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40
Subjective

What is the effect of the following processes on the bond order in $\mathrm{N}_2$ and $\mathrm{O}_2$ ?

(a) $\mathrm{N}_2 \longrightarrow \mathrm{N}_2^{+}+e^{-}$

(b) $\mathrm{O}_2 \longrightarrow \mathrm{O}_2^{+}+e^{-}$

Explanation

According to molecular orbital theory, electronic configurations and bond order of $\mathrm{N}_2, \mathrm{~N}_2^{+}, \mathrm{O}_2$ and $\mathrm{O}_2^{+}$species are as follows

$$\begin{aligned} \mathrm{N}_2\left(14 e^{-}\right) & =\sigma 1 s^2, \stackrel{\star}{\sigma} 1 s^2, \sigma 2 s^2, \stackrel{\star}{\sigma} 2 s^2,\left(\pi 2 p_x^2 \approx \pi 2 p_y^2\right), \sigma 2 p_z^2 \\ \text { Bond order } & =\frac{1}{2}\left[N_b-N_a\right]=\frac{1}{2}(10-4)=3 \\ \mathrm{~N}_2^{+}\left(13 e^{-}\right) & =\sigma 1 s^2, \stackrel{\star}{\sigma} 1 s^2, \sigma 2 s^2, \stackrel{\star}{\sigma} 2 s^2,\left(\pi 2 p_x^2 \approx \pi 2 p_y^2\right) \sigma 2 p_z^1 \\ \text { Bond order } & =\frac{1}{2}\left[N_b-N_a\right]=\frac{1}{2}(9-4)=2.5 \\ \mathrm{O}_2\left(16 e^{-}\right) & =\sigma 1 s^2,{ }^{\star} \sigma 1 s^2, \sigma 2 s^2, \stackrel{\star}{\sigma} 2 s^2, \sigma 2 p_z^2,\left(\pi 2 p_x^2 \approx \pi 2 p_y^2\right),\left({ }^{\star} \pi 2 p_x^1 \approx \stackrel{\star}{\pi} 2 p_y^1\right) \\ \text { Bond order } & =\frac{1}{2}\left[N_b-N_a\right]=\frac{1}{2}(10-6)=2 \\ \mathrm{O}_2^{+}\left(15 e^{-}\right) & =\sigma 1 s^2, \stackrel{\star}{\sigma} 1 s^2, \sigma 2 s^2, \stackrel{\star}{\sigma} 2 s^2, \sigma 2 p_z^2,\left(\pi 2 p_x^2 \approx \pi 2 p_y^2\right),\left({ }^{\star} \pi 2 p_x^1 \approx \stackrel{\star}{\pi} 2 p_y\right) \\ \text { Bond order } & =\frac{1}{2}\left[N_b-N_a\right]=\frac{1}{2}(10-5)=2.5 \end{aligned}$$

(a) $\underset{\text { B.O. }=3}{\mathrm{~N}_2} \longrightarrow \underset{\text { B.O. }=2.5}{\mathrm{~N}_2^{+}}+e^{-}$

Thus, bond order decreases.

(b) $\underset{\text { B.O = 2}}{\mathrm{O}_2} \longrightarrow \underset{\text { B.O = 2.5}}{\mathrm{O}_2^{+}}+\mathrm{e}^{-}$

Thus, bond order increases.

41
Subjective

Give reason for the following.

(a) Covalent bonds are directional bonds while ionic bonds are non-directional.

(b) Water molecule has bent structure whereas carbon dioxide molecule is linear.

(c) Ethyne molecule is linear.

Explanation

(a) A covalent bond is formed by the overlap of atomic orbitals. The direction of overlapping gives the direction of bond. In ionic bond, the electrostatic field of an ion is non-directional. Each positive ion is surrounded by a number of anions in any direction depending upon its size and vice-versa. That's why covalent bonds are directional bonds while ionic bonds are non-directional.

(b) In $\mathrm{H}_2 \mathrm{O}$, oxygen atom is $s p^3$ hybridised with two lone pairs. The four $s p^3$ hybridised orbitals acquire a tetrahedral geometry with two corners occupied by hydrogen atoms while other two by the lone pairs. The bond angle is reduced to $104.5^{\circ}$ from $109.5^{\circ}$ due to greater repulsive forces between Ip-Ip and the molecule thus acquires a V -shape or bent structure (angular structure).

In $\mathrm{CO}_2$ molecule, carbon atom is $s p$-hybridised. The two $s p$ hybrid orbitals are oriented in opposite direction forming an angle of $180^{\circ}$.

That's why $\mathrm{H}_2 \mathrm{O}$ molecule has bent structure whereas $\mathrm{CO}_2$ molecule is linear.

(c) In ethyne molecule, both the carbon atoms are $s p$ hybridised, having two unhybridised orbitals, i.e., $2 p_x$ and $2 p_y$. The two $s p$ hybrid orbitals of both the carbon atoms are oriented in opposite direction forming an angle of $180^{\circ}$.

That's why ethyne molecule is linear.

42
Subjective

What is an ionic bond? With two suitable examples the difference between an ionic and a covalent bond?

Explanation

Ionic bond The bond formed, as a result of the electrostatic attraction between the positive and negative ions was termed as the electrovalent bond or ionic bond. e.g., the formation of NaCl from sodium and chlorine can be explained as

$$\underset{[\mathrm{Ne}] 3 \mathrm{~s}^1}{\mathrm{Na}} \longrightarrow \underset{[\mathrm{Ne}]}{\mathrm{Na}^{+}}+\mathrm{e}^{-}$$

$$\underset{[\mathrm{Ne}] 3 s^2 3 p^5}{\mathrm{Cl}}+\mathrm{e} \longrightarrow \underset{[\mathrm{Ne}] 3 s^2 3 p^6 \text { or }[\mathrm{Ar}]}{\mathrm{Cl}^{-}}$$

$$\mathrm{Na}^{+}+\mathrm{Cl}^{-} \longrightarrow \mathrm{NaCl} \text { or } \mathrm{Na}^{+} \mathrm{Cl}^{-}$$

Similarly, the formation of $\mathrm{CaF}_2$ may be shown as

$$\underset{[A r] 4 \mathrm{~s}^2}{\mathrm{Ca}} \longrightarrow \underset{[A r]}{\mathrm{Ca}^{2+}}+2 e^{-}$$

$$\underset{[\mathrm{He}] 2 s^2 2 p^5}{\mathrm{~F}}+\mathrm{e}^{-} \longrightarrow \underset{[\mathrm{He}] 2 s^2 2 p^6 \text { or }[\mathrm{Ne}]}{\mathrm{F}^{-}}$$

$$\mathrm{Ca}^{2+}+2 \mathrm{~F}^{-} \longrightarrow \mathrm{CaF}_2 \text { or } \mathrm{Ca}^{2+}\left(\mathrm{F}^{-}\right)_2$$

Covalent bond The bond formed between the two atoms by mutual sharing of electrons between them is called covalent bond. e.g., the formation of chlorine molecule can be explained as

Similarly, in the formation of HCl

43
Subjective

Arrange the following bonds in order of increasing ionic character giving reason.

$$\mathrm{N}-\mathrm{H}, \quad \mathrm{F}-\mathrm{H}, \quad \mathrm{C}-\mathrm{H} \text { and } \mathrm{O}-\mathrm{H}$$

Explanation

Greater is the electronegativity difference between the two bonded atoms, greater is the ionic character.

Bond $$\mathrm{N-H}$$ $$\mathrm{F-H}$$ $$\mathrm{C-H}$$ $$\mathrm{O-H}$$
Electronegativity difference $$(3.0-2.1)=0.9$$ $$(4.0-2.1)=1.9$$ $$(2.5-2.1)=0.4$$ $$(3.5-2.1)=1.4$$

Therefore, increasing order of ionic character of the given bonds is as follows

$$\mathrm{C}-\mathrm{H}<\mathrm{N}-\mathrm{H}<\mathrm{O}-\mathrm{H}<\mathrm{F}-\mathrm{H}$$

44
Subjective

Explain why $\mathrm{CO}_3^{2-}$ ion cannot be represented by a single Lewis structure. How can it be best represented?

Explanation

A single Lewis structure of $\mathrm{CO}_3^{2-}$ ion cannot explain all the properties of this ion. It can be represented as a resonance hybrid of the following structures

If, it were represented only by one structure, there should be two types of bonds, i.e., $\mathrm{C}=\mathrm{O}$ double bond and C — O single bonds but actually all bonds are found to be identical with same bond length and same bond strength.