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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.
| 50001. |
(A): Ionic compounds tend to be non-volatile (R) : Inter ionic forces in ionic compounds are weak 2 |
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Answer» Both (A) and (R) arc true and (R) is the CORRECT explanation of (A) |
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| 50002. |
(A): Ionic compounds exhibit isomerism (R) : Ionic bond is non directional bond |
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Answer» Both (A) and (R) ARC true and (R) is the CORRECT explanation of (A) |
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| 50003. |
(A): Instead of using conventional fuels and energy systems, non-conventional fuels and non- conventional energy systems must be put into practice. (R): Non-conventional fuels and non-conventional energy systems will cause more pollution. |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT explanation'of (A) |
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| 50004. |
(A) : Instead of pesticides, herbicides like NaClO_3, Na_3AsO_3 are used in agriculture sector. (R) : The fields sprayed with herbicides are more easily attacked by insects and plant diseases |
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Answer» Both (A) and (R) are true and (R) is the CORRECT EXPLANATION of (A) |
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| 50005. |
(A): Inductive effecti is possible in chlorobutane but absent in Butane (R ): Inductive effect is possible in the organic compounds having at least one polar covalent bond |
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Answer» A and R are TRUE, R EXPLAINS A |
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| 50006. |
(A) in Victor Meyer's method, vapour density is considered to be one-half of molecular weight. |
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Answer» If both ASSERTION and reason are CORRECT and reason is the correct EXPLANATION of the assertion |
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| 50007. |
a. In the sixth period, after filling of 6p orbitals, the next electron (i.e. 57th) enters the 5d-orbital against aufbau principal and there after the filling of seven 4f-orbitals starts with cerium (Z = 71). Explain this anomalous behaviour. b. In the seventh period, after the filling of 7s-orbital, the next two electrons (i.e. 89th and 90th) enter the 6d-orbital against Aufbau principle and there after the filling of seven 5f-orbitals begins with proactinium (Pr, Z = 91) and ends up with lawrencium (Lr, Z = 103). Explain this anomalous behaviour. |
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Answer» Solution :This can be explained on the basis of GREATER stability of the xenon (inert gas) core. After barium `(Ba, Z = 56)`, the addition of the next electron (i.e. `57th)`, should occur in `4f`-orbitals lie inside the core, and will tend to destatabilise the xenon core. `Z = 54, rArr [kr]4d^(10)4f^(0)5s^(2)5p^(6)5d^(0)` Therefore, the `57th` electron prefers to enter `5d`-orbitals which lies outside the xenon core and whose energy is only sightly greater than that of `4f`-orbitals. Thus, stability of the atom due to xenon core compensates more than the SLIGHT instability caused by the addition of one electron to the higher energy `5d`-orbital instead of the lower energy `4f`-orbital. Thus, the outer electronic configuration of La `(Z = 57)` is `5d^(1)6s^(2)` rather than the expected `4f'6s^(2)`. Once `5d`-orbital has one electron, the next electron (i.e. `58th)` instead of entering the outer `5d`-orbital, enters the inner `4f`-orbnital. This is due to greater nuclear charge and thereafter the continuous filling of the `4f`-sub shell occurs till it is complete at lutetium `(Lu, Z = 71, 4f^(14)5d^(1)6s^(2))` b. The anomalous behaviour is due to: i. The SMALLER energy difference between `5f`- and `6d`-orbitals than between `4f`- and `5d`-orbitals. ii. Due to the greater stability of radon `(Rn, Z = 86)`, the next two electrons (i.e. `89th` and `90th)` after filling the `7s`-orbital prefer to enter `6d`-orbitals before filling of `5f`-orbitals begin with proactinium `(Pa, Z = 91)` and continues till it is complete with lawrencium `(Lr, Z = 103)`. |
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| 50008. |
A in purte state is a white solid of sodium which acts as an oxidizing agent. A+CO_(2)toB+C A+H_(2)SO_(4)overset(hot)rarrC+D+H_(2)O Based on above reaction sequance select incorrect statement: |
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Answer» The bond order of anionc species in A is 0.5 |
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| 50009. |
(A): In Haber's process, N_(2) and H_(2) combine in 1: 3 volume ratio (R): Gases combine in simple volume ratio |
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Answer» Both A and R are true and R is the |
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| 50010. |
(A) : In an atom the velocity of electron as it moves into higher orbits keeps on decreasing (R) : Velocity of electron is inversely proportional to the radius of orbit |
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Answer» Both (A) and (R) are TRUE and (R) is the correct explanation of (A) |
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| 50011. |
(a) In a polar solvenbt , PCl_(5) undergoes an ionization reaction as follows :2 PCl_(5)hArrPCl_(4)^(+) + PCl_(6) ^(-) Wht will be the geometrical shape of each species present In the equalilbrium maxture ? (b) Why doesPCl_(5) exist as[ PCl_(4)]^(+) [PCl_(6)]^(-) ? |
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Answer» Solution :`X = (1)/(2) [{{:(" NO of valence electrons "),("of the central atom "):}}+{{:(" NO of monovalent atoms/group "),("surrounding the central atom "):}}]` `- {{:("Charge on the cation if the given "),(" SPECIES is a polyatomic cation "):}}+{{:(" Charge on the anion if the given "),(" species is a polyatomic anion"):}}` For `PCl_(5), X = (1)/(2) [5 + 5] - 0 + 0 = 5(sp^(3) d ,`bipyramidal) For`PCl_(4_^(+) , X = (1)/(2) [ 5 + 4 - 1 + 0 ] = 4 (sp^(3), `tetrahedra) For ` PCl_(6) ^(-), X = (1)/(2) [ 5 + 6 - 0 + 1] = 6 (sp^(3) d^(2)` , octahedral) (B) ` PCl_(5)` has DIFFERENT axial and equatorial bohnd lengths and THEREFORE , has an unsymmetrical structure Which is not stable . By dissociation, it changes into tetrahedral ` ([PCl_(4)]^(+)` and octahedral`[PCl_(6)]^(-)` Which are stable . |
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| 50012. |
(A) : In a pressure cooker, the water is brought to boil. The cooker is then removed from the stove. Now on removing the lid of pressure cooker, the water starts boiling again (R) : The impurities in water bring down its boiling point. |
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Answer» If both (A) and (R) are correct and (R) is the correct EXPLANATION for (A) |
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| 50013. |
(A) In a gaseous reaction, K_(c) is unitless when Deltan = 0. (R) Unit of K_(c) = (mol L^(-1))^(Deltan). |
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Answer» Both (R) and (A) are true and reason is the CORRECT explanation of assertion. |
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| 50014. |
(A) If sulphur and nitrogen are also present in orgaic compound along with halogen then AgNO_(3) solution is added in acidified sodium fusion extract. (R) On acidification, NaCN and Na_(2)S decompose. NaCN + HNO_(3) rarr NaNO_(3)+HCNuarr Na_(2)S +2HNO_(3) rarr 2NaNO_(3)+H_(2)Suarr |
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Answer» If both assertion and REASON are correct and reason is the correct EXPLANATION of the assertion |
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| 50015. |
(a). If both (A) and (R) are correct and (R) is the correct explanation for (A). (b). If both (A) and (R) are correct and (R) is the correct explanation for (A). (c). If (A) is correct and (R) is incorrect. (d). If (A) is incorrect and (R) is correct. ltbr. (e). If both (A) and (R) are incorrect. Q. Assertion (A): In organic layer test, Cl_2 water is added to the sodium extract, which oxidises Br^(c-) and I^(c-) ions to Br_2 andI_2, respectively. Reason (R): Reduction potential of Cl_2 is greater than that of Br_2 and I_2. |
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Answer» `Cl_2+Br^(c-)rarr2Cl^(c-)+Br_2` `Cl_2+I^(c-)rarr2Cl^(c-)+I_2` Reduction POTENTIAL `E^@(Cl_2)/(CL^(c-))gtE^@(Br_2)/(Br^(c-))` or `(E^@I_2)/(I^(c-))`. |
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| 50016. |
(a). If both (A) and (R) are correct and (R) is the correct explanation for (A). (b). If both (A) and (R) are correct and (R) is the correct explanation for (A). (c). If (A) is correct and (R) is incorrect. (d). If (A) is incorrect and (R) is correct. ltbr. (e). If both (A) and (R) are incorrect. Q. Assertion (A): Dumas method is more applicable to nitrogen containing organic compounds than Kjeldahl's method. Reason (R): Kjeldahl's method does not give stisfactory results for compounds in which N is linked to O atom. |
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Answer» |
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| 50017. |
(a). If both (A) and (R) are correct and (R) is the correct explanation for (A). (b). If both (A) and (R) are correct and (R) is the correct explanation for (A). (c). If (A) is correct and (R) is incorrect. (d). If (A) is incorrect and (R) is correct. ltbr. (e). If both (A) and (R) are incorrect. Q. Assertion (A): Essential oils are volatile and are isoluble in H_2O Reason (R): Essential oils are purified by steam ditillation. |
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Answer» |
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| 50018. |
A ideal gas does work on its surroundings when it expands by 2.5 L against external pressure 2 atm. This work done is used to heat up 1 mole of water at 293 K. What would be the final temperature of water in kelvin if specific heat for water is "4.184 Jg"^(-1)"K"^(-1)? |
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Answer» 300 `W=-2xx2.5=-"5 L atm"=-"506.3 J"` Because this work is used in raising the temperature of water, so work done is equalto the HEAT supplied i.e., `w=q=m.c_(s).DeltaT` Given that, `m= 18g"(= 1 mole)", c_(s)=4.184g^(-1)K^(-1)`, `q=+506.3J, DeltaT=? `(Heat is given to water) `DeltaT=(q)/(c_(s).m)=(506.3)/(4.184xx18)=6.72` `therefore"FINAL temperature, "T_(f)=T_(i)+DeltaT=293+6.72` `=299.72K~~300K` |
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| 50019. |
(A) I_(3)^(-) ion is linear. ( R) It is not in sp hybridized state. |
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Answer» Both (A) and (R) are true and (R) is the correct EXPLANATION of (A) |
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| 50020. |
(a) (i) Radius of a cation is smaller then the parent atom. Account for the following. |
| Answer» Solution :A cation os formed by loss of one or more electrons . The NUCLEAR charge REMAINS the same but the number of electrons becomes less than the PARENT RESULTING in the increase In the effective nuclear charge per electrons. This causes decrease in size . | |
| 50021. |
A hypothetical reaction A to 2B, proceed through following sequence of steps A to C , Delta H = q _(1) , C to D, Delta H = q _(2)(1)/(2) D to B , Delta H = q _(3) The heat of reaction is |
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Answer» `q _(1) + q _(2) + 2q _(3)` |
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| 50022. |
A hypothetical electromagnetic wave is shown in Figure. Find out the wavelength of the radiation. |
| Answer» Solution :The w avelength is distance BETW EEN two successive peaks or two successive troughs of a wave. So, `lambda = 4 XX 2.16 pm = 8.64 pm` | |
| 50023. |
A hydroxy acid on heating gives a 5-membered lactone. The acid is |
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Answer» `CH_(2)OHCH_(2)CH_(2)COOH` |
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| 50024. |
A hydrogen sample is prepared in a particular excited state A. Photons of energy 2.55eV/atom gel absorbed into the sample to take some of the electrons to a further excited state B. As per the number of photons when an electron in the higher excited state returns to the ground state, which of the following are may not be possible? |
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Answer» 15 `n_A = 2 , n_B = 4` `THEREFORE ` no. of photons EMITTED `=(4xx3)/(2) = 6` |
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| 50025. |
A hydrogen molecule and helium atom are moving with the same velocity. Then the ratio of their de Brogile wavelength is |
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Answer» `3.313xx10^(-20)CM^(-1)` |
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| 50026. |
A hydrogen molecule and helium atom are moving with the same velocity. Then the ratio of their de Broglie wavelength is |
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Answer» `1:1` |
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| 50027. |
A hydrogen like atom in ground state absorbs .n. photons having the same energy nad it emits exactly .n. photons when electronic transition takes place. Then the energy of the absorbed photon may be |
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Answer» 91.8 eV |
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| 50028. |
A hydrogen like atom with atomic number 'Z' is in higher excited state of quantum number 'n'. This excited state can make a transition to the first excited state by successively emitting two photons energies 10 eV and 17 eV respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by emitting two photons of energies 4.25 eV adn 5.95 eV respectively. Calculate the values of 'n' and 'Z' |
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Answer» Solution :For the given EXCITED state, `n_(2) = n` 1st excited state means, `n_(1) =2` 2nd excited state means, `n_(1) =3` ENERGY difference between n = 2 and n = 3 will be `= (10 + 17) - (4.25 + 5.95) = 16.8 eV` `:. 16.8 = 13.6 ((1)/(2^(2)) - (1)/(3^(2))) Z^(2) = 13.6 xx (5)/(36) xx Z^(2)` or `Z^(2) = (16.8)/(13.6) xx (36)/(5) = 9 or Z = 3` Energy difference between `n_(2) = n and n_(1) = 2` will be `= 10 + 17 eV = 27 eV` `:. 27 = 13.6 ((1)/(2^(2)) - (1)/(n^(2))) xx 3^(2)` or `3 = 13.6 ((1)/(4) - (1)/(n^(2))) or (1)/(4) - (1)/(n^(2)) = (3)/(13.6) or (1)/(n^(2)) = (1)/(4) - (3)/(13.6) = (13.6 - 12)/(4 xx 13.6)` or `n^(2) = (4 xx 13.6)/(1.6) or n = 6` 
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| 50029. |
A hydrogen-like atom (atomic number Z) is in a higher excited state of quantum number n. This excited atom can make a transition to the first excited state by successively emitting two photons of energies 10.20 eV and 17.00 eV respectively.Alternatively the atom from the same excited state can make a transition to the second excited state by snccessively emitting two photons of energies 4.25 eV and 5.95 eV respectively. Determine the values of n and Z (ionization energy of hydrogen atom = 13.6 eV) |
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Answer» n=5  `E= (-2^2)/(n^2) (13.6eV)` `impliesR.hc.z^2 (1/9 - 1/n^2) = 27.2 ….. (1)` `impliesR.hc.z^2 (1/9 - 1/n^2) = 10.2 eV…(2)` `((1))/((2))= (((n^2 - 4)/(4n^2)))/(((n^2 - 9 )/(9n^2))) = (27.2)/(10.2) = 2.66` `9(n^2 - 4)xx 4 (n^2 - 9) (2.66)` `implies9n^2 - 36 = 10 . 66 n^2 - 96` `therefore 1.66 n^2 = 60 impliesn^2 =(60)/(1.66) = 36` |
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| 50030. |
(A) : Hydrogen has only one electron in its orbit but produces several spectral lines (R) : There are many excited energy levels available in a sample of Hydrogen gas |
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Answer» Both (A) and (R) are TRUE and (R) is the correct explanation of (A) |
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| 50031. |
A hydrogen atom in the ground state is excited by monochromatic ratiation of wavelength lambda Å . The resulting spectrum consists of maximum 15 different lines. What is the wavelength lambda ? (R_H = 109677.8 cm^(-1)) |
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Answer» `937.3 Å` `bar(upsilon) = 1/(lambda) = R(1/1 - 1/(36)) = (35R)/(36)` `lambda =(36)/(35R) = 937.3A^@` |
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| 50032. |
A hydrogen atom in an excited state emits a photon which has the longest wavelength of the Paschen series. Further emissions from the atom cannot include the |
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Answer» longest wavelength of the Lyman series |
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| 50033. |
A hydrogen atom emits a photon of energy 12.1 eV. By how much will the angular momentum of the electron change? |
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Answer» Solution :`E_(n_(2)) - E_(n_(1)) = 12.1 EV = 13.6 ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))` This equation is correct for transition from `n_(2) = 3 " to " n_(1) = 1` Angular MOMENTUM `= (NH)/(2pi)` `:.` Change in angular momentum `= (h)/(2pi) (n_(2) - n_(1)) = (h)/(2pi) (3 -1) = (h)/(pi) = (6.626 xx 10^(-34)Js)/(3.14)` `= 2.11 xx 10^(-14) Js` |
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| 50034. |
A hydrocatbon with formula C_8H_(18) He gives one monochloro derivative. The hydrocarbon is: |
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Answer» n- OCTANE |
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| 50035. |
A hydrocarbon (X) was found to have a molecular weight of 80-85. A 10.02 mg sample took up 8.40mL of H_(2) gas measured at 0^(@)C and 760 mm pressure. Ozonolysis of (X) yields only HCHO and OHC-CHO. What was hydrocarbon? |
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Answer» `C_(6)H_(8)` `= (8.40 xx 10^(-3))/(22.4) = 3.75 xx 10^(-4)` Moles of compound `=(10.02 xx 10^(-3))/(22.4) = 1.254 xx 10^(-4)` THUS, moles of `H_(2)` taken up by 1 mole of compound `=(3.75 xx 10^(-4))/(1.254 xx 10^(-4)) =3` Therefore, molecules has three double bonds and it may be hexa-1,3,5-triene. |
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| 50036. |
A hydrocarbon 'X' on reduction gives 2-methyl butane. On ozonolysis it gives acetone as one of the products. The hydrocarbon is |
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Answer» `CH_(3)-CH_(2)-CH=CH_(2)` |
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| 50037. |
A Hydrocarbon X (M.F.C_4H_6) produces an aldehyde Y through Hydrocarbon Oxidation and a ketone Z through Oxymercuration Demercuration. Y and Z are functional isomers. X gives P when treated with excess of HOCl and Q when treated with excess of HCl. The structure of X is : |
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Answer» `CH_3-C-=C-CH_3` |
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| 50038. |
A Hydrocarbon X (M.F.C_4H_6) produces an aldehyde Y through Hydrocarbon Oxidation and a ketone Z through Oxymercuration Demercuration. Y and Z are functional isomers. X gives P when treated with excess of HOCl and Q when treated with excess of HCl. The correct statement is : |
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Answer» <P>P and Q are POSITIONAL isomers |
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| 50039. |
A hydrocarbon 'X' have 81% of carbon. Volume of CO_(2) liberated at 298K and 76 cm of Hg when 0.55gm of 'x' undergoes combustion |
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Answer» 0.90l `0.55g --------rarr 0.4455 g` `12g rarr24.4658l (V=(nRT)/(P)=(1xx0.0821xx298)/(1))` `0.4455 g rarr 0.908 l` |
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| 50040. |
A hydrocarbon was found to contain 91.3%C. The molecular formula of the compound is |
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Answer» `C_(6)H_(6)` |
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| 50041. |
A hydrocarbon that reacts with sodium in liquid NH_3is |
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Answer» `CH_3 - CH_2 - C -= C - H` |
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| 50042. |
A hydrocarbon P has molecular formula C_8H_(16), and it does not undergo hydrogenation with raney Ni. A on free radical chlorination yields three monochloro derivatives Q, R and S (all structural isomers) with their molecular formula C_8 H_(15)Cl. Only Q can show optical isomerism R. On further chlorination produces a dichloride C_8 H_(14) Cl_2, as one of the several isomers which on heating with sodium metal in etheral solution gives bicyclo [2, 2, 2] octane. How many stereoisomers are possible for Q? |
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Answer» 2  It has three chiral CARBON , HENCE , ` 2^(3) =8` stereoisomers .
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| 50043. |
A hydrocarbon P has molecular formula C_8H_(16), and it does not undergo hydrogenation with raney Ni. A on free radical chlorination yields three monochloro derivatives Q, R and S (all structural isomers) with their molecular formula C_8 H_(15)Cl. Only Q can show optical isomerism R. On further chlorination produces a dichloride C_8 H_(14) Cl_2, as one of the several isomers which on heating with sodium metal in etheral solution gives bicyclo [2, 2, 2] octane. If S is dichlorinated, how many isomers would be produced? (exclude stereoisomers) |
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Answer» 2 
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| 50044. |
A hydrocarbon P has molecular formula C_8H_(16), and it does not undergo hydrogenation with raney Ni. A on free radical chlorination yields three monochloro derivatives Q, R and S (all structural isomers) with their molecular formula C_8 H_(15)Cl. Only Q can show optical isomerism R. On further chlorination produces a dichloride C_8 H_(14) Cl_2, as one of the several isomers which on heating with sodium metal in etheral solution gives bicyclo [2, 2, 2] octane. Which of the following satisfy the criteria for P? |
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Answer»
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| 50045. |
A hydrocarbon ono oxidation ozonolysisproduces Oxalic acid and Butanedioic acid. Its structure is |
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Answer»
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| 50046. |
A hydrocarbon of molecular formula C_(4)H_(6) decolourises bromine water and gives white ppt with Tollens reagent. This hydrocarbon on hydration gives butanone. The hydrocarbon is : |
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Answer» 1, 3 - butadiene |
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| 50047. |
A hydrocarbon has the molecular mass 70. Write all the possible structural isomers and give their IUPAC names |
| Answer» SOLUTION :The hydrocarbon having MOLECULAR MASS of 70 is `C_5H_10`. It exists in the following four structural isomers : (i)`CH_3CH_2CH_2CH=CH_2` (pent-1-ene), (ii)`(CH_3)_2CH-CH=CH_2` (3-methylbut-1-ene) , (III)`CH_3CH_2CH=CHCH_3` (pent-2-ene) and (IV)`(CH_3)_2C=CHCH_3` (2-methylbut-2-ene)] | |
| 50048. |
A hydrocarbon decolourised bromine water. On ozonolysis it gives 3-methyl butanal and acetaldehyde. Write the structure of the hydrocarbon. |
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Answer» Solution : Since the given hydrocarbon decolourises bromine WATER, it MUST be an unsaturated hydrocarbon. On ozonolysis it gives two aldehydes. Therefore, it must be an alkene. The products of ozonolysis are `UNDERSET("3-methylbutanal")(CH_3 - undersetoverset(|)(CH_3)(C)H - CH_2 - oversetunderset(|)(H)(C)= O)underset("acetaldehyde")(CH_3 - oversetunderset(|)(H)(C) = O)` The STRUCTURE of the given alkene can be obtained by joining the oxygenated carbons of the two aldehydes by a double BOND. Hence, the given hydrocarbon is `underset(5-methylhex-2-ene ")(CH_3- undersetoverset(|)(CH_3)(C)H - CH_2 - oversetunderset(|)(H)(C) = oversetunderset(|)(H)(C)- CH_3` |
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| 50049. |
A hydrocarbon contains 85.7 % C. If 42 mg of the compound contains 3.01xx10^(20)molecules, the molecular formula of the compound will be ...... |
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Answer» `C_(3)H_(6)` `:.` Mass of `6.02xx10^(23)` atoms `=84g` `:.` Molecular formula : `C_(6)H_(12)` (C of 85.7) |
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| 50050. |
A hydrocarbon contains 90% carbon by weight. What is its empirica formula? |
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Answer» |
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