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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. |
Regarding silica I) Quartz is amorphous form of silica II) Silica dissolves in NaOH III) Silica dissolves in HF The correct combination is |
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Answer» all are correct |
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| 2. |
Regarding ionization potential some statement are given a) Ionization process is an endothermic process b) The order of IP values is IP_1 lt IP_2 ltIP_3 c) With increases in the number of electrons in the shells. The I.P. values gradually increase d) Stable configuration leads to greater I.P value e) In a period the I.P. value decreases from left to right In the above statements |
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Answer» A, B, C are CORRECT |
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| 3. |
Regarding lanthanide contraction, some statements are given a) It arises because of the poor shielding effect of 5f-electrons b) The atomic radii and ionic radii steadily decrease from Ce to Lu c) The crystal structures and other properties are similar d) The seperation of lanthanides from one another is difficult from their salt solutions. The correct statements are |
| Answer» Answer :B | |
| 4. |
Regarding hybridisation which is incorrect? |
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Answer» `BF_3, C_2 H_4, C_6 H_6` INVOLVE `SP^2` HYBRIDISATION |
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| 5. |
Regarding hybridisation the correct statement is |
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Answer» Orbitals of different atoms hybridize |
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| 6. |
Regarding H_2O_2 the wrong statement is |
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Answer» it ACTS as an oxidising agent as WELL as a reducing agent |
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| 7. |
Regarding geometrical isomers which is not correct |
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Answer» both cis and trans isomer posses same m.p |
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| 8. |
Regarding electronegativity i) The element with maximum electro negativity : Fluorine ii) In Pauling scale, the reference element is : Hydrogen iii) Elements with stable configuration havehigh electronegativity |
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Answer» All are correct |
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| 9. |
Regarding dipole - dipole attractions the incorrect statement is |
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Answer» Dipole - dipole ATTRACTIONS are more if the molecules have high dipole MOMENT values. |
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| 10. |
Regarding diamond I) C-C bond length is 1.54Å II) It has least refractive index among solids III) It has a 3-dimensional structure. The correct combination is |
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Answer» all are CORRECT |
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| 11. |
Regarding Benzene some statements are given A) It is aromatic B) It burns with smoky and sooty flame C) It can't decolourise Br_2 water D) It mainly participates in electrophilic substitution reactions |
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Answer» Both 'A' and 'B' are CORRECT |
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| 12. |
Regarding Boricacid, which of the following statements is correct ? |
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Answer» Boric ACID has layered lattle structure |
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| 13. |
Regarding 'Al' the wrong statement is |
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Answer» It REACTS with both acids and bases |
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| 14. |
Refractive index of a solid is observed to have the same value along all directions. Comment on the nature of the solid. Would it show cleavage property ? |
| Answer» Solution :As the solid has same value of refractive index ALONG all directions, this means that it is isotropic and HENCE amorphous. Being an amorphous solid, it WOULD not show a clean CLEAVAGE when cut with a knife. INSTEAD, it would break into pieces with irregular surfaces | |
| 15. |
Refractive index of a solid is obseved to have the same value along alll directions. Comment on the nature of the soild. Would it show cleavage property ? |
| Answer» Solution :As the SOLID has same value of refractive index along all directions, this means that it is isotropic and hence amorphous. Being ans amorphous solid, it would not show a clean cleavage when CUT with a knife. Instead, it would break into PIECES with IRREGULAR surfaces. | |
| 16. |
Reforming is the process of converting .......... and ....... alkanes into corresponding ....... hydrocarbons. |
| Answer» SOLUTION :ACYCLIC, CYCLIC, AROMATIC | |
| 18. |
Refer to the periodic table given in your book and now answer the following questions : (a) Select the possible non metals that can show disproportionation reaction. (b) Select three metals that can show disproportionation reaction. |
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Answer» SOLUTION :Reactant participated in disproportionation reaction has at least three OXIDATION state. (a) P, CL and S, there 3 non-metals gives more than 3 oxidation state, therefore, it shows disproportionate reaction. (b) MN, Cu and GA there metals shows disproportionate reaction. There metals also shows more than 3 oxidation state. |
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| 19. |
Refer to the periodic table given in your book and now answer the following questions: (a) Select the possible non metals that can show disproportionation reaction. (b) Select three metals that can show disproportionation reaction. |
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Answer» Solution :The metals are Cu, Ga and In. Their DISPROPORTIONATION reactions are as follows. `2Cu^(+)(aq)toCu^(2+)(aq)+Cu(s)` `3Ga^(+)(aq)TOGA^(3+)(aq)+2Ga(s)` `3ln^(+)(a)toln^(3+)(aq)+2ln(s)` hese metals can EXIST in three oxidation states as shown below : `Cu:+2,0,+1` `Ga:+3,0,+1` `ln:+3,0,+1` |
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| 20. |
Refer to the periodic table given in your book and now answer the following questin (a) Select the possible non metals that can show disporportionation reaction (b) Select three metal that show disproportionation rection |
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Answer» Solution :(a) `P+_(4)(s)+3OH^(-)(aq)+3H_(2)(l)rarrPH_(3)(g)+3H_(2)PO_(2)^(-)(aq)` (II) `CI_(2)(g)+2OH^(-)overset(cold)rarrCi^(-)(aq)+CIO^(-)(aq)+H_(2)O(l)` or `3CI_(2)(g)+6OH^(-)(aq)overset(hot)rarr5CI^(-)(aq)+CIO_(3)^(-)(aq)+3H_(2)O(l)` (iii) `S_(8)(s)+12OH^(-)rarr4S^(2-)(aq)+2S_(2)O_(3)^(2-)(aq)+6H_(2)O(l)` (iii) `S_(8)(s)+12 OH^(-)rarr45S^(2-)(aq)+2S_(2)O_(3)^(2-)(aq)+6H_(2)O (l)` (b) The metals are :`Cu^(+),Ga^(+),In^(+),Mn^(3+)`, ETC `2CU^(+)(aq)rarrCu^(2+)(aq)+Cu^(s)` `3Ga^(+)(aq)rarrGa^(3+)(aq)+2Ga(s)` `3IN^(+)(aq)rarrIn^(3+)(aq)+2In(s)` |
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| 21. |
Reductive ozonolysis of but - 1, 3 - diene gives |
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Answer» `HCHO and {:(CHO),(|),(CHO):}` |
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| 22. |
Reduction potential value of A, B, C are 0.34V, -0.80V, -0.46V respectively then what will be order of strength of reducing agent ? |
| Answer» SOLUTION :`BgtCgtA` (GREATER reduction potential LOWEST reducing agent) | |
| 23. |
Reduction of methylbenzoate (C_(6)H_(5)COOCH_(3)) to benzyl alcohol (C_(6)HCH_(2)OH) can be accomplished using |
| Answer» Answer :B | |
| 24. |
Reduction of BCI,_3 with lithium aluminium hydride gives |
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Answer» Borazole |
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| 25. |
Reduction of an isonitrile gives a |
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Answer» PRIMARY amine |
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| 26. |
Reduction of 2-butyne with Lindlar's catalyst gives __________while reduction with sodium in liquid ammonia gives ________ |
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Answer» |
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| 27. |
Redox reactions play a pivotal role in chemistry and biology.The values of standard reduction potential E^(@) of two half cell reaction decide which way the reaction is expected to proceed. A simple exaple is Danie cell in which zinc goes in to solution and copper gets deposited Given below are set of half cell reaction (acidic medium) along with their E^(@) values. I_(2)+2e^(-)rarr2I^(-) ""E^(@)=0.54 CI_(2)+2e^(-)rarr2CI^(-) ""E^(@)=0.54 Mn^(3+)+e^(-)rarrMn^(2+) ""E^(@)=1.36 Fe^(3+)+e^(-)rarrMn^(2+)""E^(@)=0.77 O_(2)+4H^(+)e^(-)rarr2h_(2)O""E^(@)=1.23 Using these data , obtain the correct explanation for the following question. Sodium fusion extact obvtined from aniline. On treatment with iron (II suphate and H_(2)SO_(4)) in presence of air givers a prussian blue precipitte.The blue colour is due to the formation of |
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Answer» `Fe_(4)[Fe(CN)_(6)]^_(3)` (Sodium FUSION extract) `Fe^(2+)+6CN^(-)rarr[Fe(CN)_(6)]^(4-)` In presence of AIR, `Fe^(2+)` ions get oxidised to `Fe^(3+)` ions. `(Fe^(2+)rarrFe^(3+)+E^(-)]xx4,E^(@)=-0.77V)` `(O_(2)+4H^(+)+4e^(-)rarr2H_(2)O,E^(@)=+1.23 V)/(4Fe^(2+)+4H^(+)+O_(2)rarr4Fe^(3+)+2H_(2)O)` `Fe^(3+)` ions then combine with `[Fe(CN)_(6)]^(4-)` ion to form FERRIC ferrocyanide which has prussian blue colour `4Fe^(3+)+3[Fe(CN)_(6)]^(4-)rarrFe_(4)[Fe(CN)_(6)]_(3)` |
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| 28. |
Reductived ozonolysis of allene CH_3 - CH = C = CH_2 will give |
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Answer» only `CH_3 - CHO` 
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| 29. |
Red precipitate overset(Cu_(2)Cl_(2))underset(NH_(4)OH)(larr) P(C_(5)H_(8)) overset("Ozonolysis")to 2-Methylpropanoic acid +compound (Q) structure of P can be |
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Answer» `CH_(3)-CH_(2)-CH_(2)-C-=CH` |
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| 30. |
Red phoshphorus is …………..reactive than white phosphorus is ………… and consists of…………..of P_(4) units. |
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Answer» |
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| 32. |
Rectified spirit is a mixture of |
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Answer» 95% ethyl alcohol +5% water |
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| 34. |
Rectified spirit can be converted into absolute alcohol by |
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Answer» FRACTIONAL distillation |
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| 35. |
Rearrangement of an oxime to an amide in the presence of strong acid is called |
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Answer» CURTIUS REARRANGEMENT |
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| 36. |
Rearragne the following (I to IV) in the order of increasing masses and choose the correct answer (Atomic masses O=16, Cu=63 and N=4) I. 1 Molecule of oxygen II. 1 atom of nitrogen IIIgt 1xx 10^(-10)g atomic weight of copper |
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Answer» II lt I lt III lt IV (ii) 1. atom of N= (14)/(6.022 xx 10^(23)`g = `2.3 xx 10^(-23)`g IIIgt `10^(-19)` g MOLECULAR weight of oxygen `= (10^(-10) xx 32 = 3.2 xx 10^(-9)`g IV. `10^(-10)`gatomic weight of copper `= 10^(-10) xx 63 =6.3 xx 10^(-9)`g `therefore` ORDER of increasing mass is II lt I lt III lt IV |
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| 37. |
Real gases shows deviation than ideal gases. Explain with examples. |
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Answer» Solution :(A) According of Boyle.s Law, Explanation of graph of `pV to p` (constant) : Theoretical value is `pV to nRT`. At constant temperature and all pressure, parallel to X - axis LINE is obtained for graph `pV to p`. Because parallel to X - axid STRAIGHT line obtained. According to Boyle.s law for graph `pV to p`. But in real graph for `pV to p` is not obtained straight line. In figure (graph) against are shown ast constant temperature `pV to p`. This grpah is not straight line for ideal gas and shows deviation with ideal gas.  More over it is not parallel to X - axis as an ideal gas. Graph Type - 1 : Graph of Dihydrogen and Helium are strasight line and pV increases with increasing pressure. Graph Type - 1 : The real gas Carbon MONOXIDE (CO) and Methane `(CH_(4))` shows different type of curve. According to curve, CO and `CH_(4)` gas shows negative deviation with Ideal gas. Value of pV decreases with increasing in pressure. Vaslue of pV decreases with increasing in pressure and reached at minimum value. After that value of pV increases with increasing in pressure and it intercept line of ideal gas and deviation becomes zero. Value of pV increases with increasing in pressure and continuously positive deviation observed. So real gases do not follow ideal gas equation in each energy condition. (B) According of Boyle.s LAw, Explanation of deviation of graph of `p to V` (T constant ):Graph of `p to V` of real gas shows deviation than ideal gas is shown in graph, `p to V` curve is given according to Boyle.s Law and some experimental information.  This graph indicate that, (i) Measured volume at high pressure (volume of real gas is higher than ideal gas). (ii) Volume of real gas measured at low pressure it will be nearest thans volume of ideal gas. From result of A & B. Prove that real gases do not follow Boyle.s Law, Charle.s Law and Avogradro.s Law. |
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| 38. |
Real gases deviated from ideal behaviour due to the following two faulty assumptions of kinetic theory of gases : (i) Actual volume occupied by the gas molecule is negligible as compared to the total volume of the gases (ii) Forces of attraction and repulsion among gas molecules are negligible To explain the extent of deviation of the real gas from ideal behaviour in terms of compressibility or compression factor (z), which is the function of pressure and temperature for real gases z = (P_(0)V_(0))/(nRT) For ideal gases z = 1. for real gases either z gt 1 " or " z gt 1. When z gt 1, then it is less compressible because force of repulsion dominates over force of attraction. When z lt 1, force of attraction dominates over the repulsion and it is more compressible. Graph in between z and P is shown below on increasing the temperature, z increases and approaches to unity. Graph between z and p at different temperature are as under Answer the following questions on the basis of above write up : Which of the following statements is correct ? |
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Answer» The compressibility factor for IDEAL gases is dependent of temperature and PRESSURE |
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| 39. |
Real gases deviated from ideal behaviour due to the following two faulty assumptions of kinetic theory of gases : (i) Actual volume occupied by the gas molecule is negligible as compared to the total volume of the gases (ii) Forces of attraction and repulsion among gas molecules are negligible To explain the extent of deviation of the real gas from ideal behaviour in terms of compressibility or compression factor (z), which is the function of pressure and temperature for real gases z = (P_(0)V_(0))/(nRT) For ideal gases z = 1. for real gases either z gt 1 " or " z gt 1. When z gt 1, then it is less compressible because force of repulsion dominates over force of attraction. When z lt 1, force of attraction dominates over the repulsion and it is more compressible. Graph in between z and P is shown below on increasing the temperature, z increases and approaches to unity. Graph between z and p at different temperature are as under Answer the following questions on the basis of above write up : Which of the following statements is correct gas A having molar mass 16 g and density 0.75 g/litre at 2 atmospheric pressure and 27^(@)C temperature |
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Answer» FORCE of ATTRACTION is dominating than force of REPULSION among the GAS molecules |
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| 40. |
Real gases deviated from ideal behaviour due to the following two faulty assumptions of kinetic theory of gases : (i) Actual volume occupied by the gas molecule is negligible as compared to the total volume of the gases (ii) Forces of attraction and repulsion among gas molecules are negligible To explain the extent of deviation of the real gas from ideal behaviour in terms of compressibility or compression factor (z), which is the function of pressure and temperature for real gases z = (P_(0)V_(0))/(nRT) For ideal gases z = 1. for real gases either z gt 1 " or " z gt 1. When z gt 1, then it is less compressible because force of repulsion dominates over force of attraction. When z lt 1, force of attraction dominates over the repulsion and it is more compressible. Graph in between z and P is shown below on increasing the temperature, z increases and approaches to unity. Graph between z and p at different temperature are as under Answer the following questions on the basis of above write up : Which of the following is the correct order of temperature shown in the above graph z vs P for the same gas |
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Answer» `T_(4) gt T_(3) gt T_(2) gt T_(1)` |
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| 41. |
Real gases exert less pressure when compared with ideal gases due to the attraction between the gas molecules. Reduction in pressure of real gases prop ("Concentration of gas")^(n). What is the value of n ? |
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Answer» |
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| 42. |
Real gases deviated from ideal behaviour due to the following two faulty assumptions of kinetic theory of gases : (i) Actual volume occupied by the gas molecule is negligible as compared to the total volume of the gases (ii) Forces of attraction and repulsion among gas molecules are negligible To explain the extent of deviation of the real gas from ideal behaviour in terms of compressibility or compression factor (z), which is the function of pressure and temperature for real gases z = (P_(0)V_(0))/(nRT) For ideal gases z = 1. for real gases either z gt 1 " or " z gt 1. When z gt 1, then it is less compressible because force of repulsion dominates over force of attraction. When z lt 1, force of attraction dominates over the repulsion and it is more compressible. Graph in between z and P is shown below on increasing the temperature, z increases and approaches to unity. Graph between z and p at different temperature are as under Answer the following questions on the basis of above write up : What is correct increasing order of liquefibility of the gases shown in the above graph ? |
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Answer» `H_(2) LT N_(2) lt CH_(4) lt CO_(2)` |
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| 44. |
What is meant by Boyle temperature Boyle point? |
| Answer» Solution :Boyle point of a gas: The TEMPERATURE at which a real gas OBEYS ideal gas equation is called Boyle temperature (TB) or Boyle point. | |
| 45. |
Real gases behave ideally only at certain conditions. Density of a gas was found to be 5.5 g L at 2 bar pressure. Calculate its molar mass. [R = 0.083 bar L mol^-1 K^-1] t=250 degree C |
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Answer» SOLUTION :M = dRT/P M = `(5.5xx0.083xx300)/2` = 68.02 g/mol |
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| 46. |
Real gases behave ideally only at certain conditions. Write the expression for compressibilty factor. What is its value for an ideal gas? |
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Answer» Solution :COMPRESSIBILITY FACTOR z = PV/nRT For an ideal gas z = 1 at all temperatures and PRESSURES |
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| 47. |
Reagent(s) which can be used to bring about the following transformation is (are): |
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Answer» `LiAlH_(4)` in `(C_(2)H_(5))_(2)O` |
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| 48. |
Reagent 'X' dissolves AgCI precipitate to form a soluble complex 'Y'. On passing H_(2)S gas through Y, a black precipitate is obtained. Find the number of possibilities of 'X' from the following reagents. KCN, conc. HCI, Na_(2)S_(2)O_(3),NH_(3), dil. HNO_(3) |
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Answer» |
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| 49. |
Reagents that can be used to convert CH_3MgBr to CH_4 are |
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Answer» `H_2O` |
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