salts are high proportion in the ash of shrubs ? |
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Answer» `PO_(4)^(3-)` |
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| 33. |
In mutarotation of alpha-D Glucose hArr beta-D Glucose equilibrium constant is 1.8. At equilibrium what will be the percentage of alpha-D Glucose ? |
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Answer» 35.8 `K=alpha/(1-alpha)`=1.8 `alpha=1.8-1.8alpha` `therefore 2.8 alpha-1.8` `therefore alpha`=0.642 `therefore 1-alpha`=0.358 So, at the equilibrium of `alpha`-D Glucose =35.8% |
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| 34. |
In multi-electron atom, 4s - orbital is lower in energy than |
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Answer» 3D -orbital |
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| 35. |
In molecule, the formal charges of oxygen atoms 1,2,3 are respectively |
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Answer» `-1,0,+1` |
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| 36. |
In modern periodic table, if an element has electronic configuration 1s^(2), 2s^(2), 2p^(6), 3s^(2), 3p^(4), then what will be the atomic number of element, which is just below the element in its group ? |
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Answer» 18 Electronic configuration of the element `1S^(2) 2s^(2) 2p^(6) 3s^(2) 3P^(4)` ` :. ` TOTAL electrons in it = 16 and Z = 16 Which INDICATES element is S(Z = 16) of`III^(rd)` Period, and`16^(th)` /VIB GROUP. Element just below S in same group is Se having atomic number 34. |
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| 37. |
In Millikan's oil drop experiment, we make use of: |
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Answer» OHM's law |
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| 38. |
In Millikan's experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge in the oil drop is -1.282 xx 10^(-18)C, calculate the number of electrons present in it. |
| Answer» Solution :ELECTRONS present = `(-1.282 xx 10^(-18)C)/(-1.6022 xx 10^(-19)C) = 8` | |
| 39. |
in milikanexperimentstaticelectricchargeon theoildrops has beenobtainedby shiningX-raysif thestaticelectric chargeon theoildropis-1.282xx 10^(18) Ccalculatethe numberof electronpresencton it. |
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Answer» Solution :Staticelectricchargeon theoil drop(q ) =1.282 `XX 10^(18) C` Chargeon oneelectron(e ) = 1.6022 `xx 10^(19) C` NUMBEROF electron= `(q ) /( e )` `(1.282 xx 10^(18) C)/( 1.6022xx 10^(19) C)` 8.00 |
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| 40. |
In [Mg(H_(2)O)_(6)]^(2+) Mg hybridiasationis |
| Answer» Answer :1 | |
| 41. |
In MgH_2oxidation number of hydrogen is: |
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Answer» `-1` |
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| 42. |
In meta-aluminates, co-ordination number of aluminium is |
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Answer» 3 `UNDERSET("Meta-aluminate ion")([AL(OH)_(4)(H_(2)O)_(2)]^(-))` |
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| 43. |
In melting of ice, which one of the conditions will be more favorable ? |
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Answer» HIGH TEMPERATURE and high PRESSURE |
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| 44. |
In measurement of BOD_x, is generally taken as |
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Answer» |
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| 45. |
In long form of periodic table, an element has electronicconfiguration 1s^(2)2s^(2)2p^(6)3s^(2)3p^(3). Thus what will be the atomic number of element which is just below this element in its group ? |
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Answer» <P>23 ` :. ` TOTAL electrons = 15 ` :. ` Element is P of group 15 or VB The element just below to P is As As has ATOMIC number = 33 and As is of group 15 or VB |
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| 46. |
In long form of periodic table, an elem ent has electronic configurationIs^(2) 2s^(22)p^(6)3s^(2)3p^(3) Thus what will be the atomic no. of elem ent ? Which is just below this elem ent in its group ? |
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Answer» 23 |
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| 47. |
In lithium aluminium hydride molecule,how many hydrogen atoms are present ? |
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Answer» |
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| 48. |
In Leibig's method 0.24 g of organic compound on combustion with dry oxygen produced of 0.62 g of CO_(2) and 0.11 g of H_(2)O. Determine the percentage composition of the compound. |
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Answer» SOLUTION :Mass of organic COMPOUND `=m=0.24g` Mass of carbon dioxide formed `=0.62 g` Mass of water formed `=0.11 g` Percentage of carbon `=12/4xx0.62/0.24xx100=70.45` Percentage of hydrogen `=2/4xx0.11/0.24xx100=5.09` Percentage of oxygen `=[100-(70.5+5.0)]=24.46` |
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