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29801.

The IIUPAC name of C_(17)H_(35)COOH is :

Answer»

Heptadecanoic ACID
Octadecanoic acid
STEARIC acid
2-Methyl hexadecanoic acid.

ANSWER :B
29802.

The I.E_(1). And the I.E_(2) in kJ mol^(-1) of a few elements designated by P,Q,R,S are shown below: Based on the above information, answer the following questions. Q. Which of the above elements forms a stable binary halide of the formula MX_(2)?

Answer»

P
Q
R
S

Answer :C
29803.

The I.E_(1). And the I.E_(2) in kJ mol^(-1) of a few elements designated by P,Q,R,S are shown below: Based on the above information, answer the following questions. Q. Which represents a noble gas?

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P
Q
R
S

Answer :A
29804.

The I.E_(1). And the I.E_(2) in kJ mol^(-1) of a few elements designated by P,Q,R,S are shown below: Based on the above information, answer the following questions. Q. Which of the elements is likely to be reactive non-metal?

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P
Q
R
S

Answer :D
29805.

The I.E_(1). And the I.E_(2) in kJ mol^(-1) of a few elements designated by P,Q,R,S are shown below: Based on the above information, answer the following questions. Q. Which of the element is likely to be reactive metal?

Answer»

P
Q
R
S

Answer :B
29806.

The I.E_(1) among group 13 members follows as

Answer»

`BgtAlltGaltTl`
`BgtAlgtGAgtTl`
`BgtGagtAlgtTl`
`BgtGaltAlltTl`

Solution :The `IE_(1)` of Ga is more than that of AL because of the SMALL atomic size and greater effective NUCLEAR CHARGE of Ga.
29807.

The idea which prompted bartlett to prepare first ever compound of noble gas was:

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High BOND energy of Xe-F
Low bond energy of F-F in `F_2`
IONISATION engies of `O_2` and xenon were almost similar
None of thses

Answer :C
29808.

The O-O bond length in ozone is:

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Resonance
Presence of CARBONYL group
Presence of ALKYL group
None

Answer :A
29809.

The value of C-O-C angle in ether molecule is :

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Resonance
Presence of CARBONYL group
Presence of ALKYL group
None

Answer :A
29810.

The -I effect is shown by:

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-COOH
`-CH_3`
`-CH_2CH_3`
`-CHR_2`

ANSWER :A
29811.

The +I effect is shown by:

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`-CH_3`
-OH
-F
`-C_6H_5`

ANSWER :A
29812.

The hypothetical complex triamminediaquachloridocobalt(III) chloride can be represented as:

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`[CoCl(NH_(3))_(3)(H_(2)O_(2))]`
`[Co(NH_(3))_(3)(H_(2)O)Cl_(3)]`
`[Co(NH_(3))_(3)(H_(2)O)_(2)Cl]Cl_(2)`
`[Co(NH_(3))_(3)(H_(2)O)_(3)Cl_(3)]`

SOLUTION :N//A
29813.

The hypothetical complex chloro-diaquatriamminecobalt (III) chloride can be represented as

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`[COCL(NH_(3))_(3)(H_(2)O)_(2)]Cl_(2)`
`[CO(NH_(3))_(3)(H_(2)O)Cl_(3)]`
`[Co(NH_(3))_(3)(H_(2)O)_(2)Cl]`
`[Co(NH_(3))_(3)(H_(2)O)_(3)]Cl_(3)`

Solution :The complexchlorodiaquatriamminecobalt (III) chloride can have the structure `[COCl(NH_(3))_(3)(H_(2)O)_(2)]Cl_(2)`
29814.

The hypothetical complex chlorodiaquatriam minecobalt (III) chloride can be presented as

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`[CO(NH_3)_3(H_2O)_2Cl]Cl_2`
`[Co(NH_3)_3(H_2O)Cl_3]`
`[Co(NH_3)_3(H_2O)_2Cl]`
`[Co(NH_3)_3(H_2O)_3]Cl_3`

ANSWER :A
29815.

The hydroxyl compound that gives a precipitate immediately when treated with concentrated HCl and anhydrous ZnCl_(2) is :

Answer»

3-methyl-2-butanol
3-methyl-1-butanol
1-butanol
2-methyl-2-butanol

Solution :`underset(2-"methyl-2-butanol")(Ch_(3)-CH_(2)-underset(OH)underset(|)OVERSET(CH_(3))overset(|)C-CH+HCl) overset("ANHYD." ZnCl_(2))to CH_(3)-CH_(2)-underset(Cl)underset(|)overset(CH_(3))overset(|)C-CH_(3)`
Mechanism of this reaction involves the formation of CARBOCATION and the `3^(@)` carbocation is most stable. In the above reaction `3^(@)` carbocationis formed.Hence, when hydrozyl compound is treated with conc. HCl, the precipitate of 2-chloro-2-methylbutane is formed IMMEDIATELY.
29816.

The hydroxide with least ksp value at room temperature is:

Answer»

`MG(OH)_2`
`CA(OH)_2`
`BA(OH)_2`
`Be(OH)_2`

Answer :D
29817.

The hydroxide which is max. soluble in water is ?

Answer»

`BA(OH)_2`
`Mg(OH)_2`
`Sr(OH)_2`
`CA(OH)_2`

ANSWER :A
29818.

The hydroxide of which metal is soluble in excess of ammonia:

Answer»

Cr
Cu
Fe
Al

Answer :B
29819.

The hydrophobic end of lauryl sulphate is

Answer»

`C_(17)H_(35)`
`C_(17)H_(33)`
`C_(12)H_(25)`
`-OSO_(3)--`

Answer :C
29820.

The hydronium ion concentration in 5xx10^(-4) M aqueous solution of NaOH at 25^@C, is :

Answer»

zero
`5XX10^(-4) M`
`2XX10^(-11) M`
`2xx10^(3) M`

ANSWER :c
29821.

The hydrolysisof whichof the followinglakesthe longest time ?

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`CH_(3) COCl`
`(CH_(3)CO)_(2)O`
`CH_(3)COOC_(2)H_(5)`
`CH_(3)CONH_(2)`

Answer :D
29822.

The hydrolysis of which compound is an example of disproportionating reaction?

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`SCl_(4)`
`OF_(2)`
`S_(2)Cl_(2)`
`S_(2)Cl_(2) and OF_(2)`

Answer :C
29823.

The hydrolysis of XeF_6is not a redox reaction(R) XeF_6on complete hydrolysis will give XeO_3

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Both (A) and (R) are TRUE and (R) is the correct EXPLANATION of (A)
Both (A) and (R) are true and (R) is not the correct explanation of (A)
(A) is true but (R) is false
Both (A) and (R) are false

Answer :B
29824.

The hydrolysis of TiCl_(4) in the presenceof HCl gives :

Answer»

`TiCl_(2)`
`TiO_(2)`
`TiOCl_(2)`
Ti

ANSWER :C
29825.

The hydrolysis of the salt of weak acid and strong base is known as:

Answer»

ANIONIC hydrolysis
Cationic hydrolysis
Neutral hydrolysis
Acid hydrolysis

Answer :A
29826.

The hydrolysis of the salt of strong acid and weak base is called:

Answer»

INCREASES with concentration
decreases with concentration
Amphoteric hydrolysis
None of these

Answer :B
29827.

The hydrolysis of sucrose produce a mixture which is

Answer»

Laevorotatory
Dextrorotatory
Equally both (+) and (-)ROTATORY
OPTICALLY INACTIVE

ANSWER :A
29828.

The hydrolysis of the salt of strong acid and weak base is called

Answer»

AMPHOTERIC hydrolysis
Cationic hydrolysis
Anonic hydrolysis
None

Answer :B
29829.

The hydrolysis of starchy foods begins in the mouth by enzymes present in saliva . The enzymes are

Answer»

Amylase
Protease
Ptyalin
Maltase

Answer :C
29830.

The hydrolysis of sodium carbonate involves the reaction betweem:

Answer»

`Na^+` and water
`Na^+` and `OH^-`
`CO_3^(2-)` and water
`CO_3^(2-)` and `H^+`

Answer :C
29831.

The hydrolysis of PCl_3 produces

Answer»

`H_(3)PO_(3)+HCLO`
`H_(3)PO_(3)+HCl`
`H_(3)PO_(4)+HCl`
`PH_(3)+HClO`

Answer :B
29832.

The hydrolysis of PCl_3 produces:

Answer»

`H_3PO_3 + HCLO`
`H_3PO_3 + HCL`
`H_3PO_4 + HCl`
`PH_3 + HClO`

ANSWER :B
29833.

The hydrolysis of optically active 2- bromobutane with aqueous NaOH result in the formation of

Answer»

`(pm)` butan - 2- ol
`(pm)` butan - 1- ol
(-) butan - 2- ol
(+) butan - 2- ol

Solution :Usually `1^@` carbocation followes second order kinetics and `2^@and 3^@` carbocation follows first order kinetics.
The reaction between 2- tromobutonewith equal NaOH follows first order kinetic, i.e.,the rate of reaction depends upon the concentration of 2-brombutane (`S_N 1` MECHANISM).

Since, the 2-bromobutane is OPTICALLY active than the produt is a receive mixture. This is because carbocations are as intermadiate in `S_N1` REACTIONS. Since, carbonation being `Sp^2`-HYBRIDIZED is planar species, the attack of the nucleotide on it can occur from both the face with almost case giving a 50 : 50 mixture of two enantiomes.
29834.

The hydrolysis of optically active 2-bromobutane with aqueous NaOH results in the formation of:

Answer»

(-)butane-2-ol
`(+-)`butane-2-ol
(+)butane-2-ol
`(+-)`butane-1-ol

Solution :
RACEMIC `(+-)` butan-2-ol is FORMED.
29835.

The hydrolysis of NCl_3 by water produces :

Answer»

`NH_3 and HOCL`
`NH_2NH_2 and HCL`
`NH_4 OH and HOCl`
`NH_2Cl and HOCl`

ANSWER :C
29836.

The hydrolysis of NCl_(3) by H_(2)O produces

Answer»

`NH_(2)OH and HOCl`
`NH_(2)NH_(2) and HCl`
`NH_(4)OH + HOCl`
`NH_(2)Cl and HOCl`

Solution :`NH_(3)` THUS produced combines with `H_(2)O` to FORM `NH_(4)OH`. Thus, the products of hydrolysis are `NH_(4)OH` and HOCl
29837.

The hydrolysis of methyl formate in acid solution has rate expression, rete =k [HCOOCH_3][H]^+ the balanced equation being, HCOOCH_3+H_2O to HCOOH +CH_3OH. The rate law contains [H^+] though the balanced equation does not ontain [H^+] because

Answer»

of convenience to express the rate law
`H^+` ION is a CATALYST
`H^+` is an important constituent of any reaction all acids contains `H^+` ions
all acids contains `H^+` ions

Solution :`H^+` ions acts catalyst in the ester HYDROLYSIS.
29838.

The hydrolysis of methyl formate in acid solution has rate expression : rate =k[HCOOCH_3][H^+], the balanced equation being HCOOCH_3 +H_2O to HCOOH +CH_3OH The rate law contains [H^+]though the balanced equation does not contain [H^+] though the balanced equation does not contain [H^+] because

Answer»

more for convenience to express the rate law
`H^+` ion is a catalyst
`H^+` is an important constituent of any reaction
all acids contains `H^+` ions

Solution :`H^(+)` ions ACT as catalyst for THEGIVEN reaction
29839.

The hydrolysis of methyl cyanide in presence of acid gives :

Answer»

Methanoic ACID
ETHANOIC acid
Methylamine
Methyl alcohol

Answer :B
29840.

The hydrolysis of ethyle acetate in dilute aqueous solution in the presence of a mineral acid is represented by the equation : CH_(3) COOC_(2) H_(5)+H_(2)O overset(H^(+))(rarr) CH_(3) COOH+C_(2)H_(5)OH The reaction is :

Answer»

trimolecular
SECOND order
PSEUDO first order
ZERO order

ANSWER :C
29841.

The hydrolysis of ethyl acetate,CH_3COOC_2H_5+H_2Ooverset(H^+)rarrCH_3COOH+C_2H_5OH is:

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FIRST order
Second order
Third order
Zero order

Answer :A
29842.

The hydrolysis of ethyl acetate, R-COOR^1 + H_2O overset(H^+)to RCOOH+R^1-OH is

Answer»

first ORDER
second order
third order
ZERO order

Solution :A pseudo UNIMOLECULAR reaction ,
`R prop [ R COOR'] "" because [H_(2)O]` = constant
`THEREFORE ` It is first order reaction
29843.

The hydrolysis of ethyl acetat CH_3COOC_2H_5+H_2O overset(H^(+))to CH_3COOH + C_2H_5OH is a reaction of

Answer»

FIRST order
second order
third order
ZERO order

SOLUTION :The GIVEN reaction is of 1st order .
29844.

The hydrolysis of ester to carboxylic acid and alcohol takes place easily in presence of enzyme …………………..

Answer»

Esterase
Carboxylase
LIPASE
ALCOHOL dehydrogenase

SOLUTION :Lipase
29845.

The hydrolysis of ester was carried out separately with 0.05 N HCl and 0.05 N H_2SO_4. Which of the following will be true:

Answer»

`K_(HCI)gtK_(H_2SO_4)`
`K_(H_2SO_4)gtK_(HCL)`
`K_(H_2SO_4)=2K_(HCL)`
`K_(H_2SO_4)=K_(HCL)`

ANSWER :B
29846.

The hydrolysis of ester in the presence of alklai solution is a ………………………………………order reaction

Answer»

1
2
0
3

Answer :B
29847.

The hydrolysis of an underline ("alkyl cyanide ")always yields formic acid.

Answer»

SOLUTION :The HYDROLYSIS of an alkyl isocyanide alwaysyields FORMIC acid
29848.

The hydrolysis of an ester in acid medium is first order reaction. Half life period of a first order reaction is 20 seconds. How much time will it take to complete 90% of the reaction?

Answer»

Solution :`k=(0.693)/t_(1/2)=0.693/20=0.3465s^(-1)`
If INITIAL CONCENTRATION`[R]_0`=100,[R]=100-90=10
`thereforet=2.303/k LOG ([R]_0)/([R])2.303/0.03465 log 100/10=(2.303xx1)/0.03465=66.46`SECONDS
29849.

The hydrolysis of an ester in acid medium is first order reaction. What do you mean by a first order reaction ?

Answer»

SOLUTION :If the RATE ofa reaction is proportional to the first power of the concentration of only one rectant,it is CALLED a first order reaction `r=K[A]^1`
29850.

Write the relation between half lifet_(1/2)and rate constant (k)of a first order reaction .

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SOLUTION :For a first ORDER reaction,`t_(1/2)=0.693/k`where k is the RATE constant.