This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Borax on heating with cobalt oxide forms a blue bead of : |
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Answer» `CO(BO_2)_2` |
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| 2. |
Borax [Na_(2)B_(4)O_(7).10H_(2)O] when heated on platinum loop it gives a dark transparent glass like bead. The hot bead is dipped in the salt till it reacts with transition metal oxide. It produces characteristic bead of meta borate. {:("Colour of the bead","lon"),("(a) Blue green or light blue",Cu^(+2)),("(b) Yellow",Fe^(+2)or Fe^(+3)),("(c) Green",Cr^(+3)),("(d) Violet",Mn^(+2)),("(e) Dark blue",Co^(+2)),("(f) Brown",Ni^(+2)):} The colour of bead Ni(BO_(2))_(2) is |
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Answer» Green |
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| 3. |
Borax [Na_(2)B_(4)O_(7).10H_(2)O] when heated on platinum loop it gives a dark transparent glass like bead. The hot bead is dipped in the salt till it reacts with transition metal oxide. It produces characteristic bead of meta borate. {:("Colour of the bead","lon"),("(a) Blue green or light blue",Cu^(+2)),("(b) Yellow",Fe^(+2)or Fe^(+3)),("(c) Green",Cr^(+3)),("(d) Violet",Mn^(+2)),("(e) Dark blue",Co^(+2)),("(f) Brown",Ni^(+2)):} The flame used in Boram Bead test is |
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Answer» Reducing |
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| 4. |
Borax [Na_(2)B_(4)O_(7).10H_(2)O] when heated on platinum loop it gives a dark transparent glass like bead. The hot bead is dipped in the salt till it reacts with transition metal oxide. It produces characteristic bead of meta borate. {:("Colour of the bead","lon"),("(a) Blue green or light blue",Cu^(+2)),("(b) Yellow",Fe^(+2)or Fe^(+3)),("(c) Green",Cr^(+3)),("(d) Violet",Mn^(+2)),("(e) Dark blue",Co^(+2)),("(f) Brown",Ni^(+2)):} Glassy bead is of |
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Answer» `B_(2)O_(3)+NaBO_(2)` |
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| 5. |
Borax [Na_(2)B_(4)O_(7).10H_(2)O] when heated on platinum loop it gives a dark transparent glass like bead. The hot bead is dipped in the salt till it reacts with transition metal oxide. It produces characteristic bead of meta borate. {:("Colour of the bead","lon"),("(a) Blue green or light blue",Cu^(+2)),("(b) Yellow",Fe^(+2)or Fe^(+3)),("(c) Green",Cr^(+3)),("(d) Violet",Mn^(+2)),("(e) Dark blue",Co^(+2)),("(f) Brown",Ni^(+2)):} The hybridisation of B in Borax is |
| Answer» Answer :C | |
| 6. |
Borax is prepared by treating colemanite with : |
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Answer» `NaNO_3` |
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| 7. |
Borax is ………………. In nature. |
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Answer» basic |
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| 8. |
Borax is converted into B by the following steps: Boraxoverset(I)toH_(3)BO_(3)overset(Delta)toB_(2)O_(3)overset(II)toB B I and I reagents are |
| Answer» Answer :D | |
| 9. |
Borax is converted into B by steps Borax overset(1)to H_(3)BO_(3) overset(Delta) to B_(2)O_(3) overset(II)to B I and II reagents are |
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Answer» ACID, Al |
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| 11. |
Borax is : |
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Answer» `Na_2B_4O_7` |
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| 12. |
Borax is _____________ |
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Answer» `Na_(2)[B_(4)O_(5)(OH)_(4)].8H_(2)O` |
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| 13. |
Borax heat test is given by |
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Answer» `Co^(2+)` |
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| 14. |
Borax dissolves to give |
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Answer» `NAOH^(+) B_2 O_3` |
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| 15. |
Borax bead test is responded by : |
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Answer» DIVALENT metals |
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| 16. |
Borax bead test is not given by : |
| Answer» Answer :A | |
| 17. |
Borax bead test is given by |
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Answer» `Co^(2+)` HENCE, (A), (C) and (D) are the CORRECT answers. |
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| 18. |
Borax bead test depends upon the formation of : |
| Answer» Answer :D | |
| 19. |
Borax bead is responded generally by : |
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Answer» ALKALI METAL SALT |
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| 20. |
Borax bead cannot be performed with which of the following salts |
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Answer» `CuCl_(2)` `CrO_(2)Cl_(2)+NaOHrarrNa_(2)CrO_(4)("yellow")+NaCl` `Na_(2)CrO_(4)overset(H^(+))(rarr) Na_(2)Cr_(2)O_(7)` |
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| 21. |
Borate from green colour flame when burnt with (Conc. H_2SO_4+ ethanol). Green colour flame is obtained due to due to formation of |
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Answer» `(C_2H_5O)_3B` |
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| 22. |
Borane is an electron deficient compound. It has only six valence eletons, so the boron atom lacks an octet. Acquiring an octet is the driving force for the unusual bonding structure found in boron compounds. As an electron deficient compound, BH_(3) is a strong electrophile, capable of adding to a double bond. This hydroboration of double bond is though to oC Cur in one step, with the boron atom adding to the less highly substituted end of the double bond. In transition state, the boron atom withdraws electrons from the pi bond and the carbon at theother end of the double bond acquires a partial positive charge. This positive charge is more stable on the more highly subsituted carbon atom. The second step is the oxidation of boron atom, removing it from carbon and replacing it with hydroxyl group by using H_(2)O_(2)//OH^(bar(..)). The simultaneous addition of boron and hydrogen to the double bond leads to a syn addition. Oxidation of the trialkyl borane replaces boron with a hydroxyl group in the same stereochemical position. Thus, hydroboration of alkenen is an example of steropecific reaction, in which different steroisomers of starting compounds react to give different steroisomers of the product. Y is : |
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Answer» |
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| 23. |
Borane is an electron deficient compound. It has only six valence eletons, so the boron atom lacks an octet. Acquiring an octet is the driving force for the unusual bonding structure found in boron compounds. As an electron deficient compound, BH_(3) is a strong electrophile, capable of adding to a double bond. This hydroboration of double bond is though to oC Cur in one step, with the boron atom adding to the less highly substituted end of the double bond. In transition state, the boron atom withdraws electrons from the pi bond and the carbon at theother end of the double bond acquires a partial positive charge. This positive charge is more stable on the more highly subsituted carbon atom. The second step is the oxidation of boron atom, removing it from carbon and replacing it with hydroxyl group by using H_(2)O_(2)//OH^(bar(..)). The simultaneous addition of boron and hydrogen to the double bond leads to a syn addition. Oxidation of the trialkyl borane replaces boron with a hydroxyl group in the same stereochemical position. Thus, hydroboration of alkenen is an example of steropecific reaction, in which different steroisomers of starting compounds react to give different steroisomers of the product. underset((ii)H_(2)O_(2)//OH^(bar(..)))overset((i)BH_(3)//THF)rarr "product". The product is |
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Answer» THREO CYCLIC alchohol |
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| 24. |
Boot polish is what type of colloid. |
| Answer» SOLUTION :Boot POLISH is a LIQUID in solid i.e. gels TYPE of COLLOID | |
| 25. |
Books, periodicals, magazines and calendars are printed in large numbers. Type metal used in printing presses as alphabet letter printing contains |
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Answer» Sulphur |
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| 26. |
Bones glow in the dark. This is due to: |
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Answer» the presence of red phosphorus |
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| 27. |
Bones glow in the dark, because: |
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Answer» They contain a SHINING material |
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| 28. |
Bones glow in the dark because |
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Answer» They contains SHINING materials |
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| 29. |
Bone black is a polymorphic form of |
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Answer» PHOSPHORUS |
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| 30. |
Bond type between O and B in BH_3 larr (OC_2H_5)_2 is: |
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Answer» COORDINATE |
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| 31. |
Bond present in O_2 molecule ls |
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Answer» `ppi-ppi` |
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| 32. |
Bond present in benzene diazonium chloride are |
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Answer» only IONIC |
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| 33. |
Bond order of N_2^- anion is: |
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Answer» 3 |
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| 34. |
Bond order normally gives idea of stability of a molecular species. All the molecules viz. H_(2) Li_(2) and B_(2) have the same bond order yet they are not equally stable. Their stability order is |
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Answer» `H_(2) GT B_(2) gt Li_(2)` `H_(2) = sigma2s^(2)` (no electron anti-bonding) `Li_(2)= sigma 1s^(2) sigma ^(* *) 1 s^(2) sigma 2s^(2)` (two anti-bonding electrons) `B_(2)=sigma 1s^(2) 1s^(2) sigma^(* *) 1s^(2) sigma 2s^(2) sigma^(* *) 2s^(2){pi 2p_(y)^(1)=pi2p_(z)^(1)}` (4 anti-bonding electrons) THOUGH the bond order of all the species are same (B.O = 1) but stability is different. This is DUE to difference in the presence of no. of anti-bonding electron. Higher the no. of anti-bonding electron lower is the stability hence the correct order is `H_(2) gt Li_(2) gt B_(2)` |
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| 35. |
Bond order is a concept in the molecular orbital theory. It depends on the number of electrons in the bonding and antibonding orbitals. Which of the following statements is true about ? The bond order |
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Answer» Can have a negative quantity |
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| 36. |
Bond order for nitrogen molecular is ………………. . |
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Answer» 1 |
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| 37. |
Bond length order in various xenon fluorides is |
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Answer» `XeF_6gtXeF_4gtXeF_2` |
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| 38. |
Bond length is maximum in |
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Answer» HI |
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| 39. |
Bond length between carbon-carbon in ethylene molecule is |
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Answer» 1.54 Å |
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| 40. |
Bond formed in crystal by anion and cation is |
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Answer» ionic |
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| 41. |
Bond entnalpy of F_(2) is less than of Cl_(2). |
| Answer» Solution :This is due to relatively LARGE electronic repulsion among the lone pair of `F_(2)` molecule where they are much CLOSER to each other than in CASE of `Cl_(2)`. | |
| 42. |
Bond enthalpy of fluorine is lower than that of chlorine. Why? |
| Answer» Solution :Fluorine atom being smaller in size, electron-electron repulsions among the lone PAIRS of `F_2` MOLECULE are LARGER compared to that in `Cl_2` molecule. | |
| 43. |
Bond enthalpy of fluorine is lower than that of chlorine why ? |
| Answer» Solution :BOND enthalpy of `F - F` is smaller DUE to greater REPULSIVE interactions between the lone pair of one F ATOM with those of other. The repulsive interaction arise due to greater concentration of electron density on each F atom because of its extremely SMALL size. | |
| 44. |
Bond enthalpy of bromine is 194 kJ mol^(-). If enthalpy of vapourisation of Br_2 is +30 kJ mol^(-), electron gain enthalpy of Br is -325 kJ mol^(-1) and hydration enthalpy of bromide is -339 kJ mol^(-1) calculate the change in enthalpy for the reaction, 1/2Br_(2)(l) + e^(-) overset(aq)(rarr) Br^(-)(aq). |
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Answer» Solution :`{:((1)/(2)Br_(2)(l)rarr(1)/(2)Br_(2)(g),,Delta=+15kJmol^(-1)),((1)/(2)Br_(2)(g)rarrBr(g),,DeltaH=+97kJmol^(-1)),(Br(g)rarrBr^(-)(g),,DeltaH=-325kJmol^(-1)),(Br^(-)(g)+AQ rarr Br^(-)(aq),,DeltaH=-339kJ MOL^(-1)):}` Adding these EQUATIONS we get, `(1)/(2)Br_(2)(l)+e^(-)overset(aq)rarrBr^(-)(aq)),DeltaH=-552kJ mol^(-1)` |
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| 45. |
Bond energy of N-N is x kJ "mol"^(-1). Then bond energy of N-=N is |
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Answer» x kJ `MOL^(-1)` |
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| 46. |
Bond energy of covelent O-H bonds in water is |
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Answer» GREATER than BOND ENERGY of H-bonds |
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| 47. |
Bond energy of a molecule: |
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Answer» Is always negative |
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| 48. |
Bond energy of a molecule |
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Answer» Is always POSITIVE |
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