Chemistry NCERT Exemplar Solutions Class 11th Chapter Four

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Vishal Baghel

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Ans: All the C-Obonds in carbonate ion (CO32-) are equal in length due to the equivalent resonance of all the three carbon-oxygen bonds gets a double bond character atleast once. This type of double bond character happened throughout 3C- O skeleton, and hence all the bonds acquired equal bond length.

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Vishal Baghel

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Ans: In compound BCl3, Boron has sp2-hybridisation and the shape is Triangular Planar.

In methane CH4, Carbon has sp3 -hybridization and shape are Tetrahedral.

In carbon dioxide CO2, carbon has sp-hybridisation and shape is Linear.

In ammonia NH3, nitrogen has sp3-hybridisation and shape is Pyramidal.

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Vishal Baghel

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Ans: (i)

 

(ii)

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Vishal Baghel

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Ans: (i) According to the electronic configuration, a total of 8 electrons must be present in the valence shell of an element. Element X has 4 valence electrons, so it will share the remaining 4 electrons for the formation of the bond, the molecular formula will be XH4 . The element Y has 5 valence shell electrons, so it will form 3 bonds and the formula will be YH3 . The element Z has 7 valence shell electrons, so it will form one bond with hydrogen and has the molecular formula H-Z .

(ii) Elements X, Y and Z having, 5 and 4 7 valence electrons respectively belongs

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Vishal Baghel

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Ans: BeCl2 has a linear structure

HOCl is also non-linear in structure.

H2O has a V-shaped structure.

Cl2O has a V-shaped structure.

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Vishal Baghel

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Ans: The hybridization of Carbon 1 is sp, carbon 2 is sp, carbon 3 sp2, carbon 4 is sp3 and carbon 5 is sp2. The triple bond has 2 pie bonds and one sigma bond. Each double bond has one sigma and one pie bond. Every single bond is a sigma bond. Thus, the total number of sigma bonds is 11 and pie bonds are 4.

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Vishal Baghel

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Ans: The carbonate ion CO32- can be best represented by its resonating structures. The carbonate ion cannot be represented by a single Lewis structure because the three carbon oxygen bond lengths are the same. This cannot be shown by a single Lewis structure. For showing the similar lengths of all the carbon to oxygen bonds three hybrid structures are constructed which are in resonance with each other.

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Vishal Baghel

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N—H, F—H, C—H and O—H

Ans: The ionic character in a molecular species is decided by the electronegativity difference between the two bonded atoms. Greater the electronegativity difference between two bonded pairs, the greater will be the ionic character. The electronegativity difference of the given species is

C - F = (2.5 - 2.1) = 0.4

N -H = (3.0 - 2.1) = 0.9

O - H = (3.5 - 2.1) = 1.4

F - H = (4.0 - 2.1) = 1.9

According to the above given difference in the electronegativities, the order of increasing ionic character is:

C – H< N - H

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Vishal Baghel

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Ans: Ionic bonds are such bonds in which complete transfer of electrons from one atom to another atom. Due to complete transfer positive and negative ions are formed in this bond. The ions in this bond are held together by electrostatic force of attraction. The formation of calcium fluoride CaF2 leads to the formation of an ionic bond.

Ca→  Ca2+ + e -   The electronic configuration of  Ca= [Ar] 4s2   and Ca2+= [Ar]

F +e -→F- The electronic configuration of F= [He] 2p5  and F = [He ]2p6

Thus Ca2+ +2F- → CaF2

A covalent bond is forme

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Vishal Baghel

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Ans: (i) Due to the electrostatic forces between the two opposite charges, ionic bonds are non-directional. The bonding direction does not matter as the electrostatic field of an ion is non-directional. Whereas, the formation of covalent bonds happens with the overlap of the atomic orbitals. The direction of the bonds is given by the direction of overlapping.

 

(ii) The hybridization of the oxygen atom in water molecules is 3 sp due to the presence of two lone pairs of electrons on the oxygen atom. Tetrahedral geometry is acquired by these four 3 sp hybridized or

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