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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. |
Statethe numberof Radialnode andangularnod of 5F ? |
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Answer» SOLUTION :Redialnode `=n-l-1 ` =5-3-1=1 ANGULAR NODE =1=3 |
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| 2. |
State the Newland's law of octaves. (i) What are the two exceptions of block division in the periodic table? |
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Answer» Solution :The Law of octaves states that, "when elements are arranged in the order of INCREASING atomic weights, the properties of the eighth ELEMENT are a REPETITION of the properties of thefirst element" (ii) 1.Helium has two electrons. Its electronic configuration is `1s^(2)`?. As per the configuration, it is Supposed to be placed in .s. block, but actually placed in `18^(th)` group which belongs to .p. block. Because it has a completely filled valence shell as the other elements present in `18^(th)` group. It also resembles with `18^(th)` group elements in other properties. Hence helium is placed with other noble gases. 2. The other exception is hydrogen. It has only one s-electron and hence can be placed in group 1. It can also GAIN an electron to achieve a noble gas arrangement and hence it slovens `(17^(th))` group elements). Because of these assumptions, position of hydrogen becomes a special case. Finally, it is placed separately at the top of the periodic table. |
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| 3. |
State the numberof radialnodes in 2p , 3p ,4p ,5porbitals |
| Answer» SOLUTION :In orbitalradial NODES(n-1-1)representingnodeare 1,2,3 | |
| 4. |
state the name of method for measurement ? |
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Answer» |
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| 5. |
State the Newland's law of octaves. |
| Answer» SOLUTION :The Law of OCTAVES STATES that, "when ELEMENTS are arranged in the order of increasing atomic weights, the properties of the EIGHTH element are a repetition of the properties of the element". | |
| 6. |
State the name of components of photo chemical smog. |
| Answer» Solution :OZONE, nitric oxide, acrolein, FORMALDEHYDE and peroxyacetic NITRATE are the general components of the PHOTOCHEMICAL SMOG. | |
| 7. |
Statethe nameof orbitalsas perquantumnumber,magneticquantumnumberandnoof oribitals . (i) n=3,l=2(ii)n=4,l=3 (iii)n= 2, l = 0(iv)n=5,l=1 |
Answer» SOLUTION :
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| 8. |
State the name of scientists who contributed in discovery of periodic table and its rules. |
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Answer» SOLUTION :Dobereiner : Law of triads A.B.E de Chancourtois : Cylindrical table John ALEXANDER NEWLANDS : Law of octaves. Mendeleev : Low of period or periodic law Henry Moseley: Modern Periodic Law |
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| 9. |
State the molecules of 1 liter slight less water ? |
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Answer» 18 `=1000 ` gm moles `= (1000)/(18) = 55.55` molecules `= "moles" XX N_(A) =55.55N_(A)` |
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| 10. |
Statethe modernperiodiclaw. |
| Answer» Solution :MODERN periodiclaw states thatphysicaland CHEMICAL PROPERTIESOF theelement s area periodicfunction oftheiratomic NUMBERS. | |
| 11. |
State the mass of oxygen in 0.1 moleNa_(2)CO_(3)*10 H_(2)O. |
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Answer»
Moles of oxygen : `0.1 XX 3 = 0.3` and 1.0 mol Mass of oxygen `= "Mole " xx "Atomic mass"` `=1.3xx16` `=20.8` gm |
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| 13. |
State the limitations of Lewis theory. |
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Answer» Solution :Lewis theory is success to write the SHAPE of the atom to write the distribution of electron in bond. It can.t EXPLAIN CHEMICAL bond structure, Bond ENTHALPY difference of bond length. ALSO not explain shape of polyatomic molecule. |
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| 14. |
State the law governing entropy and temperature. |
| Answer» Solution :THIRD LAW of thermodynamics -Refer to PAGES. | |
| 15. |
State the important biological and environmental chemical equilibrium with example. |
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Answer» Solution :Chemical EQUILIBRIA are important in numerous biological and environmental processes. Ex.-1: Equilibria involving `O_2` molecules and the PROTEIN hemoglobin play a crucial role in the transport and delivery of `O_2` from our lungs to our muscles. Protein hemoglobin `hArr O_2` Ex.-2 : Biochemical equilibrium between CO molecules and hemoglobin. Hemoglobin `hArr CO_((g))`. This equilibrium account for the toxicity of CO. At the equilibrium property of the MIXTURE remain constant. Equilibrium can be obtain by forward or reverse process. Equilibrium is dynamic in nature. EXTERNAL factor like temperature, pressure, concentration affect on equilibrium. |
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| 16. |
State the II law of Thermodynamics. Give the equation that relates Gibbs energy with Enthropy and Enthalpy ofa system? |
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Answer» Solution :Second law of thermodynamics cab be stated in a number of forms: 1. All spontaneous processes or naturally occurring processes are thermodynamially irreversible. 2. Without the help of an external AGENCY, a spotaneous PROCESS cannot to REVERSED. 3. The entropy of an ISOLATED system increase if it is to be spontaneous in a particular direction. 4. In a NON -isolated systemthe total entropy of both the system and surroundings must increase or must be positive. 5. The total entropy of the universe must tend to increase in a spontaneous process. `G=H-TS` |
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| 17. |
State the harmful effects of excessive production of sulfur oxide. |
| Answer» Solution :HIGH CONCENTRATION of `SO_2` irritation to the eyes resulting in TEARS and redness. It leads to stiffness of flower buds which eventually fall off from PLANTS. | |
| 18. |
State the gas laws. |
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Answer» Solution :At constant TEMPERATURE, the pressure of given AMOUNT of gas varies inversely with its VOLUME. According to COMBINED gas law, `(P_1V_1)/T_1 = (P_2V_2)/T_2` `P_2 = (P_1V_1T_2)/(T_1V_2) = (700xx500xx388)/(308xx450)` = 727.3 mm HG |
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| 19. |
State the formaton of difference angles observed in H-C-H of ethane ? |
| Answer» SOLUTION :`120^(@)` | |
| 20. |
State the formal charge of each atom in N_(2)O |
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Answer» Solution :Here, FC = formal charge `N_(A)`= number of nonbonding electron Number of e,lectron in valence orbital of N = 5 Number of electron in valence orbital of O = 6SO, TOTAL number of electron = 16, So, FC = valence`bar(e) - (N_(A) + " number of BOND " )` Formal charge of `N_(1) = 5 - (4 + 2) =5 - 6 = - 1 ` Formal charge of `N_(2) = 5 - ( 0 + 4) = + 1 ` Formal charge of `O = 6 - (4 + 2 ) = 0 ` Thus, |
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| 21. |
State the first law of thermodynamics (ii) Calculate the enthalpy of combustion of ethylene at 300 K, at constant pressure, If its heat of combustion at constant volume (AU) is -1406 kJ. |
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Answer» Solution :The first law of thermodynamics states that the dotal energy of an inolated SYSTEM remains constant though it may change from one form to ANOTHER" (or) Energy can neither be created nor destroyed, but may be converted from one form to another (ii) The complete ethylene combustion REACTION can be written as `C_(2)H_(4(g))+3O_(2(g))to2CO_(2(g))+2H_(2)O_((1))` `DeltaU=-140kJ` `Deltan=n_(p(g))-n__(r)(g)` `Delta=2-4=-2` `DeltaH=DeltaU+RTDeltan_(g)` `/_|=-1406+(8.314xx10^(-3)xx300xx(-2))` `DeltaH=-1410.9kJ` |
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| 22. |
State the first law of thermodynamics. |
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Answer» Solution :The FIRST law of thermodymics, also KNOWN as the law of conservation of energy , states that "The total energy of an isolated SYSTEM remains constant though it may CHANGE from one FORM to another . The mathematical statement of the First Law is : `DeltaU=q+w` Where- the amount of heat supplied to the system w- work done on the system |
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| 23. |
Define first law of thermodynamics. |
| Answer» Solution :The first law of thermodynamics states that the total ENERGY of an ISOLATED system REMAINS constant though it may change from one form to another" (or) Energy can neither created nor destroyed, but may be CONVERTED from one form to another . | |
| 24. |
State the findings of modern periodic law. |
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Answer» Solution :The number of electrons increases by the same numbers as the increase in the atomic number. As the number of electrons increases , the electronic SRTUCTURE of the ATOM changes. Electrons in the outermost SHELL of an atom (valence electrons) determine the chemmical properties of the ELEMENTS. |
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| 25. |
State the factors affecting vapour pressure of solution. |
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Answer» Solution :(i) NATURE of LIQUID : At same TEMPERATURE the vapour pressure of different liquid is different. The liquid which is more volatile has more vapour pressure ether and PATROL are more volatile. So, V.P. of ether and patrol is more then water and alcohol. (ii) Temperature: As the temperature increase vapour pressure increases. |
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| 26. |
Statethe elementin whichvalenceelectronare 2p^(4) ,3s^(2) ,3p^(3) |
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Answer» Solution :`2p^(4) `Means : `1s^(2) 2S^(2) 2p^(4) ` and `Z=8` (OXYGEN ) `3s^(2) ` means `:1s^(2)2s^(2)2p^(6)3s^(2)`and `Z=12` (mg ELEMENT ) `2p^(3) ` means `: 1s^(2) 2s^(2) 2p^(3)` and `z=7 ` (Nelement ) |
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| 27. |
State the difference in Lewis dot structure. (i) BCl_(3) and BH_(3) "(ii) " BeCl_(2)and BeH_(2) |
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Answer» Solution :In `BCl_(3)` central ATOM B has incomplete octet . while a has octet In `BH_(3)` central atom B has incomplete octet and H does not possess octet. In `BeCl_(2)` Be has incomplete octet. TWO bonds are these so only 4 ELECTRONS while Cl has octet. In `BeH_(2)`Be and H both does not possess octet. In `BeCl_(2) and BeH_(2)` both Be has same electron but Cl and H has different electron. |
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| 28. |
Statethe electronconfigurationof atom having (z=16)Mentionthat howmanyp electron are there in and give the numberof halffilledand fullfilledorbitals . |
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Answer» SOLUTION :Electron CONFIGURATION `1s^(2)2S^(2)2p^(6)3s^(2) 3p_(y) 3p_(z)` totalp electron: 2p= 6 and3p =0 =10 COMPLETELY filledorbital: 1s2s, 3s three2pone3p Halffilledorbital: `3p_(x)^(1)` and `3p_(z)^(1)` |
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| 29. |
State the electron pair possessed by Xe in XeF_2,XeF_4 and XeF_6 respectively. |
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Answer» 2,1,3 |
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| 30. |
State the criteria of DeltaG for spontaneity at constant temperature and pressure. |
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Answer» SOLUTION :If `DeltaG lt 0 (-ve)`, the PROCESS is spontaneous. `DeltaG GT 0 (+ve)`, the process is NON spontaneous. |
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| 31. |
State the contribution of Nyholm & Gillespie in VSEPR theory. |
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Answer» Solution :Nyholm and Gillespie (1957) refined the VSEPR model by explaining the important difference between the lone pair and bonding pairs of electrons. While the lone pairs are localised on the central atom, each bonded pair is shared between TWO atoms. As a result, the lone pair electrons in a molecule occupy more SPACE as compared to the bonding pairs of electrons. This RESULTS in greater repulsion between lone pairs ofelectrons as compared to the lone pair - BOND j pair and bond pair - bon pair repulsions. These repulsion effects result in deviations- from idealised shapes and alterations in bond ANGLES in molecules. |
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| 32. |
State the conditions for resonance structure. |
| Answer» Solution :(i) The position of ATOM must be unchanged in each structure. (II) CHANGE the position of p or `pi` electron only. (iii) If possible them the electron pair must be on more ELECTRONEGATIVE atom. | |
| 33. |
Statethe characteristicandusesof X-rays |
| Answer» SOLUTION :X-raysare notdeflectedby theelectricand WAVELENTHS(-0.1)NM) | |
| 34. |
Statethe angularmomentum whenelectronreventorbits ? |
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Answer» Solution :ANGULARMOMENTUM inorbitis in multipleof`(H)/(2PI)= mvr = (nh)/( 2pi)` wheren=1,2,3 …accepted ORBITS |
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| 35. |
State the action to central environmental pollution. |
| Answer» Solution :(i) MANAGING waste (ii) Use system that reduce POLLUTION in EVERY day life. | |
| 36. |
State sigma and pi bond in following molecules ? (a) C_(2) H_(2) "(b) " C_(2)H_(4) |
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Answer» Solution :(a) H-C=C- H there sigma BOND & two pi BONDS In triple bond betrween two carbon one sigma bond & two `pi` bonds. Two C - H `sigma` bond. `C_(2)H_(2)` total FIVE sigma and one pi bonds. Four C - H sigma and In C = C one sigma. So total five `sigma ` bond and in double one `pi` bond. 
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| 37. |
State seconds for 4 day. |
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Answer» `= 4 xx 24 xx 60 xx 60s` `=345600s` |
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| 40. |
State-Saytzeff's rule. |
| Answer» SOLUTION :Some haloalkanes when treated with alcoholic KOH yield a mixture of olefins with diflerent amounts. It is explained by Saytzef".s RULE which STATES that in a dehydrohalogeation reaction, the preferreed product is that alkene which has more NUNBER of alkyl group attachedto the doubly bonded carbon ATOM. | |
| 41. |
State relation between K_p and K_c based on Deltan_((g))=0,Deltan_((g))= positive and Deltan_((g)) = negative. |
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Answer» Solution :`Deltan_((G))=0 K_p=K_c` `Deltan_((g)) GT 0K_p gt K_c` `Deltan_((g)) LT 0K_p lt K_c` |
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| 42. |
State Raoult law and obtain expression for lowering of vapour pressure when nonvolatile solute is dissolved in solvent. |
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Answer» Solution :Ranoult.s law: This law states that "in the case of a solution of volatile liquids the partial vapour pressure of each component (A & B) of the solution is directly proporational to its mole fraction. `P _(A)prop X _(A)` when `x _(A) =1,` then `k = P^(@)_(A) ""(P_(A)^(@)=` vapour pressure of pure component) `therefore P _(A) = P _(A)^(@) x a` ` P _(B) = P _(B) ^(@) . x _(b)` when a non volatile is dissolved in pure WATER, the vapour pressure of the pure solvent will decrease. In such solution, the vapours pressure of the solution will depend only on the solvent molecules as the solute in non-volatile. `P _("solution") prop x _(A)` `P _("solution") =k. x _(A)` when `x _(A) =1,k = p^(@)_("solvent")` `therefore P _("solution") =p_("solvent"^(@) . x _(A)` Lowering of vapour pressure `=P _("solvent")^(@)-P_("solution")` RELATIVE lowering of vapour pressure `=(P^(@)-P)/(P ^(@))=x _(B)` where `x _(B)=` Mole fraction of solute. |
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| 43. |
State Raoult law and obtain expression forlowering of vapour pressure when nonvolatile solute is dissolved in solvent. |
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Answer» <P> Solution :Roult's LAW :This law states that " in the case of a solution of volatile liquids the partial vapour pressure of each component (A & B) of the solution is directly proportional to its mole fraction "` P_(A) alpha X_A` ` (P_(A) ^(@)" is the vapour pressure of pure component") ` when ` X_A=1 , ` then k =` p_(A)^(@) ` ` P_(B) =p_B^(@) ,X_b ` When a nonvolatile solute is dissolved in a pure water the VAPOURPRESSURE of the solution will depend only the solvent molecules as the solute in non-volatile ` P_("solute ") PROP X_A ` ` P_("solution ") = k,X_A ` ` X_A =1, k =P_("solvent ")^(@)-P_("solution ") ` Relativelowering of vapour pressure ` =("P_(solvent ") ^(@) -P_("solution ") )/( P_("solvent ") ^(@) ) =X_B ` where ` X_B ` =Mole fraction of the solute. |
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| 44. |
State Pauli's exclusion principle. Give the possible values of l for n=2 |
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| 45. |
State on of covalent bond in H_(2) O , NH_(3) and "CCl"_(4). |
| Answer» Solution :In `H_(2)O , NH_(3) and "CCL"_(4)` RESPECTIVE 2, 3 and 4 COVALENT bond. | |
| 46. |
State Octet rule. |
| Answer» SOLUTION :The atoms TRANSFER or share electrons so that all the atoms involved in chemical bonding obtain eight electrons in their outer shell (VALENCE shell). It is CALLED octet RULE. | |
| 47. |
State number of covalent bond in H_(2)O and "CCl"_(4) and give bond structure and Lewis structure in H_(2)O & "CCl"_(4) |
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Answer» Solution :In WATER `(H_(2)O)` two O - H covalent one bond and in carbon tetrachloride `("CCl"_(4))` four C - Cl covalent one bond present. Covalent bond formation in `(H_(2)O)` water :  Note : • H atom attain a duplet (twoelectron) of one pair. Around O the OCTET like Ne. The VALANCE electron shown by the symbol `xx` for H and • for O. Covalent bond formation in carbon tetrachloride `("CCI"_(4))`:  In each bond one electron of Cl and one electron of C are present |
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| 48. |
State nonbonding electron on P in PCl_(5). |
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Answer» 5 |
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| 49. |
State the modern periodic law. |
| Answer» Solution :Properties of elements are PERIODIC FUNCTIONS of their atomic NUMBER. | |