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.

D -Glucose shows mutarotation because,

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it is dextrorotatory
it undergoes INTER conversion between its pyranose STRUCTURE and furanose structure
`alpha` and `BETA(+)` GLUCOPYRANOSE structures
it undergoes interconversion with D(-) fructose

Answer :C
2.

D(+) -glucose reacts with hydroxylamine and yields on oxime. The structure of the oxime would be

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SOLUTION :Optin (d) has the RIGHT orientation of OH GROUPS at `C_2,C_3 and C_5`.
3.

D-Glucose shows muta rotation between

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it is dextrarotatory
it undergoes inter CONVERSION between it’s pyranose structure and furanose structure
it undergoes INTERCONVERSION between it’s `ALPHA and beta` (+) Glucopyranose structure
it undergoes interconversion with D(-) fructose

Answer :C
4.

D(+) glucose reacts with hydroxylamine and yields an oxime. The structure of the oxime would be

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ANSWER :D
5.

D(+)glucose reacts with hydroxylamine and yields an oxime. The structure of the oxime would be :

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ANSWER :D
6.

D-glucose overset("Reagent")rarr D-gluconic acid. This reagent can be :

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Schiff's reagent <BR>FEHLING solution
`Br_(2)+H_(2)O`
`HNO_(3)`

Solution :N//A
7.

D(+) - glucose overset(5(CH_(3)CO)_(2)O)to D(+) glucose pentaccetate. Whichof the followingstatementis trueabout theproduct ?

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It will reduce Tollens' reagent.
It will reduce Fehling reagent.
It REACTS with phenylhydrazine
It UNDERGOES ALDOL CONDENSATION,

SOLUTION :
8.

D-Glucose and D-Galactose are ……………………

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C2 epimers
C4 epimers
C3 epimers
C5 epimers

Solution :C4 epimers
D-Glucose and D-Galactose differs in the POSITION of -OH group at C4 only.
9.

D-glucose & D-fructose can be differentiated by :

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FEHLING SOLUTION
TOLLEN's reagent
Benedict TEST
`Br_2//H_2O`

ANSWER :D
10.

(d) Estimation of phosphorous: A known mass of compound is heated with fuming HNO_(3) or sodiumperoxide (Na_(2)O_(2)) in Carius tube which converts phosphorrous to H_(3)PO_(4) . Magnesia mixture (MgCl_(2) + NH_(4)Cl) is then added, which gives the precipitate of magnesium ammonium phosphate (MgNH_(4).PO_(4)) which on heatinggives magnesium pyrophosphate (Mg_(2)P_(2)O_(7)) , which is weighed. 0.124 gm of an organic compound containing phosphorus gave 0.222 gm of Mg_(2)P_(2)O_(7) by the usual analysis . Calculate the percentage of phosphorous in the compound (Mg=24, P=31)

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25
75
62
50

Answer :D
11.

(d) Estimation of phosphorous: A known mass of compound is heated with fuming HNO_(3) or sodiumperoxide (Na_(2)O_(2)) in Carius tube which converts phosphorrous to H_(3)PO_(4) . Magnesia mixture (MgCL_(2) + NH_(4)Cl) is then added, which gives the precipitate of magnesium ammonium phosphate (MgNH_(4).PO_(4)) which on heatinggives magnesium pyrophosphate (Mg_(2)P_(2)O_(7)) , which is weighed. An organic compound has 6.2 % of phosphorus. On sequence of reaction, the phosphorous present in 10 gm of organic compound is converted of Mg_(2)P_(2)O_(7) . Find the weight of Mg_(2)P_(2)O_(7) formed.

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2.22 gm
10.0 gm
4.44 gm
1.11 gm

Answer :A
12.

d-block elements readily form complexes. Give reason.

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Solution :(i) Transition elements (d-block elements) have a tendency to FORM coordination compounds (complexes) with a species that has an ability to donate an electron pair to form a coordinate covalent bond.
(ii) Transition metal ions are small and highly CHARGED and they have vacant low ENERGY orbitals to accept an electron pair donated by other groups. Due to these properties, transition metals form large number of complexes.
(iii) Examples: `[Fe(CN)_(6)]^(4-), [Co(NH_(3))]^(3+)`
13.

d-block elements of periodic table consists of

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THREE SERIES
SIX series
Two series
Four series

Answer :D
14.

d-block elements generally form:

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COVALENT hydrides
Metallic hydrides
Salt-like hydrides
None

Answer :C
15.

d-block elements form complex compounds, Give reason.

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Solution :(i) Transition metal ions have smaller SIZE and greater positive charge. Hence their polarizing power is high. POLARISING power `= ("charge")/("size")`
DUE to high polarizing power, d-block metal ions can attract electrons of the ligands to form co ordinate bonds.
(II) d-block metal ions have vacant d-orbitals that can overlap with the orbital of the ligands.
16.

D-block elements form co-ordination compounds. Give reasons.

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Solution : (a) SMALL size of metal ions
(b) High IONIC charges
(c) AVAILABILITY of d-orbitals
17.

d -block elements form alloys easily because they have

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same NUMBER of shells
same ELECTRON configuration
NEARLY same ATOMIC size
same atomic weight

Answer :C
18.

d-block elements can act as catalysts due to their ability to

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EXHIBIT variable oxidation STATES
Coloured ION formation
Paramagnetic NATURE
Alloy formation

Answer :A
19.

Name two important compounds of transition elements.

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SOLUTION :`KMnO_4,K_2Cr_2O_7`
20.

d- block elements belong to groups 3-12 in the periodic table, in which the d-orbitals are progressively filledWhat is misch-metal?

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Solution :Misch METAL is an alloy of`95%` lanthanoids and` 5% `IRON and traces of S, C, CA and Al
21.

Lanthanides and actinoids are f-block elementsWhat is the common oxidation state of lanthanoids?

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Solution :The COMMON OXIDATION STATE of the metals OFTHE .ScZ series is `+2`
22.

d-block elements are called transition elements. Justify this statement.

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SOLUTION :(i) d-block elements occupy the central position of the periodic table, between s and p block elements.
(ii) Their PROPERTIES are transitional between HIGHLY reactive metals of s-block and elements of p-block which are MOSTLY non-metals. That is why d-block elements are called transitional elements.
23.

d-block elements are called inner transtion elements.,

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SOLUTION :TRANSTION ELEMENTS.
24.

d-block elements are called ul("inner-transition elements.")

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REPRESENTATIVE ELEMENTS
normal elements
inner-TRANSITION elements
transition elements.

Solution :transition elements.
25.

d-block elements are called

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SOLUTION :TRANSITION
26.

d- and l-forms of an optically active compound differ in_________.

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BOILING points
melting points
specific rotation
specific gravity

Answer :C
27.

d-block elements are also called........metals.

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SOLUTION :TRANSITION
28.

D-Alanine differs from L - Alanine with respect to

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configuration
chemical formula
number of `-NH_(` GROUPS
number of `-COOH` groups 

ANSWER :A
29.

A solution of (+)-2-chloro-2-phenyl ethane in toluene racemises slowly in the presence of small amount of SbCl_(5), due to the formation of:

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Carbanion
Carbocation
Free RADICAL
Carbene

Answer :B
30.

D-2-deoxyglucose and D-3-deoxyglucose differ in

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latter gives OSAZONE with phenyl hydrazine.
former gives only phenyl HYDRAZONE with phenyl hydrazine.
former can be CONVERTED to latter using dil. `HNO_(3)`
former can be converted to latter using `Pb(OAC)_(4)`

Answer :A::B
31.

Cytosine, Thymine and Uracil are similar with respect to

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`C=O` at 2nd POSITION in pyrimidine ring 
`NH_(2)` group at 4th carbon in pyrimidine ring 
`C=O` at 4th position in pyrimidine ring
absence of `C=O` group at 2nd position in pyrimidine ring 

ANSWER :A
32.

Cylindrical shape of an alkyne is due to

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THREE `sigma`-bonds
Three `PI`-bonds
Two `sigma` and one `pi`-bond
Two `pi` and one `sigma` -bond

Solution :In `CH-=CH` TRIPLE bond CONSISTS of one `sigma` and two `pi` bonds.
33.

Cynogen is a pseudohalogen.

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ANSWER :1
34.

Cyclopropane rearranges to form propene Delta to CH_(3) - CH = CH_(2) This follows first order kinetics . The rate constant is 2.714 xx 10^(-3) sec^(-1) . The initial concentration of cyclopropane is 0.29 M . What will be the concentration of cyclopropane after 100 sec .

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0.035 M
0.22 M
0.145 M
0.0018 M

Solution :`k = (2.303)/(t) "log" (a)/((a-x))`
(a-x) is the concentration LEFT after 100 sec .
`2.7 xx 10^(-3) = (2.303)/(100) "log" (0.29)/((a-x))`
`IMPLIES (a - x) = 0.22 M`.
35.

Cyclopropane is one of the products in which of the following reaction(s)?

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`BrCH_2CH_2CH_2Br+Znrarr`
`CH_2CH_2CH_2NH_2+HNO_2rarr`
`ClCH_2CH_2CH_2OHoverset(DELTA)RARR`

ANSWER :A::D
36.

Cyclopropane and oxygen at partial pressure of 170 and 570 torr respectively are mixed in a gas cylinder . The ratio of the number of moles of cyclopropane to the number of moles of oxygen is

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`0.190`
0.23
`0.30`
0.39

Solution :The RATIO of moles is the same as the ratio of their PARTIAL PRESSURES `:`
`( 170)/( 570) = 0.30`
37.

Cyclopentyl bromide is reacted withNaI in dry acetone gives

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cylopentyl SODIUM iodide
cylopentane
cyclopentane
cyclopentyl iodide

Answer :C
38.

Cyclopentene on treatment with alkaline KMnO_4 gives

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Cyclopentanol
TRANS 1, 2-cyclopentanediol
CIS 1, 2-cyclopentanediol
1 : 1 MIXTURE of cis and trans 1, 2-cyclopentanediol

Solution :
39.

Cycloocta -1,3,5,7- tetrane (A), a non-armatic compound reacts with 2 molpotassium to from a stable compound (B) without producing. H_(2). Write the structure of (B) and explain the reason for its stability.

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SOLUTION :Potassium metal `(K)` GIVES two `oveline(e)'s` to the cylcoo-ctatertranceto from cycloctanetrenyl dianion, which has ``10 overline(e)'s` system, FOLLOWS `(4N + 2)` RULE`(n = 2)` assumes plannarity, and becomes aromatic and hence more stable.
40.

Cyclopentanol on reaction with NaH followed by CS_(2) and CH_(3)I produces a/an

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ketone
alkene
ether
xanthate

Solution :
41.

cyclohexeneunderset("Monohalogenetion ")overset(CI_2 //hv ) to Aoverset(alc KOH ) to B underset((2)Zn //H_2 O) overset( (1)O_3) toproducts. Thefinalproductin thereactionsequenceabove

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Butandial&GLYOXAL
Propandial
Hexandial
propandial&glyoxal

SOLUTION :
42.

Cyclohexene to adipic acid

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SOLUTION :CYCLOHEXENE to ADIPIC ACID :
43.

Cyclohexene reacts with cold dilute alkaline KMnO_4 to yield

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cis-1,2-cyclohexanediol
trans-1,2-cyclohexanediol
cyclohexanone
hexane-1,6-dial

Answer :A
44.

Cyclohexene on reaction with OsO_4 followed by reaction with NaHSO_3 gives

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Cis-diol
Trans-diol
Epoxy
Alcohol

Solution :
45.

Cyclohexene on ozonolysis followed by reaction with zinc dust and water gives compound E. compound E on further treatment with aqueous KOH yields compound F. compound F is

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SOLUTION :
46.

Cyclohexene on ozonolysis followed by reaction with zinc dust and water gives compound E Compound E on further treatment with aqueous KOH yields compound F. Compound F is

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SOLUTION :
47.

Cyclohexene is best prepared from cyclohexanol by which of the following reagents:

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conc. `H_(3)PO_(4)`
conc. `HCl//ZnCl_(2)`
conc. HCl
conc. `HBr`

Solution :to suppress substitution and to favour elimination, the acid used MUST have a non-nucleophilic counter ION. The counter ion of HCl or HCl/`ZnCl_(2)` or HBr, i.e., `Cl^(-)` or `Br^(-)` ion is a strong NUCLEOPHILE and HENCE forms either CYCLOHEXYL chloride or cyclohexyl bromide by substitution reaction. in contrast, the counter ion of `H_(3)PO_(4)`, i.e., `PO_(4)^(3-)` is non-nucleophilic and hence does not give substitution product but gives only the elimination product, i.e., cyclohexene.
48.

Cyclohexene is best prepared from cyclohexanol by which of the following ?

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`"CONC." H_(3)PO_(4)`
`"Conc." HCl//ZnCl_(2)`
`"Conc." HCl`
`"Conc." HBr`

ANSWER :A
49.

Cyclohexanoyl chloride on reduction by poisoned catalyst gives

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benzaldehyde
cyclohexanol
cyclohexanone
cyclohexyl METHANAL

ANSWER :D
50.

Cyclohexanone oxime is converted into epsilon-caprolactam by treatement with

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DIL HCl
`NaOC_(2)H_(5)`
Ammonia
CONC. `H_(2)SO_(4)`

ANSWER :D