InterviewSolution
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InterviewSolution
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.
| 1. |
The configuration to second excited state of the element is (O) electronic with `O_(2)` or `P^(-)` or `Cl^(+)` isA. `[Ne]3s^(2)3p_(x)^(2)3p_(y)^(1)3p_(z)^(1)`B. `[Ne]3s^(2)`C. `[Ne]3s^(1)3p_(z)^(1)3p_(y)^(1)3p_(z)^(1)3d_(xy)^(1)3d_(yz)^(1)`D. `[Ne]3s^(2)3p_(x)^(1)3p_(z)^(1)3p_(xy)^(1)` |
| Answer» In `O_(2)` or `P^(-)` or `Cl^(+)` the electrons present are 16. The element is sulphur (Z = 16). The electronic configuration of sulphur in second excited state is given by (c ). | |
| 2. |
Which of the following has firstly organise the elements according to same trend ?A. NewlandB. MendeleefC. Lother MeyerD. Dobereiner |
| Answer» Dobereiner has firstly used tried system to organise the elements. | |
| 3. |
The electron affinity values (in `kJ mol^(-1))` of three halogens, `x,y`, and `z`are, respectively, `-349, -333`, and `-325`. Then `x, y`, and `x`, are respectively,A. `F_(2), Cl_(2)` and `Br_(2)`B. `Cl_(2), F_(2)` and `Br_(2)`C. `Cl_(2), Br_(2)` and `F_(2)`D. `Br_(2), Cl_(2)` and `F_(2)` |
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Answer» The electron affinity for halogens, `F=332.6 ~~ 333 rArr Y` `Cl =-348.5 ~~ -349 rArr X` `Br = - 324.7 ~~ 325 rArr Z` X has the highest value of electron affinity. Therefore, the correct order of electron affinity is `Cl_(2)gt F_(2) gt Br_(2)` |
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| 4. |
Calculate electronegativity of carbon at Pauling scale Given that : ` E_(H-H) =104 .2 kcal "mol"^(-1) E_(C-C) =83.1 kcal "mol"^(-1)` , `E_(C-H) =98.8 kcal "mol"^(-1)`. Electronegativity of hydrogen `=2.1`.A. `0.498`B. `0.598`C. `2.134`D. `2.597` |
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Answer» `X_(C )-X_(H)=0.208 sqrt(Delta)` where, `Delta = E_(C-H)-sqrt(E_(C-C)xxE_(H-H))` `Delta = 98.8-sqrt(83.1xx104.2)` `therefore Delta = 5.75` `X_(C )-2.1=0.208 sqrt(5.75)` `X_(C )-2.1=0.497` `X_(C )=2.597` |
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| 5. |
The atomic numbers of elements A,B,C and D are Z - 1, Z, Z + 1 and Z + 2 respectively. If B is a noble gas, choose the correct statement among the following statements : I. A has higher electron affinity. II. C exists in +2 oxidation state. III. D is an alkaline earth metal.A. I and IIB. II and IIIC. I and IIID. I, II and III |
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Answer» `A = Z - 1` `B = Z rarr` Noble gas (outer subshells are s or p) `C = Z + 1` `D = Z + 2` Noble gas with outer s-subshell = `He(1s^(2))` Noble gas with outer p-subshell `=ns^(2)np^(6)` like [Ne] In periodic table, electrons affinity is highest in chlorine of group 17 with electronic configuration of `3s^(2)p^(5)`. Noble gas `=3s^(2)3p^(6)=[Ar]` `B=[Ar]rarr Z rarr 3s^(2)3p^(6)` (Noble gas) `A = Z-1 rarr 3s^(2)3p^(5)` (Halogen family) `C=Z+1rarr 3s^(2)3p^(6),4s^(1)` (Alkali metal) `D = Z+2 rarr 3s^(2)3p^(6),4s^(2)` (Alkaline earth metal) |
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| 6. |
Which one of the following is correct increasing order of size ?A. `Mg lt Na^(+)lt F^(-)lt Al`B. `Na^(+)lt Al lt Mg lt F^(-)`C. `Na^(+)lt F^(-)lt Al lt Mg`D. `Na^(+) lt F^(-)lt Mg lt Al` |
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Answer» `Na^(+)` and `F^(-)` ions are isoelectronic, therefore `F^(-)` has the large and `Na^(+)` has the lowest size. Further, Al with higher nuclear charge has lower size than Mg. Thus, the overall order is `Na^(+)lt Al lt Mg lt F^(-)` |
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| 7. |
The one with the largest ionic size isA. `O^(2-)`B. `Mg^(2+)`C. `F^(-)`D. `Na^(+)` |
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Answer» `O^(2-), F^(-)Mg^(2+)` and `Na^(+)` are isoelectronic species (each having `10 e^(-)`). The size of isoelectronic species decreases with increase in nuclear charge (number of protons). Hence, correct order of size is `O^(2-)gt, F^(-)gt Na^(+)gt Mg^(2+)` |
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| 8. |
Among the following, the element with highest ionisation potential isA. boronB. carbonC. oxygenD. nitrogen |
| Answer» Nitrogen has stable half-filled orbital. | |
| 9. |
Which one of the following statements is false ?A. The electron affinity of chlorine is less than that of fluorineB. The electronegativity of fluorine is more than that of chlorineC. The electron affinity of bromine is less than that of chlorineD. The electronegativity of chlorine is more than that of bromine |
| Answer» The electron affinity of fulorine is less than that of chlorine, due to very small size of fluorine in which negative charge is highly concentrated and repels the incoming electrons. | |
| 10. |
The pair in which the ionisation energy of first species is less than that of second isA. N, PB. `Be, Be^(+)`C. S, PD. `N, N^(-)` |
| Answer» P has higher ionisation energy due to stable half-filled congiguration. | |
| 11. |
The lower electron affinity of fluorine than that of chlorine is due toA. Smaller sizeB. smaller unclear chargeC. difference in their electronic configurationsD. its highest reactivity |
| Answer» Because of the small size of F, electron-electron repulsions present in its relatively compact 2p-subshell, do not easily allow theaddition of an extra electron. On the other hand, Cl because of its comparatively bigger size than F, allows the addition of an extra electron more easily. Thus, the EA of Cl is higher than that of F. | |
| 12. |
The value in electron-volt per atom which represent the first ionisation energy of oxygen and nitrogen atom respectively areA. `13.6` and `14.6`B. `14.6` and `13.6`C. `14.6` and `14.6`D. `13.6` and `13.6` |
| Answer» Ionisation energy of `N gt` ionisation energy of oxygen, since, nitrogen has stable configuration due to half-filled configuration. | |
| 13. |
Which one of the following is incorrect ?A. Non-metals have strong tendency to gain electronB. Electronegativity is directly related to non- metallic properties of elementsC. Electronegativity is inversely proportional to the metallic properties of elementsD. Increase in electronegativity done the group is accompanide by a decrease in non-metallic properties |
| Answer» Decrease in electronegativity down the group is accompanied by a decrease in non-metallic properties. | |
| 14. |
Which of the following choices represent the correct order of first ionisation enthalpy ?A. `B lt C lt N lt O lt F`B. `B gt C gt N gt O gt F`C. `B lt C lt N gt O lt F`D. `B lt C lt N gt O gt F` |
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Answer» Ionisation energy is the minimum amount of energy required to removed the outermost electron from an isolaed gasseous atom. Quantitatively, it depends on the attraction between electron present on outermost shell and nucleus. Greater the interaction between outermost electron and nucleus, higher will be its ionisation enthalpy. So, correct order of first it must be `B lt C lt N lt O lt F` But due to extra stable half-filled electronic configuration of p-orbital, N has more value of first ionisation enthalpy than oxygen hence, correct order is `B lt C lt N gt O gt F` |
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| 15. |
The incorrect statements among the following isA. the firstr ionisation potential of Al is less than the first ionisation potential of MgB. the second ionisation potential of Mg is greater than the second ionisation potential of NaC. the first ionisation potential of Na is less than the first ionisation potential of MgD. the third ionisation potential of Al |
| Answer» `IE_(2)` of Mg is lower than that of Na because in case of `Mg^(2+)`, 3s electron has to be removed while in case of `Na^(+)`, an electron from the stable inert gas configuration (neon) has to be removed. | |
| 16. |
Ionisation enthalpy of Na would be same asA. value of electron affinity of `Na^(+)`B. value of electronegativity of NaC. value of ionisation potential of MgD. value of electron affinity of Na |
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Answer» `Na rarr Na^(+)+e^(-)` `Na^(+)+e^(-) rarr -X , Delta_(eg)=-IE` |
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| 17. |
The third alkaline earth metal ion contains number of electrons and protons asA. `20e^(-), 20p`B. `18e^(-), 18p`C. `18e^(-), 20p`D. `19e^(-), 20p` |
| Answer» `Ca^(2+), p=20, e^(-)=18` | |
| 18. |
How many `Cs` atoms can be converted to `Cs^+` ions by ` 1` joule energy , If `IE_1` for `Cs` si `376kJ mol^(-1)` ?A. `1.60xx10^(23)`B. `1.60xx10^(15)`C. `1.60xx10^(18)`D. `16.0xx10^(26)` |
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Answer» `376xx10^(3)J` energy produces `6.023xx10^(23)` ions. `therefore 1J` enery will produce `=(6.023xx10^(23))/(3.76xx10^(3))` `=1.60xx10^(18)Cs^(+)` ions |
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| 19. |
How many joules of energy must be absorbed to convert Li to `Li^(+)` , all the atoms present in `1.00` mg of gaseous Li ? `lE_(1)` of Li is `520.3 kJ mol^(-1)(Li=7)` .A. `0.00743 kJ`B. `0.520 kJ`C. `520 kJ`D. `0.0743 kJ` |
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Answer» `1.00 mg=1xx10^(-3)g=(1x10^(-3))/(7)` mol Li 1 mole of Li is converted by `520.3 kJ`. `therefore(1xx10^(-3))/(7)` mole is converted to `Li^(+)` by `=(520.3xx1xx10^(-3))/(7)kJ=0.0743 kJ` |
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| 20. |
There are 10 neutrons in the nucleus of the element `_(Z)M^(19)` . It belong toA. s-blockB. d-blockC. f-blockD. None of the above |
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Answer» `Z=19-10=9` `EC=1s^(2)2s^(2)2p^(5)`, here, it belongs to p-block. |
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| 21. |
Elements/ions having same number of electrons are known as isoelectronic species. Arrange the following elements incorrect order of atomic/ionic radii and choose the correct from the four choices given below `O^(2-), Na^(+), Mg^(2+), F^(-), Al^(3+)`A. `Al^(3+)lt Mg^(2+)lt Na^(+) lt F^(-)lt O^(2-)`B. `Al^(3+)lt Na^(+)lt Mg^(2+)lt F^(-)lt O^(2-)`C. `Al^(3+)gt Mg^(2+)gt Na^(+)gt O^(2-)`D. None of the above |
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Answer» Elements/ions having equal number of electrons are known as isoelectronic species. Among isoelectronic species, cation `lt` Neutral atom `lt` Anion Hence, correct choice is `Al^(3+)lt Mg^(2+)lt Na^(+)F^(-)lt O^(2-)` |
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| 22. |
Observe the following statements, I. The physical and chemical properties of elements are the periodic functions of their electronic configuration. II. Electronegativity of fluorine is less than the electronegativity of chlorine. III. Electropositive nature decreases from top to bottom in a group. The correct answer isA. I, II and IIIB. Only IC. I and IID. II and III |
| Answer» The physcial and chemical properties of elements are periodic function of their electronic configuration. Electronegativity of fluorine is more than that of chlorine. Electropositive nature increases from top to bottom in a group. | |
| 23. |
Match the Column I with Column II and select the correct answer by given codes. CodesA. `{:(A,B,C,D),(2,4,3,1):}`B. `{:(A,B,C,D),(2,4,1,3):}`C. `{:(A,B,C,D),(4,2,3,1):}`D. `{:(A,B,C,D),(4,2,1,3):}` |
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Answer» `A rarr 2, B rarr 4, C rarr 1, D rarr 3` A. `Li^(+)lt Al^(3+)lt Mg^(2+)lt K^(+)` The cation with the greater positive charge will have a smaller radius because of the greater attraction of the electrons to the nucleus. Anion with the greater negative charge will have the larger radius. Positive charge `prop (1)/("ionic radius")` Negative charge `prop` ionic radius B. Greater positive charge, increases ENC in case of isoelectronic species while for same group elements, ENC decreases down the group. C. `Cl gt F gt Br gt I` Electron affinity of Cl is highest in halogen family. D. `F gt Cl gt Br gt I` Electronegativity of fluorine (F) is higher than that of Cl, Br and I. |
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| 24. |
Which pair of elements has same chemical properties ?A. `13, 22`B. `3, 11`C. `4, 24`D. `2, 4` |
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Answer» The pair which belong to same group, i.e., in which both the elements have same outer electronic configuration has same chemical properties. `3 rArr 1s^(2), 2s^(1)` `11 rArr 1s^(2), 2s^(2)2p^(6), 3s^(1)` |
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| 25. |
The first ionisation energy of beryllium is more than that of boron becauseA. boron has higher nuclear chargeB. boron has only one electron in p-subshellC. atomic size of boron is less than that of berylliumD. atomic size of boron is more than that of beryllium |
| Answer» Boron has only one electron in p-subshell, hence it can be lost easily, while beryllium has comparatively stable configuration `(1s^(2), 2s^(2))` hence, cannot given an electron easily, consequently itas IE is greater. | |
| 26. |
The least stable ion among the following isA. `Li^(+)`B. `Be^(-)`C. `B^(-)`D. `C^(-)` |
| Answer» Be has fully filled 2s-subshell `(2s^(2))` and therefore, show least tendency to accept an electron. Thus, `Be^(-)` is least stable. | |
| 27. |
Generally, the valency of noble gases isA. twoB. threeC. oneD. zero |
| Answer» They are chemically inert (inactive). | |