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11 months ago

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Payal Gupta

Contributor-Level 10

3.42. (a) Elements of the same group have similar valence shell electronic configuration and, therefore, exhibit similar chemical properties. However, the elements of the same period have an incrementally increasing number of electrons from left to right, and, therefore, have different valences.

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alok kumar singh

Contributor-Level 10

Schottky defect: Schottky defect is basically a vacancy defect shown by ionic solids. In this defect, an equal number of cations and anions are missing to maintain electrical neutrality. It decreases the density of a substance. Significant number of Schottky defects is present in ionic solids. For example, in NaCl, there are approximately 106 Schottky pairs per cm3 at room temperature. Ionic substances containing similar-sized cations and anions show this type of defect. For example: NaCl, KCl, CsCl, AgBr, etc.
Frenkel defect: Ionic solids containing large differences in the sizes of ions show this type of defect. When the smaller ion (

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11 months ago

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Vishal Baghel

Contributor-Level 10

(i) The two substituents in the benzene ring are present at p-positions. Therefore, the sequence of reactions should be such that first an o, p-directing group, i.e., Br atom should be introduced in the benzene ring and this should be followed by nitration. Thus,

 
(ii) Here since the two substituents are at p-position w.r.t. each other, therefore, the first substituent in the benzene ring should be a o, p-directing group (i.e., CH3) and then the other group (i.e., NO2) should be introduced. Therefore, the sequence of reactions is:
 
(iii) Here since the two substituents are at m-position w.r.t. each other, therefore

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New answer posted

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Vishal Baghel

Contributor-Level 10

 

Due to the presence of a sp3-hybridized carbon, the system is not
planar. It does contain six n-electrons but the system is not fully conjugated since all the six n-electrons do not form a single cyclic electron cloud which surrounds all the atoms of the ring. Therefore, it is not an aromatic compound.
 
(ii)
Due to the presence of a sp3 -carbon, the system is not planar. Further, it contains only four n-electrons, therefore, the system is not aromatic because it does not contain planar cyclic cloud having (4n + 2) n-electrons.
 
(iii)

Cyclo-octatetraene is not planar but is tub shaped. It is, therefore, a non-planar system

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alok kumar singh

Contributor-Level 10

1.47 The energy gap between the valence band and conduction band in an insulator is very large while in a conductor, the energy gap is very small or there is overlapping between valence band and conduction band.
(ii) In a conductor, the valence band is practically filled or there is overlapping between valence band and conduction band while in semiconductor, there is always a small energy gap between them.

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New answer posted

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Vishal Baghel

Contributor-Level 10

The necessary conditions for a molecule to be aromatic are:

  • It should have a single cyclic cloud of delocalised n-electrons above and below the plane of the molecule.
  • It should be planar. This is because complete delocalization of n-electrons is possible only if the ring is planar to allow cyclic overlap of p-orbitals.
  • It should contain Huckel number of electrons, i.e., (4n + 2) n-electrons where n = 0, 1, 2, 3 etc.
    A molecule which does not satisfy any one or more of the above conditions is said to be non-aromatic.

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11 months ago

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Vishal Baghel

Contributor-Level 10

Benzene is a resonance hybrid of two canonical forms. In the resonance hybrid, all the six pi electrons are completely delocalized. This results in resonance stabilization.

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11 months ago

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alok kumar singh

Contributor-Level 10

1.45 For fcc unit cell, a=2√2?.

Here, a is the edge length and r is the atomic radius (0.144 nm).

a = 2√2 *0.144 = 0.407 nm

Hence, the length of a side of a cell is 0.407 nm.

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