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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.
| 251. |
The correct decreasing order of atomic size among the following species is: `Ar,K^(+),CI^(-),S^(2-),Ca^(2+)`A. `Ca^(2+) gt K^(+) gt Ar gt CI^(-) gt S^(2-)`B. `K^(+) gt Ca^(2+) gt CI^(-) gt Ar gt S^(2-)`C. `S^(2-) gt CI^(-) gt Ar gt K^(+) gt Ca^(2+)`D. `S^(2-) gt Ar gt CI^(-) gt Ca^(2+) gt K^(+)` |
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Answer» Correct Answer - C In isoelectronic ions, the atomic size decreases as z/e ratio increases. `S^(2-) (z)/(e) = (16)/(18), CI^(-1) (z)/(e) = (17)/(18), Ar(z)/(6) =(18)/(18), K^(+) (z)/(e) = (19)/(18), Ca^(2+) (z)/(e) = (20)/(18)` |
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| 252. |
The correct decreasing order of atomic size among the following species is : `K^(+), CI^(-), S^(2-), ca^(2+)`A. `Ca^(2+) gt K^(+) gt CI^(-) gt S^(2-)`B. `K^(+) gt Ca^(2+) gt CI^(-) gt S^(2-)`C. `S^(2-) gt CI^(-) gt K^(+) gt Ca^(2+)`D. `S^(2-) gt CI^(-) gt Ca^(2+) gt K^(+)` |
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Answer» Correct Answer - C In isoelectronic ions, the atomic size decreases as z/e ratio increases. ltbr `S^(2-)(z)/(e) = (16)/(18), CI^(-)(z)/(e) = (17)/(18),` `K^(+) (z)/(e), (19)/(18), Ca^(2+) (z)/(e) =(20)/(18)` |
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| 253. |
The first ionisation potential of `Na,Mg,Al` and `Si` are in the orderA. `Na lt Mg gt AI lt Si`B. `Na gt Mg gt AI gt Si`C. `Na lt Mg lt AI lt Si`D. `Na gt Mg gt AI gt Si` |
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Answer» Correct Answer - A Across the period (i.e. `3^(rd)` period) the size of atom decreases and nuclear charge increases. So generally the ionisation energy increases. However the ionisation energy of `Mg` is greater than `AI` because of more penetration power of `2s` sub-shell electrons of `Mg` as compared to that of the `2p` sub-shell electron of `AI`. |
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| 254. |
Arrange the following ions in the increasing order of their size: `Be^(2+), CI^(-), S^(2-), Na^(+), Mg^(2+), Br^(-1)`? |
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Answer» Correct Answer - `Be^(2+) lt Mg^(2+) lt Na^(2+) lt CI^(-) lt S^(2-) lt Br^(-)` `Be^(2+)` is smaller than `Mg^(2+)` as `Be^(2+)` has one sheel where as `Mg^(2+)` has two sheels. `Mg^(2+)` and `Na^(+)` are isoelectronic species: Ionic radius `prop 1//nuclear` charge. `CI^(-)` and `S^(2-)` are isoelectronic species: Ionic radius `prop 1//nuclear` charge. `CI^(-)` is smaller than `BI^(-)` as `CI^(-)` has three shells where as `BI^(-)` has four shells. |
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| 255. |
`Mg^(2+), O^(2-),Na^(+),F^(-),N^(3-)` (Arrange in decreasing order of ionic size) |
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Answer» Correct Answer - `[N^(-3) gt O^(-2) gt F^(-1) gt Na^(+) gt Mg^(2+)]` [In a isolectronic species, if the positive charge increases atomic size decreases. If the negative charge increases, atomic size increase] |
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| 256. |
The energy needed for `Li_(g) rarr Li_(g)^(3+) + 3e` is `1. 96 xx 10^4 kJ mol ^(-1)` . If the first ionisation energy of Li is ` 520 k J mol^(-1)` . Calcuate the second ionisation energy of `Li` . (Given : `IE_1` for ` H = 2 . 2. 18 xx 10^(-18) k J "atom"^(-1)`).A. `5270KJ "mole"^(-1)`B. `3210 KJ "mole"^(-1)`C. `7270 KJ "mole"^(-1)`D. `9290 KJ "mole"^(-1)` |
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Answer» Correct Answer - C |
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| 257. |
Which of the following represent the correct order of increasing first ionisation enthalpy for `Ca,Ba,S,Se` and `Ar`A. `Ca lt S lt Ba lt Se lt Ar`B. `S lt Se lt Ca lt Ba lt Ar`C. `Ba lt Ca lt Se lt S lt Ar`D. `Ca lt Ba lt S lt Se lt Ar` |
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Answer» Correct Answer - C |
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| 258. |
Which element has an outer electron configuration os `s^(2)p^(4)`?A. CaB. CrC. GeD. Se |
| Answer» Correct Answer - D | |
| 259. |
All elements of the same group will haveA. Same electron configurationB. Silimar outer electron configurationC. Same ionization potential valueD. Different chemical properties |
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Answer» Correct Answer - B |
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| 260. |
In the periodic table, where are non-metals located?A. Between groups IIA and IIIAB. On the lower left hand sideC. On the upper left hand sideD. On the upper right hand side |
| Answer» Correct Answer - D | |
| 261. |
Following statements regarding the periodic trends of chemical reactivity of the alkali metals and the halogens are given. Which of these statements gives the correct picture:A. In alkali metals the reactivity increases but in the halogens it decreases with increase in atomic number down the groupB. The reactivity decreases in the alkali metals but increases in the halogens with increases in atomic number down the group.C. In both the alkali metals and the halogen the chemical reactivity decreases with increases in atomic number down the groupD. Chemical reactivity increases with increases in atomic number down the group in both the alkali metals and halogens. |
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Answer» Correct Answer - A |
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| 262. |
Following statements regarding the periodic trends of chemical reactivity of the alkali metals and the halogens are given. Which of these statements gives the correct picture:A)In alkali metals the reactivity increases but in the halogens it decreases with increase in atomic number down the groupB)The reactivity decreases in the alkali metals but increases in the halogens with increases in atomic number down the group.C)In both the alkali metals and the halogen the chemical reactivity decreases with increases in atomic number down the groupD)Chemical reactivity increases with increases in atomic number down the group in both the alkali metals and halogens.A. In alkali metals the reactivity increases but in the halogens it decreases with increase in atomic number down the groupB. The reactivity decreases in the alkali metals but increases in the halogens with increases in atomic number down the group.C. In both the alkali metals and the halogen the chemical reactivity decreases with increases in atomic number down the groupD. Chemical reactivity increases with increases in atomic number down the group in both the alkali metals and halogens. |
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Answer» Correct Answer - A |
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| 263. |
In the long form of periodic all non-metals are placed inA. s- blockB. p-blockC. d-blockD. f-block |
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Answer» Correct Answer - B |
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| 264. |
Statement-1: The decreasing order of acidic character of `CO_(2),N_(2)O_(5),SiO_(2)` and `SO_(3)` is `SO_(3) gt N_(2)O_(5) gt CO_(2) gt SiO_(2)`. Statement-2: As electronegativity difference (E-O) decreases, acidic character of the oxide increases.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-2B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-2C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - A | |
| 265. |
Assertion: The first ionisation energy of `Be` is greater than that of `B`. Reason: 2p-orbital is lower in energy than 2s-orbital.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-24B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-24C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - C | |
| 266. |
Statement-1: The acidic strength order of hydra acids is `HF lt HCI lt HBr lt HI` Statement-2: The E.N. of halogens are `F gt CI gt Br gt I`.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-2B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-2C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - B | |
| 267. |
The `IE` values of `Al_((g)) = Al^(+) +e` is `577.5 kJ mol^(-)` and `DeltaH` for `Al_((g)) = Al^(3+) +3e` is `5140 kJ mol^(-1)`. If second and third IE values are in the ratio 2:3. Calculate `IE_(2)` and `IE_(3)`. |
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Answer» Correct Answer - `[IE_(2) = 1825 kJ//"mole", IE_(3) = 2737.5 kJ//"mol"]` `[Al(g)+Al^(+) +e^(-), IE_(1) = 577.5` `I.E_(1) +IE_(2) +IE_(3) = 5140` `(IE_(2))/(IE_(3)) = (2)/(3), I.E_(2) = (2)/(3),IE_(2) = (2)/(3)IE_(3)]` |
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| 268. |
The increasing order of the first ionization enthalpies of the elements B,P,S and F (lowest first) is:A. `F lt S lt P lt B`B. `P lt S lt B lt F`C. `B lt P lt S lt F`D. `B lt S lt P lt F` |
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Answer» Correct Answer - D `{:("Element":,B,S, P, F),(I.E.(kJ mol^(-1)):,801,1000,1011,1681):}` In general as we move from left to right in a period, the ionization enthalpy increases with increasing atomic number. The ionization enthalpy decreases as we move down a group. `P (1s^(2), 2s^(2),3s^(2), 3p^(3))` has a stable half filled electronic configuration than `S(1s^(2), 2s^(2), 2p^(6), 3s^(2), 3p^(4))`. For this reason, ionization enthalpy of `P` is greater than `S.` |
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| 269. |
Which will have the maximum value of electron affinity `O^(x),O^(y),O^(z)` [x,y and z respectively are `0,-1` and `-2]`?A. `O^(x)`B. `O^(y)`C. `O^(z)`D. All have equal. |
| Answer» Being neutrai atom oxygen will have higher electron affinity as there is electrostatic repulsion between additional electron and negative ion in case of `O^(-)` and `O^(2-)`. So option `(A)` is correct. | |
| 270. |
Why do halogens have high electron gain enthalpies (i.e. `-Delta_(eg)H^(o.))`? |
| Answer» The valence sheel electronic configuration of halogens is `ns^(2)np^(5)` and thus they require one electron to acquire the stable noble gas configuration `ns^(2)np^(5)`. Because of this they have strong tendency to accept an additional electron and hence have high electron gain enthalpies. | |
| 271. |
The electron gain enthalpies of halogens are as given below: `F =- 332, CI =- 349, Br =- 324, I =- 295 kJ mol^(-1)`. The less negative value for F as compared to that of CI is due to:A. string electron-electron repulsions in the compact 2-p sub shell of FB. strong electro-electron repulsion in the bigger 3-p sub shell of CIC. higher electronegativity value of CID. higher effective nuclear charge of F |
| Answer» Correct Answer - A | |
| 272. |
The formation of the oxide ion `O_((g))^(2-)` requires first an exothermic and then an endothermic step as shown below. `O_((g)) +e^(-) = O_((g))^(-) DeltaH^(@) =- 142 kJ mol^(-1)` `O_((g))^(-) + e^(-) = O_((g))^(2-) DeltaH^(@) = 844 kJ mol^(-1)` This is because of :A. `O^(-)` ion will tend to resist the addition of another electron of account of same chargeB. Oxygen has high electron affinityC. Oxygen is more electronegativeD. `O^(-)` ion has comparatively larger size than oxygen atom. |
| Answer» Correct Answer - A | |
| 273. |
Which of the following has minimum electron affinity?A. OB. SC. `Se`D. `Te` |
| Answer» Correct Answer - A | |
| 274. |
For the electron affinity of halogens (with `-ve` sign), which of the following is correct?A. `Br gt F`B. `F gtCI`C. `Br gt I`D. `F^(-) gt I` |
| Answer» Correct Answer - C | |
| 275. |
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» Correct Answer - B |
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| 276. |
The properties of the elements (atomic/ionic radii, electron gain enthalpy, ionization enthalpy, electronegativity, valence, oxidising/reducing power, acid/base character, etc.) which are directly or indirectly related to their electronic configirations are called periodic properties. These properties show a regular gradation on moving from left to right in a period or form top to bottom in a group. Down a group, the atomic/ionic radii, metallic character and reducing character increase while ionization enthalpy and electronegativity decrease. Along a period from left to right, atomic/ionic and metallic character decrease while ionization enthaloy, electronegativity, non-metallic character and oxiding power increase. However, electron gain enthalpy becomes less negative down a group butmore negative along a period. In contrast, inert gases have positive electron gain enthalpies which do not show may regular trend. The outermost electronic configuration of the most electronegative elements is:A. `ns^(2)np^(3)`B. `ns^(2)np^(4)`C. `ns^(2)np^(5)`D. `ns^(2)np^(6)` |
| Answer» Correct Answer - C | |
| 277. |
The properties of the elements (atomic/ionic radii, electron gain enthalpy, ionization enthalpy, electronegativity, valence, oxidising/reducing power, acid/base character, etc.) which are directly or indirectly related to their electronic configirations are called periodic properties. These properties show a regular gradation on moving from left to right in a period or form top to bottom in a group. Down a group, the atomic/ionic radii, metallic character and reducing character increase while ionization enthalpy and electronegativity decrease. Along a period from left to right, atomic/ionic and metallic character decrease while ionization enthaloy, electronegativity, non-metallic character and oxiding power increase. However, electron gain enthalpy becomes less negative down a group butmore negative along a period. In contrast, inert gases have positive electron gain enthalpies which do not show may regular trend. Which of the following isoelectronic ions has the lowest first ionization enthalpy?A. `K^(+)`B. `Ca^(2+)`C. `CI^(-)`D. `S^(2-)` |
| Answer» Correct Answer - D | |
| 278. |
The properties of the elements (atomic/ionic radii, electron gain enthalpy, ionization enthalpy, electronegativity, valence, oxidising/reducing power, acid/base character, etc.) which are directly or indirectly related to their electronic configirations are called periodic properties. These properties show a regular gradation on moving from left to right in a period or form top to bottom in a group. Down a group, the atomic/ionic radii, metallic character and reducing character increase while ionization enthalpy and electronegativity decrease. Along a period from left to right, atomic/ionic and metallic character decrease while ionization enthaloy, electronegativity, non-metallic character and oxiding power increase. However, electron gain enthalpy becomes less negative down a group butmore negative along a period. In contrast, inert gases have positive electron gain enthalpies which do not show may regular trend. Amomgst the following elements (whose electronic configurations are given below) the one having the highest ionization enthalpy is :A. `[Ne] 3s^(2)3p^(1)`B. `[Ne]3s^(2)3p^(3)`C. `[Ne] 3s^(2)3p^(2)`D. `[Ar] 3d^(10)4s^(2)4p^(3)` |
| Answer» Correct Answer - B | |
| 279. |
The properties of the elements (atomic/ionic radii, electron gain enthalpy, ionization enthalpy, electronegativity, valence, oxidising/reducing power, acid/base character, etc.) which are directly or indirectly related to their electronic configirations are called periodic properties. These properties show a regular gradation on moving from left to right in a period or form top to bottom in a group. Down a group, the atomic/ionic radii, metallic character and reducing character increase while ionization enthalpy and electronegativity decrease. Along a period from left to right, atomic/ionic and metallic character decrease while ionization enthaloy, electronegativity, non-metallic character and oxiding power increase. However, electron gain enthalpy becomes less negative down a group butmore negative along a period. In contrast, inert gases have positive electron gain enthalpies which do not show may regular trend. Tick the correct order of second ionization enthalpy in the following:A. `F gt O gt N gt C`B. `O gt F gt N gt C`C. `O gt N gt F gt C`D. `C gt N gt O gt F` |
| Answer» Correct Answer - B | |
| 280. |
Of the atoms listed, which has the largest third ionization energy?A. `Ca`B. `Mg`C. `AI`D. `Si` |
| Answer» Correct Answer - B | |
| 281. |
Which of these properties increase across the period from N to CI? (P) Atomic radius (Q) Density (R) ElectronegativityA. P onlyB. R onlyC. P and Q onlyD. Q and R only |
| Answer» Correct Answer - D | |
| 282. |
Which of the following will have the most negative electron gain enthalpy and which the least negative? `P, S, Cl, F`. Explain your answer.A. `P, CI`B. `CI, F`C. `CI, S`D. `CI, P` |
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Answer» Correct Answer - D In chlorine, the addition of additional electron to larger `3p`-subsheel experiences less electron-electron repulsion than smaller `2p`-subshell of fluorine. Phosphours has very low electron affinity because there is high electron repulsion when the incoming electron enters an orbital that is already half filled. |
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| 283. |
Which has the largest bond dissociation energy?A. `H-F`B. `H-CI`C. `H-Br`D. `H-I` |
| Answer» Correct Answer - A | |
| 284. |
From each set, choose the atom which has the largest ionization enthalpy and explain your answer with suitable reasons. (a) `F, O, N` (b) `Mg, P, Ar` |
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Answer» (a) Fluorine `(F)` has the largest ionization enthalpy because in moving from left to right in a period, atomic size decreases and electrons are held more tightly. Since `F` has the smallest size and maximum nuclear charge. It has the largest ionization enthalpy among these elements. (b) Argon `(Ar)` has the largest ionization enthalpy as argon has extra stable fully filled configuration. |
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| 285. |
The element having the lowest atomic number and a ground state electronic configuration of `(n-1) d^(6)ns^(2)` is placed in:A. fifth periodB. fourth periodC. sixth periodD. third period |
| Answer» Correct Answer - B | |
| 286. |
Which element has the largest atomic radius?A. BrB. KC. MgD. Na |
| Answer» Correct Answer - B | |
| 287. |
Which element has the largest atomic radius?A. LiB. KC. AsD. Br |
| Answer» Correct Answer - B | |
| 288. |
Why second ionization enthalpy is always higher than the first ionisation enthalpy for every element? |
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Answer» Correct Answer - Electron is more tightly bound by the nucleus in an cation (i.e `M^(+))` as the number of proton remains the same as in neutral atom whereas number of electron is one less than the proton. This increases the attraction between the valence sheel electrons and the nucleus `(Z_(eff)` increases). So, second ionization enthalpy is always higher than the first ionisation enthalpy for every element. Electron is more tightly bound by the nucleus in an cation (i.e `M^(+))` as the number of proton remains the same as in neutral atom whereas number of electron is one less than the proton. This increases the attraction between the valence sheel electrons and the nucleus `(Z_(eff)` increases). So, second ionization enthalpy is always higher than the first ionisation enthalpy for every element. |
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| 289. |
The elecron affinity of chlorine is `3. 7 eV`. How much energy in kcal is released when `2g` chlorine is completely converted to `cl^-` ion in a gaseous state ? `(1 eV = 23. 06 kcal "mol"^(-10))`.A. `4.8 Kcal`B. `2.4Kcal`C. `10.2Kcal`D. `14.2Kcal` |
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Answer» Correct Answer - A |
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| 290. |
The atomic number of an element is always equal toA. Number of neutrons in nucleusB. Half of the atomic weightC. Electrical charge of the nucleusD. Weight of the nucleus |
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Answer» Correct Answer - C |
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| 291. |
Which of following ions do not exist togther in aqueous solution:A. `Pb^(2+),F^(-)`B. `TI^(3+),I^(-)`C. Both `(A)` and `(B)`D. None of these |
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Answer» Correct Answer - B `TI^(3+)` gets reduced to `TI^(+)` because of `T^(-)` and then it forms the compound TII. |
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| 292. |
Atomic number of 15, 33,51 represents the following family:A. carbon familyB. nitrogen familyC. oxygen familyD. none of these |
| Answer» Correct Answer - B | |
| 293. |
The places that were left empty by Mendeleev were for:A. aluminium and siliconB. gallium and germaniumC. arsenic and antiomnyD. molybdenum and tungsten |
| Answer» Correct Answer - B | |
| 294. |
The element with atomic number `Z = 118` will be categorised as a:A. noble gasB. transition metalC. alkali metalD. alkaline earth metal |
| Answer» Correct Answer - A | |
| 295. |
Ionisation energies three hypothetical elements are given below (in kJ/mole) : `{:(,I,II,III),(X,122,340,1890),(Y,99,931,1100),(Z,118,1220,16552):}` What could be the value of the first electron affinity of `Z^(++)` in `KJ mol^(-1)`?A. 118B. 1220C. 1652D. 734 |
| Answer» Correct Answer - B | |
| 296. |
In modern periodic table, the element with atomic number `Z = 118` will be:A. Uuo, Ununoctium, alkaline earth metalB. Uno, Unniloctium, transition metalC. Uno, Unniloctium, alkali metalD. Uuo, Ununoctium, nobles gas |
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Answer» Correct Answer - D `Z = 118 [Rn]^(86)5f^(14)6d^(10)7s^(2)7p^(6)`, as last electron enters in `p`-subshelll, it belongs to `p`-block. Thus its group number will be `10 +2 +6 = 18`. Hence the element is a noble gas. |
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| 297. |
Ionisation energies three hypothetical elements are given below (in kJ/mole) : `{:(,I,II,III),(X,122,340,1890),(Y,99,931,1100),(Z,118,1220,16552):}` Which of the following pairs represents elements could be of the same group?A. Y,ZB. X,YC. X,ZD. X,Y,Z |
| Answer» Correct Answer - A | |
| 298. |
Ionisation energies three hypothetical elements are given below (in kJ/mole) : `{:(,I,II,III),(X,122,340,1890),(Y,99,931,1100),(Z,118,1220,16552):}` Which of the following is likely to be 2nd group element?A. XB. ZC. YD. X and Y |
| Answer» Correct Answer - A | |
| 299. |
The valence shell of the element X contains 1 electron in 5s subshell and below that shell, 4 electrons in 4d subshell. The element belongs to which group (IUPAC) of periodic table?A. 4th groupB. 5th groupC. 6th groupD. 7th group |
| Answer» Correct Answer - B | |
| 300. |
In the long form of the periodic table the valence shell electronic configuration of `5s^(2)5p^(4)` corresponds to the element present in:A. group 16 and period 5B. group 17 and period 6C. group 17 and period 5D. group 16 and period 6 |
| Answer» Correct Answer - A | |