Class 11th

This is a Short Answer Type Questions as classified in NCERT Exemplar

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 to the second period and group 14^th ,15^th and 17^th . The electronegativity of the elements increases on moving across the period and the dipole moment is calculated by the difference in the electronegativity of the atoms in a molecular structure. Higher the difference in the electronegativity will be the dipole moment. So, the dipole moment of H - Z will be the highest.

This is a Short Answer Type Questions as classified in NCERT Exemplar

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 formed due to mutual sharing of electrons between two atoms of a non-metal. This mutual sharing of electrons between the two non-metal species leads to the formation of a covalent bond. The making of Cl₂ leads to a formation of a covalent bond between two chloride ions.

This is a Short Answer Type Questions as classified in NCERT Exemplar

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 orbitals with two bonds of hydrogen atoms and two lone pairs of electrons. Due to the presence of lone pairs of electron the bond angle gets reduced as there is a repulsion between the LP-LP and thus the molecule acquires a bent shape. The hybridization of carbon atoms in the carbon dioxide molecule is sp. The two sp are arranged in such a way that it makes an angle of 180° . That is why carbon dioxide has a linear shape.

 

 

(iii) Both the carbon atoms in ethyne molecules are sp hybridized. Ethyne has two unhybridized 2p_x 2p_y orbitals. The two sp hybrid orbitals of both the carbon atoms are arranged such that they form an angle of 180° with each other. That is why ethyne molecules are linear in shape.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Ans: (i) N₂ → N₂⁺ + e^-

The electronic configuration of N₂ is:

σ1s² σ*1s² σ2s² σ*2s² π2p²_x = π2p²_y σ2p²_z

So, its bond order will be 3

The electronic configuration of N₂⁺ is:

σ1s² σ*1s² σ2s² σ*2s² π2p²_x = π2p²_y σ2p¹_z

Its bond order will be 2.5

Hence the bond order decreases in this reaction N₂ → N₂⁺ + e^-

(ii) O₂→ O₂⁺ + e-

The electronic configuration of O₂ is:

σ1s² σ∗1s² σ2s² σ∗2s² s2p_z² π2p²_y π2p²_x π*2p¹_y π*2p¹_x

Its bond order will be 2

The electronic configuration of O₂⁺ will be:

σ1s² σ∗1s² σ2s² σ∗2s² s2p_z², π2p²_y , π2p²_x, π*2p¹_x

Its bond order will be 2.5 .

Hence, the bond order in this reaction increases O₂→ O₂⁺ + e-

This is a Short Answer Type Questions as classified in NCERT Exemplar

Ans: The general sequence of the energy level of the molecular orbital is σ1s < σ*1s < σ2s< σ*2s < π2px = π2py < σ2pz

N₂    =   σ1s² σ*1s² σ2s² σ*2s² π2p²_x = π2p²_y σ2p²_z

N₂⁺  =   σ1s² σ*1s² σ2s² σ*2s² π2p²_x = π2p²_y σ2p¹_z

N₂^-   =   σ1s² σ*1s² σ2s² σ*2s² π2p²_x= π2p²_y σ2p²_z σ2p²_x

N₂²⁺ =   σ1s² σ*1s² σ2s² σ*2s² π2p²_x = π2p²_y

The formula for finding the bond order (B. O) for any molecular species is:

Bond order = $\frac{1}{2}$ ( N_b- N_a)

Hence, the bond orders of the given molecular species are:

N₂= $\frac{1}{2}$ ( 10-4)= 3

N₂⁺= $\frac{1}{2}$ ( 9-4)= 2.5

N₂^-= $\frac{1}{2}$ ( 10-5)= 2.5

N₂²⁺= $\frac{1}{2}$ (8-4)= 2

The order for the stability of the given molecular species thus will be:

N_2.> N₂^-> N₂⁺> N₂²⁺