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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.
| 351. |
How does `"PCl"_(5)` react with the following ? Water |
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Answer» `"PCl"_(5)` undergo hydrolysis to form phosphoric acid. `"PCl"_(5)+H_(2)Oto"POCl"_(3)+2"HCl"` `"POCl"_(3)+3"H"_(2)O toH_(3)"PO"_(4)+3" HCl"` |
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| 352. |
How does `"PCl"_(3)` react with water. |
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Answer» `"PCl"_(3)` reacts with water (hydrolysis) to form phosphorus acid. It undergo hydergo hydrolysis in presence of moisture. `"PCl"_(3)+3"H"_(2)O to H_(3)PO_(3)+3"HCl". ` |
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| 353. |
Which oxide of sulphur can act as both oxidizing and reducing agent ? Give one example each. |
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Answer» Sulphur dioxide `(SO_(2))` acts as both oxidising as well as reducing agent. `SO_(2)` as Oxidising agent : Sodium sulphide oxidises to hypo with `SO_(2)`. `2Na_(2)S+3SO_(2)to2Na_(2)S_(2)O_(3)+S` `SO_(2)` as Reducing agent : `SO_(2)` reduces `Fe^(+3)` ions to `Fe^(+2)` ions. `2Fe^(+3)+SO_(2)+2H_(2)O to2Fe^(+2)+SO_(4)^(-2)+4H^(+)` |
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| 354. |
How does `"PCl"_(5)` react with the following ? Ag |
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Answer» `PCl_(5)` reacts with Ag to form `PCl_(3)` and AgCl `PCl_(5)+2" Ag"to2" AgCl"+PCl_(3)` |
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| 355. |
How does `"PCl"_(5)` react with the following ? `"CH"_(3)"COOH"` |
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Answer» `PCl_(5)` reaccts with `CH_(3)COOH` to form acetyl chloride. `CH_(3)COOH+PCl_(5)toCH_(3)COCl+POCl_(3)+HCl` |
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| 356. |
`PCl`_(3)` can act as an oxidizing as well as a reducing agent - Justify. |
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Answer» `"PCl"_(3)` acts as reducing agent. It is evidented by the following reaction. `underset((+3))("PCl"_(3))+"Cl"_(2)tounderset((+5))("PCl"_(5))` `"PCl"_(3)` acts as oxidising agent. It is evidented by the following reaction. `underset((+3))(2"PCl"_(3("vapour")))overset(H_(2))tounderset((+2))(P_(2)"Cl"_(4))` |
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| 357. |
Which of the following are not known ? `"PCl"_(3),"AsCl"_(3),"SbCl"_(3),"NCl"_(5),"BiCl"_(5),"PH"_(5)` |
| Answer» `"NCl"_(5),"BiCl"_(5),"PH"_(5)` are not known in the given compounds. | |
| 358. |
Identify the statement that is not correct as far as structure of dibroane is concernedA. there are two bridging hydrogen atoms in diboraneB. each boron atom forms four bonds in diboraneC. all `B-H` bonds in diborane are similarD. the hydrogen atoms are not in the same plane in diborane |
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Answer» Correct Answer - C Diborane possess four `B-H` covalent bonds and two three centred (two electrons) `B-H-B` or hydrogen bridge bonds. These bonds are also known as banana bonds. |
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| 359. |
Match the species given in Column I with properties given in Column II. `{:(,"Column I",,,"Column II"),((i),"Dibroane",,(a),"Used as a flux for soldering metals"),((ii),"Gallium",,(b),"Crystalline form of silica"),((iii),"Borax",,(c ),"Banana bonds"),((iv),"Aluminosilicate",,(d),"Low melting , high boiling, useufl for measuring high temperature"),((v),"Quartz",,(e),"Used as catalyst in petrochemical industries"):}` |
| Answer» Correct Answer - `(i) rarr (c ) ; (ii) rarr (d) ; (iii) rarr (a) ; (iv) rarr (e ) ; (v) rarr (b)`. | |
| 360. |
Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of `F_(2) and Cl_(2)` |
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Answer» Correct Answer - The electrode potential of `F_(2) (+2.87 V)` is much higher than that of `Cl_(2) (+ 1.36 V)`, therefore, `F_(2)` a much stronger oxidizing agent than `Cl_(2)`.now, electrode potantial depends upon three factors : (i) bond dissociation energy (ii)electron gain enthalpy and (iii)hydration energy.Although electron gain enthalpy of fluorine is less negative `(-333 kJ mol^(-1))` than that of chlorine `(- 349 kJ mol^(-1))` the bond dissociation energy of `F-F` bond is much lower `(158.8 kJ mol^(-1))` than that of `CI-CI` bond `(242.6 kJ mol^(-1))` and hydration energy of `F^(-)` ion `(515 kJ mol^(-1))` is much higher than that of `Cl-Cl` ion `(381 kJ mol^(-1))`.The later two factors more than compensate the less negative electron gain enthalpy of `F_(2)`.As a result, electrode potantial of `F_(2)` is higher than that of `Cl_(2)` and hence `F_(2)` is a much stronger oxidizing agent than `Cl_(2)`. |
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| 361. |
Carbon does not form any complesses while slicon particlipates in the formation of complexes. Explain. |
| Answer» Carbon atom has no vacant 2d-orbitals to eaccept electron paris from electron donor species also called ligands. But silicon with vacant 3d orbitals can easily form complexes with ligands such as `F^(-),Cl^(-)` etc. | |
| 362. |
Consider the following statement . I. `XeOF_(4)` has square pyramidal structure . II. `XeF_(2)` has linear structure . Which of the above mentioned statements(s) is/are true ? Choose the correct option.A. Only IB. Only IIC. Both I and IID. Neither I nor II |
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Answer» Correct Answer - C `XeF_2` has linear structure and `XeOF_4` has square pyramidal structure . |
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| 363. |
How can you explain higher stability of `BCl_(3)` as compared to `TlCl_(3)` ? |
| Answer» Due to the poor shielding of the s-electrons of the valence shell (6s) by the 3d-,4d-,5d and 4f-electrons inert pair effect is maximum in Tl. As a result only `6 p^(1)` electrons participates in bond formation and contrast, due to the absence of d-and f-electrons , B does not show inert pair effect. In other words, all the three valence electrons (i.e, two 2s - and one 2p-) take part in bond formation and hence B shows an oxidation state of +3 and thus forms `BCl_(3)`. thus, `BCl_(3)` is more stable than `TlCl_(3)`. | |
| 364. |
Which of the following com bines with haemoglobin of the blood to form carboxyhaemoglobin?A. COB. `CO_(2)`C. `O_(2)`D. `N_(2)`. |
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Answer» Correct Answer - A CO forms carboxy haemoglobin. |
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| 365. |
Which one of the following pentafluorides cannot be formed?A. `PF_(5)`B. `AsF_(5)`C. `SbF_(5)`D. `BiF_(5)` |
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Answer» Correct Answer - D The +5 oxidation state of Bi is unstable due to inert pair effect . Thus , `BiF_(5)` cannot be formed . |
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| 366. |
Which of the following is most volatile compound?A. HIB. HClC. HBrD. HF |
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Answer» Correct Answer - B Boiling point of HF is highest due to H-bonding . For other halogen acids , boiling point increases in the order `HCl lt HBr lt HL` . Therefore , most volatile (with lower boiling point ) is HCl. |
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| 367. |
Which of the following does not exist in free stateA. `BF_(3)`B. `BCl_(3)`C. `BBr_(3)`D. `BH_(3)` |
| Answer» `BH_(3)` exists as a dimer i.e, `B_(2)H_( 6)`. | |
| 368. |
Which one of the following does not exist in the free form?A. `BF_(3)`B. `BCl_(3)`C. `BBr_(3)`D. `BH_(3)` |
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Answer» Correct Answer - D Due to back bonding `NF_(3),BCl_(3)` and `BBr_(3)` exist in free form. But `BH_(3)` does not. |
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| 369. |
An orange solid (X) on heating, gives a colourless gas (Y) and a only green residue (Z). Gas (Y) on treatment with Mg, produces a white solid substance……A. `Mg_(3)N_(2)`B. `MgO`C. `Mg_(2)O_(3)`D. `MgCl_(2)` |
| Answer» Correct Answer - A | |
| 370. |
The reaction which is consistent with the fact that `Cl_(2)O_(6)(s)` exists as `[ClO_(2)^(+)][ClO_(4)^(-)]`, would be:A. `Cl_(2)O_(6)+NaOH rarr NaClO_(3)+NaClO_(4)+H_(2)O`B. `Cl_(2)O_(6)+HF rarr ClO_(2)F+HClO_(4)`C. `2HClO_(4)+P_(2)O_(5) rarr2HPO_(3)+Cl_(2)O_(7)`D. `2ClO_(2)+2O_(3) overset(0^(@)C) rarr Cl_(2)O_(6)+2O_(2)` |
| Answer» Correct Answer - B | |
| 371. |
Write two uses of `ClO_(2)`. |
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Answer» Uses of `ClO_(2)` : `to ClO_(2)` is highly reactive oxidising agent. `to` It is used as bleaching agent for paper pulp and textiles. `to` It is used in water treatment. |
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| 372. |
`ClO_(2)` reacts with water and alkali to give:A. sodium chlorateB. sodium chloriteC. sodium chlorate and sodium chloriteD. none of the above |
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Answer» Correct Answer - C |
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| 373. |
The order of the oxidation state of the phosphours atom in `H_(3)PO_(2),H_(3)PO_(4),H_(3)PO_(3)` and `H_(4)P_(2)O_(6)` isA. `H_(3)PO_(4) gt H_(3)PO_(2) gt H_(3)PO_(3) gt H_(4)P_(2)O_(6)`B. `H_(3)PO_(4) gt H_(4)P_(2)O_(6) gt H_(3)PO_(3) gt H_(2)PO_(2)`C. `H_(3)PO_(2) gt H_(3)PO_(3) gt H_(4)P_(2)P_(6) gt H_(3)PO_(4)`D. `H_(3)PO_(3) gt H_(3)PO_(2) gt H_(3)PO_(4) gt H_(4)P_(2)O_(6)` |
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Answer» Correct Answer - B |
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| 374. |
When `H_2S` gas in passed through nitric acid, the product is :A. rhombic SB. prismatic S (colloidal )C. amorphous SD. monocilinic S |
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Answer» Correct Answer - B When `H_2S` gas is passed through nitric acid , colloidal sulphur is formed as `H_2S + 2HNO_3 to 2H_2O +2NO_2 +S ` |
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| 375. |
Nitrogen forms `N_(2)` but phosphorous when forms `P_(2)` gets readily converted into `P_(4)` becauseA. `p pi- p pi ` bonding is weakB. multiple bond is formed easillyC. `p pi - p pi` bonding is strongD. triple bond is present in phosphorus atoms |
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Answer» Correct Answer - A `p pi-p pi ` bonding is weaker in P than in N. |
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| 376. |
Why does the reactivity of nitrogen differ from phosphorus ? |
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Answer» Nitrogen gas exists as diatomic molecule. Dut to the presence of triple bond between Base N - Atoms bond dissociation energy is high (941.4 KJ / mol). Hence nitrogen is inert and unreactive. Phosphorus is a tetra atomic molecule and P-P single bond is weaker than `N-=N.P-P` bond dissociation energy is 213 KJ/mole. Hence phosphorus is more reactive than Nitrogen. |
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| 377. |
Why does nitroggen show catenation properties less than phosphorus ? |
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Answer» Explanation : `to` The single N-N bond is weaker than the single P-P bond due to high inter electronic repulsion of the non-bonding electrons in `N_(2)` because of small bond length. Therefore the catenation property is weaker in nitrogen as compared to phosphorrus. |
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| 378. |
Nitrogen exists as diatomic molecule and phosphorus as `P_(4)` - Why ? |
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Answer» Nitrogen exists as diatomic molecule : `to` Nitrogen has small size and high electronegativity and nitrogen atom forms `P pi-P pi` multiple bonds with it self (triple bond). So it exists as a discrete diatomic molecule in elementary state. Phosphorus exists as tetra atomic molecule : `to` Phosphorus has large size and less electronegative and it forms P-P single bonds. So it exists as tetra atomic i.e., `P_(4)`. |
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| 379. |
Nitrogen exists as diatomic molecule and phosphorus as `P_(4)`. Why ? |
| Answer» Nitrogen because of its small size and high electronegativity forms `ppi-ppi-` multiple bonds.Therefore, it exists as a diatomic molecule having a triple bond between the two `N`-atoms.Phosphorus, on the other hand, due to its larger size and lower electronegativity usually does not form `ppi-ppi` multiple bonds with itself.Instead it prefers to form `P-P` single bonds and hence it exists as tetrahedral `P_(4)` molecules. | |
| 380. |
The bond dissociation energy of `B-F` in `BF_(3)` is `646 kJ "mol"^(-1)` whereas that of `C-F` in `CF_(4)` is `515 kJ "mol"^(-1)`. The correct reason for higher B-F bond dissociation energy as compared to that of C-F isA. stronger `sigma`-bond between B and F in `BF_(3)` as compared to that between `C` and `F` in `CF_(4)`B. significant `ppi-ppi` interaction between B and F in `BF_(3)` whereas there is no possibility of such interaction between C and F in `CF_(4)`C. lower degree of `ppi-ppi` interaction between B and F in `BF_(3)` than that between C and F in `CF_(4)`.D. smaller size of B-atom as compared to that of C-atom |
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Answer» Correct Answer - B Due to identical size of 2p-orbitals on B and F, back donation of a lone pair of electrons from the filled 2o-orbitals of F to the empty 2p-orbital of B occurs to a considerable extent. As a result , `B-F` bond has some double character and hence bond dissociation energy of `B-F` bond in `BF_(3)` is much higher than that of C-F bond in `CF_(4)`. the reason being that C does not have an empty 2p-orbitals and hence back donation does not occur. |
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| 381. |
Which of the following oxides is amphoteric ?A. `SnO_(2)`B. `CaO`C. `SiO_(2)`D. `CO_(2)` |
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Answer» Correct Answer - A `SnO_(2)` ic amphoteric, `SiO_(2)` and `CO_(2)` are acidic while `CaO` is basic. |
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| 382. |
Litharge is chemicallyA. `PbO`B. `PbO_(2)`C. `Pb_(3)O_(4)`D. `Pb(CH_(3)COO)_(2)` |
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Answer» Correct Answer - A Lithage is `PbO`. |
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| 383. |
Among the following the maximum covalent character is shown by the compound:A. `SnCl_(2)`B. `AlCl_(3)` and `SiCl_(4)`C. `MgCl_(2)`D. `FeCl_(2)` |
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Answer» Correct Answer - B Greater the charge and smaller the size of the cation (i.e, `Al^(3+)`), greater the charge density, greater the polarization hence greater is the covalent character. i.e., `AlCl_(3)`. |
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| 384. |
`PbCl_(2)` is insoluble in cold water. Addition of HCl increases its solubility due toA. formation of soluble complex anions like `[PbCl_(3)]^(-)`B. oxidation of `Pb(II)` to `Pb(IV)`C. formation of `[Pb(H_(2)O)_(6)]^(2+)`D. formation of polymeric lead complexes |
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Answer» Correct Answer - A Addition of `Cl^(-)` ion to a suspension of `PbCl_(2)` leads to the formation of soluble complex as shown below : `underset(PbCl_(2))(PbCl_(2)) + Cl^(-) rarr underset(" Soluble complex ")([PbCl_(3)]^(-))`. |
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| 385. |
`AlCl_(3)` is a Leis acid than `BCl_(3)`. |
| Answer» Correct Answer - weaker | |
| 386. |
Assertion: Boron does not show univalent nature but unipositive nature of thallium is quite stable. Reason: Inert pait effect predominates in thallium.A. If both assertion and reason are true and the reason is the correct explanation of teh assertion.B. If both assertion and reason are true but reason in not the correct explation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - A If both assertion and reason are true and the reason is the correct explanation of the assertion. |
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| 387. |
Thallium shows different oxidation states due to :A. it is atypical metalB. it shows iner pair effectC. it is amphoteric in natureD. it has very low inoisation enthhalpy. |
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Answer» Correct Answer - B Thallium shows inert pair effect. |
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| 388. |
The most stable oxidation state of aluminum is `"……………."` while that of thallium is `"………….."`. |
| Answer» Correct Answer - `+3,+1` | |
| 389. |
The `+1` oxidation state is more stable than the +3 oxidation state for thallium. Explain why ? |
| Answer» The outer electronic configuration of thalium is `6s^(2)6p^(1)`. Therefore , it can show oxidation states of `+1` and `+3` But due to inert pair effect, the `6s^(2)` electrons are more strongly attracted by the nucleus than `6p^(1)` electron. Therefore, `6s^(2)` electrons do not participate in bond formation. Instead only `6p^(1)` electron takes part is more stable than its `+3` state. | |
| 390. |
For `BCl_(3), AlCl_(3)` and `GaCl_(3)` the increasing order of ionic character isA. `BCl_(3) lt AlCl_(3) lt GaCl_(3)`B. `GaCl_(3) ltAlCl_(3)lt BCl_(3)`C. `BCl_(3) lt GaCl_(3) lt AlCl_(3)`D. `AlCl_(3) lt BCl_(3) lt GaCl_(3)` |
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Answer» Correct Answer - C The electronegativity decreases in ordr : `B(2.0) gt Ga(1.6) gt Al (1.5)`, therefore ,ionic character of their chlorides increases in the reverse order , i.e., `BCl_(3) lt GaCl_(3) lt AlCl_(3)` and hence option (c ) is correct. |
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| 391. |
For the properties mentioned, the correct trend for the different species is inA. strength as Lewis acids - `BCl_(3) gt AlCl_(3) gt GaCl_(3)`B. inert pair effect - `Al gt Ga gt In`C. oxidising property `-Al^(3+) gt Ga gt In`D. first ionization enthalpy `-B gt Al gt Tl` |
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Answer» Correct Answer - A Only option (a) , i.e., strength of Lewis acids decreases in the order : `BCl_(3) gt AlCl_(3) gt GaCl_(3)` is correct while all other options are incorrect as explained below : Al does not show inert pair effect, `Tl^(3+)` has the highest oxidising power while ionization enthalpy of Tl is higher than that of Al. |
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| 392. |
The +1 oxidation state of thallium is more stable than its +3 oxidation state because ofA. its atomic sizeB. its ionization potentialC. inert pair effectD. diagonal relationship |
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Answer» Correct Answer - C Inert pair effect. |
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| 393. |
Acidified potassium permanganate is dropped over sodium peroxide taken in a round bottom flask at room tempreture, vigorus reaction takes place to procuce:A. hydrogen peroxideB. mixure of hydrogen and oxygenC. a colourless gas hydrogenD. a colour gas dioxygen. |
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Answer» Correct Answer - D `2KMnO_(4)+5Na_(2)O_(2)+8H_(2)SO_(4)rarrK_(2)SO_(4)+2MnSO+5Na_(2)SO_(4)+8H_(2)O+5O_(2)` |
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| 394. |
in the electrolysis method of acidified water to give `O_(2)`, the cathode used isA. graphiteB. leadC. platinumD. Nickel |
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Answer» Correct Answer - C In the electrolysis method of acidified water to give `O_(2)`, the cathote is used in plantinum |
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| 395. |
Ozone depleton due to the fomation of following compound in AntarcticaA. AcroleinB. Peroxy acety nitrateC. `SO_(2)` and `SO_(3)`D. Chlorine nitrate |
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Answer» Correct Answer - D Formation of chlorine nitrate is the main cause of ozone depletion. |
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| 396. |
Ozone depleton due to the fomation of following compound in AntarcticaA. AcrolienB. Peroxy acetyl nitrateC. `SO_2` and `SO_3 `D. Chlorine nitrate |
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Answer» Correct Answer - D Formation of chlorine nitrate is the main cause of ozone depletion |
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| 397. |
A greenish yellow gas reacts with an alkin metal hydroxide to form a halate which can be used in fireworks and saftey matches. The gas and the halate areA. `Br_2 , KBrO_3`B. `Cl_2, KClO_3`C. `I_2, NaIO_3`D. `Cl_2 , NaClO_3` |
| Answer» Correct Answer - B | |
| 398. |
A greenish yellow gas reacts with an alkin metal hydroxide to form a halate which can be used in fireworks and saftey matches. The gas and the halate areA. `Br_2, KBrO_3`B. `Cl_2, KClO_3`C. `l_2, NalO_3`D. none |
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Answer» Correct Answer - B |
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| 399. |
Silicones have the general formulaA. `SiO_(4)^(4-)`B. `Si_(2)O_(7)^(6-)`C. `(R_(2)SiO)_(n)`D. `(SiO_(3))_(n)^(2-)` |
| Answer» Correct Answer - C | |
| 400. |
Anhydrous `AlCl_(3)` is covalent but hydrated `AlCl_(3)` is electrovalent. Explain. |
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Answer» Anhydrous salt exists as a dimer `Al_(2)Cl_(6)` and is covalen t in nature. In contact with water, it gests hydrated and the hydrated salt `[Al(H_(2)O)_(6)]Cl_(3)` becomes ionic. In fact, during hydration, a large amount of hydration energy is released which helps in the inoisation of the salt. `[Al(H_(2)O)_(6)]Cl_(3)hArr[Al(H_(2)O)_(6)]^(3+)+3Cl^(-)` |
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