Chemistry

Benzene is a rich source of electrons because of the presence of an electron cloud containing 6 n-electrons above and below the plane of the ring. Consequently, it attracts the electrophiles (electron-deficient) reagents towards it and repels nucleophiles (electron- rich) reagents. As a result, benzene undergoes electrophilic substitution reactions easily and nucleophilic substitutions with difficulty.

3.44.  (d) Assertion is a wrong statement. Non-metallic elements have a strong tendency to gain electrons. Therefore, electronegativity is directly related to those non-metallic properties of elements. It can be further extended to say that the electronegativity is inversely related to the metallic properties of elements.

Thus, the increase in electronegativities across a period is accompanied by an increase in non-metallic properties (or a decrease in metallic properties) of elements

The hybridization state of carbon in these three compounds is:

Since s-electrons are closer to the nucleus, therefore, as the s-character of the orbital making the C—H bond increases, the electrons of C—H bond lie closer and closer to the carbon atom. In other words, the partial +ve charge on the H-atom and hence the acidic character increases as the s-character of the orbital increases. Thus, the acidic character decreases in the order: Ethyne > Benzene > Hexane.

1.49 NaCl is doped with 10⁻³ mol % of SrCl₂

100 moles of NaCl are doped with 0.001 moles of SrCl₂

o-Xylene may be regarded as a resonance hybrid of the following two Kekule structures. Ozonolysis of each one of these gives two products as shown below:

Thus, in all, three products are formed. Since all the three products cannot be obtained from any one of the two Kekule structures, this shows that o-xylene is a resonance hybrid of the two Kekule structures (I and II).

1.48 It is given that aluminum crystallises in a cubic closed packed structure.
Its metallic radius is 125 pm.

For cubic close-packed structure

a=2√2r=2√2*125=354 pm

Here, a is the edge length of the unit cell and r is the atomic radius.

(ii) Volume of one unit cell = a³ =(354 pm)³=4.4*10⁻²³cm³(1 pm=10⁻¹⁰cm)

Number of unit cells in 1.00cm³= 1.00 cm³ / 4.4*10^-23 cm³

= 2.27*10²²

Branching of carbon atom chain decreases the boiling point of alkane.

The expanded formula of the given compound is