Explore topic-wise InterviewSolutions in Current Affairs.

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 :

Answer»

`CO(BO_2)_2`
`CoBO_2`
`Co_3(BO_3)_2`
`Na_3Co(BO_3)_2`

ANSWER :A
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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Green
Brown
Violet
Blue

Answer :B
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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Reducing
Oxidising
Both (1) & (2)
NEITHER (1) nor (2)

ANSWER :C
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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`B_(2)O_(3)+NaBO_(2)`
`Na_(3)BO_(3)`
`Na_(2)B_(4)O_(7)`
`SiO_(2)`

Answer :A
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

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`sp^(2)`
`sp^(3)`
Both (1) & (2)
sp

Answer :C
6.

Borax is prepared by treating colemanite with :

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`NaNO_3`
`NACL`
`Na_2CO_3`
`NaHCO_3`

ANSWER :C
7.

Borax is ………………. In nature.

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basic
acidic
amphoteric
chemically inert

Answer :A
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

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ACID, AI
acid, C
acid, FE
 acid, Mg

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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ACID, Al
acid, C
acid, Fe
acid, MG

ANSWER :D
10.

Borax is a sodium salt of ……………… .

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ANSWER :tetraboric ACID
11.

Borax is :

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`Na_2B_4O_7`
`(Na_2B_4O_7).4H_2O`
`(Na_2B_4O_7).7H_2O`
`(Na_2B_4O_7).10H_2O`

ANSWER :D
12.

Borax is _____________

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`Na_(2)[B_(4)O_(5)(OH)_(4)].8H_(2)O`
`Na_(2)[B_(4)O_(5)(OH)_(6)].7H_(2)O`
`Na_(2)[B_(4)O_(3)(OH)_(8)].6H_(2)O`
`Na_(2)[B_(4)O_(2)(OH)_(10)].5H_(2)O`

Answer :A
13.

Borax heat test is given by

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`Co^(2+)`
`Zn^(2+)`
`Cu^(2+)` ions
`NI^(2+)`

Solution :White salts or ions GIVING colourless HEATED do do not give borasx BEAD test ,Salts of `Cu^(2+),Cu^(2-)` and `Ni^(2+)` give borax bead test
14.

Borax dissolves to give

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`NAOH^(+) B_2 O_3`
`NaOH^(+) B_2 H_`
`NaOH^(+) H_3 B O_3`
`NaOH^(+) HBO_2`

ANSWER :C
15.

Borax bead test is responded by :

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DIVALENT metals
Heavy metals
Light metas
Metals FORMING COLOURED metal-borates

Answer :D
16.

Borax bead test is not given by :

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An ALUMINIUM salt
A COBALT salt
A COPPER salt
A NICKEL salt

Answer :A
17.

Borax bead test is given by

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`Co^(2+)`
`ZN^(2+)`
`Cu^(2+)`
`Ni^(2+)`

Solution :White salts or IONS giving colourless beads do not GIVE borax bead test.
HENCE, (A), (C) and (D) are the CORRECT answers.
18.

Borax bead test depends upon the formation of :

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BORON oxide
metal borides
elemental boron
metal META borates.

Answer :D
19.

Borax bead is responded generally by :

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ALKALI METAL SALT
ALKALINE EARTH metals
p-block metal salt
d-block metal salt

Answer :D
20.

Borax bead cannot be performed with which of the following salts

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`CuCl_(2)`
`FeCl_(2)`
`CoS`
`AlCl_(3)`

SOLUTION :`NACL+K_(2)CrO_(7)+KSO_(4)rarrCrO_(2)Cl_(2)`(orange red )
`CrO_(2)Cl_(2)+NaOHrarrNa_(2)CrO_(4)("yellow")+NaCl`
`Na_(2)CrO_(4)overset(H^(+))(rarr) Na_(2)Cr_(2)O_(7)`
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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`(C_2H_5O)_3B`
`(C_2H_5)_2BO_3`
`(C_2H_5)_3B`
A and C are CORRECT

ANSWER :C
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 :

Answer»



both
None of these

SOLUTION :N//A
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

Answer»

THREO CYCLIC alchohol
Erythreo cyclic alcohol
Optically active alchohol
Both (B) and (C)

SOLUTION :N//A
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

Answer»

Sulphur
chromium
lead
copper

Answer :C
26.

Bones glow in the dark. This is due to:

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the presence of red phosphorus
conversion of WHITE P into red P
slow combustion of white P in CONTACT with air
conversion of red P into white P

Answer :C
27.

Bones glow in the dark, because:

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They contain a SHINING material
They contain red PHOSPHORUS
White phosphorus changes into red phosphorus
White phosphorus undergoes SLOW COMBUSTION with air.

Answer :D
28.

Bones glow in the dark because

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They contains SHINING materials
They contain red phosphorus
WHITE phosphorus undergoes slow combustion in contact with air
White phosphorus changes into red FORM

Answer :C
29.

Bone black is a polymorphic form of

Answer»

PHOSPHORUS
Sulphur
Carbon
Nitrogen

Solution :Bone BLACK is the POLYMORPHIC form of phosphorus.
30.

Bond type between O and B in BH_3 larr (OC_2H_5)_2 is:

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COORDINATE
covalent
ionic
hydrogen

Answer :A
31.

Bond present in O_2 molecule ls

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`ppi-ppi`
`ppi-dpi`
`dpi-dpi`
`dpi-Ppi`

ANSWER :A
32.

Bond present in benzene diazonium chloride are

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only IONIC
ionic, COVALENT and co-ordinate
only covalent
ionic and covalent

Answer :B
33.

Bond order of N_2^- anion is:

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3
2
2.5
1.5

Answer :C
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

Answer»

`H_(2) GT B_(2) gt Li_(2)`
`H_(2) gt Li_(2) gt B_(2)`
`Li_(2) gt B_(2) gtH_(2)`
`B_(2) gt H_(2) gt Li_(2)`

Solution :The molecular orbital configuration of the given molecules is
`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)`
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

Answer»

Can have a negative quantity
Has always an integral VALUE
can ASSUME any positive or integral or FRACTIONAL value INCLUDING ZERO
Is a non zero quantity

Answer :C
36.

Bond order for nitrogen molecular is ………………. .

Answer»

1
2
3
0

Solution :3
37.

Bond length order in various xenon fluorides is

Answer»

`XeF_6gtXeF_4gtXeF_2`
`XeF_2=XeF_4=XeF_6`
`XeF_2gtXeF_4gtXeF_6`
cannot be predicted

Answer :C
38.

Bond length is maximum in

Answer»

HI
HBr
HCl
HF

Answer :A
39.

Bond length between carbon-carbon in ethylene molecule is

Answer»

1.54 Å
1.35 Å
1.19 Å
2.4 Å

Answer :B
40.

Bond formed in crystal by anion and cation is

Answer»

ionic
metallic
covalent
dipole

SOLUTION :GENERALLY cationand anion form ionic BOND.
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).

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)`
45.

Bond energy of N-N is x kJ "mol"^(-1). Then bond energy of N-=N is

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x kJ `MOL^(-1)`
`lt kJ mol^(-1)`
3x kJ `mol^(-1)`
`GT 3x kJ mol^(-1)`

Answer :B
46.

Bond energy of covelent O-H bonds in water is

Answer»

GREATER than BOND ENERGY of H-bonds
Equal to bond energy of H-bonds
Less than bond energy of H-bond
NONE of these

ANSWER :A
47.

Bond energy of a molecule:

Answer»

Is always negative
Is always POSITIVE
Either positive or negative
Depends UPON the PHYSICAL STATE of the system

Answer :B
48.

Bond energy of a molecule

Answer»

Is always POSITIVE
EITHER positive or NEGATIVE
Is always negative
Depends UPON the PHYSICAL state of the system

Answer :A
49.

Bond energy is highest for which of the following?

Answer»

`F_2`
`Cl_2`
`Br_2`
`I_2`

Solution :Cl-Cl BOND is STRONGEST and hence `Cl_2` has HIGHEST bond ENERGY
50.

Bond energy is highest for :

Answer»

Sn-Sn
Ge-Ge
`C-C`
`Si-Si`

ANSWER :C