Answer

(b)

(a) $ NF_3$ is pyramidal whereas $BF_3$ is planar triangular.

(b) $ BF_4^-$and $ NH_4^+ $ion both are tetrahedral and $s p^3$ hybridisation.

(c) $ BCl_3$ is triangular planar whereas $ BrCl_3$ is $T$ shaped.

(d) $NH_3$ is pyramidal whereas $\mathrm{NO}_{3}^{-}$is triangular planar.

• (a) $NF_3$ is pyramidal whereas $BF_3$ is planar triangular.
• (c) $BCl_3$ is triangular planar whereas $BrCl_3$ is T-shaped.
• (d) $NH_3$ is pyramidal whereas $NO_3^-$ is triangular planar.

Answer

(c) $\mathrm{CO}_{2}$ being symmetrical has zero dipole moment

$$ \begin{gathered} \mathrm{O} \leftrightarrows \mathrm{C} \stackrel{\leftarrow}{=}\mathrm{O} \\ \mu=0 \end{gathered} $$

Among $ HI, SO_2$ and $ H_2 O$ dipole moment is highest for $ H_2 O$ as in it the central atom contains 2 lone pairs.

$$ \begin{gathered} \mathrm{H} \vec{\longrightarrow} \mathrm{I}\\ \mu=0.38 \mathrm{D} \end{gathered} $$

$ \mu=1.84 \mathrm{D} \quad \quad \quad \quad \mu=1.62 \mathrm{D} $

• (a) $\mathrm{CO}_{2}$: $\mathrm{CO}_{2}$ is a linear molecule with a symmetrical structure. The dipole moments of the two $\mathrm{C=O}$ bonds are equal in magnitude but opposite in direction, canceling each other out, resulting in a net dipole moment of zero.

• (b) $\mathrm{HI}$: $\mathrm{HI}$ has a dipole moment due to the difference in electronegativity between hydrogen and iodine. However, its dipole moment (0.38 D) is lower compared to $\mathrm{H}_{2} \mathrm{O}$ because the electronegativity difference and the bond polarity are not as significant.

• (d) $\mathrm{SO} _{2}$: $\mathrm{SO} _{2}$ is a bent molecule with a dipole moment due to the difference in electronegativity between sulfur and oxygen. However, its dipole moment (1.62 D) is lower than that of $\mathrm{H} _{2} \mathrm{O}$ because the molecular geometry and the distribution of electron density result in a smaller net dipole moment compared to $\mathrm{H} _{2} \mathrm{O}$.

Answer

(b) The type of hybrid orbitals of nitrogen can be decided by using VSEPR theory counting $b p$ and as $1 p$ in

$\mathrm{NO}_{2}^{+}=2 \mathrm{bp}+0 lp=$ linear $=s p$ hybridised

$\mathrm{NO}_{3}^{-}=3 b p+0 lp \Rightarrow s p^{2}$ hybridised

$\mathrm{NH}_{4}^{+}=4 \mathrm{bp}+0 lp \Rightarrow s p^{3}$ hybridised

• Option (a) is incorrect because $\mathrm{NO} _{3}^{-}$ is $sp^2$ hybridized, not $sp^3$.
• Option (c) is incorrect because $\mathrm{NO} _{2}^{+}$ is $sp$ hybridized, not $sp^2$.
• Option (d) is incorrect because $\mathrm{NO} _{3}^{-}$ is $sp^2$ hybridized, not $sp^3$, and $\mathrm{NH} _{4}^{+}$ is $sp^3$ hybridized, not $sp$.

Answer

(b) Strength of $\mathrm{H}$-bond is in the order $\mathrm{H} \ldots . . \mathrm{F}>\mathrm{H} \ldots . . \mathrm{O}>\mathrm{H} \ldots . . . \mathrm{N}$.

But each $ H_2 O $ molecule is linked to four other $ H_2 O $ molecules through $ H $-bonds whereas each $ HF $ molecule is linked only to two other $ HF$ molecules.

Hence, b.p of $ H_2 O >$ b. p of $ HF >$ b.p. of $ NH_3$

• Option (a) $HF > H_2 O > NH_3$:

  • This option is incorrect because, although the strength of the hydrogen bond in HF is stronger than in H₂O, each H₂O molecule can form four hydrogen bonds (two as a donor and two as an acceptor), whereas each HF molecule can form only two hydrogen bonds (one as a donor and one as an acceptor). This extensive hydrogen bonding network in H₂O leads to a higher boiling point for H₂O compared to HF.

• Option (c) $NH_3 > HF > H_2 O$:

  • This option is incorrect because NH₃ has the weakest hydrogen bonds among the three compounds due to the lower electronegativity of nitrogen compared to oxygen and fluorine. Additionally, NH₃ can form fewer hydrogen bonds (each NH₃ molecule can form only one hydrogen bond as a donor and one as an acceptor) compared to H₂O and HF. Therefore, NH₃ has a lower boiling point than both HF and H₂O.

• Option (d) $NH_3 > H_2 O > HF$:

  • This option is incorrect for similar reasons as option (c). NH₃ has the weakest hydrogen bonds and can form fewer hydrogen bonds compared to H₂O and HF. Consequently, NH₃ has the lowest boiling point among the three compounds, not the highest.

Answer

(c) In $\mathrm{PO}_{4}^{3-}$ ion, formal charge on each $\mathrm{O}$-atom of $\mathrm{P}-\mathrm{O}$ bond

$$ =\frac{\text { total charge }}{\text { Number of O-atom }}=-\frac{3}{4}=-0.75 $$

• Option (a) +1: This is incorrect because the formal charge on an oxygen atom in the $\mathrm{PO}_{4}^{3-}$ ion cannot be positive. Oxygen is more electronegative than phosphorus and typically carries a negative formal charge in such compounds.

• Option (b) -1: This is incorrect because if each oxygen atom in the $\mathrm{PO}_{4}^{3-}$ ion had a formal charge of -1, the total charge on the ion would be -4 (since there are four oxygen atoms), which does not match the given charge of -3 for the ion.

• Option (d) +0.75: This is incorrect because a positive formal charge on an oxygen atom in the $\mathrm{PO}_{4}^{3-}$ ion is not consistent with the typical electronegativity and bonding patterns of oxygen. Additionally, the calculation of formal charge should result in a negative value for oxygen in this context.

Thinking Process

To solve this question, we must know the structure of $\mathrm{NO}_{3}^{-}$ion i.e.,

$$ \left[\begin{array}{c} \underset{..}{\ddot{O}}=N- \underset{..}{\ddot{O:}} \\ \underset{..}{\underset{:O:}{\mid}}\\ \end{array}\right]^- $$

Then, count the bond pairs and lone pairs of electron on nitrogen.

Answer

(d) In $\mathrm{N}$-atom, number of valence electrons $=5$

Due to the presence of one negative charge, number of valence electrons $=5+1=6$ one O-atom forms two bond (= bond) and two O-atom shared with two electrons of $\mathrm{N}$-atom

Thus, $3 \mathrm{O}$-atoms shared with 8 electrons of $\mathrm{N}$-atom.

$\therefore$ Number of bond pairs (or shared pairs) $=4$

Number of lone pairs $=0$

• Option (a) 2,2: This option is incorrect because it suggests that there are 2 bond pairs and 2 lone pairs on the nitrogen atom. However, in the $\mathrm{NO}_{3}^{-}$ ion, the nitrogen atom forms 4 bonds with the oxygen atoms and has no lone pairs. Therefore, the number of bond pairs is 4, not 2, and the number of lone pairs is 0, not 2.

• Option (b) 3,1: This option is incorrect because it suggests that there are 3 bond pairs and 1 lone pair on the nitrogen atom. In reality, the nitrogen atom in the $\mathrm{NO}_{3}^{-}$ ion forms 4 bonds with the oxygen atoms and has no lone pairs. Thus, the number of bond pairs is 4, not 3, and the number of lone pairs is 0, not 1.

• Option (c) 1,3: This option is incorrect because it suggests that there is 1 bond pair and 3 lone pairs on the nitrogen atom. However, in the $\mathrm{NO}_{3}^{-}$ ion, the nitrogen atom forms 4 bonds with the oxygen atoms and has no lone pairs. Therefore, the number of bond pairs is 4, not 1, and the number of lone pairs is 0, not 3.

Answer

(a) $\mathrm{BH}_{4}^{-} \Rightarrow 4$ bond pairs +0 lone pair $\Rightarrow s p^{3}$ hybridised $=$ tetrahedral geometry

$\mathrm{NH}_{2}^{-}=\mathrm{V}$ - shape

$\mathrm{CO}_{3}^{2-}=$ triangular planar

$\mathrm{H}_{3} \mathrm{O}^{+}=$pyramidal

• $\mathrm{NH}_{2}^{-}$: This species has 2 bond pairs and 2 lone pairs on the nitrogen atom, leading to an angular or bent (V-shaped) geometry due to the repulsion between lone pairs and bond pairs.

• $\mathrm{CO}_{3}^{2-}$: This species has 3 bond pairs and no lone pairs on the central carbon atom, resulting in a trigonal planar geometry.

• $\mathrm{H}_{3} \mathrm{O}^{+}$: This species has 3 bond pairs and 1 lone pair on the oxygen atom, leading to a pyramidal geometry due to the lone pair-bond pair repulsion.

Answer

(a) The given compound will have the correct structure as

There are $5 \pi$-bonds and $8 \mathrm{C}-\mathrm{H}+11 \mathrm{C}-\mathrm{C} \sigma$-bonds, i.e., $19 \sigma$-bonds are present in the above molecule.

• Option (a) 6,19: This option is incorrect because the given structure has only 5 π-bonds, not 6. The count of σ-bonds is correct at 19, but the number of π-bonds is overestimated.

• Option (b) 4,20: This option is incorrect because the given structure has 5 π-bonds, not 4. Additionally, the number of σ-bonds is 19, not 20. Both the π-bonds and σ-bonds counts are incorrect in this option.

• Option (d) 5,20: This option is incorrect because, although the number of π-bonds is correctly identified as 5, the number of σ-bonds is overestimated. The correct number of σ-bonds is 19, not 20.

Answer

(c) The electronic configuration of the given molecules are

$ N_2^+=\sigma 1 s^2, \sigma^{\star} 1 s^2, \sigma 2 s^2, \sigma^{\star} 2 s^2, \pi 2 p_x^2=\pi p_y^2, \sigma 2 p_z^1$

It has one unpaired electron.

$ O_2=\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \approx \pi 2 p_y^2, \pi * 2 p_x^1 \approx \pi * 2 p_y^1$

$\mathrm{O}_{2}$ has two unpaired electrons.

$ O_2^{2-}=\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \approx \pi 2 p_y^2, \pi * 2 p_x^2 \approx \pi * 2 p_y^2$

Thus, $\mathrm{O}_{2}^{2-}$ has no unpaired electrons.

$ B_2=\sigma 1 s^2, \sigma^{\star} 1 s^2, \sigma 2 s^2, \sigma^{\star} 2 s^2, \pi 2 p_x^1 \approx \pi_2 p_y^1$

Thus, $\mathrm{B}_{2}$ has two unpaired electrons.

• (a) $\mathrm{N}_{2}^{+}$: The electronic configuration of $\mathrm{N}_{2}^{+}$ is $\sigma 1 s^2, \sigma^{\star} 1 s^2, \sigma 2 s^2, \sigma^{\star} 2 s^2, \pi 2 p_x^2=\pi p_y^2, \sigma 2 p_z^1$. It has one unpaired electron, making it incorrect.

• (b) $\mathrm{O}_{2}$: The electronic configuration of $\mathrm{O}_{2}$ is $\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \approx \pi 2 p_y^2, \pi * 2 p_x^1 \approx \pi * 2 p_y^1$. It has two unpaired electrons, making it incorrect.

• (d) $\mathrm{B}_{2}$: The electronic configuration of $\mathrm{B}_{2}$ is $\sigma 1 s^2, \sigma^{\star} 1 s^2, \sigma 2 s^2, \sigma^{\star} 2 s^2, \pi 2 p_x^1 \approx \pi_2 p_y^1$. It has two unpaired electrons, making it incorrect.

Answer

(c) $\mathrm{XeF}_{4} \Rightarrow 4 b p+2 l p \Rightarrow$ square planar $\Rightarrow$ all bonds are equal

$\mathrm{BF}_{4}^{-} \Rightarrow 4 \mathrm{bp}+0 lp \Rightarrow$ tetrahedral (all bonds are equal)

$ C_2 H_4 \Rightarrow _H^H > C = C <_H^H \Rightarrow C= C $ bond is not equal to $ C- H $ bond

$\mathrm{SiF}_{4} \Rightarrow 4 b p+0 lp \Rightarrow$ tetrahedral (all bonds are equal)

Thus, in $ C_2 H_4 $ all the bonds are not equal.

• (a) $\mathrm{XeF}_{4}$: The molecule $\mathrm{XeF}_{4}$ has a square planar geometry with 4 bonding pairs and 2 lone pairs of electrons. In this geometry, all the $\mathrm{Xe-F}$ bonds are equal.

• (b) $\mathrm{BF}_{4}^{-}$: The ion $\mathrm{BF}_{4}^{-}$ has a tetrahedral geometry with 4 bonding pairs and no lone pairs of electrons. In this geometry, all the $\mathrm{B-F}$ bonds are equal.

• (d) $\mathrm{SiF}_{4}$: The molecule $\mathrm{SiF}_{4}$ has a tetrahedral geometry with 4 bonding pairs and no lone pairs of electrons. In this geometry, all the $\mathrm{Si-F}$ bonds are equal.

Answer

(b) $ HCl, HI $ and $ H_2 S$ do not from $ H $-bonds. Only $ H_2 O$ forms hydrogen bonds. One $ H_2 O $ molecule forms four $ H $-bonding.

• (a) $\mathrm{HCl}$: Hydrogen chloride (HCl) does not form hydrogen bonds because the electronegativity difference between hydrogen and chlorine is not sufficient to create the strong dipole necessary for hydrogen bonding. Additionally, chlorine is not small enough to allow for effective hydrogen bonding.

• (c) $\mathrm{HI}$: Hydrogen iodide (HI) does not form hydrogen bonds because iodine is much less electronegative compared to oxygen, nitrogen, or fluorine, and it is also much larger in size. This results in a weaker dipole and insufficient conditions for hydrogen bonding.

• (d) $\mathrm{H}_{2} \mathrm{~S}$: Hydrogen sulfide (H$_2$S) does not form hydrogen bonds because sulfur is less electronegative than oxygen, and the size of the sulfur atom is larger, which makes the dipole interactions weaker and insufficient for hydrogen bonding.

Answer

(d) The given electronic configuration shows that an element is vanadium $(Z=22)$. It belongs to $d$-block of the periodic table. In transition elements i.e., $d$-block elements, electrons of $n s$ and $(n-1) d$ subshell take part in bond formation.

• (a) $3 p^{6}$: The $3 p^{6}$ electrons are part of a completely filled subshell and are not typically involved in chemical bonding for transition elements. These electrons are more stable and less likely to participate in bond formation compared to the outer $4 s$ and $3 d$ electrons.

• (b) $3 p^{6}, 4 s^{2}$: While the $4 s^{2}$ electrons can be involved in bonding, the $3 p^{6}$ electrons are not. The $3 p^{6}$ electrons are part of a filled inner shell and are not available for bonding in transition elements.

• (c) $3 p^{6}, 3 d^{2}$: Similar to the previous options, the $3 p^{6}$ electrons are part of a filled inner shell and do not participate in bonding. The $3 d^{2}$ electrons can be involved in bonding, but the $3 p^{6}$ electrons are not.

Answer

(b) For $s p^{2}$ hybridisation, the geometry is generally triangular planar.

Thus, bond angle is $120^{\circ}$.

• (a) $90^{\circ}$: This angle is typically associated with $dsp^2$ or $d^2sp^3$ hybridisation, not $sp^2$ hybridisation. In $sp^2$ hybridisation, the bond angles are approximately $120^{\circ}$ due to the trigonal planar geometry.

• (c) $180^{\circ}$: This angle corresponds to $sp$ hybridisation, where the geometry is linear. In $sp^2$ hybridisation, the bond angles are approximately $120^{\circ}$ due to the trigonal planar geometry.

• (d) $109^{\circ}$: This angle is characteristic of $sp^3$ hybridisation, where the geometry is tetrahedral. In $sp^2$ hybridisation, the bond angles are approximately $120^{\circ}$ due to the trigonal planar geometry.

Answer

(a) The given electronic configuration shows that $A$ represents noble gas because the octet is complete. $A$ is neon which has 10 atomic number.

• Option (b) $A_{2}$: Noble gases are chemically inert and do not form diatomic molecules under normal conditions. Therefore, $A_{2}$ is not a stable form for a noble gas like neon.

• Option (c) $A_{3}$: Noble gases do not form triatomic molecules. The electronic configuration of neon indicates a complete octet, making it highly stable and unlikely to form $A_{3}$.

• Option (d) $A_{4}$: Noble gases do not form tetra-atomic molecules. The stable form of a noble gas like neon is as a single atom, not as a molecule with four atoms.

Answer

(b) The electronic configuration of $C$ represent chlorine. Its stable form is dichlorine $(Cl_2 )$ i.e., $C_2$.

• Option (a) $\mathrm{C}$: This option suggests that the stable form of chlorine is a single chlorine atom. However, chlorine is a diatomic molecule in its stable form, meaning it naturally exists as $\mathrm{Cl}_2$ rather than a single $\mathrm{Cl}$ atom.

• Option (c) $\mathrm{C}_{3}$: This option suggests that the stable form of chlorine is a triatomic molecule. Chlorine does not naturally form $\mathrm{Cl}_3$ molecules; its stable form is diatomic, $\mathrm{Cl}_2$.

• Option (d) $\mathrm{C}_{4}$: This option suggests that the stable form of chlorine is a tetraatomic molecule. Chlorine does not naturally form $\mathrm{Cl}_4$ molecules; its stable form is diatomic, $\mathrm{Cl}_2$.

Answer

(d) The electronic configuration show that $B$ represents phosphorus and $C$ represents chlorine. The stable compound formed is $PCl_3$ i.e., $BC_3$.

• Option (a) $BC$: This option is incorrect because phosphorus (B) typically forms three bonds with chlorine (C) due to its valence of 5, needing three more electrons to complete its octet. Therefore, a 1:1 ratio does not satisfy the valence requirements of phosphorus.

• Option (b) $B_{2}C$: This option is incorrect because it suggests a 2:1 ratio of phosphorus to chlorine. Phosphorus does not typically form compounds with such a ratio with chlorine. The common valence of phosphorus leads to the formation of $PCl_3$ or $PCl_5$, not a compound with a 2:1 ratio.

• Option (c) $BC_{2}$: This option is incorrect because it suggests a 1:2 ratio of phosphorus to chlorine. Phosphorus typically forms three bonds with chlorine, resulting in a 1:3 ratio, not 1:2. Therefore, $BC_2$ does not satisfy the valence requirements of phosphorus.

Answer

(b) The bond between $B$ and $C$ will be covalent. Both $B$ and $C$ are non-metal atoms. $B$ represents phosphorus and $C$ represent chlorine.

• Ionic: An ionic bond typically forms between a metal and a non-metal, where one atom donates electrons and the other accepts them. Since both $B$ (phosphorus) and $C$ (chlorine) are non-metals, they are more likely to share electrons rather than transfer them, making an ionic bond unlikely.

• Hydrogen: A hydrogen bond is a type of weak interaction that occurs between a hydrogen atom, which is covalently bonded to a more electronegative atom (like oxygen, nitrogen, or fluorine), and another electronegative atom. Since neither $B$ (phosphorus) nor $C$ (chlorine) involves hydrogen in their bonding, a hydrogen bond is not applicable here.

• Coordinate: A coordinate bond (or dative covalent bond) involves one atom providing both electrons for the bond. This type of bond typically occurs in complex molecules or ions where one atom has a lone pair of electrons and the other has an empty orbital. In the case of $B$ (phosphorus) and $C$ (chlorine), they are more likely to form a regular covalent bond by sharing electrons rather than forming a coordinate bond.

Answer

(a) The correct increasing order of energies of molecular orbitals of $\mathrm{N}_{2}$ is given below

$\sigma 1 s<\sigma^* 1 s<\sigma 2 s<\sigma^* 2 s<(\pi 2 p_x \approx \pi 2 p_y)<\sigma 2 p_z<(\pi^* 2 p_x \approx \pi^* 2 p_y)<\sigma^* 2 p_z$

• Option (b) is incorrect because it suggests that the energy of the $\pi 2p_y$ orbital is greater than that of the $\sigma 2p_z$ orbital, which is not true for $\mathrm{N}_2$. In $\mathrm{N}_2$, the $\pi 2p_x$ and $\pi 2p_y$ orbitals have lower energy than the $\sigma 2p_z$ orbital.

• Option (c) is incorrect because it repeats the same order as option (a) but is labeled incorrectly. The correct order is already given in option (a).

• Option (d) is incorrect because it suggests that the energy of the $\pi 2p_y$ orbital is greater than that of the $\sigma 2p_z$ orbital, which is not true for $\mathrm{N}_2$. In $\mathrm{N}_2$, the $\pi 2p_x$ and $\pi 2p_y$ orbitals have lower energy than the $\sigma 2p_z$ orbital.

Answer

(d) Existance of molecule, bonding nature and energy order of molecular orbitals can be explained on the basis of molecular orbital theory as follows

(i) Molecules having zero bond order never exists while molecular having non-zero bond order is either exists or expected to exist.

(ii) Higher the value of bond order, higher will be its bond strength.

Electrons present in bonding molecular orbital are known as bonding electrons $\left(N_{b}\right)$ and electrons present on anti-bonding molecular orbital are known as anti-bonding electrons $\left(N_{\mathrm{a}}\right)$ and half of their difference is known as bond order i.e.,

(a) $ Be_2(4+4=8)=\sigma 1 s^2, \sigma^* 1 s^2, \sigma 1 s^2, \sigma^* 2 s^2$

Bond order $(\mathrm{BO})=\frac{1}{2}$

[Number of bonding electrons $\left(N_{b}\right)$ - Number of anti-bonding electrons $N_{a}$ ]

$$ =\frac{4-4}{2}=0 $$

Here, bond order of $\mathrm{Be}_{2}$ is zero. Thus, it does not exist.

(b) $\mathrm{He}_{2}(2+2=4)=\sigma 1 s^{2}, \sigma^{*} 1 s^{2}$

$$ \mathrm{BO}=\frac{2-2}{2}=0 $$

Here, bond order of $\mathrm{Be}_{2}$ is zero. Hence, it does not exist.

$$ \begin{gathered} \mathrm{He}_{2}^{+}(2+2-1=3)=\sigma 1 s^{2}, \sigma^{*} 1 s^{1} \\ \mathrm{BO}=\frac{2-1}{2}=0.5 \end{gathered} $$

Since, the bond order is not zero, this molecule is expected to exist.

(c) $N_2(7+7=14)=\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^2 \approx \pi 2 p_y^2, \sigma 2 p_z^2$

$$ \mathrm{BO}=\frac{10-4}{2}=3 $$

Thus, dinitrogen $(N_2)$ molecule contain triple bond and no any molecule of second period have more than double bond. Hence, bond strength of $N_2$ is maximum amongst the homonuclear diatomic molecules belonging to the second period.

(d) It is incorrect. The correct order of energies of molecular orbitals in $\mathrm{N}_{2}$ molecule is

$\sigma 2 s<\sigma^{\star} 2 s<\left(\pi 2 p_{x} \simeq \pi 2 p_{y}\right)<\sigma 2 p_{z}<\pi^{\star} 2 p_{x} \approx \pi^{\star} 2 p_{y}<\sigma^{\star} 2 p_{z}$

• (a) $\mathrm{Be}_{2}$ is not a stable molecule: This statement is correct. The bond order of $\mathrm{Be}_{2}$ is zero, indicating that it does not exist as a stable molecule.

• (b) $He_2$ is not stable but $He_2^+$ is expected to exist: This statement is correct. The bond order of $He_2$ is zero, indicating that it does not exist as a stable molecule. However, $He_2^+$ has a bond order of 0.5, suggesting that it can exist.

• (c) Bond strength of $\mathrm{N}_{2}$ is maximum amongst the homonuclear diatomic molecules belonging to the second period: This statement is correct. The bond order of $\mathrm{N}_{2}$ is 3, indicating a triple bond, which is the highest bond order among homonuclear diatomic molecules in the second period, resulting in maximum bond strength.

Thinking Process

To calculate bond order, write the molecular orbital configuration of particular species and afterwards using the formula.

Bond order $=\frac{1}{2}\left[\right.$ Number of bonding electrons $\left(N_{b}\right)-$ Number of anti-bonding electrons

$\left.\left(N_{a}\right)\right]$

Answer

(b) Electronic configuration of $\mathrm{O}_{2}$ (16 electrons)

$$ =\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \approx \pi 2 p_y^2, \pi^* 2 p_x^1 \approx \pi^* 2 p_y^1 $$

Bond order $=\frac{1}{2}\left(N_{b}-N_{a}\right)=\frac{1}{2}(10-6)=2$

Electronic confiquration of $\mathrm{O}_{2}^{+}$(15 electrons)

$$ =\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \approx \pi 2 p_y^2, \pi^* 2 p_x^1 \approx \pi^* 2 p_y^0 $$

Bond order $=\frac{1}{2}\left(N_{b}-N_{a}\right)=\frac{1}{2}(10-5)=2.5$

Electronic confiquration of $\mathrm{O}_{2}^{-}$(17 electrons)

$$ =\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \approx \pi 2 p_y^2, \pi^* 2 p_x^2 \approx \pi^* 2 p_y^1 $$

Bond order $=\frac{1}{2}\left(N_{b}-N_{a}\right)=\frac{1}{2}(10-7)=1.5$

Thus, the order of bond order is $O_2^-<O_2<O_2^+$

• Option (a) $O_2^->O_2>O_2^+$:

  • This option is incorrect because it suggests that the bond order of $O_2^-$ is greater than that of $O_2$, and the bond order of $O_2$ is greater than that of $O_2^+$. However, the bond orders are calculated as follows:
    • $O_2^-$ has a bond order of 1.5.
    • $O_2$ has a bond order of 2.
    • $O_2^+$ has a bond order of 2.5.
  • Therefore, the correct order should be $O_2^- < O_2 < O_2^+$, not $O_2^- > O_2 > O_2^+$.

• Option (c) $O_2^->O_2<O_2^+$:

  • This option is incorrect because it suggests that the bond order of $O_2^-$ is greater than that of $O_2$, but the bond order of $O_2$ is less than that of $O_2^+$. However, the bond orders are:
    • $O_2^-$ has a bond order of 1.5.
    • $O_2$ has a bond order of 2.
    • $O_2^+$ has a bond order of 2.5.
  • Therefore, the correct order should be $O_2^- < O_2 < O_2^+$, not $O_2^- > O_2 < O_2^+$.

• Option (d) $O_2^-<O_2>O_2^+$:

  • This option is incorrect because it suggests that the bond order of $O_2$ is greater than both $O_2^-$ and $O_2^+$. However, the bond orders are:
    • $O_2^-$ has a bond order of 1.5.
    • $O_2$ has a bond order of 2.
    • $O_2^+$ has a bond order of 2.5.
  • Therefore, the correct order should be $O_2^- < O_2 < O_2^+$, not $O_2^- < O_2 > O_2^+$.

Answer

(a) The electronic configuration represents

$$ \begin{aligned} & 2 s^{2} 2 p^{5}=\text { fluorine }=\text { most electronegative element } \\ & 3 s^{2} 3 p^{5}=\text { chlorine } \\ & 4 s^{2} 4 p^{5}=\text { bromine } \\ & 5 s^{2} 5 p^{5}=\text { iodine } \end{aligned} $$

• (b) $3 s^{2} 3 p^{5}$: This electronic configuration represents chlorine, which is less electronegative than fluorine.
• (c) $4 s^{2} 4 p^{5}$: This electronic configuration represents bromine, which is less electronegative than fluorine.
• (d) $5 s^{2} 5 p^{5}$: This electronic configuration represents iodine, which is less electronegative than fluorine.

Answer

(b) The electronic configuration of options (b) and (d) have exactly half-filled $3 p$ orbitals (b) represents phosphorus and (c) represents arsenic but (b) is smaller in size than (d). Hence, (b) has highest ionisation enthalpy. Ionisation enthalpy increases left to right in the periodic table as the size decreases.

• Option (a) is incorrect because the electronic configuration $[\mathrm{Ne}] 3 s^{2} 3 p^{1}$ represents aluminum, which has a lower ionization enthalpy compared to phosphorus due to its larger atomic size and less stable electron configuration.

• Option (c) is incorrect because the electronic configuration $[\mathrm{Ne}] 3 s^{2} 3 p^{2}$ represents silicon, which has a lower ionization enthalpy compared to phosphorus. Although silicon has a relatively stable configuration, it is not as stable as the half-filled $3p$ orbitals of phosphorus.

• Option (d) is incorrect because the electronic configuration $[\mathrm{Ar}] 3 d^{10} 4 s^{2} 4 p^{3}$ represents arsenic, which, despite having a half-filled $4p$ orbital, is larger in size compared to phosphorus. The larger atomic size of arsenic results in a lower ionization enthalpy compared to phosphorus.

Answer

$(a, b)$

$\mathrm{CN}^{-}$(number of electrons $=6+7+1=14$ )

$\mathrm{NO}^{+}$(number of electrons $=7+8-1=14$ )

$\mathrm{O}_{2}^{-}$(number of electrons $=8+8+1=17$ )

$\mathrm{O}_{2}^{2-}$ (number of electrons $=8+8+2=18$ )

Thus, $\mathrm{CN}^{-}$and $\mathrm{NO}^{+}$because of the presence of same number of electrons, have same bond order.

• $\mathrm{O}_{2}^{-}$ has 17 electrons, which results in a different bond order compared to $\mathrm{CN}^{-}$ and $\mathrm{NO}^{+}$, which both have 14 electrons.
• $\mathrm{O}_{2}^{2-}$ has 18 electrons, which results in a different bond order compared to $\mathrm{CN}^{-}$ and $\mathrm{NO}^{+}$, which both have 14 electrons.

Answer

$(a, d)$

$ BeCl_2( Cl - Be - Cl)$ and $ CS_2( S = C= S)$ both are linear. $ NCO^+$is non-linear. However, [remember that $ ^- NCO(- N = C = O)$ is linear because it is isoelectronic with $ CO_2$ ]. $ NO_2$ is angular with bond angle $132^{\circ}$ and each $ O- N$ bond length of $1.20$ $\mathrm{A}$ (intermediate between single and double bond).

• NCO⁺: This ion is non-linear. Although the neutral molecule NCO⁻ is linear because it is isoelectronic with CO₂, the positive ion NCO⁺ does not maintain this linear structure.

• NO₂: This molecule is angular with a bond angle of 132° and each O-N bond length of 1.20 Å, which is intermediate between a single and double bond. Therefore, it does not have a linear structure.

Thinking Process

Isoelectronic species are those species have same number of electrons but different nuclear charge.

Answer

$(a, b)$

Electrons present in $\mathrm{CO}=6+8=14$

Then, $\quad$ In $\mathrm{NO}^{+}=7+8-1=14$

In $\quad \mathrm{N}_{2}=7+7=14$

In $\mathrm{SnCl}_{2}=$ (very high) $50+17 \times 2=50+34=84$.

In $\mathrm{NO}_{2}^{-}=7+16+1=24$

• Option (c) $\mathrm{SnCl}_{2}$: The total number of electrons in $\mathrm{SnCl} _{2}$ is 84, which is significantly higher than the 14 electrons in $\mathrm{CO}$. Therefore, $\mathrm{SnCl} _{2}$ is not isoelectronic with $\mathrm{CO}$.

• Option (d) $\mathrm{NO} _{2}^{-}$: The total number of electrons in $\mathrm{NO} _{2}^{-}$ is 24, which is also higher than the 14 electrons in $\mathrm{CO}$. Thus, $\mathrm{NO} _{2}^{-}$ is not isoelectronic with $\mathrm{CO}$.

Answer

(c, $d)$

The shape of following species are

$$ \begin{aligned} CO_{2} & =\text { linear } \\ CCl_{4} & =\text { tetrahedral } \\ O_{3} & =\text { bent } \\ NO_{2}^{-} & =\text {bent } \end{aligned} $$

• (a) $\mathrm{CO}_{2}$: This species is linear because it has a central carbon atom with two double-bonded oxygen atoms, resulting in a bond angle of 180 degrees.

• (b) $\mathrm{CCl}_{4}$: This species is tetrahedral because it has a central carbon atom with four single-bonded chlorine atoms, resulting in bond angles of approximately 109.5 degrees.

• (c) $\mathrm{O}_{3}$: This species is bent because it has a central oxygen atom with one double-bonded oxygen and one single-bonded oxygen, along with a lone pair, resulting in a bond angle of less than 120 degrees.

• (d) $\mathrm{NO}_{2}^{-}$: This species is bent because it has a central nitrogen atom with two oxygen atoms and one lone pair, resulting in a bond angle of less than 120 degrees.

Answer

$(c, d)$

The hybridisation of central atom in $\mathrm{CO}_{3}^{2-}$ is $s p^{2}$. Hence, (a) is wrong.

Due to resonance all $\mathrm{C}-\mathrm{O}$ bond lengths are equal.

Formal charge on each $\mathrm{O}$-atom $=\frac{\text { total charge }}{\text { Number of } \mathrm{O}-\text { atoms }}=\frac{-2}{3}=-0.67$ units.

All $\mathrm{C}-\mathrm{O}$ bond lengths are equal as mentioned above.

• The hybridisation of central atom in $\mathrm{CO}_{3}^{2-}$ is $sp^2$, not $sp^3$. Hence, option (a) is incorrect.

• The resonance structure of $\mathrm{CO}_{3}^{2-}$ does not have one $\mathrm{C}-\mathrm{O}$ single bond and two $\mathrm{C}=\mathrm{O}$ double bonds. Instead, it has three equivalent resonance structures where the double bond character is delocalized over all three $\mathrm{C}-\mathrm{O}$ bonds, making them all equivalent. Hence, option (b) is incorrect.

Answer

$(a, d)$

(a) Electronic configuration of $N_2=\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x ^2 \approx \pi 2 p_y ^2, \sigma 2 p_z ^2$.

It has no unpaired electron indicates diamagnetic species.

(b) Electronic configuration of $N_2^2-$ ion $=\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x ^2 \approx \pi p_y ^2, \sigma 2 p_z^2$,

$$ \pi^* 2 p_x^1 \approx \pi^* 2 p_y^1 $$

It has two unpaired electrons, paramagnetic in nature.

(c) Electronic configuration of $O_2=\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \approx \pi 2 p_y^2$,

$$ \pi^* 2 p_x^1 \approx \pi^* 2 p_y^1 $$

The presence of two unpaired electrons shows its paramagnetic nature.

(d) Electronic configuration of $O_2^2-$ ion $=\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \approx \pi 2 p_y^2$,

$$ \pi^* 2 p_x^2 \approx \pi^* 2 p_y^2 $$

It contains no unpaired electron, therefore, it is diamagnetic in nature.

• (b) $\mathrm{N}_{2}^{2-}$: The electronic configuration of $\mathrm{N}_{2}^{2-}$ is $\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^2 \approx \pi 2 p_y^2, \sigma 2 p_z^2, \pi^* 2 p_x^1 \approx \pi^* 2 p_y^1$. It has two unpaired electrons, which makes it paramagnetic in nature.

• (c) $\mathrm{O}_{2}$: The electronic configuration of $\mathrm{O}_{2}$ is $\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \approx \pi 2 p_y^2, \pi^* 2 p_x^1 \approx \pi^* 2 p_y^1$. The presence of two unpaired electrons shows its paramagnetic nature.

Answer

(c, $d)$

Bond order of the following species are calculated using molecular orbital electronic configuration and found as

$$ \begin{aligned} & N_{2}=3 \\ & ~N_{2}^{-}=2.5 \\ & ~F_{2}^{+}=1.5 \\ & O_{2}^{-}=1.5 \end{aligned} $$

• $\mathrm{N}_{2}$: The bond order of $\mathrm{N}{2}$ is 3, which is different from the bond order of $\mathrm{F}{2}^{+}$ and $\mathrm{O}_{2}^{-}$, both of which have a bond order of 1.5.

• $\mathrm{N}_{2}^{-}$: The bond order of $\mathrm{N}{2}^{-}$ is 2.5, which is different from the bond order of $\mathrm{F}{2}^{+}$ and $\mathrm{O}_{2}^{-}$, both of which have a bond order of 1.5.

Answer

$(a, b)$

(a) $\mathrm{NaCl}$ is a bad conductor of electricity in solid due to the absence of free ions.

(b) Canonical structures differ in the arrangement of electrons, not in the arrangement of atoms.

(c) Hybrid orbitals form stronger bonds than pure orbitals because they have better overlap, leading to more stable and stronger bonds.

(d) VSEPR theory can explain the square planar geometry of $\mathrm{XeF}_{4}$ because it accounts for the repulsion between electron pairs, including lone pairs, which leads to the observed geometry.

Answer

Central atom of $H_{2}$ is $S$. There are 6 electrons in its valence shell $( _{16} ~S =2,8,6)$. Two electrons are shared with two $H$-atoms and the remaining four electrons are present as two lone pairs.

Hence, total pairs of electrons are four (2 bond pairs and 2 lone pairs). Due to the presence of 2 lone pairs the shape becomes distorted tetrahedral or angular or bent (non-linear).

$PCl_{3}-$ Central atom is phosphorus. There are 5 electrons in its valence shell $\left( _{15} P =2,8,5\right)$. Three electrons are shared with three $Cl$-atoms and the remaining two electrons are present as one lone pair.

Hence, total pairs of electrons are four ( 1 lone pair and 3 bond pairs). Due to the presence of one lone pair, the shape becomes pyramidal (non-planar).

Answer

According to molecular orbital theory electronic configurations of $O_{2}^{+}$and $O_{2}^{-}$species are as follows

$ O_{2}^{+}:(\sigma 1 s)^{2}(\sigma^{*} 1 s)^{2}(\sigma 2 s)^{2}(\sigma^{\star} 2 s)^{2}(\sigma 2 p_{z})^{2}(\pi 2 p_{x}^{2}, \pi 2 p_{y}^{2})(\pi * 2 p_{x}^{1})$

Bond order of $\mathrm{O}_{2}^{+}=\frac{10-5}{2}=\frac{5}{2}=2.5$

$ O_{2}^{-}:(\sigma 1 s)^{2}(\sigma^{\star} 1 s^{2})(\sigma 2 s^{2})(\sigma^{\star} 2 s^{2})(\sigma 2 p_{z})^{2}(\pi 2 p_{x}^{2}, \pi 2 p_{y}^{2})(\pi^{\star} 2 p_{x}^{2}, \pi^{*} 2 p_{y}^{1})$

Bond order of $\mathrm{O}_{2}^{-}=\frac{10-7}{2}=\frac{3}{2}=1.5$

Higher bond order of $ O_{2}^{+}$shows that it is more stable than $\mathrm{O}_{2}^{-}$. Both the species have unpaired electrons. So, both are paramagnetic in nature.

Answer

The central atom $\mathrm{Br}$ has seven electrons in the valence shell. Five of these will form bonds with five fluorine atoms and the remaining two electrons are present as one lone pair.

Hence, total pairs of electrons are six (5 bond pairs and 1 lone pair). To minimize repulsion between lone pairs and bond pairs, the shape becomes square pyramidal.

Answer

(a) Compound (I) will form intramolecular $\mathrm{H}$-bonding. Intramolecular $\mathrm{H}$-bonding is formed when $\mathrm{H}$-atom, in between the two highly electronegative atoms, is present within the same molecule. In ortho-nitrophenol (compound I), $\mathrm{H}$-atom is in between the two oxygen atoms.

(I)

Compound (II) forms intermolecular $\mathrm{H}$-bonding. In para-nitrophenol (II) there is a gap between $\mathrm{NO}_{2}$ and $\mathrm{OH}$ group. So, $\mathrm{H}$-bond exists between $\mathrm{H}$-atom of one molecule and $\mathrm{O}$-atom of another molecule as depicted below.

(II)

(b) Compound (II) will have higher melting point because large number of molecules are joined together by $\mathrm{H}$-bonds.

(c) Due to intramolecular $\mathrm{H}$-bonding, compound (I) is not able to form $\mathrm{H}$-bond with water, so it is less soluble in water. While molecules of compound II form $\mathrm{H}$-bonding with $\mathrm{H}_{2} \mathrm{O}$ easily, so it is soluble in water.

Answer

In the figure (I), area of ++ overlap is equal to +- overlap, so net overlap is zero, while in figure (II), there is no overlap due to different symmetry.

Answer

$\mathrm{PCl}_{5}$-The ground state and the excited state outer electronic configurations of phosphorus $(Z=15)$ are represented below

$\mathrm{P}$ (ground state)

P(excited state)

In $PCl_{5}, P $ is $s p^{3} d$ hybridised, therefore, its shape is trigonal bipyramidal.

$\mathrm{IF}_{5}-$ The ground state and the excited state outer electronic configurations of iodine $(Z=53)$ are represented below.

In $IF_{5}$, I is $s p^{3} d^{2}$ hybridised, therefore, shape of $ IF_{5}$ is square pyramidal.

Answer

Dimethyl ether has greater bond angle than that of water, however in both the molecules central atom oxygen is $s p^{3}$ hybridised with two lone pairs. In dimethyl ether, bond angle is greater $\left(111.7^{\circ}\right)$ due to the greater repulsive interaction between the two bulky alkyl (methyl) groups than that between two $\mathrm{H}$-atoms.

Dimethyl ether

Actually $\mathrm{C}$ of $\mathrm{CH}_{3}$ group is attached to three $\mathrm{H}$-atoms through $\sigma$-bonds. These three $\mathrm{C}-\mathrm{H}$ bond pair of electrons increases the electronic charge density on carbon atom.

Answer

The Lewis structure of the following compounds and formal charge on each atom are as

(i) $HNO_3$

Formal charge on an atom in a Lewis structure

$=$ [total number of valence electrons in free atom]

-[total number of non-bonding (lone pairs) electrons]

$-\frac{1}{2}$ [total number of bonding or shared electrons]

Formal charge on $\mathrm{H}=1-0-\frac{1}{2} \times 2=0$

Formal charge on $\mathrm{N}=5-0-\frac{1}{2} \times 8=1$

Formal charge on $\mathrm{O}(1)=6-4-\frac{1}{2} \times 4=0$

Formal charge on $\mathrm{O}(2)=6-4-\frac{1}{2} \times 4=0$

Formal charge on $\mathrm{O}(3)=6-6-\frac{1}{2} \times 2=-1$

(ii) $NO_{2}$

Formal charge on $\mathrm{O}(1)=6-4-\frac{1}{2} \times 4=0$

Formal charge on $\mathrm{N}=5-1-\frac{1}{2} \times 6=+1$

Formal charge on $\mathrm{O}(2)=6-6-\frac{1}{2} \times 2=-1$

(iii) $H_{2} SO_{4}$

Formal charge on $\mathrm{H}(1)$ or $\mathrm{H}(2)=1-0-\frac{1}{2} \times 2=0$

Formal charge on $\mathrm{O}(1)$ or $\mathrm{O}(3)=6-4-\frac{1}{2} \times 4=0$

Formal charge on $\mathrm{O}(2)$ or $\mathrm{O}(4)=6-6-\frac{1}{2} \times 2=-1$

Formal charge on $\mathrm{S}=6-0-\frac{1}{2} \times 8=+2$

Answer

Electronic configuration of $\mathrm{N}$-atom $(Z=7)$ is $1 s^{2} 2 s^{2} 2 p_{x}^{1} 2 p_{y}^{1} 2 p_{z}^{1}$. Total number of electrons present in $N_{2}$ molecule is 14,7 from each $N$-atom. From the view of various rules for filling of molecular orbitals, the electronic configuration of $N_{2}$ molecule will be

$$ \sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^2 \approx \pi 2 p_y^2, \sigma 2 p_z^2 $$

Comparative study of the relative stability and the magnetic behaviour of the following species

(i) $N_{2}$ molecule $\sigma 1 s^2, \sigma^* s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi p_x^2 \approx \pi 2 p_y^2, \sigma 2 p_z^2$

Here, $N_{b}=10, N_{a}=4$.

Hence, Bond order $=\frac{1}{2}\left(N_{b}-N_{a}\right)=\frac{1}{2}(10-4)=3$

Hence, presence of no unpaired electron indicates it to be diamagnetic.

(ii) $N_2^+$ions $\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^2 \approx \pi 2 p_y^2, \sigma 2 p_z^1$

Here, $N_{b}=9, N_{a}=4$ so that $\mathrm{BO}=\frac{1}{2}(9-4)=\frac{5}{2}=2.5$

Further, as $N_2^+$ion has one unpaired electron in the $\sigma\left(2 p_{2}\right)$ orbital, therefore, it is paramagnetic in nature.

(iii) $N_2^-$ions $\sigma 1 s^2, \sigma^* s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^2 \approx \pi 2 p_y^2, \sigma 2 p_z^2, \pi^* 2 p_x^1$

Here, $N_{b}=10, N_{a}=5$ so that $\mathrm{BO}=\frac{1}{2}(10-5)=\frac{5}{2}=2.5$

Again, as it has one unpaired electron in the $\pi^{*}\left(2 p_{x}\right)$ orbital, therefore, it is paramagnetic.

(iv) $N_2^{2+}$ ions $\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^2 \approx \pi 2 p_y^2$

Here, $N_{b}=8, N_{a}=4$. Hence, $B O=\frac{1}{2}(8-4)=2$

Presence of no unpaired electron indicates it to be diamagnetic in nature.

As bond dissociation energies are directly proportional to the bond orders, therefore, the dissociation energies of these molecular species in the order.

$$ N_{2}>N_{2}^{-} = N_{2}^{+}> N_{2}^{2+} $$

As greater the bond dissociation energy, greater is the stability, the stability of these species is also in the above order.

Answer

According to molecular orbital theory, electronic configurations and bond order of $N_{2}, N_{2}^{+}, O_{2}$ and $O_{2}^{+}$species are as follows

$$ N_2 (14 e^-) =\sigma 1 s^2, \stackrel{\star}{\sigma} 1 s^2, \sigma 2 s^2, \stackrel{\star}{\sigma} 2 s^2,(\pi 2 p_x^2 \approx \pi 2 p_y^2), \sigma 2 p_z^2 $$

$$ \text { Bond order } =\frac{1}{2}[N_{b}-N_{a}]=\frac{1}{2}(10-4)=3 $$

$$ ~N_{2}^{+}(13 e^{-}) =\sigma 1 s^{2}, \stackrel{*}{\sigma} 1 s^{2}, \sigma 2 s^{2}, \stackrel{\star}{\sigma} 2 s^{2},(\pi 2 p_{x}^{2} \approx \pi 2 p_{y}^{2}) \sigma 2 p_{z}^{1} $$

$$ \text { Bond order } =\frac{1}{2}[N_{b}-N_{a}]=\frac{1}{2}(9-4)=2.5 $$

$$ O_2 (16 e^-) =\sigma 1 s^2, \stackrel*{\sigma} 1 s^2, \sigma 2 s^2, \stackrel{\star}{\sigma} 2 s^2, \sigma 2 p_z^2,(\pi 2 p_x^2 \approx \pi 2 p_y^2),(\stackrel* \pi 2 p_x^1 \approx * \frac{\star}{\pi} 2 p_y^1) $$

$$ \text { Bond order } =\frac{1}{2}[N_{b}-N_{a}]=\frac{1}{2}(10-6)=2 $$

$$ O_2^+ (15 e^-) =\sigma 1 s^2, \stackrel*{\sigma} 1 s^2, \sigma 2 s^2, \stackrel*{\sigma} 2 s^2, \sigma 2 p_z^2,(\pi 2 p_x^2 \approx \pi 2 p_y^2),(\stackrel \star \pi 2 p_x^1 \approx^* \pi 2 p_y) $$

$$ \text { Bond order } =\frac{1}{2}[N_b - N_a]=\frac{1}{2}(10-5)=2.5 $$

(a) $\underset{\text { B.O. }=3}{N_2} \longrightarrow \underset{\text { B.O. }=2.5}{ N_2^+} +e^-$

Thus, bond order decreases.

(b) $\underset{\text { B.O. }=2}{O_2} \longrightarrow \underset{\text { B.O. }=2.5}{ O_2^+} +e^-$

Thus, bond order increases.

Answer

(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 $\mathrm{lp}-\mathrm{lp}$ and the molecule thus acquires a $\mathrm{V}$-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 $H_2 O $ molecule has bent structure whereas $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.

Answer

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 $\mathrm{NaCl}$ from sodium and chlorine can be explained as

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

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

Similarly, the formation of $\mathrm{CaF}_{2}$ may be shown as

$$ \begin{aligned} & \underset{[\mathrm{Ar}] 4 \mathrm{~s}^{2}}{\mathrm{Ca}} \longrightarrow \underset{[\mathrm{Ar}]}{\mathrm{Ca}^{2+}}+2 \mathrm{e}^{-} \end{aligned} $$

$$ \begin{aligned} & Ca^{2+}+2 F^- \longrightarrow CaF_2 \text { or } Ca^{2+}(F^-)_{2} \end{aligned} $$

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 $\mathrm{HCl}$

Answer

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

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

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

Answer

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 $\mathrm{C}-\mathrm{O}$ single bonds but actually all bonds are found to be identical with same bond length and same bond strength.

Answer

The hybridisation and type of bonds of each carbon in the molecule given below

Answer

The structure of the given molecules are

Therefore, only $BeCl_2$ is linear and rest of the molecules are non-linear

Answer (a)

(b) $Z$ has seven electrons in its valence shell. It is the most electronegative element. Therefore, $\mathrm{HZ}$ will have the highest dipole moment.

Answer

(a) The resonating structure of ozone molecule may be written as

(b) The resonating structure of nitrate ion ($NO_3^-$) is

Answer

The shape of $\mathrm{CH}_{4}$ is tetrahedral due to $s p^{3}$ hybridisation.

$\mathrm{CO}_{2}$ show linear shape because of $s p$ hybridisation.

$$: \ddot O=C=\ddot O:$$

The geometry of $\mathrm{NH}_{3}$ is pyramidal shape and has $s p^{3}$ hybridisation.

Ammonia, $\mathrm{NH}_{3}$

Thinking Process

To explain the reason of equal in length of $\mathrm{C}-\mathrm{O}$ bonds, it should keep in mind about the resonance. As a result of resonance, the bond length in a molecule become equal.

Answer

Carbonate ion $\left(\mathrm{CO}_{3}^{2-}\right)=3$ bond pair +1 lone pair $\Rightarrow$ trigonal planar

Due to resonance all $\mathrm{C}-\mathrm{O}$ bond length are equal.

Answer

All the similar bonds in a molecule do not have the same bond enthalpies. e.g., in $\mathrm{H}_{2} \mathrm{O}(\mathrm{H}-\mathrm{O}-\mathrm{H})$ molecule after the breaking of first $\mathrm{O}-\mathrm{H}$ bond, the second $\mathrm{O}-\mathrm{H}$ bond undergoes some chanae because of chanaed chemical environment.

Therefore, in polyatomic molecules the term mean or average bond enthalpy is used. It is obtained by dividing total bond dissociation enthalpy by the number of bonds broken.

$$e.g., \quad\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{H}(\mathrm{g})+\mathrm{OH}(\mathrm{g});$$

$$\Delta_{a} H_{1}^{\circ}=52 \mathrm{~kJ} \mathrm{~mol}^{-1} \mathrm{OH}(g) \longrightarrow \mathrm{H}+\mathrm{O}(\mathrm{g});$$

$$ \Delta_{a} H_{2}^{\circ}=427 \mathrm{~kJ} \mathrm{~mol}^{-1} $$

Average $\mathrm{O}-\mathrm{H}$ bond enthalpy $=\frac{502+427}{2}=464.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$

The bond enthalpies of $O-H$ bond in $C_{2} H_{5} OH$ and $H_{2} O$ are different because of the different chemical (electronic) environment around oxygen atom.

Answer

A. $\rightarrow(3)$

B. $\rightarrow(1)$

C. $\rightarrow(5)$

D. $\rightarrow(4)$

A. $\mathrm{SF}_{4}=$ number of $b p(4)+$ number of $lp(1)$

$=s p^{3} d$ hybridisation

B. $\mathrm{IF}_{5}=$ number of $b p(5)+$ number of $lp(1)$

$=s p^{3} d^{2}$ hybridisation

C. $\mathrm{NO}_{2}^{+}=$number of $b p(2)+$ number of $lp(0)$

$=s p$ hybridisation

D. $\mathrm{NH}_{4}^{+}=$number of $b p(4)+$ number of $lp(0)$

$=s p^{3}$ hybridisation.

Answer

A. $\rightarrow(5)$

B. $\rightarrow(1)$

C. $\rightarrow(2)$

$\mathrm{D} \rightarrow(3)$

A. $\mathrm{H}_{3} \mathrm{O}^{+}=3 b p+1 lp$ pyramidal shape

B. $\mathrm{HC} \equiv \mathrm{CH} \Rightarrow$ linear as sphybridised shape

C. $\mathrm{ClO}_{2}^{-}=2 \mathrm{bp}+2 lp \Rightarrow$ angular shape

D. $\mathrm{NH}_{4}^{+}=4 b p+0 lp \Rightarrow$ tetrahedral shape

Answer

A. $\rightarrow(3)$

B. $\rightarrow(4)$

C. $\rightarrow(1)$

D. $\rightarrow(2)$

A. $\mathrm{NO}(7+8=15)=\sigma 1 s^{2}, \sigma^{\star} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \sigma 2 p_{z}^{2}, \pi 2 p_{x}^{2} \approx \pi 2 p_{y}^{2}, \pi \star 2 p_{x}^{1}$

Bond order $=\frac{1}{2}\left(N_{b}-N_{a}\right)=\frac{10-5}{2}=2.5$

B. $CO(6+8=14)=\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \simeq \pi 2 p_y^2$

Bond order $=\frac{10-4}{2}=3$

C. $O_2^-(8+8+1=17)=\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^{\star} 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \approx \pi 2 p_y^2, \pi * 2 p_x^2 \approx \pi * 2 p_y^1$

Bond order $=\frac{10-7}{2}=1.5$

D. $O_2(8+8=16)=\sigma 1 s^2, \sigma^{\star} 1 s^2, \sigma 2 s^2, \sigma^{\star} 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \approx \pi 2 p_y^2, \pi * 2 p_x^1 \approx \pi * 2 p_y^1$

Bond order $=\frac{10-6}{2}=2$

Answer

A. $\rightarrow(4)$

B. $\rightarrow(5)$

C. $\rightarrow(2)$

D. $\rightarrow(1)$

A. Hydrogen bond $\rightarrow \mathrm{HF}$

B. Resonance $\rightarrow \mathrm{O}_{3}$

C. Ionic bond $\rightarrow$ LiF

D. Covalent solid $\rightarrow \mathrm{C}$

Answer

A. $\rightarrow(3)$

B. $\rightarrow(1)$

C. $\rightarrow(2)$

A. Tetrahedral shape $-s p^{3}$ hybridisation

B. Trigonal shape $-s p^{2}$ hybridisation

C. Linear shape - sp hybridisation

Assertion and Reason

In the following questions, a statement of Assertion (A) followed by a statement of Reason $(\mathrm{R})$ is given. Choose the correct option out of the choices given below in each question.

Answer

(a) Assertion and reason both are correct and reason is the correct explanation of assertion

$$\underset{(2,8,1)}{Na} + \underset{(2,8,7)}{Cl} \longrightarrow \underset{\underset{(2,8,8)}{(2,8,8)}}{NaCl}$$

Here both $\mathrm{Na}^{+}$and $\mathrm{Cl}^{-}$have complete octet hence $\mathrm{NaCl}$ is stable.

Answer

(a) Assertion and reason both are correct and reason is the correct explanation of assertion.

$sp^3 -hybridised \quad \quad \quad sp^3 -hybridised$

Answer

(d) Correct assertion The bond enthalpies of the two $\mathrm{O}-\mathrm{H}$ bonds in $\mathrm{H}-\mathrm{O}-\mathrm{H}$ are not equal.

Correct reason This is because electronic environment around $\mathrm{O}$ is not same after breakage of one $\mathrm{O}-\mathrm{H}$ bond.

Answer

(a) The applications of dipole moment are

(i) The dipole moment helps to predict whether a molecule is polar or non-polar. As $\mu=q \times d$, greater is the magnitude of dipole moment, higher will be the polarity of the bond. For non-polar molecules, the dipole moment is zero.

(ii) The percentage of ionic character can be calculated as

Percentage of ionic character $=\frac{\mu_{\text {observed }}}{\mu_{\text {ionic }}} \times 100$

(iii) Symmetrical molecules have zero dipole moment although they have two or more polar bonds (in determination of symmetry).

(iv) It helps to distinguish between cis and trans isomers. Usually cis-isomer has higher dipole moment than trans isomer.

(v) It helps to distinguish between ortho, meta and para isomers. Dipole moment of para isomer is zero. Dipole moment of ortho isomer is greater than that of meta isomer.

(b)

Answer

Formation of $N_2$ molecule Electronic configuration of $N$ - atom $ _7 N=1 s^2, 2 s^2, 2 p_x^1, 2 p_y^1, 2 p_z^1$ $N_2$ molecule $=\sigma 1 s^2, \sigma^\star 1 s^2, \sigma 2 s^2, \sigma^\star 2 s^2, \pi 2 p_x^2 \approx \pi 2 p_y^2, \sigma 2 p_z^2$

Bond order $=\frac{1}{2}\left[N_{b}-N_{a}\right]=\frac{1}{2}(10-4)=3$.

Bond order value of 3 means that $\mathrm{N}_{2}$ contains a triple bond.

Formation of $F_2$ molecule, $ _9 F =1 s^2, 2 s^2, 2 p_x^2, 2 p_y^2, 2 p_z^1$

$F_2$ molecule $=\sigma 1 s^2, \sigma^\star 1 s^2, \sigma 2 s^2, \sigma^\star 2 s^2, \sigma 2 p_z ^2, \pi 2 p_x ^2 \approx \pi 2 p_y ^2, \pi * 2 p^2 _x \approx \pi * 2 p^2 _y$

Bond order $=\frac{1}{2}\left[N_{b}-N_{a}\right]=\frac{1}{2}(10-8)=1$

Bond order value 1 means that $\mathrm{F}_{2}$ contains single bond.

Formation of $Ne_{2}$ molecule $ _10 Ne=1 s^2, 2 s^2, 2 p_x^2, 2 p_y^2, 2 p_z^2$

$Ne_2$ molecule $=\sigma 1 s^2, \sigma^\star 1 s^2, \sigma 2 s^2, \sigma^\star 2 s^2, \sigma 2 p_z ^2, \pi 2 p_x ^2 \approx \pi 2 p_y ^2, \pi * 2 p^2 x$

$\approx \pi * 2 p^{2} y, \sigma^{*} 2 p_{z}^{2}$

Molecular orbitals of $\mathrm{Ne}$, molecule

Bond order $=\frac{1}{2}\left[N_{b}-N_{a}\right]=\frac{1}{2}(10-10)=0$

Bond order value zero means that there is no formation of bond between two $\mathrm{Ne}$-atoms.

Hence, $\mathrm{Ne}_{2}$ molecule does not exist.

Answer

Valence bond theory (VBT) was introduced by Heitler and London (1927) and developed further by Pauling and other. VBT is based on the knowledge of atomic orbitals, electronic configurations of elements, the overlap criteria of atomic orbitals, the hybridisation of atomic orbitals and the principles of variation and superposition.

Consider two hydrogen atoms $A$ and $B$ approaching each other having nuclei $N_{A}$ and $N_{B}$ and electrons present in them are represented by $e_{A}$ and $e_{B}$. When the two atoms are at large distance from each other, there is no interaction between them.

As these two atoms approach each other, new attractive and repulsive forces begin to operate.

Attractive forces arise between

(i) nucleus of one atom and its own electron

$$ \text { i.e., } \quad H_{A}-e_{A} \text { and } H_{B}-e_{B} $$

(ii) nucleus of one atom and electron of other atom

$$ \text { i.e., } \quad H_{A}-e_{B}, H_{B}-e_{A} $$

Similarly, repulsive forces arise between

(i) electrons of two atoms like $e_{A}-e_{B}$

(ii) nuclei of two atoms like $H_{A}-H_{B}$

Attractive forces tend to bring the two atoms close to each other whereas repulsive forces tend to push them apart.

Experimentally, we have been found that the magnitude of new attractive force is more than the new repulsive forces. As a result two atoms approach each other and potential energy decreases.

Hence, a stage is reached where the net force of attraction balances the force of repulsion and system acquires minimum energy. At this stage, two $\mathrm{H}$-atoms are said to be bonded together to form a stable molecule having the bond length of $74 \mathrm{pm}$.

Attractive forces

Repulsive forces

Since, the energy gets released when the bond is formed between two hydrogen atoms, the hydrogen molecule is more stable than that of isolated hydrogen atoms.

The energy so released is called as bond enthalpy, which is corresponding to minimum in the curve depicted in the given figure. Conversely $435.8 \mathrm{~kJ}$ of energy is required to dissociate one mole of $\mathrm{H}_{2}$ molecule.

The potential energy curve for the formation of $H_{2}$ molecule as a function of internuclear distance of the $H$-atoms. The minimum in the curve corresponds to the most stable state or $H_{2}$.

Answer

Formation of $\mathrm{PCl}_{5}$

Electronic configuration of ${ }_{15} \mathrm{P}$ (ground state)

Electronic configuration of ${ }_{15} \mathrm{P}$ (excited state)

$s p^{3} d$ hybridisation

In $\mathrm{PCl}_{5}$, phosphorus is $s p^{3} d$ hybridised to produce a set of five $s p^{3} d$ hybrid orbitals which are directed towards the five corners of a trigonal bipyramidal. These five $s p^{3} d$ hybrid orbitals overlap with singly occupied $p$-orbitals of $\mathrm{Cl}$-atoms to form five $\mathrm{P}-\mathrm{Cl}$ sigma bonds.

(Trigonal bipyramidal)

$\mathrm{PCl}_{5}$

Three $\mathrm{P}-\mathrm{Cl}$ bonds lie in one plane and make an angle of $120^{\circ}$ with each other. These bonds are called equatorial bonds. The remaining two $\mathrm{P}-\mathrm{Cl}$ bonds one lying above and other lying below the plane make an angle of $90^{\circ}$ with the equatorial plane.

These bonds are called axial bonds. Axial bonds are slightly longer than equatorial bonds because axial bond pairs suffer more repulsive interaction from the equatorial bond pairs.

Formation of $\mathrm{SF}_{6}$

Electronic configuration of ${ }_{16} \mathrm{~S}$ (ground state)

S(excited state)

In $\mathrm{SF}_{6}$, sulphur is $s p^{3} d^{2}$ hybridised to produce a set of six $s p^{3} d^{2}$ hybrid orbitals which are directed towards the six corners of a regular octahedron. These six $s p^{3} d^{2}$ hybrid orbitals overlap with singly occupied orbitals of fluorine atoms to form six S-F sigma bonds.

Thus, $\mathrm{SF}_{6}$ molecule has a regular octahedral geometry and all $\mathrm{S}-\mathrm{F}$ bonds have same bond length.

Answer

Hybridisation It can be defined as the process of intermixing of the orbitals of slightly different energy or of same energy to produce entirely new orbitals of equivalent energy, identical shapes and symmetrically disposed in plane. New orbitals formed are called hybrid orbitals.

Only the orbitals of an isolated single atom can undergo hybridisation. The hybrid orbitals generated are equal in number to that of the pure atomic orbitals which mix up.

Hybrid orbitals do not make $\pi$, pi-bonds. If there are $\pi$-bonds, equal number of atomic orbitals must be left unhybridised for $\pi$-bonding.

Like atomic orbitals, hybrid orbitals cannot have more than two electrons of opposite spins. Types of hybridisation in carbon atoms

(a) (i) Diagonal or sp-hybridisation All compounds of carbon containing $C \equiv C$ triple bond like ethyne $\left(C_{2} H_{2}\right)$.

(ii) Trigonal or $sp^{2}$-hybridisation All compounds of carbon containing $C=C$ (double bond) like ethene $\left(C_{2} H_{4}\right)$

(iii) Tetrahedral or $sp^{3}$-hybridisation All compounds of carbon containing $C-C$ single bonds only like ethane $\left(C_{2} H_{6}\right)$.

Molecular orbitals are formed by the overlap of atomic orbitals. Two atomic orbitals combine to form two molecular orbitals called bonding molecular orbital (BMO) and anti-bonding molecular orbital (ABMO). Energy of anti-bonding orbital is raised above the parent atomic orbitals that have combined and the energy of the bonding orbital is lowered than the parent atomic orbitals.

Energies of various molecular orbitals for elements hydrogen to nitrogen increase in the order

$$ \sigma 1 s<\sigma^* 1 s<\sigma 2 s<\sigma^* 2 s<(\pi 2 p_x \approx \pi 2 p_y)<\sigma 2 p_z< (\pi^* 2 p_{x} \approx \pi^* 2 p_y)<\sigma^* 2 p_z \text { and } $$

For oxygen and fluorine order of energy of molecular orbitals is given below

$$ \sigma 1 s<\sigma^* 1 s<\sigma 2 s<\sigma^* 2 s<\sigma p_z<\left(\pi 2 p_x \approx \pi 2 p_y \right)<\left(\pi^* 2 p_x \approx \pi^* 2 p_y\right)<\sigma^* 2 p_z $$

Different atomic orbitals of one atom combine with those atomic orbitals of the second atom which have comparable energies and proper orientation.

Further, if the overlapping is head on, the molecular orbital is called ‘sigma’, $(\sigma)$ and if the overlap is lateral, the molecular orbital is called ‘pi’, $(\pi)$. The molecular orbitals are filled with electrons according to the same rules as followed for filling of atomic orbitals.

However, the order for filling is not the same for all molecules or their ions. Bond order is one of the most important parameters to compare the strength of bonds.

Answer

(a) In the formation of dioxygen from oxygen atoms, ten molecular orbitals will be formed.

$$ O_{2}=\frac{\sigma 1 s^2}{1} \frac{\sigma^* 1 s^2}{2} \frac{\sigma 2 s^2}{3} \frac{\sigma^* 2 s^2}{4} \frac{\sigma_{2} p_z^2}{5} \frac{\pi 2 p_x^2}{6} \frac{\pi 2 p_y^2}{7} \frac{\pi^* 2 p_x^1}{8} \frac{\pi^* 2 p_y^1}{9} \frac{\sigma^* 2 p_z^0}{10} $$

Answer

(d) Nodal plane are $\sigma^* 1 s=1, \sigma^* 2 p_z=1, \pi 2 p_x=1, \pi^* 2 p_y=2$

$\quad\quad \quad\quad$ 1s $ \quad\quad \quad\quad \quad \quad \quad\quad $ 1s

By subtraction

Anti-bonding molecular orbital

Answer

(b) On the basec of molecular orbetal therory we can calculate bond order of molecules ions as

$$ \mathrm{BO}=\frac{1}{2}\left(N_{b}-N_{a}\right) $$

Molecular orbital electronic configuration (MOEC) of $\mathrm{N}_{2}$ is

$$ \sigma 1 s^{2}, \sigma^{\star} 1 s^{2}, \sigma 2 s^{2}, \sigma^{\star} 2 s^{2}, \pi 2 p_{x}^{2} \simeq \pi 2 p_{y}^{2}, \sigma 2 p_{x}^{2} $$

Bond order of $\mathrm{N}_{2}=\frac{1}{2}(10-4)=3$

MO electronic configuration of $N_2 ^+=\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^2 \approx \pi 2 p_y^2, \sigma 2 p_2$

$\mathrm{BO}$ of $\mathrm{N}_{2}^{+}=\frac{1}{2}(9-4)=2.5$

MO electronic configuration of $N_2 ^-=\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^2 \simeq \pi 2 p_y^2, \sigma 2 p_z^2, \pi * 2 p_x \simeq^1 \approx * 2 p_y$

$\mathrm{BO}$ of $\mathrm{N}_{2}^{-}=\frac{1}{2}(10-5)=2.5$

MO electronic configuration of $O_2=\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \simeq \pi 2 p_y^2, \pi * 2 p_x^1 \simeq \pi * 2 p_y^1$

$\mathrm{BO}$ of $\mathrm{O}_{2}=\frac{1}{2}(10-6)=2$

$$ \text {MO electronic configuration of } O_2^-=\sigma 1 s^2, \sigma^\star 1 s^2, \sigma 2 s^2, \sigma^\star 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2=\pi 2 p_y^2, \pi * 2 p_x^2 \simeq \pi * 2 p_y^1 $$

$$ \text { BO of } O_2^-=\frac12(10-7)=1.5 $$

$$ \text {MO electronic configuration of } O_{2}^{+}=\sigma 1 s^{2}, \sigma^{\star} 1 s^{2}, \sigma 2 s^{2}, \sigma^{\star} 2 s^{2}, \sigma 2 p_{z}^{2}, \pi 2 p_{x}^{2} \simeq \pi 2 p_{y}^{2}, \pi * 2 p_{x}^{2} \simeq \pi * 2 p_{y} $$

$$ \text { BO of } O_{2}^{+}=\frac{1}{2}(10-5)=2.5 $$

(a) Bond order of $O_{2}$ and $N_{2}$ are 2 and 3 , respectively.

(b) Bond order of both $O_{2}^{+}$and $N_{2}^{-}$are 2.5.

(c) Bond order of $O_{2}^{-}$and $N_{2}^{+}$are 1.5 and 2.5 , respectively.

(d) Bond order of $O_{2}^{-}$and $N_{2}^{-}$are 1.5 and 2.5 respectively.

Answer

(c) Total number of electrons present in $\mathrm{N}_{2}$ molecule is 14 .

The electronic configuration of $\mathrm{N}_{2}$ molecule will be

$$ \sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \pi 2 p_x^2 \approx \pi 2 p_y^2 \sigma 2 p_z^2 $$

Note The increasing order of energies of various molecular orbitals for $O_2$ and $F_2$ is given below $01 s<\sigma^* 1 s<\sigma 2 s<\sigma^* 2 s<\sigma 2 p_z<\left(\pi 2 p_x \approx \pi 2 p_y\right)<\left(\pi^* 2 p_x \approx \pi^* 2 p_y\right)<\sigma^* 2 p_z$

However, this sequence of energy levels of $MO$ is not correct for the remaining molecules such as $Li_{2}, Be_{2}, ~B_{2}, C_{2}$ and $N_{2}$. For these molecules, the increasing order of energies of various $MO$ is

$\sigma 1 s<\sigma^* 1 s<\sigma 2 s<\sigma^* 2 s<\left(\pi 2 p_x \approx \pi 2 p_y\right)<\sigma 2 p_z<\left(\pi^* 2 p_x \approx \pi^* 2 p_y\right)<\sigma^* 2 p_z$