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New answer posted

a year ago

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A
alok kumar singh

Contributor-Level 10

7.2

Nitrogen atom can bond with another nitrogen atom by strong p–p overlap resulting in NN. The triple bond in N2 has high bond strength resulting in high bond dissociation energy. Phosphorous do not show this property of p–p overlap. Hence, nitrogen is less reactive than phosphorous.

The p-orbitals which interact to form two? bonds in N2 molecule are shown by joining them with a curve line.

New answer posted

a year ago

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V
Vishal Baghel

Contributor-Level 10

The given reaction is a nucleophillic reaction:

The given reaction is an SN2 reaction. In this reaction, CN acts as the stronger nucleophile and attacks the carbon atom to which Br is attached in nBuBr.

And, CN- ion is an ambident nucleophile and can attack through both C and N. In this case, it attacks through the C-atom.

The mechanism is shown below:

New answer posted

a year ago

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A
alok kumar singh

Contributor-Level 10

7.1

The general characteristics of Group 15 elements are:

Electronic configuration: All Group 15 elements have 5 electrons in their valence The general electronic configuration of these elements is ns2 np3

Oxidation state: Group 15 elements have 5 valence electrons and they require 3 more electrons to complete their However, the gaining of 3 electrons is difficult

Atomic size: Atomic size increases as we move down the group due to increase in the number of

Ionisation enthalpy: Ionisation enthalpy decreases as we move down the group because of increase in atomic

Electronegativity: Electronegativity decreases on moving down the group due to in

...more

New answer posted

a year ago

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Vishal Baghel

Contributor-Level 10

It is a simple substitution reaction with Cl being replaced by iodide ion.

Since ethanol is a alcohol, so in presence of alcoholic KOH, alkyl chloride undergo elimination reaction, which results in the removal of proton and being substituted by a bromide ion.

It is also a simple substitution reaction in which under the presence of aqueous NaOH, bromide ion is replaced by hydroxide ion as it is a better leaving group than hydroxide ion.

 

It is a simple substitution reaction in which a better leaving group leaves and is being substituted by an ion.

CH3CH2Br + KCN - CH3CH2CN + kBr

5-C6H5ONa + C2H5Cl

It is a simple substitution reaction in

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New answer posted

a year ago

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A
alok kumar singh

Contributor-Level 10

7.34

It is difficult to study the chemistry of radon because it is a radioactive substance having a half-life (the time period to decompose the substance half to its initial concentration) of only 3.82 days.

Radon belongs to the 18 group elements with chemical formula as Rn. In general, the elements at the bottom o periodic table are radioactive, they are very dangerous to study as they emit harmful radiations.

Also, compounds of radon such as RnF2 have not been isolated, they are still in the phase of discovery. They have only been identified by radiotracer technique and no any further properties have been determined.

New answer posted

a year ago

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Vishal Baghel

Contributor-Level 10

(i) Freon 12:- the chlorofluorocarbon compounds of methane and ethane are collectively known as freons. They are extremely stable, unreactive, non-toxic and non-corrosive. Example of Freon 12 is CCl2F2. It is mainly used in refrigeration and eventually makes its way into the atmosphere where it diffuses unchanged into the stratosphere. In stratosphere, Freon is able to initiate the radical chain reactions that can upset the natural ozone balance.

  • DDT: - it stands for p, p'-Dichlorodiphenyltrichloroethane (DDT). It is mainly used as an
  • Iodoform: - it was earlier used as an antiseptic but the antiseptic properties are due the liberation o
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New answer posted

a year ago

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Vishal Baghel

Contributor-Level 10

Let us explain each question from Haloalkane and Haloarenes chapter one by one. Students who are currently in school and plan to take the CBSE board exam for class 12th soon, need to prepare all these questions. Let us get started.

(i) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.

Sp2 hybrid carbon (s-character=33.33%) in chlorobenzene is more electronegative than a sp3-hybrid carbon (s-character=25%) in cyclohexylchloride, due to greater s-character. Thus, Carbon atom of chlorobenzene has less tendency to release electrons to Cl than carbon atom of cyclohexylchloride.

Chlorine atom in chlorobenzene is atta

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New answer posted

a year ago

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A
alok kumar singh

Contributor-Level 10

7.33

XeF6 + H2O  XeO2F2 + HF

Balanced equation: XeF6 + 2H2O XeO2F2 +4HF. The steps for balancing the reaction are as follows:

1. The main element is First, check if the atoms of Xe on both sides are balanced. YES, they are.

2. Then take another element i.e. F. It is not balanced on the right side, there are 3 atoms of F missing. So, to balance it multiply HF by 4.

3. Now check the other secondary atoms which are oxygen and hydrogen. Oxygen is not balanced on the left side as there are 2 O atoms so multiply H2O by

4. Now check for the hydrogen They are balanced on both sides.

New answer posted

a year ago

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Vishal Baghel

Contributor-Level 10

(i) Since the conversion of ethanol to but-1-yne involves the addition the two extra carbons that is why the conversion is carried out in 3 steps.

  1. In the first step ethanol is treated with thionyl chloride (SOCl2) in the presence of pyridine to give chloroethane
  • In the second step the acetylene is treated with sodamide(NaNH2) in the presence of liquid ammonia to give sodium acetylide
  • The last step involves the reaction between chloroethane and sodium acetylide to give the final product as the But-1-yne and NaCl as the by-product.
  • The bromination of ethane (Br2 in the presence of light at 520-670K) gives bromoethane which on further treatme

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New answer posted

a year ago

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Vishal Baghel

Contributor-Level 10

Carbon due to its chemical property of catenation, tetravalency and the ability to form stable covalent bonds form hydrocarbon and halo-alkanes easily.

The process of removal of hydrogen and halogen from adjacent carbon atoms of haloalkens is called dehydrohalogenation. NaOEt ( Sodium ethoxide) used in Dehydrohalogenation.  It is a strong base and helps remove a β-hydrogen from the haloalkane.

The chemical reaction for

Reaction:

R–CH2 –CHX–R′+NaOEt   ethanol > R–CH=CHR'+NaX+EtOH

(i) 1-Bromo-1-methylcyclohexane: 

      Br
       |
 CH3–C1–cyclohexane&

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