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. |
According to VSEPR theory, the repulsion between different parts of electrons obey the order . |
|
Answer» `l.p - l.p GT b.p - b.p gt l.p - b.p` |
|
| 2. |
According to V.B.T. a covalent bond is formed by the overlapping of half filled atomic orbitals with electron spinning in the opposite direction. The strength of a bond depends on extent of overlapping. Double bond energy is not twice that of single bond energy because a double bond is made up of a weak a bond and strong o bond.Bond energy is highest in |
|
Answer» `F_2 ` |
|
| 3. |
According to V.B.T. a covalent bond is formed by the overlapping of half filled atomic orbitals with electron spinning in the opposite direction. The strength of a bond depends on extent of overlapping. Double bond energy is not twice that of single bond energy because a double bond is made up of a weak a bond and strong o bond.More energy is required to break the bond between the carbon atoms in |
|
Answer» `c_2 H_6 ` |
|
| 4. |
According to V.B.T. a covalent bond is formed by the overlapping of half filled atomic orbitals with electron spinning in the opposite direction. The strength of a bond depends on extent of overlapping. Double bond energy is not twice that of single bond energy because a double bond is made up of a weak a bond and strong bond energy The expected bond angle in NH_3 molecule according to V.B.T is |
| Answer» Answer :A | |
| 5. |
According to V.B.T. a covalent bond is formed by the overlapping of half filled atomic orbitals with electron spinning in the opposite direction. The strength of a bond depends on extent of overlapping. Double bond energy is not twice that of single bond energy because a double bond is made up of a weak a bond and strong o bond.According to V.B.T, in the formation of CH_4 molecule the type of overlappings observed are |
|
Answer» `p-s ,s-s` |
|
| 6. |
According to V.B. Theory, the bonds in methane are formed due to the overlapping |
|
Answer» `1sigma s-s,3sigmas-p` |
|
| 7. |
According to van't Hoff equation, the value of osmotic pressure pi is equal to …………………… |
| Answer» SOLUTION :`PI =CRT` | |
| 8. |
According to Valence bond theory, a bond between two atoms is formed when |
|
Answer» FULLY FILLED ATOMIC orbitals overlap |
|
| 9. |
According to valence bond theory, water molecule has |
|
Answer» TWO DATIVE BONDS and BOND ANGLE`90^(@)` |
|
| 10. |
According to their electronic configuration which of the following elements belongs to some group. (i) 1s^(2) 2s^(2) "" (ii) s^(2) 2s^(2) 2p^(5) (iii) 1s^(2) 2s^(2) 2p^(6) 3s^(2)3p^(5) "" (iv) 1s^(2) 2s^(2) 2p^(6) 3s^(10) 4s^(1) |
| Answer» Solution :(ii) and (III) both are in `17^(TH)` group. | |
| 11. |
As per the modern periodic law, the physical and chemical properties of elements are periodic functions of their |
|
Answer» NUCLEAR MASSES |
|
| 12. |
According to the molecular formula, benzene appears to be highly unsaturated. But bezene mainly participatesin substitutionreactions. It is because benzene is |
|
Answer» UNSATURATED |
|
| 13. |
According to the kinetic theory of gases, in an ideal gas, between two successive collisions, a gas molecule travels |
|
Answer» in a circular path THUS, `E._k prop T` Hence, `(E._k(313K))/(E._k(293K))=313/293` |
|
| 14. |
According to the kinetic theory of gases, |
|
Answer» the PRESSURE EXERTED by a gas is proportional to mean square velocity of the molecules `:.` K.E. depends on temperature |
|
| 15. |
According to the kinetic molecular theory, the collision between gas molecules is perfectly elastic. |
|
Answer» |
|
| 16. |
According to the following equation, K_(2)Cr_(2)O_(7)+4H_(2)SO_(4) to K_(2)SO_(4)+Cr_(2)(SO_(4))_(3)+4H_(2)O+3[O] the equivalent mass of K_(2)Cr_(2)O_(7) is : |
|
Answer» `"mol. MASS"//3` |
|
| 17. |
According to the first law of thermodynamics, which of the following shows change in state function ? |
|
Answer» `q_("REV") ` |
|
| 18. |
Accordingto theelectromagnetictheoryofmaxwellwhathappenwhen chargeparticlesaccelerated ? |
|
Answer» SOLUTION :JamesMaxwellwas the FIRST to GIVE acomprehensiveexplanationabout theinteractionbetweenthe chargebodiesand thebehavior of ELECTRICAL andmagneticfields. Whenelectrically chargedparticlemovesunderaccelerationaltermatingelectricalandmagneticfieldareproduced andtransmitted . thesefieldare transmittedin the formsofwaves calledelectromagneticwavesofelectromagneticradiation. |
|
| 19. |
According to the Bohr Theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon? |
|
Answer» N = 6 to n = 1 |
|
| 20. |
According to the Bohr Theory, the third from the red end corresponds to which one of the following transitions in the hydrogen atom will give rise to the least energetic photon ? |
|
Answer» N = 6 to n = 1 `(1)/(1^(2)) - (1)/(6^(2)) = 1 - (1)/(36) = (35)/(36), (1)/(4^(2)) - (1)/(5^(2)) = (1)/(16) - (1)/(25) = (9)/(400)` `(1)/(5^(2)) - (1)/(6^(2)) = (1)/(25) - (1)/(36) = (11)/(36 xx 25)` `(1)/(3^(2)) - (1)/(5^(2)) = (1)/(9) - (1)/(25) =(16)/(225)` Thus `Delta E` will be MINIMUM for the TRANSITION n = 6 to n = 5 |
|
| 21. |
According to the Bohr theory for the atomic spectrum of hydrogen the energy levels of the proton -electron system depends, on the quantum number n. In an electron transition from a higher quantum level , n_(2) to the lower level n_(1), radiation is emitted. The frequency, v of the radiation is given by h v = (E_(n_(2)) - E_(n_(1))) where, h is Planck.s constant and E_(n_(2)), E_(n_(1)) are the energy level values for the quantum number n_(2) and n_(1). A useful formula for the wavelength, lambda = c//v is given by = lambda(Å) = (912)/(Z^(2)) xx ((n_(2)^(2)n_(1)^(2))/(n_(2)^(2)-n_(1)^(2))) where, Z = atomic number of any one electron species viz H, He^(+), Li^(2+), Be^(3+),..... For hydrogen Z=1 Calculating n_(2) for the series limit = 22800 Å. Calculate lambda for the transition (n_(2) + 1) to (n_(1) - 1) |
|
Answer» 26266 Å |
|
| 22. |
According to the Bohr theory for the atomic spectrum of hydrogen the energy levels of the proton -electron system depends, on the quantum number n. In an electron transition from a higher quantum level , n_(2) to the lower level n_(1), radiation is emitted. The frequency, v of the radiation is given by h v = (E_(n_(2)) - E_(n_(1))) where, h is Planck.s constant and E_(n_(2)), E_(n_(1)) are the energy level values for the quantum number n_(2) and n_(1). A useful formula for the wavelength, lambda = c//v is given by = lambda(Å) = (912)/(Z^(2)) xx ((n_(2)^(2)n_(1)^(2))/(n_(2)^(2)-n_(1)^(2))) where, Z = atomic number of any one electron species viz H, He^(+), Li^(2+), Be^(3+),..... For hydrogen Z=1 In the above formula, when n_(2) rarr oo, we have lambda = (912)/(1^(2)) xx n_(1)^(2). This values of lambda is known as the series limit for the given value of n_(1). Calculate the value of n_(1) for the series limit lambda = 8208 Å |
|
Answer» 5 |
|
| 23. |
According to the Bohr model, the energy levels of hydrogen atom can be found by making two assumptions. (i). The electrons move in circulr orbit and (ii) the angular momentum of the electron in the n^(th) energy level is quantized to have a value n(h)/(2pi). The levels calculated with nuclear charge Ze, with a single electron in the orbit, are called Hydrogenic levels. Assume that the two electrons in the ground state of a helium atom occupy the corresponding lowest Hydrogenic level. Q. if the hydrogen atom ionizatioin temperature is T, the temperature at which He atoms ionise completely (both electron having left the atom) would be: |
|
Answer» 8T `=2xx54.4eV`-repulsive energy `=108.8eV-27.2eV` `=6XX`Ionization energy of HYDROGEN `therefore` Ionisation TEMPERATURE of Helium `=6xx`Ionisation temperature of hydrogen ltBrgt `=6T` |
|
| 24. |
According to the Bohr model, the energy levels of hydrogen atom can be found by making two assumptions. (i). The electrons move in circulr orbit and (ii) the angular momentum of the electron in the n^(th) energy level is quantized to have a value n(h)/(2pi). The levels calculated with nuclear charge Ze, with a single electron in the orbit, are called Hydrogenic levels. Assume that the two electrons in the ground state of a helium atom occupy the corresponding lowest Hydrogenic level. Q. The minimum repulsive energy between the two electrons would then be: |
|
Answer» 3.4 eV `((-Ze^(2))/(2R))=-54.4eV` `(-2e^(2))/(2r)=-54.4eV` `(e^(2))/(2r)=27.2eV` `therefore` Minimum REPULSIVE energy =27.2eV (Two ELECTRONS will lie diametrically OPPOSITE). |
|
| 25. |
According to the adsorption theory of catalysis, the speed of the reaction increases because |
|
Answer» adsorption lower the activation energy of the reaction |
|
| 26. |
According to Schrodinger model, nature of electron in an atom is as |
|
Answer» PARTICLES only |
|
| 27. |
According to Raoults law, the relative lowering of vapour pressure for a solution is equal to … |
|
Answer» 1) mole FRACTION of solvent |
|
| 28. |
According to Raoult's law the relative lowering of vapour pressure for a solution is equal to |
|
Answer» mole fraction of SOLVENT |
|
| 29. |
According to qauntum mechanical model of H-like species, and electron can be represented by a wave function (psi) which contain all dynamic information about the electron. The nature of wave function depends on the type of the orbital to which the electron belongs. For an orbital psi=[sqrt(2)/(81sqrt(3pi))]((1)/(a_(0)))^(3//2)(27-18sigma+2sigma^(2))e^((sigma)/(3)) Where, sigma =((Zr)/(a_(0))),r = radial distance from nucleous, a_(0)=52.9pm The orbital could possibly be |
| Answer» Solution :`THEREFORE` It MUST be a 3s-orbital. | |
| 30. |
According to Planck's Quantun theory, the correct statements are (a) The vibrating particle in the black body does not emit continuously (b) Radiation is emitted in the form of small packets called Quanta (c) Energy associated with emmitted radiations is inversely proportional to frequency. (d) The emitted radiant energy is propagated in the form of waves. |
| Answer» Answer :C | |
| 31. |
According to periodic law of elements the variation in properties of elements is related to their. |
|
Answer» Nuclear neutron-proton number ratio |
|
| 32. |
AccordingtoPeriodicLawof element sthevariationin properties of elements isrelatedto their |
|
Answer» nuclearneutron- protonnumber RADIUS |
|
| 33. |
According to octet rule SO_3 contains _______ number of dative bonds |
|
Answer» 1 |
|
| 34. |
According to ____no two electrons in an atom will have all the four quantum numbers ____ |
| Answer» SOLUTION :PAULI EXCLUSION PRINCIPLE, same | |
| 35. |
According to (n + l) rule, after np level is filled with electrons, the next electron will enter into |
|
Answer» <P>`(N + 1) p` |
|
| 36. |
According to molecule orbital theory, which of the following will not be a viable molecule- |
|
Answer» `H_(2)^(-)` `therefore`BOND order=`(2-2)/(2)=0` Since, bond order of `H_(2)^(2-)` is 0, it does not exist. |
|
| 37. |
According to molecular orbital theory, which of the following will not be a viable molecule? |
|
Answer» `He_2^(2+) `
|
|
| 38. |
According to molecular orbital theory, which of the following will not be a viable molecule ? |
|
Answer» `He_(2)^(2-)` ` H_(2)^(2+) (3e^(-)):sigma 1 s^(2) sigma^(**) 1 s^(1) , B.O. = (2-1)/(2) = 0.5` `H_(2)^(-) (3e^(-)): sigma 1 s^(2) sigma^(**)1 s^(1) , B.O. = (2-1)/(2) = 5.0` `H_(2)^(2-) (4e^(-)): sigma 1 s^(2) sigma^(**)1 s^(2) , B.O. = (2-2)/(2) = 0` THUS , `H_(2)^(2-)` connot exist as it has zero BOND order |
|
| 39. |
According to molecular orbital theory, atomic orbitals of nearly same energy of different atoms are combined and form molecule. In homodiatomic molecule two atomic orbitals which have wave function psi_(A) and psi_(B) are combined in two ways (psi_(A)pm psi_(B)) . They have two type of electron density, one is bonding [(psi_(A) pm psi_(B))^(2)] where e- density increase between nucleus. Another is antibonding [(psi_(A)pm psi_(B))^(2)] where e^(-) density decreases between nucleus. These molecular orbitals are filled according to Hund, Pauli, afbau principle. Existence of molecule depends on bond order. where bond order = ("Number of bonding "e^(-)"s- number of antibonding "e^(-)s)/(2) lf the molecule have unpaired electron it is paramagnetic otherwise diamagnetic Statement- 1: Removal of two e^(-) from N^(2), magnetic behaviour does not change while addition of two electrons in O_(2) magnetic behaviour changes. Which of the following correct trends of bond energy is possible to explain by following statement. "Successive filling of antibonding orbital reduce the bond energy between two atoms"? (I ) N_(2) gtgt O_(2) gtgtF_(2)(II) O_(2)^(+) gt O_(2) gt O_(2)^(-) (III) C_(2)^(2) gt C_(2) (IV) CI_(2) gt Br_(2) gt F_(2) |
|
Answer» all of these |
|
| 40. |
According to molecular orbital theory, atomic orbitals of nearly same energy of different atoms are combined and form molecule. In homodiatomic molecule two atomic orbitals which have wave function psi_(A) and psi_(B) are combined in two ways (psi_(A)pm psi_(B)) . They have two type of electron density, one is bonding [(psi_(A) pm psi_(B))^(2)] where e- density increase between nucleus. Another is antibonding [(psi_(A)pm psi_(B))^(2)] where e^(-) density decreases between nucleus. These molecular orbitals are filled according to Hund, Pauli, afbau principle. Existence of molecule depends on bond order. where bond order = ("Number of bonding "e^(-)"s- number of antibonding "e^(-)s)/(2) lf the molecule have unpaired electron it is paramagnetic otherwise diamagnetic Statement- 1: Removal of two e^(-) from N^(2), magnetic behaviour does not change while addition of two electrons in O_(2) magnetic behaviour changes. Statement-2: In N_(2) removal of electron talces place from bonding orbital while addition of e^(-)in O_(2) is in antibonding orbital |
|
Answer» STATEMENT-1 is true, Statement-2 is true, Statement-2 is a correct EXPLANATION for Statement - 1 |
|
| 41. |
According to Molecular orbital theory, a molecule of H_(2) has two electrons in |
|
Answer» `rho_(1S)` |
|
| 42. |
According to molecular orbital theory |
|
Answer» `C_(2)^(2-)` is expected to be diamgnetic Hence , it is diamagnetic `O_(2)^(2+)` has bond order = 3 ,`O_(2)` has bond order = 2.5 . Hence , bond length in `O_(2)^(2+)` is LESS than that in `O_(2)` . Bond order of `N_(2)^(+)` = Bond order of `N_(2)^(-) = 2.5` Bond order of `He_(2)^(+) = (1)/(2)` ,i.e., it exists . Hence , some energy is relesed when `He_(2)^(+)` is formed from two isolated He atoms. |
|
| 43. |
Accordingto Mendeleev's periodic table , the physical and chemical properties of elements are periodicfunction of their __________ |
| Answer» SOLUTION :ATOMIC NUMBERS | |
| 44. |
According to MO theory which of the following combination between the orbitals are not possible? (1) 2p_(z),2p_(z) (2) 2s,2p_(y) (3) 1s,2s (4) 2p_(x),2p_(x) |
| Answer» SOLUTION :(2) and (3) | |
| 45. |
According to Maxwell distribution, the area under the curve is equal to the total number of molecules in collision. On increasing temperature, fraction of molecules having speed equal to u_(mp) .(most probable speed) decreases. The speed distribution also depends on the mass of the molecules along with the temperature. In general, the distribution depends upon the value of M/T (where M is molar mass and Tis temperature in kelvin). If at T_1 K is u, then value of u_(rms) at T_2 is |
|
Answer» `3sqrt((T_2)/(T_1)) U` |
|
| 46. |
According to Maxwell distribution, the area under the curve is equal to the total number of molecules in collision. On increasing temperature, fraction of molecules having speed equal to u_(mp) .(most probable speed) decreases. The speed distribution also depends on the mass of the molecules along with the temperature. In general, the distribution depends upon the value of M/T(where M is molar mass and Tis temperature in kelvin). Relationship between T_1 and T_2 is given as |
|
Answer» `T_1= T_2` |
|
| 47. |
According to Maxwell distribution, the area under the curve is equal to the total number of molecules in collision. On increasing temperature, fraction of molecules having speed equal to u_(mp) .(most probable speed) decreases. The speed distribution also depends on the mass of the molecules along with the temperature. In general, the distribution depends upon the value of M/T (where M is molar mass and Tis temperature in kelvin). On increasing the temperature, number of molecules having speed between u_1 and u_2 |
|
Answer» increases |
|
| 48. |
According to lewis what is acid and base ? |
| Answer» SOLUTION :G.N.Lewis in 1923 proposed a acid -BASE concept ACCORDING to it, "An acid is defined as SUBSTANCE which is capable of accepting a pair of electrons and a base is defined as a substance which is capable of donating an UNSHARED (lone) pair of electrons. | |
| 49. |
According to Le Chateier principle, when an equilibrium is subjected to any external change, the equilibrium/reaction shifts to comensate the effect of the change. This principle helps in shifting the reaction towards appropriate diections so as to increase % yield of any reaction. Which of the following changes cannot cause an incresase in extent of dissociation of CH_(3)COOH in its aqueous solution as per the reaction? If is given that conversion of graphite to diamond in an endothermic reaction and the conversion C_("graphite")hArrC_("diamond") attains equilibria at 1.5xx10^(9) Pa at 300K then comment at what pressure equilibria can be attained at 500 K? |
|
Answer» <P>`P GT 1.5xx10^(9)PA` |
|
| 50. |
According to Le Chateier principle, when an equilibrium is subjected to any external change, the equilibrium/reaction shifts to comensate the effect of the change. This principle helps in shifting the reaction towards appropriate diections so as to increase % yield of any reaction. Which of the following changes cannot cause an incresase in extent of dissociation of CH_(3)COOH in its aqueous solution as per the reaction? A,B,C and D are in equilibrium in a 2 litre container at 400 K and their moles are respectively 4,5,8 and 6. If the reaction involved is 3A(g)+2B(g)hArrC(g)+5D(g) then calculate equilibrium concentration of C when volume is increased to 10 litre. |
|
Answer» 4M |
|