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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.
| 1501. |
Three moles of `B_(2)H_(6)` are completely reacted with methanol. The number of moles of boron containing product formed is |
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Answer» Correct Answer - 6 Six : `3B_(2)H_(6) + 8 MeOH rarr underset("Trimethylborate")(6B(Ome)_(3))+ 18H_(2)` |
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| 1502. |
The heavier elements of groups 13 and 14 besides the group oxidation state exhibit another oxidation state which is two units lower than the group oxidation state and the stability of this lower oxidation state increases down the group. This concept which is commonly called inert pair effect has been used to explain many physical and chemical properties of the element of these groups. The strongest reductant among the following isA. `GeCl_(2)`B. `SnCl_(2)`C. `PbCl_(2)`D. `SnCl_(4)` |
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Answer» Correct Answer - A `+4` oxidation state of Ge is more stable than that of Sn, therefore, strongest reducant is GeCl_(2)` |
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| 1503. |
The ease of liquefaction of noble gases increases in the orderA. `HeltXeltArltKrltNe`B. `XeltKrltArltHeltNe`C. `HeltNeltArltKrltXe`D. `XeltHeltNeltArltKr` |
| Answer» Correct Answer - C | |
| 1504. |
Gradual addition of electronic shells in the noble gases causes a decrease in theirA. ionisation energyB. densityC. boiling pointD. atomic radius |
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Answer» Correct Answer - A As the number of shells increases , size increases and the effective nuclear charge on the outermost electron decreases . Thus , IE decreases . |
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| 1505. |
Which of the following is electron deficient ?A. `PH_(3)`B. `(CH_(3))_(2)`C. `(SiH_(3))_(2)`D. `(BH_(3))_(2)` |
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Answer» Correct Answer - D `(BH_(3))_(2)` or `B_(2)H_(6)` (diborane) is the and electron deficient compound. |
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| 1506. |
The oxidation state of phosphorus in cyclotrimetaphosphoric acid isA. `+3`B. `+5`C. `-3`D. `+2` |
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Answer» Correct Answer - B O.S. of P in `(HPO_(3))_(3)" is "(+1+x-6)_(3)=0orx=+5` |
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| 1507. |
Give chemical reaction in support of the statement that all the bonds in `PCl` molecule are not equivalent. |
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Answer» Due to greater bond pair-bond repulsions, the two axial P-Cl bonds are less stable than the three equatorial P-Cl bonds. It is because of this reason tht when `PCl_(5)` is heated, the less stable axial bonds are broken to form `PCl_(3)` and `Cl_(2)`. `PCl_(5)overset(Delta)toPCl_(3)+Cl` |
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| 1508. |
Which of the following statemnets regarding sulpher is incorrect?A. `S_(2)` moleucle is paramagnetic.B. The vapour at `200^(circ)C` consists mostly of `S_(8)` rings.C. At `600^(circ)C` the gas mainly consists of `S_(2)` moleucles.D. The oxidation state of sulphur is never less than `+4` in its compounds. |
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Answer» Correct Answer - `4` Sulpher exhibit `+2,+4,+6` oxidation states but +4 and +6 are more common. |
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| 1509. |
Which of the following products is/are obtained in the following reaction `KBrO_3+F_2+KOH to` productsA. `KBrO_4`B. KFC. HOFD. `Br_2` |
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Answer» Correct Answer - A,B `KBrO_3+F_2+2KOH to KBrO_4+2KF+H_2O`. |
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| 1510. |
Which one of the following has highest Lewis acid strength ?A. `BI_(3)`B. `BBr_(3)`C. `B_(3)`D. `BCl_(3)` |
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Answer» Correct Answer - A `BI_(3)` is the strongest Lewis acid. |
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| 1511. |
In which species is the oxidation number for hydrogen different from those in the other three?A. `AlH_(3)`B. `H_(3)AsO_(4)`C. `H_(3)PO_(3)`D. `NH_(3)` |
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Answer» Correct Answer - A `AlH_(3)^((-1))` |
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| 1512. |
Which acid should be stored in plastic containers rather than in glass ones?A. Hydrofluoric acidB. Nitric acidC. Phosphoric acidD. Sulphuric acid |
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Answer» Correct Answer - A HF reacts with galss. |
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| 1513. |
Conc. `HNO_(3)` is stored in containers made up ofA. `Cu`B. `Zn`C. `Al`D. `Sn`. |
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Answer» Correct Answer - C Al is rendered passive by formation of a thin impervious layer of `Al_(2)O_(3)` on its surface when brought in contact with conc. `HNO_(3)`. |
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| 1514. |
`Al_(2)O_(3)` can be converted to anhydrous `AlCl_(3)` by heatingA. a mixture of `Al_(2)O_(3)` and carbon in dry `Cl_(2)` gasB. `Al_(2)O_(3)` with `Cl_(2)` gasC. `Al_(2)O_(3)` with HCl gasD. `Al_(2)O_(3)` with NaCl in solid state |
| Answer» (a) `Al_(2)O_(3)+3Cl_(2)("dry")+3Cto2AlCl_(3)+3CO` | |
| 1515. |
`Al_(2)O_(3)` becomes anhydrous `AlCl_(3)` upon heatingA. with `NaCl`B. with dry `Cl_(2)` and CC. with `Cl_(2)`D. with dry `HCl` gas |
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Answer» Correct Answer - B `Al_(2)O_(3) + 3Cl_(2) + 3C overset("Heat")rarr2AlCl_(3) + 3 CO`. |
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| 1516. |
Heating an aqueous solution of aluminium chloride to dryness will giveA. `Al_(2)O_(3)`B. `Al_(2)Cl_(6)`C. `AlCl_(3)`D. `Al(OH)Cl_(2)` |
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Answer» Correct Answer - A On heating `AlCl_(3)` (aq) to dryness, `Al_(2)O_(3)` is formed. `{:(2AlCl_(3)+6H_(2)Oto2Al(OH)_(3)+6HCl),(Al(OH)_(3)toAl_(2)O_(3)+3H_(2)O):}` |
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| 1517. |
Heating an aqueous solution of aluminium chloride to dryness will giveA. `AlCl_(3)`B. `A l_(2)Cl_(6)`C. `Al_(2)O_(3)`D. `Al(OH) Cl_(2)` |
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Answer» Correct Answer - A `Al_(2)Cl_(6)` will be formed. |
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| 1518. |
Complete the following reactions : (i) `P_(4)+SOCl_(2)to` (ii) `CH_(3)COOH+PCl_(5)to` |
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Answer» (i) White phosphorus `(P_(4))` reacts with thionyl chloride `(SOCl_(2))` to form phosphorus trichloride `(PCl_(3))` `P_(4)+8SOCl_(2)to4PCl_(2)+4SO_(4)+2S_(2)Cl_(2)` (ii) Acetic acid reacts with phosphorus pentachloride `(PCl_(5))` to form acetly chloride `underset("Acetic acid")(CH_(3)COOH)+PCl_(5)tounderset("Acetyl chloride")(CH_(3)COCl)+POCl_(3)+HCl` |
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| 1519. |
In which of following ions, the hybrid state of halogen atom is `sp^3` ?A. `ClO^-`B. `ClO_4^-`C. `ClO_3^-`D. In all |
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Answer» Correct Answer - D Cl is sp3 hybridised in `ClO^(-), ClO_3^(-), ClO_4^(-)` |
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| 1520. |
The hybrid state of halogen atom is `sp^(3)` inA. `ClO_(4)^(-)`B. `ClO^(-)`C. `ClO_(3)^(-)`D. All of the above |
| Answer» Correct Answer - D | |
| 1521. |
Interhalogen compounds areA. covalent moleculesB. diamagnetic in natureC. volatile solids/liquids at 298 K except CIFD. All of the above |
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Answer» Correct Answer - D Interhalogen compounds are covalent molecules and diamagnetic in nature . These are volatile solids or liquids at 298 K except CIF. |
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| 1522. |
The stability of interhalogen compounds follows the orderA. `ClF_(3)gtBrF_(3)gtIF_(3)`B. `BrF_(3)gtIF_(3)gtClF_(3)`C. `IF_(3)gtBrF_(3)gtClF_(3)`D. `ClF_(3)gtIF_(3)gtBrF_(3)` |
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Answer» Correct Answer - C Electropositive character of halogen is in order `IgtBrgtCl`. Central atom is bigger in size, more electropositive nature form stable interhalogen compound. |
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| 1523. |
When chlorine is passed through concentrated solution of KOH, the compound formed is ____________ .A. KCIB. `KClO_3`C. `KClO_2`D. `KClO_4` |
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Answer» Correct Answer - B `6KOH + 3Cl_2 to 5KCl + KClO_3 + 3H_2O` |
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| 1524. |
As the atomic number of halogens increases. The halogensA. Lose the outermost electrons less readilyB. `Become lighter in colourC. Become less denserD. Gain electrons less readily |
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Answer» Correct Answer - D As the atomic number increase electronegativity decreases. Hence, tendency to gain electron decreases. |
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| 1525. |
mark the element which displaces three halogens from their compoundsA. `F`B. `Cl`C. `Br`D. `I` |
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Answer» Correct Answer - A `F_(2)+2Cl^(-)rarrCl_(2)+2F^(-)` `F_(2)+2Br^(-)rarrBr_(2)+2F^(-)` `F_(2)+2I^(-)rarrI_(2)+2F^(-)` |
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| 1526. |
Ammonium dichromate on heating givesA. Chromium oxide and ammoniaB. Chromium acid and nitrogenC. Chromic acid and ammoniaD. Chromic acid and ammonia |
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Answer» Correct Answer - C `(NH_(4))_(2)Cr_(2)O_(7)overset(Delta)rarrCr_(2)O_(3)+N_(2)+4H_(2)O` |
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| 1527. |
On heating ammonium dichromate, the gas evolved is :A. oxygenB. ammoniaC. nitrous oxideD. nitrogen |
| Answer» Correct Answer - D | |
| 1528. |
When boron is fused with potassium hydroxide which pair of species are formed?A. `K_(2)O+B_(2)O_(3)`B. `KO_(2)+B_(2)H_(6)`C. `K_(3)BO_(3)+H_(2)`D. `K_(3)BO_(3)+H_(2)O` |
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Answer» Correct Answer - C `2B(s)+6KOH(s)overset(Fuse)to2K_(3)BO_(3)+3H_(2)` |
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| 1529. |
Iron sulphide on combustion forms an oxide of sulphur which is dissolved in water to give an acid. The basicity of this acid isA. 2B. 3C. 1D. zero |
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Answer» Correct Answer - A `4FeS_2 + 11O_2 to 2Fe_2O_3 + 8SO_2` `2SO_2 + O_2 to 2SO_3 ` `SO_3 + H_2O to H_2SO_4 ` Dibasic acid |
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| 1530. |
Which one of the hydracid does not form any precipiate with `AgNO_(3)` ?A. HFB. HClC. HBrD. HI |
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Answer» Correct Answer - A |
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| 1531. |
Elemental silicon does not form a graphtie like structure. Explain. |
| Answer» The size of silicon atom is more than that of carbon. Therefore `p pi-p pi` bonding is not so effective in case of silicon atom. However, this is present in carbon atoms in graphtie because of smaller size. Thus, silicon cannot have graphtie like structure. Rather, it resembles diamond and is very hard as well as poor conductor of electricity. | |
| 1532. |
Select the false statements from the following and try to justify your assertion: (a) Like carbon, silicon also existas in the free state. (b) `B(OH)_(3)` is acidic in nature. (c) Boron dissovles in hydrochloric acid. (d) The composition of common glass is `Na_(2)O,CaO.6SiO_(2)`. (e) Silica contains `SiO_(2)` molecules with O=Si=O bonds. (f) `BF_(3)` is a strnger Lewis acid than `BCl_(3)` (f) Water glass is sodium silicate. (h) diborane cannot have ethane like structure because it is not a covalent molecule. |
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Answer» (a) Silicon does not exist in free state becaue it is more reactive than carbon due to bigger size andalso lesser ionisation enthalpy. (c) Boron dissolves only in oxidising acids like `H_(2)SO_(4)` and `HNO_(3)` to form boric acid. As HCl is not strong oxidising agent, boron does not dissolve in HCl. (e) In silica O=Si=O cannot be present due to the different in the energy state of 2p orbital of oxygen and 3p orbial of silicon which are to participitate in sidewise overlapping. (f) `BF_(3)` is a weaker Lewis acid than `BCl_(3)` because of greater magnitude of `p pi- p pi` back bonding, resulting in more electron density on the boron atom. (h) dibronae cannot have a ethane like structure becasue boron atom is trivalent and more than three hydrogen atoms cannot be linked to it. thus `H_(3)B-BH_(3)` structure is not possible in this case. |
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| 1533. |
In how many of the following non-metal showns -3 oxidation state ? `Ca_(3)N_(2),Ca_(3)P_(2),Na_(3)As,Zn_(3)Sb_(2),Mg_(3)Bi_(2),NaN_(3),Ba(N_(3))_(2),Mg_(3)N_(2)` |
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Answer» Correct Answer - `0005` `Ca_(3)N_(2),Ca_(3)P_(2),Zn_(3)Sb_(2),Mg_(3)Bi_(2),Mg_(3)N_(2)` |
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| 1534. |
The number of lone pairs on central atom in `XeF_2, XeF_4 and XeF_6` are:A. (A) 1,2,3B. (B) 3,2,1C. (C) 2,2,1D. (D) 1,3,2 |
| Answer» Correct Answer - B | |
| 1535. |
What happen when a solution of potassium chromate is treated with an excess of dil. Nitic acid?A. `Cr^(3+)` and `Cr_2O_7^(2-)` are formedB. `Cr_2O_7^(2-)` and `H_2O` are formedC. `CrO_4^(2-)` is reduced to +3 state of CrD. `CrO_4^(2-)` is oxidized to +7 state of Cr |
| Answer» Correct Answer - B | |
| 1536. |
A gas P is obtained at anode during the electrolysis of brine. The gas P when treated with excess of `NH_3` released a diatomic gas Q. Find the value of (x - y) where x & y are the molar mass of P and Q. |
| Answer» Correct Answer - 43 | |
| 1537. |
The ionisation energy of dioxygen (`O_2`) is very close to that of Xenon. Also F and O have the highest electronegativity and consequently can oxidise Xe among rare gases. So Xe forms a large number of compounds with F and O. Xe and `F_2` are mixed and reacted at different temperatures to give `XeF_2, XeF_4 and XeF_6`. Xe also forms an unstable gaseous `XeO_4` and soild `XeO_3` which Is a very powerful explosive at higher temperatures .Some of the rare gases form clathrates or cage compounds by being entrapped in the cages of cystals laiitice of water, phenol of quinols. Helium can form intersitital compound with transition metals. Bigger members of rare gases do not form such compounds because of their large size. He and Ne do not from any clathrates because :A. (A) He and Ne are very large in size .B. (B) Being neutral they cannot from any polar bounds with the host molecules.C. (C) Being too small , they cannot be entrapped in the cages of water,phenol or quinol.D. (D) clathrates with He and Ne are highly expolsive. |
| Answer» Correct Answer - C | |
| 1538. |
The ionisation energy of dioxygen (`O_2`) is very close to that of Xenon. Also F and O have the highest electronegativity and consequently can oxidise Xe among rare gases. So Xe forms a large number of compounds with F and O. Xe and `F_2` are mixed and reacted at different temperatures to give `XeF_2, XeF_4 and XeF_6`. Xe also forms an unstable gaseous `XeO_4` and soild `XeO_3` which Is a very powerful explosive at higher temperatures .Some of the rare gases form clathrates or cage compounds by being entrapped in the cages of cystals laiitice of water, phenol of quinols. Helium can form intersitital compound with transition metals. Bigger members of rare gases do not form such compounds because of their large size. `XeF_6` cannot be prepard by the method :A. `Xe+3F_2 underset"50 atm"overset"475-532 K"to XeF_6`B. `XeF_2+2F_2 overset"500 K"to XeF_6`C. `XeF_4 + F_2 overset"475 K"to XeF_6`D. `XeO_3+6HF underset(Delta)overset("475 K")to XeF_6+3H_2O` |
| Answer» Correct Answer - D | |
| 1539. |
The number of cornpounds.’elements oxidised by `XeF_2` among following is: HF, HBr, HCl, HI, `NH_3, CrF_2`, Pt, `S_8` |
| Answer» Correct Answer - 7 | |
| 1540. |
The compound that cannot be formed by xenon isA. `XeO_3`B. `XeF_4`C. `XeCl_4`D. `XeOF_4` |
| Answer» Correct Answer - C | |
| 1541. |
The ionisation energy of dioxygen (`O_2`) is very close to that of Xenon. Also F and O have the highest electronegativity and consequently can oxidise Xe among rare gases. So Xe forms a large number of compounds with F and O. Xe and `F_2` are mixed and reacted at different temperatures to give `XeF_2, XeF_4 and XeF_6`. Xe also forms an unstable gaseous `XeO_4` and soild `XeO_3` which Is a very powerful explosive at higher temperatures .Some of the rare gases form clathrates or cage compounds by being entrapped in the cages of cystals laiitice of water, phenol of quinols. Helium can form intersitital compound with transition metals. Bigger members of rare gases do not form such compounds because of their large size. Xenon from the larges number of compounds only with oxygen and fuorine because: (i) oxygen and fluorine have very high electronegativity. (ii) lonisation energy of Xe is the largest among rare gases. (iii) lonisation energy of Xe is low compared to those of other rare gases. (iv) low disscoiation energy of fluorine molecule compared to those of `CI_2` and `Br_2`A. (A) (I) (ll) (Ill)B. (B) (I), (Ill), (lV)C. (C) (Ill), (lV)D. (D) (I), (lV) |
| Answer» Correct Answer - B | |
| 1542. |
The order of the oxidation state of the phosphorus atom in `H_(3)PO_(2),H_(3)PO_(4),H_(3)PO_(3),andH_(4)P_(2)O_(6)` isA. `H_(3)PO_(4)gtH_(3)PO_(2)gtH_(3)PO_(3)gtH_(4)O_(2)O_(6)`B. `H_(3)PO_(2)gtH_(3)PO_(3)gtH_(4)P_(2)O_(6)gtH_(3)PO_(4)`C. `H_(3)PO_(3)gtH_(3)PO_(2)gtH_(3)PO_(4)gtH_(4)P_(2)O_(6)`D. `H_(3)PO_(4)gtH_(4)P_(2)O_(6)gtH__(3)PO_(3)gtH_(3)PO_(2)` |
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Answer» Correct Answer - D `H_(3)overset(+5)PO_(4)gtH_(4)overset(+4)P_(2)O_(6)gtH_(3)overset(+3)PO_(3)gtH_(3)overset(+1)PO_(2)` |
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| 1543. |
Identify A and B In following reaction, `H_2SO_4 + HBr to A + B + H_2O `A. (A) `Br_2, SO_3`B. (B) `Br_2, S`C. (C) `BrO_3^(-) , SO_3`D. (D) `Br_2 , SO_2` |
| Answer» Correct Answer - D | |
| 1544. |
`K_2S_2O_8` , acidic `K_2S_2O_8` and acidic `MnO_2` oxidise `I^(-), Br^(-) , Cl^(-)` to `I_2 , Br_2 and Cl_2` , respectively. From the given data the sequence that represents the correct order of increasing oxiding ability isA. `I_2 gt K_2S_2O_8 gt Br_2`B. Acidic `MnO_2 gt K_2S_2O_8 gt Cl_2`C. `K_2S_2O_8 gt I_2 gt Br_2`D. `Cl_2 gt K_2 S_2O_8 gt Br_2` |
| Answer» Correct Answer - D | |
| 1545. |
`Cl_2 + NH_3` (excess) `to A+B`A. (A) One of the product Is also obtained by decomposition of `(NH_4)_2Cr_2O_7`B. (B) Bond order in one of the product is 3C. (C) Both products contain chlorine.D. If `Br_2` is used instead of `Cl_2`, one of product remain same |
| Answer» Correct Answer - A::B::D | |
| 1546. |
`underset"Conc."(HCl) + underset"Conc."(HNO_3)to` In this reaction change in oxidation number of N Is ______ |
| Answer» Correct Answer - 2 | |
| 1547. |
How many of the following reactions would have HCl as one of the products ? (a)`CH_4+CI_2 to ` (b) `FeSO_4 + H_2SO_4 + Cl_2 to`(c) `l_2+Cl_2+H_2O to ` (d) `Cl_2+H_2O` (e)`H_2O+SO_2+Cl_2 to` (f) `SO_3+Cl_2 to` (g) `NaCI(aq)overset("Electrolysis")to` (h) `Cl_2O_7+H_2O to` (i)`Cl_2+NaOH(conc) to` |
| Answer» Correct Answer - 5 | |
| 1548. |
The following are some statements related to VA group hydrides. I. Reducing property increases from `NH_(3)` to `BiH_(3)`. II. Tendency to donate lone pair decreases from `NH_(3)" to "BiH_(3)`. III. Thermal stability of hydrides decreases from `NH_(3)" to "BiH_(3)`. IV. Bond angle of hydrides decreases from `NH_(3)" to "BiH_(3)`. The correct statements areA. I, II, III and IVB. I, III and IVC. I, II and IVD. I and IV |
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Answer» Correct Answer - A All statements are correct, i.e., option (a) is correct. |
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| 1549. |
When chlorine reacts with cold and dilute solution of sodium hydroxide, the products obtained areA. (1) `ClO_2^-` and `ClO_3^-`B. `Cl^-` and `ClO^-`C. `Cl^-` and `ClO_2^-`D. `ClO^-` and `ClO_3^-` |
| Answer» Correct Answer - B | |
| 1550. |
Chlorine on reaction with hot and concentrated sodium hydroxide gives:A. `ClO_3^-` and `ClO_2^-`B. `Cl^-` and `ClO^-`C. `Cl^-` and `ClO_3^-`D. `Cl^-` and `ClO_2^-` |
| Answer» Correct Answer - C | |