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.

NH_4 OH is a weak base because

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it has low VAPOUR pressure
it is only partially ionised
it is COMPLETELY ionised
it has low density

Answer :B
2.

NH_(4)^(+) is kept in group zero because

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Its salts are highly solublr in WATER
The `K_(sp)` of salts of AMMONIUM is high
`K_(sp)` of salts of `NH_(4)^(+)` are LOW
Ammonium salts are insoluble in water

Solution :(a)and (b) are correct because `NH_(4)^(+)` salts are ionic and can from H-bonds with water, therefore they are highly solublr in water. HIGHER the solubility, higher will be `K_(sp).`
`therefore` © and (d) are not correct.
3.

NH_4 CNS can be used to test one or more out of Fe^(3+), Co^(2+), Cu^(2+)e

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`Fe^(3+)` only
`Co^(2+),Cu^(2+)`
`Fe^(3+),Cu^(2+)`
All

Solution :`CoCl_(2)+4NH_(4)CNS OVERSET("ETHER")tounderset("in ethereal layer")underset("Blue colour")((NH_(4))_(2)[Co(CNS)_(4)])+2NH_(4)Cl`
`FeCl_(3) +3NH_(4)CNS to underset("Blood red colour")(Fe(CNS)_(3))+3NH_(4)Cl`
4.

NH_(3)C"OO"NH_(4) overset(Delta)to 2NH_(3) + CO_(2) If 6 moles of NH_(3) is produced then find moles of NH_(2) C"OO"NH_(4) intially taken.

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ANSWER : `(##ALN_NC_CHM_MC_E01_019_A01##)`
3
5.

NH_(3),CO_(2) are readily adsorbed where as, H_(2),N_(2) are slowly adsorbed. Give reason.

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Solution :The NATURE of adsorbate can influece the adsorption. Gases like `NH_(3)" "CO_(2)` are EASILY liquefiable as have greater Van der Waals forces of attraction and hence readily adsorbed due to high critical TEMPERATURE.
But permanent gases like `H_( 2),N_(2)` can not be easily liquefied and having low critical temperature and adsorbed slowly .
6.

{:([(NH_(3))_(5)Co-O-O-Co(NH_(3))_(5)]^(4+)underset("oxidise")overset([S_(2)O_(8)]^(2-))to [(NH_(3))_(5)Co-O-O-Co(NH_(3))_(5)]^(5+)),("""(Brown)""""(Green)"):} The magnetic moment of green complex is 1.7 BM and for brown complexes magnetic moment is zero. (O-O) is same in all respect in both the complexes. The oxidation state of Co in brown complex and green complex respectively are :

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`{:(III" "III" and "IV" "III),(""BROWN""""GREEN"):}`
`{:(III" "II" and "III" "III),(""brown""""green"):}`
`{:(III" "III" and "III" "II),(""brown""""green"):}`
`{:(III" "IV" and "III" "III),(""brown""""green"):}`

ANSWER :A
7.

NH_(3)+underset("(excess)")(X_(2)) rarr NX_(3) + HX rarr"---------------------"1 NH_(3)+underset("(excess)")(X_(2)) rarr N_(2) + NH_(4) X "------2 list out the halogen which reacts according 1 & 2

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`F_(2)`
`Cl_(2)`
`Br_(2)`
`I_(2)`

Solution :`Cl_(2)` & `Br_(2)` SHOW similar reactivity with `NH_(3)`
8.

NH_(3) solutionwas added to four semple solution in difference test tube and found the following observation about the precipitate. a. White ppt which is solution in oxcess of NH_(3) solution b.On heating whichis white in cold but yellow on heating c. The cationpresent in (b) forms white ppt , with hypo solutionwhich give black ppt on heating d. The cation present in (c ) forms soluble complex withexcess of NH_(3) solution White ppt in (c) and the soluble complex fromwhite ppt withthe type solution is//are

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`Pb(OH)_(2),[Pb(S_(2)O_(3))_(2)]^(2)`
`Ag_(2)O, [Ag(S_(2)O_(3))_(2)]^(2)`
`HgO.HG(NH_(2))NO_(3),[Hg(S_(2)O_(3))_(2)]^(2-)`
NONE of these

Solution :N//A
9.

NH_(3) solutionwas added to four semple solution in difference test tube and found the following observation about the precipitate. a. White ppt which is solution in oxcess of NH_(3) solution b. On heating whichis white in cold but yellow on heating c. The cationpresent in (b) forms white ppt , with hypo solutionwhich give black ppt on heating d. The cation present in (c ) forms soluble complex withexcess of NH_(3) solution The solution initialy present in (a) + H_(2)S (basic medium) gives ppt , then (a) may have

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`Zn^(2+)`
`CD^(2+)`
`Co^(2+)`
`Ni^(2+)`

SOLUTION :N//A
10.

NH_(3) solutionwas added to four semple solution in difference test tube and found the following observation about the precipitate. a. White ppt which is solution in oxcess of NH_(3) solution b.On heating whichis white in cold but yellow on heating c. The cationpresent in (b) forms white ppt , with hypo solutionwhich give black ppt on heating d. The cation present in (c ) forms soluble complex withexcess of NH_(3) solution White ppts in (a),(b) and (c ) respectively obtained are

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`Zn(OH)_(2),Zn(OH)_(2),HgOHg(NH_(2))NO_(3)`
`CD(OH)_(2),Zn(OH)_(2),HgOHg(NH_(2))NO_(3)`
`HgOHg(NH_(2))NO_(3)Zn(OH)_(2),Cd(OH)_(2)`
`AI(OH)_(2),Zn(OH)_(2),Pb(OH)_(2)`

Solution :
11.

NH_3 reacts with a gas fumes produced when methanol is passed through Cu tube at 573 K The compound obtained is polycyclic. How many rings does it contain = [X]. The number of Nitrogen atoms it has = [Y] ?

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SOLUTION :
12.

NH_(3) reacts with Cl_(2) to give N_(2) gas . How many number of moles of Cl_(2) are required ?

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Solution :`2NH_(3) + 3Cl_(2) RARR N_(2) + 6 HCL`. No of moles of `Cl_(2)` required in .3..
13.

NH_(3)+O_(2)underset(Delta) overset(Pt)to A+H_(2)O, A+O_(2)to B, B+O_(2)+H_(2)O to C Substances A, B and C of the above sequence of reactions are

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`N_(2)O, NO_(2) and HNO_(3)`
`NO, NO_(2) and HNO_(3)`
`NO_(2), NO and HNO_(3)`
`N_(2)O, NO and HNO_(3)`

ANSWER :2
14.

NH_(3)+NaOCl → NH_(2)-NH_(2)+NH_(4)Cl To obtain this product in large amount, we should use

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Tap water as medium
Glue or GELATINE in medium
Heavy metal ION in solution
by takin NAOCL in EXCESS amount

Answer :B
15.

NH_3 molecule can enter into complex formation through:

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IONIC bond
Covalent bond
Co-ordinate bond
Electron DEFICIENT bond

Answer :C
16.

NH_(3) is liquefied more easily than N_(2) Hence.

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`a and b" of "NH_(3) gt" that of "N_(2)`
`a(NH_(3)) gt a(N_(2))" but b "(NH_(3)) LT b(N_(2))`
`a(NH_(3)) lt a(N_(2))" but b"(NH_(3)) gt b(N_(2))`
NONE of the above

Answer :B
17.

NH_3is an example of:

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MOLECULAR hydride
Polymeric hydride
Metallic hydride
Inerstitial hydride

Answer :A
18.

NH_(3) has a much higher bp than PH_(3) because

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`NH_(3)` has a larger molecular weight
`NH_(3)` UNDERGOES UMBRELLA inversion
`NH_(3)` FORMS hydrogen BOND
`NH_(3)` contains ionic bonds wheras `PH_(3)` containscovalent bonds

Solution :It contain intermolecular hydrogen bonding
19.

NH_3 has a much higher boiling point than PH_3 because:

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`NH_3` has a higher MOLECULAR weight
`NH_3` undergoes umbrella INVERSION
`NH_3` forms HYDROGEN BOND
`NH_3` contains IONIC bonds whereas `PH_3` contains covalent bonds.

Answer :C
20.

NH_(3) has a higher proton affinity than PH_(3). Explain. Or NH_(3) is more basic than PH_(3).

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Solution :Due the presence of a lone pair of ELECTRONS on N and P, both `NH_(3) and PH_(3)` act as Lewis bases and accept a proton to form an additional N-H and P-H bonds respectively

However, due to smaller size of N over P, N-H bond thus formed is MUCH stronger than the P-H bond. Therefore, `NH_(3)` has HIGHER proton AFFINITY than `PH_(3)`. In other words, `NH_(3)` is more basic than `PH_(3)`.
21.

NH_3 gas is dried over:

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ANHYDROUS `CaCl_2`
`P_2O_5`
QUICK lime
Conc. `H_2SO_4`

ANSWER :C
22.

NH_3 forms complex with:

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`CuSO_4`
`CdSO_4`
AgCl
All

Answer :D
23.

NH_(3)(excess)+3CltoA+N_(2)uarr , the bonds present in compound A is

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IONIC, COVALENT and dative
Ionic and covalent
Dative and ionic
Dative only

ANSWER :A
24.

NH_(3) ,CO_(2) are readily adsorbed where as H_(2), N_(2)are slowly adsorbed. Give reason.

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Solution :• The nature of adsorbate can influence the adsorption. Gases like `NH_(3) CO_(2)` are easily liquefiable as have greater van der Waals forces of ATTRACTION and HENCE readily adsorbed due to high critical temperature.
• But PERMANENT gases like `H_(2) ,N_(2)`can not be easily liquefied and having low critical temperature and adsorbed slowly.
25.

NH_(3) does not form complex with

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AgBr
AgCl
AgI
None of these

Answer :C
26.

NH_3 does not from complex with:

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Agl
AgBr
AgCl
None

Answer :A
27.

NH_3 can be collected by the displacement of:

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Mercury
Water
Brine
Conc. `H_2SO_4`

ANSWER :A
28.

NH_(3) and HCl gas are introduced simultaneously from the two ends of a long tube. A white ring of NH_(3)Cl appears first

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A) nearer to the HCL end
B) at the centre of the tube
C) throughout the tube
D) nearer to the `NH_(3)` end

Solution :According to Graham.s law of diffusion
`(r_(1))/(r_(2))=SQRT((d_(2))/(d_(1)))=sqrt((m_(2))/(m_(1))),(r_(NH_(3)))/(r_(HCl))=sqrt((36.7)/(17))`
HENCE rate of diffusion of `NH_(3)` is more than HCl. This means `NH_(3)` reach first to HCl end. hence white ring of `NH_(4)Cl` appears first at HCl end.
29.

NH_(3)and BF_(3) form adduct readily because they form :

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IONIC BOND
COVALENT bond
coordinate bond
hydrogen bond

Answer :C
30.

NH_3 + 3F_2 to A+ 3HF The correct statement regarding A is

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`SP^(3)`, Tetra hedral , I.P. =0
Pyramidal, I.P. =1 `SP^(3)`
SP, linear, I.P. =0
`SP^(3)`, angular I.P. =2

ANSWER :B
31.

NH_(2) - underset(CH_(3))underset(|)CH-overset(O) overset(||)C-NH-CH_(2)-CO_(2)H Identify the amino acid obtianed by hydrolysis of the above compound :

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GLYCINE
Alanine
Both (1) and (2)
NONE of these

SOLUTION :
32.

NH_2NH_2 serves as :

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MONODENTATE LIGAND
Chelating ligand
Chelating ligand
(a) and (C) both

Answer :D
33.

NH overset(KOH)rarrAoverset(C_(2)H_(5)Br)rarrBunderset(H^(+))overset(HOH)rarrC+D C and D in the above sequence are

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BENZOIC acid+aniline
PHTHALIC acid+ethylamine
Phthalic acid+aniline
Benzoic acid+ethylamine

Answer :B
34.

NF_(3), is possible, but NF_(5), is not. Why?

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Solution :(1) Nitrogen has electronic configuration,
`""_(7)N 1s^(2) 2s^(2)2p_(x)^(1) 2p_(y)^(1)2p_(z)^(1)`
(2) Since N has three unpaired ELECTRONS, it forms three N-F BONDS.(3) Nitrogen does not contain d-orbital to expand and HENCE cannot accommodate 10 electrons of five N-F bonds. Hence it does not form `NF_(5)`
35.

Newman projection of Butane is given, C-2 is rotated by 120^(º) along C_(2)-C_(3) bond in anticloclwise direction, the conformation formed is :

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anti
fully eclipsed
gauche
partially eclipsed

Answer :C
36.

Newly shaped glass articles when cooled suddenly becomebrittle, therefore these are cooled slowly , this process in known as :

Answer»

Tempering
Annealing
Quenching
Galvanising

Answer :B
37.

New industrial plants for acetic acid react liquid methanol with carbon monoxide in the presence of a catalyst.CH_3OH + CO to CH_3COOHIn the experiment, 15 g of methanol and 10 g of carbon monoxide were placed in a reaction vessel. What is the theoretical yield of acetic acid? If the actual yield is 19.1 g, what is the percentage yield?

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SOLUTION :21.4 G, 89.1%
38.

New carbon - carbon bond formation does not take place in

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Gatterman - KOCH reaction
Kolbe's reaction
Reimer-Teimann reaction
ROSENMUND REDUCTION

SOLUTION :Rosenmund reaction
39.

Neutron scattering experiments have shown that the radius of the nucleus of an atom is directly proporitonal to the cube root of the number of nucleons in the nucleus. From _(3)^(7)Li to _(76)^(189)Os the radius is:

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Halved
the same
Doubled
Tripled

Solution :`R_(OS)/R_(LI)=(189/7)^(1//3)=3`
40.

Neutrons are obtained by

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Bombardment of RA with `beta-`particles
Bombardment of Be and `ALPHA`-particles
Radioactive disintegration of uranium
None of these

Solution :`._(4)Be^(9) + ._(2)He^(4) rarr ._(6)C^(12) + ._(0)N^(1)`
41.

Neutron was discovered by:

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J.J Thomson
Chadwick
Rutherford
Millikan

Answer :B
42.

Neutrino has

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Charge +1, MASS 1
Charge 0, mass 0
Charge -1, mass 1
Charge 0, mass 1

Solution :Neutrino has no mass and no charge and are THUS known as GHOST particles
43.

Neutralization of an acid with base invariably results in the production of

Answer»

`H_(3)O^(+)`
`OH^(-)`
`H_(2)O`
`H^(+)` and `OH^(-)`

SOLUTION :`NAOH + HCl underset("Reaction")overset("Neutralization")HARR underset("Salt")(NACL) + H_(2)O`
44.

Neutrino can be detected during the emission of:

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`ALPHA`-RAYS
`BETA`-particle
protons
X-rays.

Answer :B
45.

...............neutralise the acid in the stomach that causes acidity.

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SOLUTION :ANTACIDS
46.

Neutralisation of 30 gm of a mixture of acetic acid and phenol solutions required 100 ml of 2M sodium hydroxide solution. When the same mixture was treated with bromine water, 33.1 gm of precipitate was formed. Determine the mass percentage of acetic acid and phenol in the given solution.

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Solution :i.
331 gm of (B) is obtained form 94 gm of (A).
`33.1` gm (B) is obtained form `= (94)/(331)xx33.1`
`= 9.4` gm of PHENOL
Weight of phenol `=9.4` gm
`= (9.4)/(9.4)= 0.1` mol
ii.NaOH will react with both `CH_(3)COOH` and phenol. Total MOLAR equivalent of NaOH `= 100xx2`
`= 200` mEq
`= (200)/(1000)= 0.2` Eq. .of NaOH
`= 0.2` mol of NaOH
ACID + Phenol `= 0.2` mol
Acid + `0.1` mol `= 0.2` mol
`:.` Acid `= 0.2 - 0.1 = 0.1` mol
`=0.1` Eq.
1 Eq. of `CH_(3)COOH = 60` gm
`0.1` Eq. of `CH_(3)COOH = 6` gm
Weight of acid = 6 gm
Weight of phenol `= 9.4` gm
Mass percentage of acid `= (6)/(30)xx100 = 20%`
Mass percentage of phenol `= (9.4)/(30)xx100=31.3%`
47.

Neutralwater with pH about 7 becomes slightly acidic when aerated this is because

Answer»

it is a redox REACTION
metalic iron acts as a reducing agent
`O_(2)` acts as an OXIDISING agent
metallic iron is reduced to `FE^(3+)`

Solution :`4Fe+3O_(2)rarr4Fe^(3+)+6O^(2-)` ltrbgt This reaction is a redox reaction in this reaction metallic fe is oxidlised to `Fe^(3+)` and `O_(2)` reduced to `O^(2-)`
48.

Neutral ferric chloride is added to the aqueous solution of acetate. The blood red colour is obtained, it is due to the compound

Answer»

`Fe(OH)_(2)`
`Fe(OH)_(3)`
`Fe(CH_(3)COO)_(3)`
`Fe(OH)_(2)(CH_(3)COO)`

Solution :`3CH_(2)COONa+FeCl_(3)tounderset("blood RED PPT.")(Fe(CH_(3)COO)+3NaCl`
49.

Neutral refractory material used in furnaces is

Answer»

Graphite
CaO
`SiO_(2)`
MgO

Answer :A
50.

Neutral complex among the following

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`CuSO_4. 4NH_3`
`[Co(NH_3)_6]Cl_3`
`NI(CO)_4`
`[Pt(NH_3)_2]Cl_2`

Answer :C