Chemistry NCERT Exemplar Solutions Class 11th Chapter Four

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (iii) and (iv)

The formula of bond order is:

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

Number of bonding and antibonding electrons can be found from the molecular electronic configuration of the species.

(i) Total number of electrons in N2 is 14. So, its molecular electronic configuration will be

σ1s² σ∗1s² σ2s² σ∗2s², π2p²_x , π2p²_y, s2p_z²

The bond order will be:

$\frac{1}{2}$ ( 10-4)= 3

(ii) The total number of electrons in N₂ ^- is 15. So, its molecular electronic configuration will be:

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

The bond order will be:

$\frac{1}{2}$ ( 10-5)= 2.5

(iii) The total electrons in F₂⁺ is 17. So, its electronic configuration will be:

σ1s² σ∗1s² σ2s² σ∗2s² s2p_z², π2p²_y , π2p²_x, π*2p²_y π*2p¹_x

The bond order will be:

$\frac{1}{2}$ ( 10-7)= 1.5

(iv) The total electrons in O₂^- is 17 So, its electronic configuration will be

σ1s² σ∗1s² σ2s² σ∗2s² s2p_z², π2p²_y , π2p²_x, π*2p²_y π*2p¹_x

$\frac{1}{2}$ ( 10-7)= 1.5

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (i) and (iv)

(i) N₂ will have 14 electrons in its structure. So, its electronic configuration will be:

σ1s² σ∗1s² σ2s² σ∗2s², π2p²_x , π2p²_y, s2p_z²

According to the molecular electronic configuration of N₂ it does not contain any unpaired electrons, it will be diamagnetic in nature.

(ii) N₂^2– will have 16 electrons in its structure. So, its molecular electronic configuration will be:

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

According to the molecular electronic configuration of 2 N₂ ^-, it has two unpaired electrons in π*2p¹_x ,   π*2p¹_y  sub- shell. Hence, it will be paramagnetic in nature.

(iii) O₂ will have 16 electrons in its structure. So, its molecular electronic configuration will be:

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

According to the molecular electronic configuration of O₂, it has two unpaired electrons in π2p¹_y π*2p¹_x  sub-shell. Hence, it will be paramagnetic in nature.

(iv) O₂^2- will have 18 electrons in its structure. So, its molecular electronic configuration will be:

σ1s² σ∗1s² σ2s² σ∗2s² s2p_z² π2p²_y π2p²_x π*2p²_y π*2p²_x

According to the molecular electronic configuration of O₂^2- it does not contain any unpaired electrons, it will be diamagnetic in nature.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (iii) and (iv)

(i) The hybridization of the central carbon atom in CO₃ ^2- is 2 sp as the carbon is bonded with three oxygen atoms. So, the given statement is incorrect.

(ii) The resonance structure of CO₃^2- has one C= O double bonds and two C- O single bonds. So, the statement is incorrect.

(iii) The net charge on three oxygen atoms is -2, hence the average formal charge on each oxygen atom is 0.67 units. So, the given statement is correct.

(iv) As it can be seen from the resonating structures that the structures of the bonds are not fixed. Hence, all the C- O bond lengths are equal. So, the given statement is correct.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (iii) and (iv)

(i) The number of valence electrons in CO₂ is 14. The oxygen atom attached to the carbon atom forms a double bond. Therefore, the hybridization of the structure is sp². Thus, carbon dioxide is linear in shape.

(ii) The number of valence electrons in CCl₄ is 32. The structure of CCl4 is arranged in a tetrahedral manner. The chlorine atoms are arranged such that the hybridization is sp³.

(iii) The number of valence electrons in O₃ is 24. The hybridization of the molecule is sp² and due to the presence of lone pair of electrons on the central oxygen atom and due to the effect of resonance, ozone molecule has a bent shape.

(iv) The number of valence electrons in NO₂^- is 24. The hybridization of the molecule is sp² and due to the presence of lone pair on the nitrogen atom it attains a bent shape.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (i) and (ii)

The species having the same number of electrons are known as isoelectronic species.

The number of electrons in CO species is 14; 8 electrons from oxygen atom and 6 electrons from carbon atom.

The number of electrons in NO⁺ is 14; 8 electrons from oxygen atom, 7 electrons from nitrogen atom and 1 electron will be subtracted due to the positive charge on the species.

The number of electrons in N₂ is 14 ; 7 from each nitrogen atom.

The number of electrons in SnCl₂ is 84 ;50 from the tin atom and 17 from each chlorine atom.

The number of electrons in NO₂ ^- is 24 ; 7 from nitrogen atom, 8 from each oxygen atom and 1 from the negative charge on the species.

Hence, NO₊ and N₂ are the isoelectronic species of CO.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (i) and option (ii)

Bond order depends on the electronic configuration of the molecular species, and the electronic configuration depends on the number of electrons a particular atom is contributing in its molecular structure.

The CN^- species has a total of 14 electrons; 6 electrons from carbon atom, 7 electrons from nitrogen atom and 1 electron for the negative charge.

The NO⁺ species has a total of 14 electrons; 8 from oxygen atoms, 7 from nitrogen atom and 1 electron will be subtracted due to the positive charge on the species.

The O₂ ^- species has a total of 17 electrons; 8 from each oxygen atom and 1 for the negative charge on the species.

The O₂^2- species has a total of 18 electrons; 8 from each oxygen atom and 2 for the negative charge on the species. Hence, the molecular species / ion having the same bond order will be CN^- and NO⁺ .

This is a Multiple Choice Questions as classified in NCERT Exemplar

Ans: Option (ii)

Aluminium has [Ne]3s² 3p¹  electronic configuration. Phosphorus has [Ne]3s² 3p³. The electronic configuration of silicon is [Ne]3s² 3p² and the electronic configuration of arsenic is   [Ar]3d¹⁰4s²4p³.

Ionisation energy increases as the atomic number in a period increases and decreases on moving down the group. It is also known that the ionisation enthalpy of group 15 elements is more than group 16 elements as group 15 has half-filled p-subshells which gives extra stability.Thus, the electronic configuration having the highest ionization enthalpy is [Ne]3s² 3p³