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In the following structure, the double bonds are marked as I, II, III and IV Geometrical isomerism is not possible at site (s) :
Option: 1  III
Option: 2  I
Option: 3  I and II
Option: 4  III and IV  
 

Geometrical isomerism is not possible at Site I as two identical methyl groups are attached to the same carbon bearing the double bond.

Hence, the answer is Option (2)

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vishal kumar

 Which of the following ions does not liberate hydrogen gas on reaction with dilute acids ?
Option: 1 Ti2+
Option: 2 V2+
Option: 3  Cr2+
Option: 4  Mn2+  
 

The reactivity of transition metals decreases from Ti?????2+ to V??????2+ Ito Cr?????2+ and finally Mn2+. Therefore, the reactivity of Mn2+ is very less. Therefore, Mn?????2+ does not liberate hydrogen gas on reaction with dilute acids. 

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vishal kumar

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 The number of P−OH bonds and the oxidation state of phosphorus atom in pyrophosphoric acid (H4P2O7) respectively are :
Option: 1  four and four
Option: 2 five and four
Option: 3  five and five
Option: 4 four and five  
 

The number of P−OH bonds in pyrophosphoric acid (H4P2O??7) is four.

The oxidation state of the phosphorus atom in H??????4??P2O7

   ⇒  4+2x−14=0

           x = +5

 

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vishal kumar

XeF6 on partial hydrolysis with water produces a compound ‘X’.  The same compound ‘X’ is formed when XeF6 reacts with silica.  The compound ‘X’ is :
Option: 1 XeF2
Option: 2 XeF4
Option: 3  XeOF4
Option: 4 XeO3
 

 

Partial hydrolysis of  XeF????6 gives XeOF4  (compound X)
XeF????6+H2O→XeOF4+2HF


XeF????6 reacts with silica SiO????2 to form  XeOF???4 (compound X)
2XeF6+SiO2→2XeOF4+SiF????4 

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vishal kumar

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 Which one of the following is an oxide ?
Option: 1  KO2
Option: 2  BaO2
Option: 3  SiO2
Option: 4 CsO2  
 

SiO2  - oxide

KO???2? , CsO????2  -  superoxides

BaO????2 -  peroxide

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vishal kumar

 The electronic configuration with the highest ionization enthalpy is :
Option: 1  [Ne] 3s2 3p1
Option: 2  [Ne] 3s2 3p2
Option: 3  [Ne] 3s2 3p3
Option: 4 [Ar] 3d10 4s2 4p3
 

The electronic configuration with the highest ionization enthalpy is [Ne]3s23p3. On moving down the group, the  ionization enthalpy decreases. In a period, on moving from left tor fight, the  ionization enthalpy increases. 

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vishal kumar

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The most abundant elements by mass in the body of a healthy human adult are : Oxygen (61.4%); Carbon (22.9%), Hydrogen (10.0%); and Nitrogen (2.6%). The weight (in kg) which a 75 kg person would gain if all ^{1}H atoms are replaced by ^{2}H atoms is :  
Option: 1 7.5
Option: 2 10
Option: 3 15
Option: 4 37.5
 

Given that

Mass of the person = 75 kg

Mass of 1H1 present in person = 10% of 75 kg = 7.5 kg

Since Mass of 1H2 is double the Mass of 1H1

So, Mass of 1H2 will be in person = 2 X 7.5 kg =15 kg

Thus, increase in weight = 15 - 7.5 = 7.5 kg

Therefore, Option (1) is correct

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vishal kumar

The correct order of catenation is: 


Option: 1 C > Sn > Si \approx Ge
Option: 2 C > Si > Ge \approx Sn
Option: 3 Si > Sn > C > Ge
Option: 4 Ge > Sn > Si > C

Catenation/ Self Linkage -

Property of elements to form long chains or rings by self linking of their own atoms through covalent bonds. In the carbon family, it decreases down the group. Only carbon atom form double or triple bong involving p^{\pi } -p^{\pi }multiple bond with itself.

The homo atomic bond energies are as follows :

(i) C-C = 83 kcal / mol

(ii) Si - Si = 54 kcal / mol

(iii) Ge - Ge = 40 kcal / mol

(iv) Sn - Sn = 37 kcal / mol

* Very large difference exists between the bond energies of (C-C) & (Si-Si)

but negligible difference is there for (Ge - Ge ) & (Sn-Sn).

Thus, the correct order is:

C > Si > Ge \approx Sn

Therefore, option (2) is correct.

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Ritika Jonwal

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The correct order of the atomic radii of C, Cs, Al and S is :
Option: 1 S<C<Cs<Al
Option: 2 C<S<Al<Cs
Option: 3 C<S<Cs<Al
Option: 4 S<C<Al<Cs
 

 

Periodicity of atomic radius and ionic radius in period -

In a period from left to right the effective nuclear charge increases because the next electron fills in the same shell. So the atomic size decrease.

- wherein

Li>Be>B>C>N>O>F

 

 

 

Electronegativity and atomic radius -

The attraction between the outer electrons and the nucleus increases as the atomic radius decreases in a period.

- wherein

Electronegativity\propto\frac{1}{atomic\:radius}

 

 

 

Size of atom and ion in a group -

In a group moving from top to the bottom the number of shell increases.So the atomic size increases.

- wherein

Li<Na<K<Rb<Cs

As we know that

From Left to right in a period size decreases and when going down the group size increases

C< S< Al< Cs

Therefore, Option(2) is correct

  

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Ritika Jonwal

Which of the following compounds will show the maximum 'enol' content?
Option: 1 CH_{3}COCH_{2}COOC_{2}H_{5}
Option: 2 CH_{3}COCH_{2}COCH_{3}
Option: 3 CH_{3}COCH_{3}
Option: 4 CH_{3}COCH_{2}CONH_{2}
 

Difinition of Tautomerism -

Tautomers are isomers of a compound that differ only in the position of the protons and electrons. The carbon skeleton of the compound is unchanged. A reaction that involves simple proton transfer in an intramolecular fashion is called tautomerism.

The stability of enol depends on the factors

(1) Resonance

(2) Hydrogen-bonding

(3) Hyperconjugation

(4) Hydrogen which is removed from \alpha- carbon should be acidic in Nature for enol formation.

CH2 is present between 2- electron-withdrawing group,  It is having acid 'H'

                        

Over the resonance effect decreases because of cross conjugation. So the stability of the enol form decreases.

The two -dicarbonyl compounds have a higher enol content than the two mono carbonyl compounds because hydrogen bonding and conjugation stabilize their enols. The enol content in C (a mono aldehyde) is higher than D because of the reasons outlined above. 

Therefore, option (2) is correct.

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Ritika Jonwal

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