Answer

(b)

(a) $N{F}_3$ is pyramidal whereas $B{F}_3$ is planar triangular.

(b) $B{F}_4^{-}$and $N{H}_4^{+}$ion both are tetrahedral and $s{p}^3$ hybridisation.

(c) $BC{l}_3$ is triangular planar whereas $BrC{l}_3$ is $T$ shaped.

(d) $N{H}_3$ is pyramidal whereas ${\mathrm{NO}}_3^{-}$is triangular planar.

• (a) $N{F}_3$ is pyramidal whereas $B{F}_3$ is planar triangular.
• (c) $BC{l}_3$ is triangular planar whereas $BrC{l}_3$ is T-shaped.
• (d) $N{H}_3$ is pyramidal whereas $N{O}_3^{-}$ is triangular planar.

Answer

(c) ${\mathrm{CO}}_2$ being symmetrical has zero dipole moment

$$\begin{array}{c}\mathrm{O}\leftrightarrows \mathrm{C}\overleftarrow{=}\mathrm{O}\\ \mu =0\end{array}$$

Among $HI,S{O}_2$ and $H_{2}O$ dipole moment is highest for $H_{2}O$ as in it the central atom contains 2 lone pairs.

$$\begin{array}{c}\mathrm{H}\overrightarrow{⟶}\mathrm{I}\\ \mu =0.38\mathrm{D}\end{array}$$

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

• (a) ${\mathrm{CO}}_2$: ${\mathrm{CO}}_2$ is a linear molecule with a symmetrical structure. The dipole moments of the two $\mathrm{C}=\mathrm{O}$ 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 $bp$ and as $1p$ in

${\mathrm{NO}}_2^{+}=2\mathrm{bp}+0lp=$ linear $=sp$ hybridised

${\mathrm{NO}}_3^{-}=3bp+0lp\Rightarrow s{p}^2$ hybridised

${\mathrm{NH}}_4^{+}=4\mathrm{bp}+0lp\Rightarrow s{p}^3$ hybridised

• Option (a) is incorrect because ${\mathrm{NO}}_3^{-}$ is $s{p}^2$ hybridized, not $s{p}^3$.
• Option (c) is incorrect because ${\mathrm{NO}}_2^{+}$ is $sp$ hybridized, not $s{p}^2$.
• Option (d) is incorrect because ${\mathrm{NO}}_3^{-}$ is $s{p}^2$ hybridized, not $s{p}^3$, and ${\mathrm{NH}}_4^{+}$ is $s{p}^3$ hybridized, not $sp$.

Answer

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

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 $N{H}_3$

• Option (a) $HF>H_{2}O>N{H}_3$:

  • 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) $N{H}_3>HF>H_{2}O$:

  • 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) $N{H}_3>H_{2}O>HF$:

  • 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

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

It has one unpaired electron.

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

${\mathrm{O}}_2$ has two unpaired electrons.

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

Thus, ${\mathrm{O}}_2^{2-}$ has no unpaired electrons.

$B_2=\sigma 1s^2,{\sigma }^{\star }1s^2,\sigma 2s^2,{\sigma }^{\star }2s^2,\pi 2p_x^1\approx {\pi }_2{p}_y^1$

Thus, ${\mathrm{B}}_2$ has two unpaired electrons.

• (a) ${\mathrm{N}}_2^{+}$: The electronic configuration of ${\mathrm{N}}_2^{+}$ is $\sigma 1s^2,{\sigma }^{\star }1s^2,\sigma 2s^2,{\sigma }^{\star }2s^2,\pi 2p_x^2=\pi p_y^2,\sigma 2p_z^1$. It has one unpaired electron, making it incorrect.

• (b) ${\mathrm{O}}_2$: The electronic configuration of ${\mathrm{O}}_2$ is $\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\sigma 2p_z^2,\pi 2p_x^2\approx \pi 2p_y^2,\pi \ast 2p_x^1\approx \pi \ast 2p_y^1$. It has two unpaired electrons, making it incorrect.

• (d) ${\mathrm{B}}_2$: The electronic configuration of ${\mathrm{B}}_2$ is $\sigma 1s^2,{\sigma }^{\star }1s^2,\sigma 2s^2,{\sigma }^{\star }2s^2,\pi 2p_x^1\approx {\pi }_2{p}_y^1$. It has two unpaired electrons, making it incorrect.

Answer

(c) ${\mathrm{XeF}}_4\Rightarrow 4bp+2lp\Rightarrow$ square planar $\Rightarrow$ all bonds are equal

${\mathrm{BF}}_4^{-}\Rightarrow 4\mathrm{bp}+0lp\Rightarrow$ 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

${\mathrm{SiF}}_4\Rightarrow 4bp+0lp\Rightarrow$ 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 $\mathrm{Xe}-\mathrm{F}$ 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 $\mathrm{B}-\mathrm{F}$ 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 $\mathrm{Si}-\mathrm{F}$ 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 $ns$ and $(n-1)d$ subshell take part in bond formation.

• (a) $3p^6$: The $3p^6$ 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 $4s$ and $3d$ electrons.

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

• (c) $3p^6,3d^2$: Similar to the previous options, the $3p^6$ electrons are part of a filled inner shell and do not participate in bonding. The $3d^2$ electrons can be involved in bonding, but the $3p^6$ 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 $ds{p}^2$ or $d^{2}s{p}^3$ hybridisation, not $s{p}^2$ hybridisation. In $s{p}^2$ 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 $s{p}^2$ hybridisation, the bond angles are approximately ${120}^{\circ }$ due to the trigonal planar geometry.

• (d) ${109}^{\circ }$: This angle is characteristic of $s{p}^3$ hybridisation, where the geometry is tetrahedral. In $s{p}^2$ 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 $(C{l}_2)$ 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 $PC{l}_3$ i.e., $B{C}_3$.

• 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 $PC{l}_3$ or $PC{l}_5$, not a compound with a 2:1 ratio.

• Option (c) $B{C}_2$: 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, $B{C}_2$ 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

$\sigma 1s<{\sigma }^{\ast }1s<\sigma 2s<{\sigma }^{\ast }2s<(\pi 2p_x\approx \pi 2p_y)<\sigma 2p_z<({\pi }^{\ast }2p_x\approx {\pi }^{\ast }2p_y)<{\sigma }^{\ast }2p_z$

• 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) $B{e}_2(4+4=8)=\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 1s^2,{\sigma }^{\ast }2s^2$

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) ${\mathrm{He}}_2(2+2=4)=\sigma 1s^2,{\sigma }^{\ast }1s^2$

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

Here, bond order of ${\mathrm{Be}}_2$ is zero. Hence, it does not exist.

$$\begin{array}{c}{\mathrm{He}}_2^{+}(2+2-1=3)=\sigma 1s^2,{\sigma }^{\ast }1s^1\\ \mathrm{BO}=\frac{2-1}{2}=0.5\end{array}$$

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

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

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

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

$\sigma 2s<{\sigma }^{\star }2s<(\pi 2p_x\simeq \pi 2p_y)<\sigma 2p_z<{\pi }^{\star }2p_x\approx {\pi }^{\star }2p_y<{\sigma }^{\star }2p_z$

• (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) $H{e}_2$ is not stable but $H{e}_2^{+}$ is expected to exist: This statement is correct. The bond order of $H{e}_2$ is zero, indicating that it does not exist as a stable molecule. However, $H{e}_2^{+}$ 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 $=\frac{1}{2}[$ Number of bonding electrons $\left(N_b\right)-$ Number of anti-bonding electrons

$\left(N_a\right)]$

Answer

(b) Electronic configuration of ${\mathrm{O}}_2$ (16 electrons)

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

Bond order $=\frac{1}{2}(N_b-N_a)=\frac{1}{2}(10-6)=2$

Electronic confiquration of ${\mathrm{O}}_2^{+}$(15 electrons)

$$=\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\sigma 2p_z^2,\pi 2p_x^2\approx \pi 2p_y^2,{\pi }^{\ast }2p_x^1\approx {\pi }^{\ast }2p_y^0$$

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

Electronic confiquration of ${\mathrm{O}}_2^{-}$(17 electrons)

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

Bond order $=\frac{1}{2}(N_b-N_a)=\frac{1}{2}(10-7)=1.5$

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

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

• (b) $3s^{2}3p^5$: This electronic configuration represents chlorine, which is less electronegative than fluorine.
• (c) $4s^{2}4p^5$: This electronic configuration represents bromine, which is less electronegative than fluorine.
• (d) $5s^{2}5p^5$: 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 $3p$ 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 $[\mathrm{Ne}]3s^{2}3p^1$ 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 $[\mathrm{Ne}]3s^{2}3p^2$ 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 $[\mathrm{Ar}]3d^{10}4s^{2}4p^3$ 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

$(a,d)$

$BeC{l}_2(Cl-Be-Cl)$ and $C{S}_2(S=C=S)$ both are linear. $NC{O}^{+}$is non-linear. However, [remember that ${}^{-}NCO(-N=C=O)$ is linear because it is isoelectronic with $C{O}_2$ ]. $N{O}_2$ 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, $$ In ${\mathrm{NO}}^{+}=7+8-1=14$

In ${\mathrm{N}}_2=7+7=14$

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

$$\begin{array}{rl}C{O}_2& =\text{ linear }\\ CC{l}_4& =\text{ tetrahedral }\\ O_3& =\text{ bent }\\ N{O}_2^{-}& =\text{bent }\end{array}$$

• (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 $s{p}^2$, not $s{p}^3$. 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 $N_2=\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\pi 2p_x^2\approx \pi 2p_y^2,\sigma 2p_z^2$.

It has no unpaired electron indicates diamagnetic species.

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

$${\pi }^{\ast }2p_x^1\approx {\pi }^{\ast }2p_y^1$$

It has two unpaired electrons, paramagnetic in nature.

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

$${\pi }^{\ast }2p_x^1\approx {\pi }^{\ast }2p_y^1$$

The presence of two unpaired electrons shows its paramagnetic nature.

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

$${\pi }^{\ast }2p_x^2\approx {\pi }^{\ast }2p_y^2$$

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 $\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\pi 2p_x^2\approx \pi 2p_y^2,\sigma 2p_z^2,{\pi }^{\ast }2p_x^1\approx {\pi }^{\ast }2p_y^1$. It has two unpaired electrons, which makes it paramagnetic in nature.

• (c) ${\mathrm{O}}_2$: The electronic configuration of ${\mathrm{O}}_2$ is $\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\sigma 2p_z^2,\pi 2p_x^2\approx \pi 2p_y^2,{\pi }^{\ast }2p_x^1\approx {\pi }^{\ast }2p_y^1$. 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

$$\begin{array}{rl}& N_2=3\\ & N_2^{-}=2.5\\ & F_2^{+}=1.5\\ & O_2^{-}=1.5\end{array}$$

• ${\mathrm{N}}_2$: The bond order of $\mathrm{N}{2}$is3,whichisdifferentfromthebondorderof$\mathrm{F}{2}^{+}$and$\mathrm{O}_{2}^{-}$, both of which have a bond order of 1.5.

• ${\mathrm{N}}_2^{-}$: The bond order of $\mathrm{N}{2}^{-}$is2.5,whichisdifferentfromthebondorderof$\mathrm{F}{2}^{+}$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).

$PC{l}_3-$ Central atom is phosphorus. There are 5 electrons in its valence shell $({}_{15}P=2,8,5)$. 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

$O_2^{+}:(\sigma 1s{)}^2({\sigma }^{\ast }1s{)}^2(\sigma 2s{)}^2({\sigma }^{\star }2s{)}^2(\sigma 2p_z{)}^2(\pi 2p_x^2,\pi 2p_y^2)(\pi \ast 2p_x^1)$

Bond order of ${\mathrm{O}}_2^{+}=\frac{10-5}{2}=\frac{5}{2}=2.5$

$O_2^{-}:(\sigma 1s{)}^2({\sigma }^{\star }1s^2)(\sigma 2s^2)({\sigma }^{\star }2s^2)(\sigma 2p_z{)}^2(\pi 2p_x^2,\pi 2p_y^2)({\pi }^{\star }2p_x^2,{\pi }^{\ast }2p_y^1)$

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 $PC{l}_5,P$ 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 $I{F}_5$, I is $s{p}^3{d}^2$ hybridised, therefore, shape of $I{F}_5$ is square pyramidal.