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.

Specify the oxidation numbers of the metals in the following coordination entities : [Cr(NH_(3))_(3)Cl_(3)]

Answer»

Solution :OXIDATION NUMBERS can be obtained by adding the oxidation numbers of constituents and equating them with the overall charge on the ion, or ZERO, if it is a molecular complex.
Assuming x as the oxidation NUMBER of the METAL, we have
`x + 0 + 3(-1) = 0` or `x = +3`
2.

Specify the coordination of geometry and hybrization of N and B atoms in 1: 1 comples of BF_(3) and NH_(3)

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N:tetrahedral, `sp^(3)`, B:tetrahedral, `sp^(3)`.
N:pyramidal, `sp^(3)`, B:tetrahedral, `sp^(3)`.
N:pyramidal, `sp^(3)`, B:Planar, `sp^(3)`.
N:pyramidal, `sp^(3),`B:tetrahedral, `sp^(3)`.

ANSWER :A
3.

Specify the oxidation numbers of the metals in the following coordination entities : [CoBr_(2)(en)_(2)]^(+)

Answer»

SOLUTION :Oxidation NUMBERS can be obtained by ADDING the oxidation numbers of constituents and equating them with the overall charge on the ion, or zero, if it is a molecular complex.
Assuming x as the oxidation NUMBER of the metal, we have
`x + 2(-1) + 0 = +1` or `x = +3`
4.

Specify the coordination geometry around and hybridisation of N and B atoms in a 1 : 1 complex of BF_(3) and NH_(3)

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N : tetrahedral, `sp^(3)` , B : tetrahedral, `sp^(3)`
N : pyramidal, `sp^(3)` , B : pyramidal, `sp^(3)`
N : pyramidal, `sp^(3)` , B : PLANAR, `sp^(2)`
N : pyramidal, `sp^(3)` , B : tetrahedral, `sp^(3)`

Solution :In the 1 : 1 complex, `H_(3)N rarr BF_(3)`, both N and B have `sp^(3)` hybridisation and tetrahedral geometry.
5.

Specify the coordination geometry and hybridisation around N and B atoms in a 1:1 complex of BF_(3) and NH_(3)

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N: TETRAHEDRAL, `sp^(3): B:` tetrahedral , `sp^(3)`
N:PYRAMIDAL,`SP_(3)^(3-)`, B: Pyramidal, `sp^(3)`
N: pyramidal, `sp^(3)`, B: PLANAR, `sp^(2)`
N: pyramidal, `sp^(3)`, B: tetrahedral, `sp^(3)`

ANSWER :A
6.

Specificity of enzyme is due to

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the SEQUENCE of amino acids
SECONDARY STRUCTURE
TERTIARY structure
all of the above

Solution :all of the above
7.

Specify the coordination geometry around and the hybridisation of N and B atoms in 1 : 1 complex of BF_(3) and NH_(3).

Answer»

N : tetrahedral, `sp^(3),B:` tetrahedral, `sp^(3)`
N : PYRAMIDAL, `sp^(3),B:` tetrahedral, `sp^(3)`
N : pyramidal, `sp^(3),B:` planar, `sp^(2)`
N : pyramidal, `sp^(3)`, B : pyramidal, `sp^(3)`

ANSWER :A
8.

Specific impluse of a rocket and the critical temperature of the fuel reacted in the motor of the rocket has the relationship

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`I_(s) prop sqrt((1//T_(c)))`
`I_(s) prop 1//T_(c)`
`I_(s) prop sqrt((T_(c)))`
`I_(s) prop T_(c)`

SOLUTION :The energy of propellant is measurement in terms of a specific impulse INSIDE the motor of the ROCKET. It depend upon the flame temperature and average molecular mass of the gaseous products coming OT of the rocket nozzle. The specific impulse is given by the following relationship with the average molecular mass M and the flame temperature T, `I_(s) = sqrt(T_(c)//M)`.
9.

Specific reagent for decarboxylation is:

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`CaCO_3`
CaO+NaOH
NaOH
`CaCI_2`

ANSWER :D
10.

Specific heat of aluminium is 0.214 cal/g^(@)C. Heat required to raise the temperature of 40.0 g of Al from 20^(@) to 30^(@)C is

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85.6 cal
3.2 cal
171.2 cal
342.4 cal

Answer :A
11.

Specific conductivity of N/35 KCl at 298 K is 0.002768 ohm^(-1) cm^(-1) and it has resistance of 520 ohm. A N/25 solution of a salt kept in the same cell was found to have a resistance of 300 ohm at 298 K. Calculate equivalent conductance of the solution.

Answer»


SOLUTION :Cell constant = `kappa xx R`
`= 0.002768 xx 520 = 1.43936 CM^(-1)` specific conductivity for salt solution , `kappa ` = Cell constant `xx` conductance
`= 1.43936 xx (1)/(300)`
`= 0.00478 ohm^(-1) cm^(-1)`
`LAMBDA = (0.00478 xx 1000)/(1//25)`
`= 119.5S cm^(2) "EQUIV"^(-1)`
12.

Specific gravity of sulphuric acid in lead accumulator is

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1
1.5
1.28
1.17

Solution :DESCRIPTION of LEAD ACCUMULATOR.
13.

Specific gravity of slag is :

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ALWAYS HIGHER than molten METAL
Always higher than molten metal
Always LESS than molten metal
Same as that of molten metal

Answer :B
14.

Specific conductivity of 0.5 mol/dm^(3)" "AgNO_(3) solution at 298 K temperature is 5.76xx10^(-3)" S "cm^(-1), then molar conductivity is. . .

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`0.086`
`28.8`
`2.88`
`11.52`

SOLUTION :`lamda_(m)=(1000xxK)/(M)`
`=(5.76xx10^(-3))/(0.5)=11.52" S "CM^(2)`/mol
15.

Specific conductivity of a solution

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INCREASES with dilution
decreases with dilution
remains UNCHANGED with dilution
depends on MASS of electrolyte

ANSWER :B
16.

Specific conductivity of 0.1 M NaCl aqueous solution is 1.06xx10^(-2)ohm^(-1)cm^(-1), then what is its molar conductivity in ohm^(-1)cm^(2)mol^(-1) ?

Answer»

`1.06xx10^(2)`
`1.06xx10^(3)`
`1.06xx10^(4)`
53

Solution :`Lamda_(m)=(Kxx1000)/(M)=(1.06xx10xx^(-2)xx1000)/(0.1)`
`=1.06xx10^(2)ohm^(-1)cm^(2)mol^(-1)`.
17.

Specific conductance of pure water at 25^(@)C is 0.58 xx 10^(-7) "mho cm"^(-1). Calculate ionic product of water (K_(w)) if ionic conductances of H^(+) and OH^(+) ions at infinite dilution are 350 and 198 mho cm^(2) respectively at 25^(@)C.

Answer»

SOLUTION :`1 XX 10^(-14) ("mole/litre")^(2)`
18.

Specific conductance of saturated solution of AgBr is 8.486xx10^(-7)ohm^(-1)cm^(-1) at 25^(@). Specific conductance of pure water at 25^(@)C is 0.75xx10^(-6)ohm^(-1)cm^(-1)." "._m^(^^oo)for KBr, AgNO_3 and KNO_3 are 137.4,133,131 (S cm^2 mol^(-1)) respectively. Calculate the solubility of AgBr in gm/litre

Answer»


ANSWER :A::C::D
19.

Specific conductance of "1 M KNO"_(3) solution is observed to be 5.55 xx 10^(-3)" mho.cm"^(-1). What is the equivalent conductance of KNO_(3) when one litre of the solution is used ?

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SOLUTION :`"5.55 mho.cm"^(2)."GM. EQUIV"^(-1)`
20.

Specific conductance of 0.1 M sodium chloride solution is 1.06xx10^(-2)ohm^(-1)cm^(-1).Its molar conductance in ohm^(-1)cm^(2)mol^(-1) is

Answer»

`1.06xx10^(2)`
`1.06xx19^(3)`
`1.06xx10^(4)`
53

Answer :A
21.

Specific conductance of 0.1 M Nitric acid is 6.3 xx 10^(-2) "ohm"^(-1) cm^(-1) "mol"^(-1) The molar conductance of the solution is

Answer»

`630" OHM"^(-1) CM^(2) " mol"^(-1)`
`315 ohm^(-1) cm^(2) " mol"^(-1)`
`100 ohm^(-1) cm^(2) "mol"^(-1)`
`6300 ohm^(-1) cm^(2) "mol"^(-1)`

ANSWER :A
22.

Specific conductance =molar conductance xx cell constant. Is it true or false?

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SOLUTION :OBSERVED CONDUCTANCE
23.

Specific conductance for strong electrolyte on dilution ________

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SOLUTION :DECREASES RAPIDLY.
24.

Specific conductance=________x cell constant.

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SOLUTION :OBSERVED CONDUCTANCE
25.

Specific conductivity of a 0.12 normal solution of an electrolyte is 0.023 ohm^(-1)cm^(-1). Determine its equivalent conductivity.

Answer»


Solution :`wedge_(eq)=kappaxx(1000)/("Normality")=0.024Omega^(-1)cm^(-1)xx(1000CM^(3)L^(-1))/(0.12g" "eqL^(-1))=200OMEGA^(-1)cm^(2)eq^(-1)`.
26.

How does specific conductance vary with dilution?

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SOLUTION :DECREASES RAPIDLY.
27.

Species present in acidic aq. solution are :

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`H_(3)O^(+)`
`H_(5)O_(2)^(+)`
`H_(7)O_(3)^(+)`
All

SOLUTION :`H_(2) O +H^(+) to H_(3)O^(+) OVERSET(H_(2)O)to H_(5)O_(2)^(+) overset(H_(2)O)to H_(7)O_(3)^(+)`
28.

Species which are isoelectronic with OF_(2) is/are

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`CLO^(-)`
CLF
`NH_(2)CL`
`ClO_(2)`

Solution :`ClO^(-)` (17 + 8 + 1 = 26), ClF (17 + 9 = 26) and `NH_(2)Cl` (7 + 2 + 17 = 26) all the THREE are isoelectronic. Only `ClO_(2)` (17 + 2 `xx` 8 = 33) is not isoelectronic.
29.

Species which represent maximum crystal field stablisation energy :-

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`[CO(CN)_(6)]^(-3)`
`[Co(C_(2)O_(4))_(3)]^(-3)`
`[Ni(H_(2)O)_(6)]^(+2)`
Both (1) and (2)

ANSWER :A
30.

Species acting both as Bronsted acid base is

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`HSO_(4)^(-)`
`Na_(2)CO_(3)`
`NH_(3)`
`OH^(-)`

ANSWER :C
31.

Species containing carbon with three bonds and an electron are called:

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Carbenes
Carbonions
Carbonium ions
Free radicals

Answer :D
32.

sp^(3)d- hybridization is possible in

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`SCl_(2)`
`SCl_(4)`
`SF_(6)`
`SeCl_(6)`

Answer :B
33.

Sp^2 -hybridisation is shown by :

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`BeCl_2`
`BF_3`
` NH_3`
`XeF_2`

ANSWER :B
34.

sp^3-hybridisation is important in describing the bonding in :

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`NH_4^+`
`C Cl_4`
`H_3O^+`
All of these

Answer :D
35.

sp^(2) - Hybridisation is involved in the molecule of

Answer»

CO
`CO_(2)`
`SO_(2)`
`SO_(3)`

SOLUTION :`CO=SP, CO_(2)=sp, SO_(2)=sp^(3), SO_(3)=sp^(2)`
36.

Sources of inert gases are

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Air
Natural gas
Minerals
All the above

Answer :D
37.

Sorption is the term used when

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ADSORPTION takes PLACE
absorption take place
both take place
DESORPTION takes place

Answer :C
38.

Sometimes yellow turbidity appears whitepassing H_(2)S gas even in the group II radicals .This is because of

Answer»

Sulphate ispresent in themixtureas limpurity
GROUP IV radicals areprecipitate assulphides
The oxidation of `H_(2)S`GAS by someacidradicals
GroupII radicals areprecipitateas hydroxides

Answer :C
39.

Sometimes yellow turbidity appears while passing H_(2)S gas even in the absence of second group radicals. This is because:

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Sulphur is PRESENT in the mixture as an IMPURITY
The fourth group radicals get precipitated as sulphides
Of the oxidation of `H_(2)S` to sulphur by some radicals ACTING as oxidant
The third group radicals get precipitated as hydroxides.

Answer :C
40.

Sometimes yellow turbidity appears while passing H_2 S gas even in the absence of II group radicals. This is because

Answer»

Sulphur is present in the MIXTURE as impurity
IV group radicals are precipitated as sulphides
of the oxidation of `H_(2)S` gas by some acid RADICAL
III group radicals are precipitated as hydroxide.

Solution :`H_2 S + UNDERSET("redical")underset("oxidising")underset("From some")([O]) to underset("sulphur")underset("Colloidal")(Sdarr) _H_(2)O`
41.

Sometimes yellow turbidity appears while passing H_2 S gas even in the absence of group II radicals. This is because of

Answer»

sulphur is PRESENT in the mixture as impurity
group IV radicals are precipitated as sulphides
the OXIDATION of `H_2 S` gas by some acid radicals
group III radicals are precipitated as hydroxides

Solution :TURBIDITY is due to colloidal sulphur FORMED as a result of the oxidation of `H_2S ` by some OXIDISING agent (radicals) present in the mixture .
42.

Sometimes yellow turbidity appears o passing H_(2)S gas even in the absence of the second group radicals. This happens because

Answer»

Sulphur is present in the MIXTURE as an impurity
The FOURTH group radicals are precipitated as SULPHIDES
The `H_(2)S` is oxidized by some acid radicals
The THIRD group radicals are precipitated

SOLUTION :This is due to the precipitation of fourth group radical as sulphides due to high concentration of `S^(2-)` in the solution as a result yellow turbidity is obtained.
43.

Sometimesvesselsmade of copperorbronzeshowtracesof greencolur. Thisis due tothe formationof _______

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`Cu(OH)_(2)`
`CuCO_(3)`
`Cu(OH)_(2)CuCO_(3)`
`Cu(OH)_(2) CuSO_(4)`

Answer :C
44.

Sometimes the behaviour of a compound is explained by assuming that it exists in a world between two or more different possible structures. The phenomenon is called:

Answer»

Isomerism
Resonance
rotation
Allotropism

Answer :B
45.

Sometimes no precipitate is obtained even if group V radicals are present why ?

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Solution :No, precipitate is obtained when the solutio is dilute in dilute solution `(NH_(4))_(2)CO_(3)` hydrolyses to given carbonic acid which is a WEAK ELECTROLYTE. Its ionisation does not produce sufficient `CO_(3)^(2-)` ions concentration to precipitate Ca, Sr and Ba as carbonates
`(NH_(4))_(2)CO_(3)+2H_(2)Oto2NH_(4)OH+H_(2)CO_(3)`
46.

Sometimes , the colour of photochemical smog becomes brown . The reason for this brown appearance is the excess of

Answer»

`NO_2`
`SO_2`
PAN
`CH_4`

ANSWER :A
47.

Sometimes, it is possible to separate two sulphide ores by using……………….

Answer»

SOLUTION :DEPRESSANTS
48.

Sometimes it is possible to separate two sulphide ores by adjusting the proportion of oil to water or by using depressant NaCN is added to an ore conaining ZnS and PbS, what is the correct observation?

Answer»

NACN PREVENTS PbS from coming to the froth but allows ZnS to come with froth
NaCN prevents ZnS from coming to the froth but allows PbS to come with froth
NaCN prevents FROTHING of both ZnS and PbS, HENCE no froth is formed
NaCN does not act as depressant hence a mixture of PbS and ZnS is found in froth

Answer :B
49.

Sometimes it happen that when H_(2)S gas is pased into solution in dil.HCI yellowish white turbidity appears what do you conclude ? What precaution are taken to check this turbidity?

Answer»

SOLUTION :Thisyellowish white turbidity is due to colloidal sulphur , formed by OXIDATION of `H_(2)S` by theprersence of `NO_(3)^(-)` or `NO_(2)^(-)` or `SO_(3)^(2-)`
`H_(2)S+HNO_(3)tounderset(colloidal)(S)`
Before passing `H_(2)S GAS solution is boiled so as to decompose these RADICALS.
50.

Sometimes indigestion is caused by the secretion of too much hydrochloricx acid in the stomach to eaase the pain caused a table can tbe taken that reacts to reduce the amount of acid present which one amog the folllwoing wuld be inapprapriatefor amanufacturer to inlcude as a major reactant in the tablet?

Answer»

OXIDIZING agents are themselves REDUCED
Reducing agents gains or appear to gain electrons
Reducing agents oxidized
oxidizing agents oxidize other substances

SOLUTION :Here the reducing agent `FeBr_(3)` GETS oxidized and truns into aqueous solution of `FeCI_(3)`