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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.
| 86701. |
A transition metal exists in its highest oxidation state. It is expected to behave as |
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Answer» A chelating AGENT |
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| 86702. |
A transition metal A has 'spin-only' magnetic moment value of 1.8 Bm. When it is reacted with dilute sulphuric acid in the presence of air, its compound B is formed. B reacts with compound C to give compound D with the liberation of iodine. Then the metal A and compounds B, C and D are respectively |
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Answer» `TI,TiSO_(4), KI and TiI_(2)` |
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| 86703. |
Transition elements show catalytic property. Give two reasons. |
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Answer» Solution :(a) (i) They show variable oxidation states (ii) Large surface area (iii) presence of incompletely FILLED or partially filled d-orbitals or vacant d-orbital (iv) Formation of intermediate compounds. (b) Zinc (ZN) or SCANDIUM (Sc) (a) Two reasons, because transition elements show catalytic property. (i) Presence of vacant d-orbitals. (ii) Ability to exhibit variable VALENCES. (b) Zinc (Zn) is the element of 3d-series that does not show variable valency (oxidation state) due to as electronic configuration. Electronic configuration of `Zn : [Ar] 3d^(10) 4s^(2)`. |
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| 86704. |
(A): Transition elements form alloys very easily. (R): They exhibit variable oxidation states. |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
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| 86705. |
(A): Transition elements form alloys easily (R) : Transition elements have high melting points |
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Answer» Both (A) and (R) are true and (R) is thecorrect EXPLANATION of (A) |
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| 86706. |
A transition element X has the configuration [Ar]d^(4) in its +3 oxidation state. Its atomic number is |
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Answer» 25 |
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| 86707. |
A transition element with atomic number 24 will have its magnetic moment : |
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Answer» 5.92 B.M Configuration : [AR] `4s^(1)3d^(5)` No. of unpaired electrons= 6 `THEREFORE mu = sqrt(6(6+2)) = sqrt(48) = 6.93` B.M. |
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| 86708. |
(A) Towards S_N2 reaction, order of reactivity is CH_3Br > CH_3CH_2Br > (CH_3)_2CHBr > (CH_3)_3 C Br. (R) Greater the stability of carbocation, greater will be its ease of formation from alkyl halide and faster will be the rate of S_N1 reaction. |
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Answer» Both A & R are TRUE, R is the CORRECT explanation of A |
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| 86709. |
A total of 69500 C of electricity was required to reduced 37.7 g of M^(3+) to the metal. What is M ? |
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Answer» |
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| 86710. |
(A) :Tolene on chlorination, followed by hydrolysis gives benzaldehyde (R) : Benzal chloride is formed by chlorination of benzene . |
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Answer» Both A & R are TRUE, R is the CORRECT EXPLANATION of A |
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| 86711. |
(A) To prepare ter.butylethylether, the reactants used are ter.butylbromide and ethoxide solution(R) Williamson's synthesis is an electrophilic substitution reaction of RX with alkoxide ion |
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Answer» Both (A) and (R) are TRUE and (R) is the correct EXPLANATION of (A) |
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| 86712. |
(A) : To prepare aldehydes from primary alcohols , acidifiedKMnO_4 " or" K_(2) Cr_(2)O_7 is not used.(R) : Strong oxidising agents like acidified KMnO_(4) " or" K_(2)Cr_(2)O_(7)oxidise aldehydes to carboxylic acids . |
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Answer» Both A & R are true, R is the CORRECT EXPLANATION of A |
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| 86713. |
A to F alkenes with minimum possible carbon. (i). Aunderset(Delta)overset(H^(o+)//KMnO_(4))toMeCOOH as the only product (iii). Cunderset(Delta)overset(H^(o+)//KMnO_(4))toMeCH_(2)COOH as the only organic product. (iv). (v). Eunderset(Delta)overset(H^(o+)//KMnO_(4))toHOOC-C-C-C-overset(O)overset(||)(C)-underset(O)underset(||)(C)-C-C-C-COOH (vi). Funderset(Delta)overset(H^(o+)//KMnO_(4))to acetone+ethanoic acid |
Answer» SOLUTION :
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| 86714. |
A to B , K_1= 0.693 sec^(-1) C to D, K_2 = 0.693 "min"^(-1) If t_1 and t_2are half lives of two reactions, then |
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Answer» `t_1 = t_2` |
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| 86715. |
A to B is a first order reaction with rate 6.6 xx 10^(-5) m.s^(-1). When [A] is 0.6 m, rate constant of the reaction is- |
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Answer» `1.1 XX 10^(-5)s^(-1)` |
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| 86716. |
A toB and C toD are first order reactions. Ratio of t_(99.9%)values is 4:1, then, ratio of rate constants K_1to K_2is |
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Answer» `4:1` |
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| 86717. |
A to 2B + C 3B to 2D Find moles of D produced if initially 3 moles of A are taken. |
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Answer» 4 |
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| 86718. |
(A): Ti(IV) in aqueous solutions is colour less. (R): Ti(IV) has no electrons in d-subshell. |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
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| 86719. |
(a) [Ti(H_(2)O)_(6)]^(3+) absorbs light of wavelength 5000Å. Name one ligand which would form a titanium (III) complex absorbing light to lower wavelength than 5000Å and one ligand which would form a complex absorbing light of wavelength higher than 5000 Å. (b) Calculate the magnetic moments (spin only ) of the following complexes (i)[PtCl_(6)]^(2-) "" (ii) [Cr(CO)_(6)] "" (iii) [Ir(NH_(3))_(6)]^(3+)"" (iv) [Pd(en)_(2)]^(2+) |
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Answer» (B) (i) 0 (ii)0 (iii)0 (IV)0 |
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| 86720. |
A thixotropic gel is characterised by |
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Answer» reversible TRANSFORMATION of GEL into sol on heating. |
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| 86721. |
(A) Thionyl chloride reacts with primary alcohols to form pure alkyl halides in the presence of pyridine. (R) In the reaction between SOCl_2 and R-OH, SO_2 escapes from the reaction mixture and HCI is absorbed by pyridine. |
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Answer» Both A & R are TRUE, R is the correct explanation of A |
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| 86722. |
A third group is least likely to enter between two groups in the meta relationship. This is the result of steric hindrance and increases in importance with the size of the groups on the ring and with the size of the attacking species. When a Meat-directing group is meta to an ortho-para directing group, the incoming group primarily goes ortho th the meta directing group rather than para. |
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Answer»
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| 86723. |
A thermometer which can be used only accurate measurement of small differences in temperature is know as a : |
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Answer» BECKMANN's thermometer |
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| 86724. |
A Thermometer designed by Beckman is useful to measure the absolute values of |
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Answer» B.P |
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| 86725. |
A thermodynamics state function is a quantity |
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Answer» used to determine HEAT changes |
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| 86726. |
(A): Thermodynamically all solids possess a tendency to acquire defects (R) : During defects the entropy of the system increases in solids. |
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Answer» Both (A) and (R) are true and (R) is the CORRECT explanation of (A) |
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| 86727. |
A thermally isolated gaseous system can exchange energy with the surroundings. The mode of transference of energy can be |
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Answer» heat |
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| 86728. |
A thermally isolated gaseous system can exchange energy with the surrounding. The mode of transference of energy can be: |
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Answer» HEAT |
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| 86729. |
A thermally insulated container initially holds N_0 molecules of an ideal monoatomicgas at an absolute temperature T_0. Molecules escape from the container through a small hole in the wall, and it can be shown that in such a process at a temperature T, the average kinetic energy of the escaping molecules is 2 KT. How many molecules remain in the container when the temperature has fallen to T_0// 2? |
| Answer» SOLUTION :`N_0// 8` | |
| 86730. |
(A) : Thermal stability of the hydrides of VIA group elements decreases from H_(2)O to H_(2)Po (R) : The heats of dissociation of M - H bond of hydrides of VIA group decreases doun the group |
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Answer» Both (A) and (R) are TRUE and (R) is the correct explanation of (A) |
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| 86731. |
(A): Thermal stability of VA group hydrides decreases from NH_(3) to BiH_(3) (R): The dissociation energy of M - H bond! increases down the group regularly |
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Answer» Both (A) and (R ) are TRUE and (R ) is the correct explanation of (A) |
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| 86732. |
(A): Thermal stability of the hydrides of VIA group elements decreases from H_(2)O to H_(2)Po (R): The heats of dissociation of M-H bond of hydrides of VIA group decreases down the group |
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Answer» Both (A) and (R ) are TRUEAND (R ) is the correct explanation of (A) |
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| 86733. |
(A) Thermal stability of 16 ^(th) group hydrides decreases from H _(2) O to H_(2) PO (R ) From H _(2)O to H_(2)PObond dissociation energy increases. |
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Answer» Both (A) and (R ) are TRUE and (R ) is the CORRECT explanation of (A) |
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| 86734. |
(A) There is no sulphur sulphur bond in thiosulphate (R ) All sulphur sulphur bonded compounds are persulphates |
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Answer» Both (A) and (R ) are true and (R ) is the CORRECT EXPLANATION of (A) |
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| 86735. |
A: There is very less gap between the value of radii of 4d and 5d elements R: Size of 5d is more than 4d subshell |
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Answer» If both Assertion & REASON are true and the reason is the correct explanation of the assertion, then MARK (1) |
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| 86736. |
(A) The volume of 1 mole of an ideal gas at 1 bar pressure at 25^(@)C is 24.78 litre. (R) 1 bar=0987atm. |
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Answer» If both (A) and (R) are CORRECT and (R) is the correct EXPLANATION of (A). |
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| 86737. |
(A) The two oxygen-oxygen bond lengths in the ozone molecule are identical (128 pm) (R ) Ozone exhibit resonance. |
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Answer» Both (A) and (R ) are true and (R ) is the CORRECT EXPLANATION of (A) |
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| 86738. |
(a) The wave function of an electron in 2s orbital in hydrogen atom is given below: psi_(2s)=1/(4(2pi)^(1//2))(z/a_(0))^(3//2)(2-r/a_(0))exp(-r//2a_(0)) where a_(0) is the radius. This wave function has a radial node at r=r_(0). Express r_(0) in terms of a_(0). (b) Calculate the wavelength of a ball of mass 100 g moving with a velocity of 100 ms^(-1). (c) _(92)X.^(238)underset(-6beta)overset(-8alpha)(rarr) Y. Find out atomic number, mass number of Y and identify it. |
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Answer» `Psi_("at node")^(2)=0` (probability of finding an electron is zero at node) for node at `r=r_(0), PSI^(2)=0` So, `Psi^(2)=0=(1)/(4sqrt(2pi))[1/a_(0)]^(3), [2-r_(0)/a_(0)]xxe^(r_(0)//2a_(0))RARR [2-r_(0)/a_(0)]=0` or `2=r/a_(0)` `rArr r=2a_(0)` (b) The wavelength can be calculate with the help of de-Broglie's FORMULA i.e., `lambda=h/(mv)=(6.626xx10^(-34))/(100xx100xx10^(-3))=(6.626xx10^(-34))/(10,000xx10^(-3))=6.626xx10^(-35) m` or `6.626xx10^(-25) Å` (c) (i) The atomic mass of element reduces by `4` and atomic number by `2` on emission of an `alpha`-particle. (ii) the atomic mass of an element remains unchanged and atomic increases by `1` on emission of a `beta`-particle. Thus change in atomic mass on emission of `8alpha`-particles will be `8xx4=32` New atomic mass = old atomic mass `-32=238-32=206` similarly change in atomic number on emission of `8alpha`-particle will be: `8xx2=16` i.e., New atomic number = old atomic `-16=92-16=76` On emission of `6beta`-particles the atomic mass remains unchanged thus, atomic mass of the new element will be `206`. The atomic number increases by `6` unit thus new atomic number will be `76+6=82` Thus, the equation looks like: `._(92)X.^(238) underset(-6beta)overset(-8alpha)(rarr)._(82)Y^(206)` |
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| 86739. |
(a) The vapour pressures of benzene and toluene at 293 Kare 75 mm Hg and 22 mm Hg respectively. 23.4 g of benzene and 64.4 g of toluene are mixed. If the two form an ideal solution, calculate the mole fraction of benzene in the vapour phase assuming that the vapour pressures are in equilibrium with the liquid mixture at this temperature. (b) What is meant by +ve and -ve deviations from Raoult'slaw and how is the sign of H solution related to +ve and -ve deviations from Raoult's law ? |
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Answer» SOLUTION : (a) Calculation of the total vapour pressure Moles of benzene `(N_(A))` ` = ("Mass of Benzene")/("molar mass of benzene " (C_(6) H_(6)))` =` ((23.4 g))/((78 "g mol"^(-1))) = 0.3 ` mol Moles of toluene `(N_(B))` = `(" mass of toluene")/("Molar mass of toluene " (C_(7) H_(8)))` `= ((64.6 g))/((92g "mol"^(-1))) ` = 0.7 mol Mole fraction of benzene `(X_(A))` `= (n_(A))/(n_(A) + n_(B))` = `((0.3 "mole"))/((0.3 " mole " + 0.7 "mol")) = 0.3` M.ole fraction of toluene`(X_(B))` `= (n_(B))/(n_(A) + n_(B))` = `((0.7 "mole"))/((0.3 " mole" + 0.7 "mol")) = 0.7` Vapour pressure of toluene in solution`(P_(A))` `P_(A)^(0) xx X_(A) = (75 mm) xx 0.3` = 22.5 mm Vapour pressure of toluene in solution` (P_(B))` `P_(B)^(0) xx X_(B)` = (22 mm) `xx 0.7` = 15.4 mm Total vapour pressure of solution = `P_(A) + P_(B)` = 22.5 + 15.4 = 37.9 mm Calculation of mole fraction of benzene in the vapour phase. Mole fraction of benzene in vapour phase `= ("Partial vapur pressure of benzene ")/("Total vapour pressure ")` = `((22.5 mm))/((37.9 " mm")) = 0.591` (b) SOLUTIONS showing positive deviations, the partial vapour pressure of each component of solution is greater than the vapour pressure as expected according to Raoult.s low. Solutions showing negative deviations, the partial vapour pressure of each component of solution is LESS than vapour pressure as expected according to Raoult.s law. For posivitve DEVIATION `Delta H_("mixing ") = + `ve For negative deviation `Delta_("mixing")` = -ve |
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| 86740. |
(A) Thethermodynamical reaction proceeds forward if Delta G is negative in the equation Delta G = Delta H - T Delta S (R) If Delta S and Delta Hare positive and with increasing temperature,T Delta Sincreases |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
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| 86741. |
a. The treatment of RX with aqueous KOH leads to the information of alcohols but in the presence of alcholic KOH or NaOH, alkenes are themajor products. Expalin why. b. CHF_(3) is less acidic than CHCl_(3). Explain. c. Wurtz reactionin case of tert-alkyl halide fails. Explain. d. Dipole moment of C_(6) H_(5) Cl is lower thanthat of C_(6) H_(11) Cl (cuclohexul choride). Expalin why. e. Why should Grignardreagent be preparedunder anthydrous conditions? f. Why does p- dichlorobenzene have higher melting point and lower solubility than o- and m- isomers ? |
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Answer» Solution :a. `NaOH` or `KOH` is COMPLETELY ionised in aquioes solution to give `Oh` ions, which acts as a strongnuclophile, so `SN` reactiontakes placewith `RX` to give alcholos. Moreover, `OH` ions are hydrarded or solvated in aquires soltion which reduces the basic character of `overset(ᶱ)(O)H` that cannot abstact acidic `beta-H` atom of `EX` to from alkene. In case of alcholic solution of `NaOH` or `KOH, overset(ᶱ)(O)H`reacts with `ROH` to fromRO` (alkoxide ion) which is a stronger base than `overset((-))(O)H`, and `RO^((-))` can abstract acidic `beta-H` atom of `RX` easily to form alkense. d `-I` effect of `FgtCl`. According to `-I` effect, `CHF_(3)~ should be more acidic than `CHCl_(23)`. but this is not observed. the conjugate base `overset(..)overset((-))(C)Cl_(3)` is resonance stabilised due to the presence of `d`-orbital in `Cl[2p(C)-3d(Cl)]` overlap. But conjugate base `overset(..)overset((-))(C)F_(3)` is not resonance stabilised due to the absence of `d`-orbital in `F`.  (c) `Me_(3)C-X+2Natounderset("t-Butylsodium")Me_(3)overset((-))(C)-overset(o+)(Na)+NaX` `t`-ALKYL halides undergo dehydrohalogenation in the presence of storng base such as `Na` metal rather thanWurtz reaction.  Therefore, `1^(@)` and `2^(@) RX` undergo Wurtz reaction, while `3^(@) RX` undergo dehydroghalogenation to give alkenes.  (i) `sp^(2)` -HYBRID `C` atom is more `EN` than `sp^(3)` -hybrid `C` atom . (ii) Therefore `sp^(2)` -hybrid `C` atom of `(C-Cl)` bond in (i) has less tendency to release electrons to `Cl` than ans `sp^(3)` -hybrid `C` of `(II)`. (iii) Therefore,`(C-Cl)` bond in `(I)` less polar than in (II), i.e., magnitude opf `delta^((-))` charge is less on `Cl` atom of `(I)` than that of (II) (iv) Due top resonance in (I), the `(C-Cl)` bond in `(I)` acquire some double bond character and is shorter than that in (II). (v). Since `mu=qxxd` (charge `xx`distance), therefore, `(I)` has lower `mu` than `(II)`. (e) Grignard reagents are very reactive and they react with moisture present INT the reaction mixture, therefore they must be prepared under anhydrous conditions.  (f) Due to symmetrical packing in its crystal lattice of `p`-isomer, they have stronger intermolecular force of attraction than `o-` and `m-` isomers. Therefore, large amount of energy is required for its melting than for the `o-` and `m-`isomers. The enthalpy of intermolecular force of attraction in `p`-isomer is greater than the enthalpy of dissolution. hence `p-`isomer has lower solubility than its isomers. |
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| 86742. |
(A) The study of radon is difficult among noble gases(R) Radon is radio active in nature |
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Answer» Both (A) and (R) are true and (R) is the CORRECT explanation of (A) |
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| 86743. |
(A): The spin only magnetic moment of Sc^(+3) 1.73 BM (R) : The spin only magnetic moment (in BM) of an ion is equal to sqrt(n(n+2)) |
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Answer» Both (A) and (R) are true and (R) is the correct EXPLANATION of (A) |
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| 86744. |
(A) : The solubility of maleic acid in water is higher than that of fumaric acid . (R) : Maleic acid and fumaric acids are cistrans isomers. |
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Answer» Both A & R are true , R is the correct EXPLANATION of A |
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| 86745. |
(A) The simplest alpha -amino acid is optically inactive. (R) Simplest alpha amino acid has no chiral carbon centre. |
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Answer» Both A & R are true, R is the correct explanation of A |
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| 86746. |
A : The S-S-S bond angle in S_8 molecule is 105° R : S_8 has a V-shape. |
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Answer» a |
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| 86747. |
(a) The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500Omega. What is the cell constant if the conductivity of 0.11 M KCl solution at 298 K is 0.146xx10^(-3)"S "cm^(-1)? (b) Predict the products of electrolysis of the solution of H_(2)SO_(4) with platinum electrodes. |
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Answer» |
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| 86748. |
(a) The resistance of 1/10 M solution is found to be 2.5 xx 10^(3) ohms. Calculate molar conductance (Given Cell constant = 1.15 cm). (b) Mention two general methods for prevention of corrosion. |
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Answer» Solution : (a) `k=1/Axx1/R=1.15xx1/(25xx10^(3))=4.60xx10^(-4)`ohm cm ` wedge_(m)=(1000xxk)/M=(1000xx4.60xx10^(-4))/0.1=4.60 ohm^(-1) cm^(2) mol^(-1)` (i) painting(ii) Electrochemical METHOD |
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| 86749. |
(A) The reduction of metal oxide is easier if the metal is in liquid state at temp ofreduction (R) Delta Sis -ve and Delta G is + ve |
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Answer» Both (A) and (R) are TRUE and (R) is the correct explanation of (A) |
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| 86750. |
(A) The reduction of Cr_(2) O_(3) with Al posses Delta G^(0)= - ve but does not take place at room 'T' (R) Certain amount of E_(a)is required and hence heating is required . |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT explanation of (A) |
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