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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.
| 151. |
The size of any species depends on various factors such as nature of charge, magnitude of charge/oxidation state, effective nuclear charge, electronic configuration etc. Select the order of size which are correct?A. `O gt O^(-) gt O^(2-)`B. `B gt Be gt Li`C. `Mg gt Mg^(+) gt Mg^(2+)`D. `Sc lt Ti` |
| Answer» Correct Answer - C | |
| 152. |
Variable valency is exhibited byA. Normal elementsB. Metallic elementsC. Transitional elementsD. Non-metallic elements |
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Answer» Correct Answer - C |
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| 153. |
The first five inization energies of an element are `9.1,16.2,24.5,35` and `205.7eV` respectively then, number of valence electrons in the atom is:A. 2B. 3C. 4D. 5 |
| Answer» Correct Answer - C | |
| 154. |
The element having 18 electrons in its outer most shell is:A. `._(28)Ni`B. `._(46)Pd`C. `._(29)Cu`D. None of these |
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Answer» Correct Answer - B |
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| 155. |
Identify the group (in Modern Peridic Table) and valency of a hypothetical elemt having atomic number 11.9. if group number is `x` and valency is `y`. Given the the value of `x+y`. |
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Answer» Correct Answer - 2 `8s^(1)` `x = 1, y = 1` `1+ 1 = 2`. |
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| 156. |
Nitrogen can form many oxides with oxygen, and thus is said to exhibit variable valency. Similary, sulphur, phosphours and carbon can exhibit variable valency. Which sequence of compounds is according to the increasing order of the oxidation state of chlorine?A. `CI_(2),CI_(2)O,CI_(2)O_(3),CI_(2)O_(5),CI_(2)O_(7),NH_(4)CI`B. `CI_(2),CI_(2)O,CI_(2)O_(3),CI_(2)O_(7),CI_(2)O_(5)`C. `CI_(2),CI_(2)O,CI_(2)O_(3),CI_(2)O_(5),CI_(2)O_(7)`D. `CI_(2),CI_(2)O_(3),CI_(2)O_(7),CI_(2)` |
| Answer» Correct Answer - C | |
| 157. |
Nitrogen can form many oxides with oxygen, and thus is said to exhibit variable valency. Similary, sulphur, phosphours and carbon can exhibit variable valency. Amongst the following, select those which are most likely to be neutral.A. `CO, N_(2)O, NO`B. `CO_(2),NO_(2),N_(2)O_(3)`C. `SO_(2),CO,N_(2)O_(4)`D. `SO_(3),N_(2)O_(5),CO_(2)` |
| Answer» Correct Answer - A | |
| 158. |
Atom loses electrons and becomesA. Only cationB. Only anionC. Either cation or anionD. Neither cation nor anion |
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Answer» Correct Answer - A |
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| 159. |
`Al^(3+)` has low ionic radius than `Mg^(2+)` becauseA. `Al^(3+)` has high nuclear charge than `Mg^(2+)`B. `Mg` atom has less no. of neutrons than `Al` atomC. `Mg` and `Al` differ in electronegativity valuesD. `Al` atom has low `I_(1)` value than `Mg` atom |
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Answer» Correct Answer - A |
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| 160. |
The oxidation state and valency of `Al` in `[AlCl (H_(2)O)_(5)]^(2+)`A. `+ 6` & 3B. `+3` & 6C. `+ 6, 6`D. `+3`, & 3 |
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Answer» Correct Answer - B |
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| 161. |
The maximum number of valency electrons possible for atom in the second period of the periodic table is :A. 18B. 10C. 8D. 2 |
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Answer» Correct Answer - C |
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| 162. |
Match Column-I (atomic number of elements) with Column-II (position of element in periodic table)A. `{:(Column-I,Column-II),(19,(P)p-block):}`B. `{:(Column-I,Column-II),(22,(Q)f-block):}`C. `{:(Column-I,Column-II),(32,(R)d-block):}`D. `{:(Column-I,Column-II),(64,(S)s-block):}` |
| Answer» Correct Answer - A::B::C::D | |
| 163. |
The position of element with `Z=24` in the periodic table isA. V A group and 4 periodB. VI B group and 4 periodC. IV A group and 3 periodD. III B group and 3 period |
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Answer» Correct Answer - B |
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| 164. |
The most common oxidation states of cerium areA. `+2,+3`B. `+2,+4`C. `+3,+4`D. `+3,+5` |
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Answer» Correct Answer - C |
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| 165. |
Using the data given below,predict the nature of heat changes for the reaction . `Mg_(g) + 2F_(g) rarr Mg_(g)^(2+) +2F_(g)^-` `IE_1` and `IE_2` of `Mg_(g) `are `737.7 ` and `451 kJ "mol"^(-1)` . `EA_1 ` for `F_(g)` is `-328 kJ "mol"^(-1)`.A. `1232.4 KJ "mole"^(-1)`B. `+1532.7 KJ "mole"^(-1)`C. `-1232.4 KJ "mole"^(-1)`D. `-1532.7 KJ "mole"^(-1)` |
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Answer» Correct Answer - B |
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| 166. |
What is the `DeltaH` of the following affinity. `Mg_(g)+2F_(g) rarr Mg_((g))^(2+)+2F_((g))^(-)`. If the electron affinity of `F_(g) = -328 kJ " mol"^(-1)` and first ionisation energy of `Mg = 737.7kJ " mol"^(-1)` and second ionisation energy of `Mg = 1451kJ " mol"^(-1)`A. `1532.7 kJ " mol"^(-1)`B. `1860.7kJ " mol"^(-1)`C. `2516.7kJ " mol"^(-1)`D. `2844.7kJ " mol"^(-1)` |
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Answer» Correct Answer - A |
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| 167. |
Which oxide is more basic, `MgO` or `BaO`? Whay? |
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Answer» Correct Answer - `[BaO]` [Among metal oxide, down the group basic character increases.] |
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| 168. |
The elements which exhibit both vertical and horizontal similarities are:A. inert gas elementsB. representative elementsC. trantition elementsD. none of these |
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Answer» Correct Answer - C This is a characteristic feature of transition metals. |
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| 169. |
Statement-1: `Sc(Z = 21)` is placed as d-block element. Statement-2: Last filling electron goes into 3d-subshell.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-12B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-12C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - A | |
| 170. |
The correct statement is:A. F has more electron affinity than `O^(o+)`B. `Ai^(3o+)` ion has more hydration energy than `Mg^(2o+)` ionC. noble gases have positive electron affinityD. all of the above |
| Answer» Correct Answer - B | |
| 171. |
`Be` and `Ne` have positive values of electron gain enthalpy against the general trend in their period in modern periodic table. Explain. |
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Answer» Correct Answer - In `Be`, the extra electron is to be addedx in `2p` orbital because `2s` orbital is completely filled and in `Ne`, it is to be added to a noble gas configuration. Since full-fulled orbitals and noble gas configuration are more stable, reluctancy in accepting the electron is found. So, they have positive values of electron gain enthalpy. In `Be`, the extra electron is to be addedx in `2p` orbital because `2s` orbital is completely filled and in `Ne`, it is to be added to a noble gas configuration. Since full-fulled orbitals and noble gas configuration are more stable, reluctancy in accepting the electron is found. So, they have positive values of electron gain enthalpy. |
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| 172. |
Period number of `Sc = x` Modern periodic table group number of `TI = x` (according to 1 to 18 convention) Find the value of y-x |
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Answer» Correct Answer - `009` `x = 4, y = 13` |
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| 173. |
Assertion: Electron gain enthalpy value of the 3rd peirod p-block elements of the mordern periodic table are generally more negative than the 2nd period element of the same group. Reason: Due to smaller atomic size of the 2nd peirod element, its electron density is high which eases the addition of electron.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - C | |
| 174. |
The number of elements which should be theoretically present in `8^(th)` period of the modern long from of periodic table, isA. 32B. 40C. 50D. 48 |
| Answer» Correct Answer - C | |
| 175. |
The correct order of electronegativity is:A. `F gt CI gt Br gt I`B. `F gt O gt N gt C`C. `S lt O lt Se lt Te`D. All of these |
| Answer» Correct Answer - A,B | |
| 176. |
Statement-1: The `5^(th)` period of periodic table contains 18 elements are 32. Statement-1: `n = 5, l = 0,1,2,3`. The order in which the energy of available orbitals `4d,5s` and 5p increases is `5s lt 4d lt 5p` and the total number of orbitals avaiable are 9 and thus 18 electrons can be accommodated.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-4B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-4C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - A | |
| 177. |
Which is a true statement?A. Larger is the value of ionisation energy easier is the formation of cationB. Larger is the value of electron affinity easier is the formation of anionC. Larger is the value of ionisation energy as well as electron affinity the smaller is the electronegativity of atomD. Larger is the `Z_(eff)` larger is the size of atom |
| Answer» Correct Answer - B | |
| 178. |
Assertion: the `4f^(-)` and `5f^(-)` inner transition series of elements are placed separately at the bottom of the modern periodic table. Reason : (i) Position of f-block elements prevents the undue expansion of the mordern periodic table, ie, maintains its structure. (ii) Position of f-block elements preserves the principle of classification by keeping elements with similar properties in a single column.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-5B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-5C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - A | |
| 179. |
As one moves down the group from top to bottom then which of the following will not be observed?A. Ionisation energy increasesB. Electron affinity decreasesC. Electronegativity decreasesD. Atomic radii increase |
| Answer» Correct Answer - A | |
| 180. |
The elements present in d-block areA. (a) metals and non-metalsB. (b) only metalsC. (c) only non-metalsD. (d) metals, metalloids and non-metals |
| Answer» All d-block elements are mentals. | |
| 181. |
Which one of the following is correct order of second ionisation potential of `Na, Ne Mg` and `Al`?A. `Al lt Na lt Mg lt Ne`B. `Ne lt Al lt Na lt Mg`C. `Na lt Mg lt Ne lt Al`D. `Mg lt Al lt Ne lt Na` |
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Answer» Correct Answer - D |
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| 182. |
The ionization potential of elements in any group decreases from top to bottom. This is due toA. Increase in size of atomB. Increase in atomic numberC. Increase in screening effectD. Both increase in size of atom and increase in screening effect |
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Answer» Correct Answer - D |
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| 183. |
The `I_(1)` of potassium is `4.339 eV//"atom"`. The `I_(1)` of sodiumA. 4.339B. 2.21C. 5.138D. 1.002 |
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Answer» Correct Answer - C |
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| 184. |
Two elements A and B belonging to same group have electrones affinity as `(EA)_(A)` and `(EA)_(B)` and ionistation potentials `I_(A)` and `I_(B)`. If A is more electronegative element than B, then, there must beA. `(EA)_(A) gt (EA)_(B)` and `I_(A) = I_(B)`B. `(EA)_(A) + I_(A) lt (EA)_(B) + I_(B)`C. `(EA)_(A) + I_(A) gt (EA)_(B) + I_(B)`D. `(EA)_(A) gt (EA)_(B) ` but `I_(A) lt I_(B)` |
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Answer» Correct Answer - C |
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| 185. |
Which one of the following represents the electronic configuration of the most electropositive element?A. `[He] 2s^(1)`B. `[He] 2s^(2)`C. `[Xe] 6s^(1)`D. `[Xe]6s^(2)` |
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Answer» Correct Answer - C |
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| 186. |
Which one of the following electronic configuration corresponds to the element with maximum electropositive character?A. (a) `[Kr]5s^(1)`B. (b) `[Ne]3s^(1)`C. (c) `[Ar]4s^(1)`D. (d) `[Xe]6s^(1)` |
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Answer» Electropositive character increases from top to bottom in a group. Hence, the element with electronic configuration [Xe] `6s^(1)` has the maximum electro- positive character. |
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| 187. |
The ionisation potential of "X" ion is equal toA. The electron affinity of "X" atomB. The electronegativity of "X" atomC. The ionisation energy of "X" atomD. The electron affinity of "`X^(2+)`" ion |
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Answer» Correct Answer - B |
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| 188. |
The `I_(1)` value of potassium is less than the `I_(1)` value of sodium. This is due toA. Large size of potassium atomB. Small size of potassium atomC. Low density of potassiumD. Univalent nature of potassium |
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Answer» Correct Answer - A |
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| 189. |
The first, second and third ionisation potentials (E1, E2, and E3) for an element are 7eV, 12.5eV and 142.3ev respectively. The most stable oxidation state of the element will beA. `+1`B. `+2`C. `+3`D. `+4` |
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Answer» Correct Answer - B |
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| 190. |
The correct values of ionization enthalpies(in KJ `"mol"^(-1)`) of Si, P, Cl, and S respectively are:A. `786, 1012, 999, 1256`B. `1012, 786, 999, 1256`C. `786, 1012, 1256, 999`D. `786, 999, 1012, 1256` |
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Answer» Correct Answer - C |
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| 191. |
The correct values of ionization enthalpies(in KJ `"mol"^(-1)`) of Si, P, Cl, and S respectively are:a)`786, 1012, 999, 1256`b)`1012, 786, 999, 1256`c)`786, 1012, 1256, 999`d)`786, 999, 1012, 1256`A. `786, 1012, 999, 1256`B. `1012, 786, 999, 1256`C. `786, 1012, 1256, 999`D. `786, 999, 1012, 1256` |
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Answer» Correct Answer - C |
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| 192. |
The correct values of ionization enthalpies(in KJ `"mol"^(-1)`) of Si, P, Cl, and S respectively are:a)`786, 1012, 999, 1256`b)`1012, 786, 999, 1256`c)`786, 1012, 1256, 999`d)`786, 999, 1012, 1256`A. `786,1012,999,1256`B. `1012,786,999,1256`C. `786,1012,1256,999`D. `786,999,1012,1256` |
| Answer» Correct Answer - C | |
| 193. |
The `I_(1),I_(2),I_(3),I_(4)` values of an element "M" are 120 kJ/mole, `600 kJ//"mole", 1000 kJ//"mole"` and `8000 kJ//"mole"`. Then the formula of its sulphate isA. `MSO_(4)`B. `M_(2)(SO_(4))_(3)`C. `M_(2)SO_(4)`D. `M_(3)(SO_(4))_(2)` |
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Answer» Correct Answer - B |
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| 194. |
The element with high electron affinity isA. NitrogenB. OxygenC. SulphurD. Phosphorous |
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Answer» Correct Answer - C |
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| 195. |
Which of the following is not an anomalous pair?A. `s, Cl`B. `Te, I`C. `Co, Ni`D. `Ar, k` |
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Answer» Correct Answer - A |
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| 196. |
The properties of the following elements were predicted by Mendeleeff before their isolation areA. `Co` and `Ni`B. `I` and `Te`C. `Sc, Ga` and `Ge`D. `Cl, Ar` and `K` |
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Answer» Correct Answer - C |
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| 197. |
Mendeleff corrected the atomic weight ofA. BeB. NC. OD. Cl |
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Answer» Correct Answer - A |
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| 198. |
In Mendeleef table, the triad of VIII group isA. `Ru, Rh, Pd`B. `Cu, Ag, Au`C. `N, O, F`D. `Tl, Pb, Bi` |
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Answer» Correct Answer - A |
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| 199. |
Eka silicon is now called asA. GalliumB. ScandiumC. GermaniumD. Indium |
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Answer» Correct Answer - C |
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| 200. |
The atomic weights of "Be" and "In" were correctly by Mendeleef using for formulaA. `sqrt(v) = a(Z-b)`B. `mvr = (nh)/(2pi)`C. Atomic weight = Equivalent weight `xx` valencyD. Equivalent weight = Atomic weight `xx` valency |
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Answer» Correct Answer - C |
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