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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. |
`XeF_(6)` on complete hydrolysis givesA. `XeO_(3)`B. `XeO`C. `XeO_(2)`D. `Xe` |
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Answer» Correct Answer - a `XeF_(6) + 3H_(2)O rarr XeO_(3) + 6HF` |
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| 152. |
Why zero group element do not form compound under ordinary conditions? |
| Answer» Zero group elements have general electronic configuration `ns^(2)np^(6) (except He- Is^(2))` hence they do not have any lendensy to lose or gain electron or share electron with other atoms . Hence z to group element do not from compound under ordinary conditions | |
| 153. |
The lightest, non-inflammable gas isA. `H_(2)`B. `He`C. `N_(2)`D. Ar |
| Answer» Correct Answer - d | |
| 154. |
Helium is used in gas balloons instead of hydrogen becauseA. it is higher than `H_(2)`B. it is none-combustibleC. it is more abundant than `H_(2)`D. its linkage can be detected easily |
| Answer» Correct Answer - b | |
| 155. |
In the clathrates of xenon with water , the nature of bonding between xenon and water molecule is _________.A. covalentB. hydrogen bondingC. coordinateD. dipole-induced dipole |
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Answer» Correct Answer - d Clathrate formation involves dipole induced dipole interaction. |
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| 156. |
The lifting power of helium is______of hydrogen. |
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Answer» Correct Answer - `92%` `92%` |
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| 157. |
Assertion : He and Ne do no form any clathrates Reason : Both He and Ne are very small in sizeA. If both (A) and (R ) are correct and (R ) is the correct explanation of (A)B. If both (A) and (R ) are correct,but (R ) is the correct explanation of (A)C. If (A) is correct but (R ) is incorrectD. If (A) is incorrect but (R ) is correct |
| Answer» Correct Answer - a | |
| 158. |
Radon is obtained from the decay of radium. |
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Answer» Correct Answer - T 1 |
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| 159. |
A radioactive element `X` decays to give two inert gases`X` isA. `._(92)^(238)U`B. `._(88)^(226)Ra`C. `._(90)^(232)Th`D. `._(89)^(227)Ac` |
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Answer» Correct Answer - b `._(88)^(226)Ra rarr ._(2)^(4)He+._(86)^(22)Rn` |
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| 160. |
In the clathrates of xenon with water , the nature of bonding between xenon and water molecule is _________. |
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Answer» Correct Answer - Dipole-indaced dipole interaction Dipole-indaced dipole interaction |
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| 161. |
On moving along the period , the atomic radii decreases |
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Answer» Correct Answer - F From halogens to noble gases atomic radii increases as amongst noble gas atoms, it is van der waals forces of attraction which are holding them together .Hence,atomic radii of noble gases is large than corresponding halogon. |
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| 162. |
Complete the following reactions (a)`XeF_(4) + H_(2)O rarr` (b)`XeF_(6) + SiO_(2) rarr` (c )`XeF_(6) + H_(2) rarr` (d)`XeF_(6) + H_(2)O rarr` (e )`XeF_(6) + SbF_(5) rarr` (f)`XeF_(6) + NH_(3) rarr` |
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Answer» a.`2XeF_(4) + 3H_(2)O rarr Xe + XeO_(3) + 6HF + F_(2)` b `2XeF_(6) + SiO_(2) rarr 2XeOF_(4) + SiF_(4)` `2XeOF_(4) + SiO_(2) rarr 2XeO_(2)F_(2) + SiF_(4)` `2XeO_(2)F_(2) + SiO_(2) rarr 2XeO_(4) + SiF_(4)` c. `XeF_(2) + H_(2) rarr 2HF + Xe` d.`XeF_(6) + 3H_(2)O rarr XeO_(3) + HF` c.`XeF_(6) + SbF_(5) rarr [2XeF]^(Theta)[SbF_(6)]^(Theta)` `XeF_(6) + 6NH_(3) rarr Xe + 6NH_(4)F + N_(2)` |
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| 163. |
The value of `n` in the molecular fromula ` Be_n Al_2Si_6 O_(18)`. |
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Answer» Correct Answer - `(3)` The value of `n` in the molecular formula `Be_(n) Al_(2) Si_(6) O_(18)` is `3`. `2n + (2 xx 3) +(4 xx 6) +(18 xx -2) = 0` or `n = 3`. Beryl is `Be_(3)Al_(2)Si_(6) O_(18)`. |
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| 164. |
Ozone reacts with dry iodine to form an oxide having ____oxygen atoms in its molecules. |
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Answer» Correct Answer - `(9)` `9O^(3) + underset(dry) (2I_(2) rarr I_(4) O_(9) + 9O_(2)`. |
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| 165. |
Sulphur and rest of the elements of group `16` are less electronegative than oxygen, Therefore, their atoms cannot take electrons easily. They can acquire `ns^2 np^6` configuration by sharing two electrons with the atoms of other elements and thus, exhibit `+2` oxidation state in their compounds. In addition to this, their atoms have vacant d-orbitals in their valence shell to which electrons can be promoted from the `p` and s-orbitals of the same shell. As a result, they can show `+4` and `+ 6` oxidation states. Oxygen exhibits `+2` oxidation state inA. `H_2 O`B. `OF_2`C. `Cl_2 O`D. `H_2 O_2` |
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Answer» Correct Answer - B Let the oxidation state of `O` in `OF_2` be `x` `x + (-1) xx 2 = 0 rArr x = + 2`. |
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| 166. |
Sulphur and rest of the elements of group `16` are less electronegative than oxygen, Therefore, their atoms cannot take electrons easily. They can acquire `ns^2 np^6` configuration by sharing two electrons with the atoms of other elements and thus, exhibit `+2` oxidation state in their compounds. In addition to this, their atoms have vacant d-orbitals in their valence shell to which electrons can be promoted from the `p` and s-orbitals of the same shell. As a result, they can show `+4` and `+ 6` oxidation states. The nature of the compounds of sulphur having `+4` oxidation state isA. Act as oxidising agentsB. Acts as reducing agentsC. Act as oxidising as well as reducing agentsD. Cannot be predicted |
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Answer» Correct Answer - C Compound of sulphur in oxidation state `+4` can reduce as well as oxidise and hence behave as oxidising agent and reducing agent. |
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| 167. |
a. `XeF_(6) + 4NaOH + 8H_(2)O rarr _______` b. `XeF_(6) + 7OH^(Θ)(strongly basic) rarr ______`- 3H_(2)O + 6F^(Θ)` c. `2HXeO_(4)^(Θ) + 2OH^(Θ) rarr _____ Xe + 2H_(2)O +O_(2)` d. `XeF_(6) + H^(Θ)(acidic) rarr _______XeOF_(4) +_____` e. `6XeF_(4) + 12H_(2)O rarr _______+24HF + 4Xe + 3O_(2)` f. `2XeF_(2) + 2H_(2)O rarr ____ + 4HF + 3O_(2)` g. `XeF_(6) + 3H_(2) rarr 6HF + _____` |
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Answer» Correct Answer - a.`NaXeO_(6)8H_(2)O or [XeO_(6)]^(4-)` (sodium peraxemate octabydram in percente ion)`(Xe-+8)` b,HXeO_(4)^(Theta) (Xenate ion , Xe= +6)` c.`XeO_(6)^(4-)` d.`H_(6)XeO_(6)` (Xenic acid ) or Xe(OH)_(6)` `e. 2XeO_(3)` f.`2Xe` g. Xe` a.`NaXeO_(6)8H_(2)O or [XeO_(6)]^(4-)` (sodium peraxemate octabydram in percente ion) `(Xe-+8)` b,`HXeO_(4)^(Theta) ("Xenate ion", Xe= +6)` c.`XeO_(6)^(4-)` d.`H_(6)XeO_(6)` (Xenic acid ) or `Xe(OH)_(6)` e. `2XeO_(3)` f.`2Xe` g. `Xe` |
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| 168. |
Sulphur and rest of the elements of group `16` are less electronegative than oxygen, Therefore, their atoms cannot take electrons easily. They can acquire `ns^2 np^6` configuration by sharing two electrons with the atoms of other elements and thus, exhibit `+2` oxidation state in their compounds. In addition to this, their atoms have vacant d-orbitals in their valence shell to which electrons can be promoted from the `p` and s-orbitals of the same shell. As a result, they can show `+4` and `+ 6` oxidation states. The oxidation state of sulphur in `Na_2 S_4 O_6` isA. `2//3`B. `3//2`C. `3//5`D. `5//2` |
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Answer» Correct Answer - D Let oxidation state of sulphur in `Na_2 S_4 O_6` is `x` `(+1) xx 2 + 4x + (-2) xx 6 = 0` `0 = + (10)/(4) =(5)/(2)`. |
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| 169. |
Helium is used to fill gas ballons instead of hydrogen because it is lighter and non-inflammable |
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Answer» Correct Answer - F Helium is heavier than hydrogen |
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| 170. |
Why helium and neon do not form clathrates with quinol? |
| Answer» The size of the cavities formed during crystallisatiohn of quinol is larger thum the size of helium and neon .Hence, He and Ne escapes from the cavities easily and do not from clathrates with quinol. | |
| 171. |
Neon is obtained during radioactive disintegration |
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Answer» Correct Answer - F He and Rn are obtained during radioactive disintegration |
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| 172. |
Neon in extensively used inA. cold stronge unitB. organic compoundsC. medicinesD. coloured electric discharge lamps |
| Answer» Correct Answer - d | |
| 173. |
Why neon is used in warning signal illuminations? |
| Answer» Neon lights are visible from long distance .The light is visible even during fog and mist condition .Hence neon is used in waring signal illumination . | |
| 174. |
Under ambient condition , the total number of gases released products in the final step of the reaction scheme shown below is |
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Answer» Correct Answer - c `XeF_(6) + 3H_(2)O rarr XeO_(3) + 6HF` (P) `xeO_(3) + OH Theta rarr HXeO_(4)^(Theta)` (P) (Q) `2HXeO_(4)^(Theta) + 2OH^(Theta) rarr XeF_(6)^(+) Xe_(4) + 2H_(2)O+O_(2g)` (Q) Gases are` Xe` and`O_(2)` |
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| 175. |
Why helium and neon do not form compounds with florine? |
| Answer» Atomic size of He and Ne is very small and possese very high inoisation enthalpies emthalpies.Both do not have d-orbitals in their valence shells. Thas the paired electrons present in valence shell cannot be unpaired in absence of simgly occupied orbitals, no sharting is possinble with other atoms .Hence , no compounds are formed by he and Ne. | |
| 176. |
A list of species having the formula `XZ_(4)` is given below `XeF_(4),SF_(4),SiF_(4),BF_(4)^(Θ), [Cu(NH_(3))_(4)]^(2+),[FeCl_(4)]^(2),[CoCl_(4)]^(2-)` and `[PtCl_(4)]^(2-)` Defining shape on the basis of the location of X and Z atoms , the total number of species having a square planar shape is |
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Answer» Correct Answer - `(4)` `XeF_(4), BeF_(4)^(Ө). [Cu(NH_(3))_(4)]^(2+), [PtCl_(4)]^(2-)`. |
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| 177. |
A list of species having the formula `XZ_(4)` is given below `XeF_(4),SF_(4),SiF_(4),BF_(4)^(Θ), [Cu(NH_(3))_4]^(2+),[FeCl_(4)]^(2-),[CoCl_(4)]^(2-)` and `[PtCl_(4)]^(2-)` Defining shape on the basis of the location of X and Z atoms , the total number of species having a square planar shape is |
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Answer» Correct Answer - 4 XeF_(4),BrF_(4)^(*Theta)[Cu(NH_(3))]^(-2),[PICI_(2)]^(2)` |
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| 178. |
The idea which prompted bartlett to prepare first ever compound of noble gas wasA. Low bond dissociation enthalpy of F-F in `F_(2)` moleculeB. High bond energy of `Xe-F`C. Ionisation exthalpies of `O_(2)` and `Xe` are almost sameD. none of the above |
| Answer» Correct Answer - c | |
| 179. |
The gaseous mixture used by deep sea divers for respiration isA. Nitrogens is much less soluble in blood than heliumB. Helium is much less soluble in blood than nitrogenC. Nitrogen is highly soluble in waterD. Due to high pressure deep under the sea nitrogen and oxygen react to give poisonous nitric oxide |
| Answer» Correct Answer - b | |
| 180. |
Noble gases are also known as aerogens becauseA. They occur in airB. They are rarely found in atmosphereC. They are most rarely found in atmosphereD. none of the above |
| Answer» Correct Answer - a | |
| 181. |
The mixed anhydride of nitrous and nitric acid is.A. `N_2 O`B. `NO_2`C. `NO`D. `N_2 O_5` |
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Answer» Correct Answer - B `HNO_2 + HNO_3 rarr NO_2 + H_2 O` Hence `NO_2` is mixed anhydride of `HNO_2` and `HNO_3`. |
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| 182. |
Of the following the most acidic is.A. `H_3 PO_4`B. `H_3 AsO_4`C. `H_3 SbO_4`D. `H_3 BiO_4` |
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Answer» Correct Answer - A When oxidation state of central atom is same, with the decrease in electronegativity of central atom, acidic character decreases. Hence `H_3 PO_4` is most acidic. `H_3 PO_4 gt H_3 AsO_4 gt H_3 SbO_4 gt H_3 BiO_4`. |
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| 183. |
Which elements of group `15` are metalloids ?A. NB. AsC. SbD. Bi |
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Answer» Correct Answer - B::C A strong continuous layer of oxide is formed on `Fe, Cr` and `Al` when they react with `conc HNO_3`, hence they become passive. |
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| 184. |
Which of the group `15` elelments are non-metallic ?A. NB. PC. AsD. Sb |
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Answer» Correct Answer - A::B `underset ((A)) (NH_4NO_3) + NaOH rarr underset ("Non-inflammabl e gas") (NaNO_3) + H_2 O` `NaNO_3 + 4Zn + 7NaOH rarr 4 Na_2 ZnO_2 + NH_3 + 2H_2 O` `NH_4 NO_2 (A) + NaOH rarr NH_3 + NaNO_2 + H_2 O` `3 Zn + 5 NaOH + NaNO_2 rarr 3 Na ZnO_2 + NH_3 + H_2 O`. |
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| 185. |
How many `P-O-P` bonds are present in `P_4 O_8` ? |
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Answer» Correct Answer - `(6)` There are `6 P-O-P "bonds in" P_4 O_8` |
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| 186. |
Nitrogen forms how many oxides |
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Answer» Correct Answer - `(-5)` `N_2 O, NO, N_2 O_3, NO_2 (or N_2 O_4) and N_2 O_5`. |
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| 187. |
In group `15` elements, the number of unpaired electrons in valence shell is____. |
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Answer» Correct Answer - `(3)` Valence shell electronic configuration of group `15` elements is `np^2 np^3`. Hence, there are `3` unpaired electrons in the valence shell. |
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| 188. |
White phosphorus on reaction with lime water gives calcium salt of an acid `(A)` along with a gas `(X)`. Which of the following is correct ?A. `(A)` on heating gives `(X)` and `O_2`B. The bond angle in `(X)` is less than that in case of ammonia.C. `(A)` is a disbasic acid.D. `(X)` is more basic than ammonia. |
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Answer» Correct Answer - B White phosphorous + Lime water rarr Calcium salt of an acid `(A) + Gas (X)` `2 P_4 + 8 Ca(OH)_2 + 6 H_2 O rarr 3 Ca(H_2 PO_2)_2 + 2 PH_3 (X)` `(A)` is `H_3 PO_2`, hypophosphorous acid. `(A)` is monobasic acid. `2 H_3PO_2 overset (Delta) rarr H_3 PO_4 + PH_3`. |
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| 189. |
Give the names of formulae of the compounds described below : (a) A compound of `N, H` and `O` which on heating gives laughing gas. (b) A compound of `N,H` and `O` which on heating gives nitrogen gas. ( c) A compound of `Ca, P` and `O` which is found in bones. (d) A compound of `N` and `H` which is used as refrigerant. ( e) A compound of `N` and `H` which behaves like an acid. (f) A compound of `N` and `H` which is used as a rocket fuel. (g) A compound of `N, H, S` and `O` which is used as a fertiliser. (h) A compound of `N, H, S` and `O` which is used for making oximes. (i) Two neutral oxides of nitrogen. (j) The oxyacid of phosphorus is used for the preparation of `HBr` and `H I` from bromides and iodides respectively. |
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Answer» Ammonium nitrate, `NH_4 NO_3` (b) Ammonium nitrate, `NH_4 NO_2` ( c) Calcium phosphate, `Ca_3(PO_4)_2` (d) Ammonia, `NH_3` ( e) Hydrazine acid, `N_3 H` (f) Hydrazine, `N_2 H_4` (g) Ammonium sulphate, `(NH_4)_2 SO_4` (h) Hydroxylamine, `NH_2 OH` (i) Nitrous oxide, `N_2 O` and nitric oxide, `NO` (j) Orthophosphoric acid, `H_3 PO_4`. |
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| 190. |
Extra pure `N_2` can be obtained by heatingA. `NH_3 with CuO`B. `NH_4 NO_3`C. `(NH_4)_2 Cr_2 O_7`D. `Ba(N_3)_2` |
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Answer» Correct Answer - D (a) `NH_3 + Cu O overset (Delta)rarr N_2 (g) + Cu(s) + H_2 O` (b) `NH_4 NO_3 overset (Delta) rarr N_2 O(g) + H_2 O (g)` ( c) `(NH_4)_2 Cr_2 O_7 overset (Delta) rarr N_2 (g) + H_2 O(g) + Cr_2 O_3 (s)` (d) `Ba(N_3)_2 overset (Delta) rarr Ba_3 N_2 + N_2 (g)`. |
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| 191. |
Nitrogen is an inert molecule due to low bond dissociation enthalpy. |
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Answer» Correct Answer - F Nitrogen is an inert molecule due to high bond dissociation enthalpy. |
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| 192. |
`N_2 O_3` is known as laughing gas. |
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Answer» Correct Answer - F Laughing gas is `N_2 O`. |
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| 193. |
`NCl_3` is a stable compound. |
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Answer» Correct Answer - F `NCl_3` is an explosive compound. |
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| 194. |
Yellow phosphorous is polymeric in nature. |
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Answer» Correct Answer - F Yellow phosphorous exist ad discrete `P_4` unit. |
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| 195. |
Catalytic oxidation of `NH_3` (passing a mixture of `NH_3` and air over heated `Pt` gauge) gives.A. `NO`B. `N_2 O`C. `N_2 O_3`D. `N_2 O_5` |
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Answer» Correct Answer - A `2 NH_3 + 5//2 O_2 underset (Delta) overset (Pt)rarr 2 NO + 3 H_2 O`. |
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| 196. |
Write the balanced equations for the reactions of the following compounds with `H_2 O`. (i) `CaNCN` (ii) `NCl_3`. |
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Answer» (i) With calcuim cynamide, calcuim carbonate is precipitated. `CaNCN + 5H_2O rarr CaCO_3 darr + 2 NH_4 OH` (ii) With nitrogen trichloride, ammonia and hypochlorous acid are formed. `NCl_3 + 3 H_2 O rarr NH_3 + 3HOCl`. |
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| 197. |
Nitrate `(I)` oxide is product byA. thermal decomposition of ammonium nitrateB. disproportionation of `N_2 O_4`C. thermal decomposition of ammonia nitrateD. interaction of hydroxylamine and nitrous acid. |
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Answer» Correct Answer - A::D Following two method are used for the preparation of nitrous oxide : `NH_4 NO_3 rarr N_2 O + 2 H_2 O` `NH_2 OH.HCl + NaNO_2 rarr NaCl + 2 N_2 O + 2 H_2 O`. |
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| 198. |
When silver nitrate is heated, the product areA. Oxygen and metal nitrateB. Nitrogen dioxide, `O_2` and metallic oxide.C. Nitrogen dioxide. `O_2` and metalD. Nitrogen dioxide and metal oxide. |
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Answer» Correct Answer - C `2 AgNO_3 overset (Delta) rarr 2 Ag + 2 NO_2 + O_2`. |
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| 199. |
Chlorine reacts with excess of ammonia to form.A. `NH_4 Cl`B. `N_2 + HCl`C. `N_2 + NH_4 Cl`D. `N_2 + NCl_3` |
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Answer» Correct Answer - C `Cl_2 + 4NH_3(Express) rarr N_2 + 2 NH_4 Cl`. |
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| 200. |
Which of the following statement is incorrect ?A. `NO` is heavier than `O_2`B. The formula of heavy water in `D_2 O`.C. Nitrogen diffuses faster than oxygen through an orifice.D. `NH_3` can be used as a refrigerant. |
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Answer» Correct Answer - A Molecular weight `(Mw) NO = 14 + 16 = 30` Vapour density `= (Mw)/(2) = (30)/(2) = 15` Molecular weight `(Mw) of O_2 = 16 + 16 = 32` Vapour density `= (Mw)/(2) = (32)/(2) = 16` Therefore, `NO` is lighter than `O_2`. Hence statement (a) is incorrect. ( c) `Mw of N_2 = 28, Mw of O_2 = 32`, Rate of diffusion or effusion in inversely proportional to the square root of their molecular weight or vapour densities. Since the `Mw` of `N_2` is less than `Mw` of `O_2`, so `N_2` diffuses faster than `O_2`. |
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