Chemistry Ncert Exemplar 2019 Solutions for Class 11 Science Chemistry Chapter 4 - Chemical Bonding And Molecular Structure

Answer:

Both ${\mathrm{BF}}_4^{-}$ and ${\mathrm{NH}}_4^{+}$ have sp³ hybridisation. Therefore, they have same structure i.e tetrahedral.

Hence, the correct answer is option B.

Answer:

H₂O has highest value of dipole moment. Oxygen being more electronegative, the direction of dipole moment is towards the two hydrogen atoms which add up to give a net dipole moment in the downward direction.

Hence, the correct answer is option C.

Answer:

Hybridisation of ${\mathrm{NO}}_2^{+}, {\mathrm{NO}}_3^{-}$and ${\mathrm{NH}}_4^{+}$ is sp, sp² and sp³ respectively.

Hence, the correct answer is option B.

Answer:

Among H₂O, HF and NH₃ , H₂O has the highest boiling point due to more hydrogen bonding per molecule of water than per molecule of HF. Due to less electronegativity of nitrogen in NH₃ as compared to oxygen in H₂O and fluorine in HF, NH₃ will have the lowest boiling point. Therefore, the correct order of decreasing boiling point is H₂O > HF > NH₃.

Hence, the correct answer is option B.

Answer:

The formal charge is calculated as:
$$\begin{gathered} \mathrm{Formal} \mathrm{Charge}=\left[\mathrm{Valence} \mathrm{electron} \mathrm{in} \mathrm{free} \mathrm{oxygen} \mathrm{atom}-\left(\mathrm{number} \mathrm{of} \mathrm{unshared} \mathrm{electron}+\frac{1}{2}\times \mathrm{number} \mathrm{of} \mathrm{shared} \mathrm{electron}\right)\right] \\ =\left[6-\left(6+\frac{1}{2}\times 2\right)\right] \\ = -1 \end{gathered}$$
                           

Hence, the correct answer is option B.

Answer:

The number of bond pair on nitrogen atom: 4
The number of lone pair on nitrogen atom: 0

Hence, the correct answer is option D.

Answer:

Answer:

In the given strucuture:

Number of sigma bonds = 19
Number of pi bond = 5

Hence, the correct answer is option C.

Answer:

$$\begin{gathered} {\mathrm{N}}_2^{+} =\mathrm{\sigma }1{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}1{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}2{\mathrm{s}}^2 \left[\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2=\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2\right] \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^1 \\ {\mathrm{N}}_2^{+} \mathrm{has} \mathrm{one} \mathrm{unpaired} \mathrm{electron}. \\ {\mathrm{O}}_2 =\mathrm{\sigma }1{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}1{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}2{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}2{\mathrm{p}}_{\mathrm{z}}^2 \left[\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2=\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2\right] \left[\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^1=\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^1\right] \\ {\mathrm{O}}_2 \mathrm{has} \mathrm{two} \mathrm{unpaired} \mathrm{electron}. \\ {\mathrm{O}}_2^{2-} =\mathrm{\sigma }1{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}1{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}2{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}2{\mathrm{p}}_{\mathrm{z}}^2 \left[\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2=\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2\right] \left[\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2=\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2\right] \\ {\mathrm{O}}_2^{2-} \mathrm{has} \mathrm{zero} \mathrm{unpaired} \mathrm{electron}. \\ {\mathrm{B}}_2 =\mathrm{\sigma }1{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}1{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}2{\mathrm{s}}^2 \left[\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^1=\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^1\right] \\ {\mathrm{B}}_2 \mathrm{has} \mathrm{two} \mathrm{unpaired} \mathrm{electrons}. \end{gathered}$$

Hence, the correct answer is option C.

Answer:

Ethene molecule have both sigma and pi bond. Therefore, in ethene all bonds are not equal. 



Hence, the correct answer is option C.
 

Answer:

Oxygen in H₂O is more electronegative. Therefore, hydrogen bonding in water is strongest.

Hence, the correct answer is option B.

Answer:

Valence electron will be involved in the bond formation. Therefore, 3d², 4s² will participate in bond formation.

Hence, the correct answer is option D.

Answer:

In sp² hybridisation the bond angle is 120°. 

Hence, the correct answer is option B.

Answer:

A has more stable full filled p orbital configuration. Therefore, most stable state is A. 

Hence, the correct answer is option A.

Answer:

$C_2 = \sigma 1s^2{\sigma }^{*}1s^2 \sigma 2s^2 {\sigma }^{*}2s^2 \left[\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2 =\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2\right]$

$\mathrm{Bond} \mathrm{Order} = \frac{1}{2}\left(8-4\right) =2$

C₂ is the most stable form.

Hence, the correct answer is option B.

Answer:

The electronic configuration of B and C is corresponding to phosphorous and chlorine. Therefore, the molecular formula of the compound represented by B and C is BC₃. 

Hence, the correct answer is option D.

Answer:

The bond between B and C is formed by sharing of electron. Therefore, it will be a covalent bond.

Hence, the correct answer is option B.

Answer:

$\mathrm{The} \mathrm{electronic} \mathrm{configuration} \mathrm{of} \mathrm{nitrogen}: \mathrm{\sigma }1{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}1{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}2{\mathrm{s}}^2 \left[\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2 = \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2\right] \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2 \left[{\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{x}}^0 = {\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{y}}^0\right] {\mathrm{\sigma }}^{*}2{\mathrm{p}}_{\mathrm{z}}^0$

Hence, the correct answer is option A.

Answer:

$\mathrm{The} \mathrm{electronic} \mathrm{configuration} \mathrm{of} \mathrm{nitrogen}: \mathrm{\sigma }1{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}1{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}2{\mathrm{s}}^2 \left[\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2 = \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2\right] \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2 \left[{\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{x}}^0 = {\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{y}}^0\right] {\mathrm{\sigma }}^{*}2{\mathrm{p}}_{\mathrm{z}}^0$
Therefore, order of energies for nitrogen molecule given in option D is incorrect.

Hence, the correct answer is option D.

Answer:

$$\begin{gathered} \mathrm{The} \mathrm{electronic} \mathrm{configuration} \mathrm{of} \mathrm{oxygen} \left({\mathrm{O}}_2\right): \mathrm{\sigma }1{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}1{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}2{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2 \left[\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2 = \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2\right] \left[{\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{x}}^1 = {\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{y}}^1\right] {\mathrm{\sigma }}^{*}2{\mathrm{p}}_{\mathrm{z}}^0 \\ \mathrm{Bond} \mathrm{Order}: \frac{{\mathrm{N}}_{\mathrm{b}} - {\mathrm{N}}_{\mathrm{a}}}{2} = \frac{10-6}{2} = 2 \\ \mathrm{The} \mathrm{electronic} \mathrm{configuration} \mathrm{of} \mathrm{oxygen} \mathrm{ion} \left({\mathrm{O}}_2^{+}\right): \mathrm{\sigma }1{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}1{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}2{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2 \left[\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2 = \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2\right] \left[{\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{x}}^1 = {\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{y}}^0\right] {\mathrm{\sigma }}^{*}2{\mathrm{p}}_{\mathrm{z}}^0 \\ \mathrm{Bond} \mathrm{Order}: \frac{{\mathrm{N}}_{\mathrm{b}} - {\mathrm{N}}_{\mathrm{a}}}{2} = \frac{10-5}{2} = 2.5 \\ \mathrm{The} \mathrm{electronic} \mathrm{configuration} \mathrm{of} \mathrm{oxygen} \mathrm{ion} \left({\mathrm{O}}_2^{-}\right): \mathrm{\sigma }1{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}1{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}2{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2 \left[\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2 = \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2\right] \left[{\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{x}}^2 = {\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{y}}^1\right] {\mathrm{\sigma }}^{*}2{\mathrm{p}}_{\mathrm{z}}^0 \\ \mathrm{Bond} \mathrm{Order}: \frac{{\mathrm{N}}_{\mathrm{b}} - {\mathrm{N}}_{\mathrm{a}}}{2} = \frac{10-7}{2} = 1.5 \\ \mathrm{Therefore}, \mathrm{the} \mathrm{correct} \mathrm{bond} \mathrm{order} \mathrm{is} {\mathrm{O}}_2^{-} < {\mathrm{O}}_2 < {\mathrm{O}}_2^{+} \end{gathered}$$

Hence, the correct answer is option B.

Answer:

Fluorine is most electronegative atom and it's electronic configuration is 1s² 2s² 2p⁵. 

Hence, the correct answer is option A.

Answer:

Ionisation enthalpy is inversally proportional to atomic size. The energy required to remove the electron from extra stable half filled p orbital is exceptionally high. Therefore, phosphorous have highest value of ionisation enthalpy.

Hence, the correct answer is option B.

Answer:

Both CN^- and NO⁺ are isoelectronic. Therefore, they have identical bond order.

Hence, the correct answers are option A and B

Answer:

In case of BeCl₂ and CS₂ central atom is sp hybridised. Therefore, both BeCl₂ and CS₂ are linear in shape.

Hence, the correct answer is option A and D.

Answer:

Both NO⁺ and N₂ has 14 electron each. Therefore, NO⁺ and N₂ are isoelectronic with CO. 

Hence, the correct answer is option A and B.

Answer:

Both O₃ and ${\mathrm{NO}}_2^{-}$ have two bond pair and one lone pair at the central atom. Therefore, they have same hybridisation i.e. sp² and shape.


Hence, the correct answers are options C and D.

Answer:

In  ${\mathrm{CO}}_3^{2-}$ the average formal charge on each oxygen atom is 0.67 units and due to resonance all C – O bond lengths are equal. 

Hence, the correct answers are options C and D.

Answer:

$$\begin{gathered} \mathrm{The} \mathrm{electronic} \mathrm{configuration} \mathrm{of} \mathrm{nitrogen} \left({\mathrm{N}}_2\right): \mathrm{\sigma }1{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}1{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}2{\mathrm{s}}^2 \left[\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2 = \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2\right] \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2 \left[{\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{x}}^0 = {\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{y}}^0\right] {\mathrm{\sigma }}^{*}2{\mathrm{p}}_{\mathrm{z}}^0 \\ \mathrm{The} \mathrm{electronic} \mathrm{configuration} \mathrm{of} \mathrm{nitrogen} \mathrm{ion} \left({\mathrm{N}}_2^{2-}\right): \mathrm{\sigma }1{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}1{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}2{\mathrm{s}}^2 \left[\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2 = \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2\right] \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2 \left[{\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{x}}^1 = {\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{y}}^1\right] {\mathrm{\sigma }}^{*}2{\mathrm{p}}_{\mathrm{z}}^0 \\ \mathrm{The} \mathrm{electronic} \mathrm{configuration} \mathrm{of} \mathrm{oxygen} \left({\mathrm{O}}_2\right): \mathrm{\sigma }1{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}1{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}2{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2 \left[\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2 = \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2\right] \left[{\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{x}}^1 = {\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{y}}^1\right] {\mathrm{\sigma }}^{*}2{\mathrm{p}}_{\mathrm{z}}^0 \\ \mathrm{The} \mathrm{electronic} \mathrm{configuration} \mathrm{of} \mathrm{roxygen} \mathrm{ion} \left({\mathrm{O}}_2^{2-}\right): \mathrm{\sigma }1{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}1{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}2{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2 \left[\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2 = \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2\right] \left[{\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{x}}^2 = {\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{y}}^2\right] {\mathrm{\sigma }}^{*}2{\mathrm{p}}_{\mathrm{z}}^0 \end{gathered}$$

Both N₂ and ${\mathrm{O}}_2^{2-}$  are diamagnetic.

Hence, the correct answers are options A and D.

 

Answer:

$$\begin{gathered} \mathrm{The} \mathrm{electronic} \mathrm{configuration} \mathrm{of} \mathrm{nitrogen} \left({\mathrm{N}}_2\right): \mathrm{\sigma }1{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}1{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}2{\mathrm{s}}^2 \left[\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2 = \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2\right] \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2 \left[{\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{x}}^0 = {\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{y}}^0\right] {\mathrm{\sigma }}^{*}2{\mathrm{p}}_{\mathrm{z}}^0 \\ \mathrm{Bond} \mathrm{Order} = \frac{{\mathrm{N}}_{\mathrm{b}} - {\mathrm{N}}_{\mathrm{a}}}{2} = \frac{10 - 4}{2} = 3 \\ \mathrm{The} \mathrm{electronic} \mathrm{configuration} \mathrm{of} \mathrm{nitrogen} \mathrm{ion} \left({\mathrm{N}}_2^{-}\right): \mathrm{\sigma }1{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}1{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}2{\mathrm{s}}^2 \left[\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2 = \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2\right] \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2 \left[{\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{x}}^1 = {\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{y}}^0\right] {\mathrm{\sigma }}^{*}2{\mathrm{p}}_{\mathrm{z}}^0 \\ \mathrm{Bond} \mathrm{Order} = \frac{{\mathrm{N}}_{\mathrm{b}} - {\mathrm{N}}_{\mathrm{a}}}{2} = \frac{10 - 5}{2} = 2.5 \\ \mathrm{The} \mathrm{electronic} \mathrm{configuration} \mathrm{of} \mathrm{oxygen} \left({\mathrm{F}}_2^{+}\right): \mathrm{\sigma }1{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}1{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}2{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2 \left[\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2 = \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2\right] \left[{\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{x}}^2 = {\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{y}}^1\right] {\mathrm{\sigma }}^{*}2{\mathrm{p}}_{\mathrm{z}}^0 \\ \mathrm{Bond} \mathrm{Order} = \frac{{\mathrm{N}}_{\mathrm{b}} - {\mathrm{N}}_{\mathrm{a}}}{2} = \frac{10 - 7}{2} = 1.5 \\ \mathrm{The} \mathrm{electronic} \mathrm{configuration} \mathrm{of} \mathrm{oxygen} \mathrm{ion} \left({\mathrm{O}}_2^{-}\right): \mathrm{\sigma }1{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}1{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}2{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2 \left[\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2 = \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2\right] \left[{\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{x}}^2 = {\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{y}}^1\right] {\mathrm{\sigma }}^{*}2{\mathrm{p}}_{\mathrm{z}}^0 \\ \mathrm{Bond} \mathrm{Order} = \frac{{\mathrm{N}}_{\mathrm{b}} - {\mathrm{N}}_{\mathrm{a}}}{2} = \frac{10 - 7}{2} = 1.5 \end{gathered}$$

Hence, the correct answer is C and D.

Answer:

NaCl being an ionic compound is a bad conductor of electricity in the solid state and in canonical structures the arrangement of atoms remains same. 

Hence, the correct answers are options A and B.

Answer:

In H₂S the central atom sulphur has two bond pair and one lone pair. The presence of two lone pairs brings distortion in the molecule due to repulsion by two bond pairs. Therefore, it has bent shape. Its structure is as follows:

In PCl₃ the central atom phosphorous has one lone pair and 3 bond pair. Due to the presence of lone pair it has a distorted tetrahedral shape. Therefore, it has a triagonal pyramidal shape. Its structure is as follows:

Answer:

$$\begin{gathered} \mathrm{The} \mathrm{electronic} \mathrm{configuration} \mathrm{of} \mathrm{oxygen} \mathrm{ion} \left({\mathrm{O}}_2^{-}\right): \mathrm{\sigma }1{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}1{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}2{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2 \left[\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2 = \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2\right] \left[{\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{x}}^2 = {\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{y}}^1\right] {\mathrm{\sigma }}^{*}2{\mathrm{p}}_{\mathrm{z}}^0 \\ \mathrm{Due} \mathrm{to} \mathrm{presence} \mathrm{of} \mathrm{one} \mathrm{unpaired} \mathrm{electron} \mathrm{in} {\mathrm{O}}_2^{-}, \mathrm{it} \mathrm{is} \mathrm{paramagnetic}. \\ \mathrm{Bond} \mathrm{Order} = \frac{{\mathrm{N}}_{\mathrm{b}} - {\mathrm{N}}_{\mathrm{a}}}{2} = \frac{10 - 7}{2} = 1.5 \\ \mathrm{The} \mathrm{electronic} \mathrm{configuration} \mathrm{of} \mathrm{oxygen} \mathrm{ion} \left({\mathrm{O}}_2^{+}\right): \mathrm{\sigma }1{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}1{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}2{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2 \left[\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2 = \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2\right] \left[{\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{x}}^1 = {\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{y}}^1\right] {\mathrm{\sigma }}^{*}2{\mathrm{p}}_{\mathrm{z}}^0 \\ \mathrm{Due} \mathrm{to} \mathrm{presence} \mathrm{of} \mathrm{two} \mathrm{unpaired} \mathrm{electron} \mathrm{in} {\mathrm{O}}_2^{+}, \mathrm{it} \mathrm{is} \mathrm{paramagnetic}. \\ \mathrm{Bond} \mathrm{Order} = \frac{{\mathrm{N}}_{\mathrm{b}} - {\mathrm{N}}_{\mathrm{a}}}{2} = \frac{10 - 5}{2} = 2.5 \\ \mathrm{Due} \mathrm{to} \mathrm{more} \mathrm{bond} \mathrm{order} \mathrm{of} {\mathrm{O}}_2^{+}, \mathrm{it} \mathrm{will} \mathrm{have} \mathrm{more} \mathrm{bond} \mathrm{energy}. \end{gathered}$$

Answer:

In BrF₅ the central atom bromine has one lone pair and 5 bond pair. Therefore, it has square pyramid shape and octahedral geometry.

Answer:

(a) Compound I has intramolecular hydrogen bonding and compound II has intermolecular hydrogen bonding.
(b) Compound II has intermolecular hydrogen bonding, hence it will have more boiling point.
(c) Due to intermolecular hydrogen bonding in compound II, it will form more stronger hydrogen bonding with water molecule. Hence, it is more soluble.

Answer:

Both figure represents zero overlap i.e out of phase overlap due to different orientation or direction of approach. Therefore, both will not leads to bond formation.

Answer:

PCl₅ has 5 bond pairs and zero lone pair of electrons whereas IF₅ has 5 bond pairs and one lone pair of electrons. Hence, repulsion between bond pair and lone pair takes place in IF₅ which leads to distortion in geometry. Therefore, the shape of PCl₅ and IF₅ is trigonal bipyramidal and square pyramidal respectively.

Answer:

Dimethyl ether has greater bond angle because there will be more repulsion between bond pairs of CH₃ groups attached in ether than between bond pairs of hydrogen atoms attached to oxygen in water. This is because the carbon of CH₃ in ether is attached to three hydrogen atoms through σ bonds and electron pairs of these bonds add to the electronic charge density on carbon atom.

Answer:

The Lewis structures of the following compounds with formal charge on each atom are shown below:
(i) HNO₃


(ii) NO₂

(iii) H₂SO₄

 

Answer:

The complete sequence of energy levels in the increasing order of energy in nitrogen molecule is as follows:
σ1s< σ*1s< σ2s< σ*2s< (π2p_x= π2p_y) < σ2p_z< (π*2p_x= π*2p_y) < σ*2p_z

The order of  relative stability depends on bond order.
The bond order of the following species is as follows:
$$\begin{gathered} {\mathrm{N}}_2: 3 \\ {\mathrm{N}}_2^{+}: 2.5 \\ {\mathrm{N}}_2^{-}: 2.5 \\ {\mathrm{N}}_2^{2+}: 2 \end{gathered}$$
The order of  relative stability of the following species is as follows:
${\mathrm{N}}_2 > {\mathrm{N}}_2^{+} \approx {\mathrm{N}}_2^{-} > {\mathrm{N}}_2^{2+}$

The magnetic behaviour of the following species is as follows:
$$\begin{gathered} {\mathrm{N}}_2: \mathrm{Diamagnetic} \\ {\mathrm{N}}_2^{+}: \mathrm{Paramagnetic} \\ {\mathrm{N}}_2^{-}: \mathrm{Paramagnetic} \\ {\mathrm{N}}_2^{2+}: \mathrm{Diamagnetic} \end{gathered}$$

Answer:

(i) The bond order decreases from 3 in N₂ to 2.5 in N₂⁺.
(i) The bond order increases from 2 in O₂ to 2.5 in O₂⁺.

Answer:

(i) The covalent bond is formed by overlapping of atomic orbitals and these orbitals overlap in a specific orientation. Therefore, covalent bonds are directional bonds. On the other hand, ionic bond is formed by electrostatic attraction between cation and anion which is same in all directions. Therefore, ionic bonds are non-directional bonds.

(ii) Water molecule is bent because the lone pair-bond pair repulsion is much more as compared to the bond pair-bond pair repulsion whereas there is no lone pair of electrons present on carbon atom and hence, the carbon dioxide molecule is linear.

(iii) The carbon atoms of ethyne molecule are sp hybridized and hybridized orbitals are directed in opposite directions at an angle of 180^o. Therefore, ethyne molecule is linear.

Answer:

Ionic bond is the bond formed by the transfer of electrons between a cation and an anion due to electrostatic forces of attraction.
For example, NaCl is an ionic compound formed by the transfer of an electron from sodium (Na) to chlorine (Cl). Sodium loses one electron to form Na⁺ and chlorine gains that electron to form Cl⁻. While CH₄ molecule is formed by sharing of electrons between carbon and hydrogen atoms. Thus, there are four covalent bonds in CH₄ molecule.
 

Answer:

The ionic character of a bond depends on the electronegativity difference between the two atoms involved in the bond formation. More is the electronegativity difference, more will the ionic character. 
The order of electronegativity of the following atoms is as follows:
C < N < O < F
Hence, the order of increasing ionic character will be:
C – H < N – H < O – H < F – H

Answer:

The single Lewis structure based on the presence of two single bonds and one double bond between carbon and oxygen atoms is inadequate to represent the molecule accurately as it represents unequal bonds. All carbon to oxygen bonds in ${\mathrm{CO}}_3^{2-}$ ion are equivalent. Therefore the carbonate ion is best described as a resonance hybrid of the canonical forms I, II, and III shown below.

Answer:

The hybridization of carbons 1, 2, 3, 4 and 5 are sp, sp, sp², sp³ and sp² respectively.
The total number of sigma and pi bonds in this molecule are 11 and 4 respectively.

Answer:

Linear molecule: BeCl₂
Non-linear molecules: H₂O, HOCl, Cl₂O

Answer:

(i) The valencies of elements X, Y and Z are 4, 3 and 1 respectively. Therefore, the molecular formulae are as follows:
XH₄
YH₃
HZ

(ii) Since element Z has 7 valence electrons, it belongs to halogen family and is the most electronegative among the given elements. So, the bond H−Z will be highly polar and the compound HZ will have the highest dipole moment.
 

Answer:

(i) The resonating structures of ozone molecule are drawn below:



(ii) The resonating structures of nitrate ion are drawn below:

 

Answer:

$$\begin{gathered} \mathrm{Number} \mathrm{of} \mathrm{hybridized} \mathrm{orbitals}=\frac{1}{2}\left[\mathrm{Number} \mathrm{of} \mathrm{valence} \mathrm{electrons} \mathrm{on} \mathrm{central} \mathrm{atom}+\mathrm{Number} \mathrm{of} \mathrm{atoms} \mathrm{joined} \mathrm{to} \mathrm{central} \mathrm{atom} \mathrm{by} \mathrm{single} \mathrm{bond}\right] \\ \mathrm{For} {\mathrm{BCl}}_{3, }\mathrm{number} \mathrm{of} \mathrm{hybridized} \mathrm{orbitals}=\frac{1}{2}\left[3+3\right]=3 \\ \therefore \mathrm{Hybridization} \mathrm{is} s{p}^2 \mathrm{and} \mathrm{shape} \mathrm{is} \mathrm{trigonal} \mathrm{planar}. \\ \mathrm{For} {\mathrm{CH}}_4, \mathrm{number} \mathrm{of} \mathrm{hybridized} \mathrm{orbitals}=\frac{1}{2}\left[4+4\right]=4 \\ \therefore \mathrm{Hybridization} \mathrm{is} s{p}^3 \mathrm{and} \mathrm{shape} \mathrm{is} \mathrm{tetrahedral}. \\ \mathrm{For} {\mathrm{CO}}_2, \mathrm{number} \mathrm{of} \mathrm{hybridized} \mathrm{orbitals}=\frac{1}{2}\left[4+0\right]=2 \\ \therefore \mathrm{Hybridization} \mathrm{is} sp \mathrm{and} \mathrm{shape} \mathrm{is} \mathrm{linear}. \\ \mathrm{For} {\mathrm{NH}}_3, \mathrm{number} \mathrm{of} \mathrm{hybridized} \mathrm{orbitals}=\frac{1}{2}\left[5+3\right]=4 \\ \therefore \mathrm{Hybridization} \mathrm{is} s{p}^3 \mathrm{and} \mathrm{shape} \mathrm{is} \mathrm{trigonal} \mathrm{pyramidal} \mathrm{and} \mathrm{not} \mathrm{tetrahedral} \mathrm{due} \mathrm{to} \mathrm{lone} \mathrm{pair} \mathrm{of} \mathrm{electrons} \mathrm{on} \mathrm{nitrogen}. \end{gathered}$$

Answer:

All the C – O bonds in carbonate ion $\left({\mathrm{CO}}_3^{2-}\right)$ are equal in length due to resonance. A single Lewis structure based on the presence of two single bonds and one double bond between carbon and oxygen atoms is inadequate to represent the molecule accurately as it represents unequal bonds. So, the carbonate ion is best described as a resonance hybrid of the canonical forms I, II, and III shown below due to which all the C – O bonds become equal.

Answer:

The term average bond enthalpy is used for polyatomic molecules. It is obtained by dividing total bond dissociation enthalpy by the number of bonds broken.
The  enthalpy needed to break the two O – H bonds is not the same as second O – H bond undergoes some change because of changed chemical environment. This is the reason for some difference in energy of the same O – H bond in ethanol (C₂H₅OH) and water.

Answer:

Answer:

Answer:

Answer:

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Answer:

Sodium chloride formed by the action of chlorine gas on sodium metal is a stable compound. The sum of the electron gain enthalpy of chlorine to form chloride ion and the ionization enthalpy of sodium to form sodium ion may be positive but still the crystal structure gets stabilized due to the energy released in the formation of the crystal lattice.
So, A and R both are correct, and R is the correct explanation of A.

Hence, the correct answer is option A.

Answer:

The nitrogen atom has one lone pair and oxygen atom has two lone pairs. So, lone pair-bond pair repulsion is more in H₂O than NH₃. Therefore, the H – N – H bond angle in NH₃ is greater than that of H – O – H in H₂O.
So, A and R both are correct, and R is the correct explanation of A.

Hence, the correct answer is option A.

Answer:

In case of H₂O molecule, the enthalpy needed to break the two O – H bonds is not the same. This is because the second O – H bond undergoes some change because of changed chemical environment.
So, A and R both are false.

Hence, the correct answer is option (iv).

Answer:

(i) Dipole moment is a vector quantity. It determines the polarity of a molecule. In case of polyatomic molecules, the dipole moment is the vector sum of the dipole moments of various bonds.
For example in H₂O molecule, which has a bent structure, the two O–H bonds are oriented at an angle of 104.5^o. Net dipole moment is the resultant of the dipole moments of two O–H bonds which is a non-zero value. 

Hence, the molecule is polar.

For example, in BF₃ molecule, the B – F bonds are oriented at an angle of 120^o to one another, the three bond moments give a net sum of zero as the resultant of any two is equal and opposite to the third. 

Hence, the molecule is non-polar.

(ii) 



 

Answer:

Molecular Orbital Energy level diagram of N₂
 
                   N                                       N₂                                        N
Bond order of N₂ = = $\frac{1}{2}\left(6-0\right)=3$

Molecular Orbital Energy level diagram of F₂

                     F                                               F₂                                            F
Bond order of F₂ = = $\frac{1}{2}\left(6-4\right)=1$

Molecular Orbital Energy level diagram of Ne₂

                      Ne                                            Ne₂                                         Ne
Bond order of Ne₂ = = $\frac{1}{2}\left(6-6\right)=0$

Answer:

Valence Bond Theory

Introduced by Heitler and London (1927); developed further by Pauling and others

Theory:

• Consider two hydrogen atoms A and B approaching each other to form a covalent bond. N_A, N_B, e_A, and e_B represent the nucleus and electrons of atoms A and B respectively.

• When the atoms are at large distance from each other, there is no force of attraction between them.

• As the distance reduces, the following forces start developing in them:

Attractive forces −

1. Between the nucleus and the electron of the same atom (N_A- e_A, N_B - e_B)

2. Between the nucleus and the electron of different atoms (N_A- e_B, N_B - e_A)

Repulsive forces −

1. Between the electrons of the two atoms (e_A - e_B)

2. Between the nuclei of the two atoms (N_A - N_B)

• Experimentally, the attractive forces are found to be stronger than the repulsive ones. Therefore, the two atoms approach each other, and the potential energy also drops during the process.

• When the two forces balance each other, a state of minimum energy is attained. At this point, the two atoms are said to be bonded. The distance between them is called the bond length (in case of H-atoms, the bond length is equal to 74 pm).

• Also, during bond formation, energy is released. This released energy is called bond enthalpy. The minimum of the given curve represents bond energy, which is 435.8 kJ for hydrogen.

• It can be observed in the above figure that when the two H-atoms form a bond, they overlap each other slightly. This represents overlapping of atomic orbitals. (Hence, this theory introduces the concept of atomic orbitals, stating that covalent bonds are formed by the overlapping of singly filled atomic orbitals containing electrons having opposite spins.)

• More the extent of the overlapping, stronger will be the bond.

Answer:

Formation of PCl₅

• The five sp³d hybrid orbitals orient themselves in a trigonal bipyramidal manner, which is the geometry of PCl₅.

• All the P−Cl bonds in the molecule are not equivalent. Three bonds that are present in one plane are called equatorial bonds and have an angle of 120° between them. The other two bonds are called axial bonds and are at 90° with the equatorial plane. As the axial bond pairs suffer more repulsive interaction from the equatorial bond pairs, therefore axial bonds have been found to be slightly longer and hence slightly weaker than the equatorial bonds.

Formation of SF₆

• The six sp³d² hybrid orbitals are projected towards the six corners of a regular octahedron in SF₆.
• The hybrid orbitals overlap with singly occupied orbitals of fluorine atoms to form six S–F sigma bonds. Thus SF₆ molecule has a regular octahedral geometry and all bonds have the same bond length.

Answer:

(i) Hybridisation

It states that atomic orbitals of slightly different energies combine to form a new set of equivalent orbitals (having the same energy and shape) known as hybrid orbitals.

These hybrid orbitals are more stable than atomic orbitals, and participate in bond formation.

Salient features:

1. Number of hybrid orbitals obtained = Number of atomic orbitals undergoing hybridisation

2. Hybrid orbitals are always identical in all respects.

3. To minimise electronic repulsion, they orient themselves in a preferred manner. Thus, the type of hybridisation indicates the geometry of a molecule.

Conditions required for hybridisation:

1. Orbitals in the valence shell should be hybridised.

2. Orbitals undergoing hybridisation should have almost equal energy.

3. Promotion of electrons is not essential prior to hybridisation.

4. Half as well as fully filled orbitals of the valence shell can take part in hybridisation.

Types of hybridisation in a carbon atom are:

• sp
• sp²
• sp³

Answer:

The electronic configuration of oxygen atom is 1s² 2s² 2p⁴. It has 5 atomic orbitals. So, in the formation of dioxygen from oxygen atoms 10 atomic orbitals will combine to form 10 molecular orbitals because the number of molecular orbitals formed is equal to the number of combining atomic orbitals.

Hence, the correct answer is option (i).

Answer:

The nodal plane is the plane where the probability of finding the electron density is zero. The nodal planes of the following molecular orbitals are drawn below:


Hence, the correct answer is option D.
 

Answer:

The electronic configuration of O₂⁺ is as follows:
σ1s² σ*1s² σ2s² σ*2s² σ2p_z² (π­2p_x² â‰ƒ π2p_y²) (­π* 2p_x¹ â‰ƒ π* 2p_y).

The electronic configuration of  N₂⁻ is as follows:
σ1s² σ*1s² σ2s² σ*2s² (π­2p_x² â‰ƒ π2p_y²) σ2p_z² (­π* 2p_x¹ â‰ƒ π* 2p_y).

The bond order for O₂⁺ as well as N₂⁻ is 2.5.

Hence, the correct answer is option B.

Answer:

Energies of various molecular orbitals for elements hydrogen to nitrogen increase in the order :
σ1s < σ* 1s < σ2s < σ* 2s < (π­2p_x ≈ π­2p_y) < σ2p_z < (­π* 2p_x ≈­ π* 2p_y) < σ* 2p_z 

Hence, the correct answer is option C.