Chemical Bonding And Molecular Structure – JEE Main ChemistryPractice Questions & PYQs

The $\mathrm{XeO_2F_2}$ molecule adopts a see-saw molecular geometry, which is a distorted trigonal bipyramidal structure. In this arrangement, the central xenon atom has a steric number of 5, consisting of 4 bonding domains (two double bonds to oxygen and two single bonds to fluorine) and 1 lone pair, which occupies an equatorial position to minimize repulsion. The two fluorine atoms occupy the axial positions, resulting in an $\mathrm{F}-\mathrm{Xe}-\mathrm{F}$ bond angle close to $180^{\circ}$, while the two oxygen atoms and the lone pair occupy the equatorial positions, leading to an $\mathrm{O}-\mathrm{Xe}-\mathrm{O}$ bond angle significantly less than $180^{\circ}$ (approximately $106^{\circ}$).

Regarding the electron count, each double bond consists of 2 electron pairs and each single bond consists of 1 electron pair, which, when added to the 1 lone pair, results in a total of 7 electron pairs ($2 \times 2 + 2 \times 1 + 1 = 7$) in the valence shell of xenon. Consequently, the valence shell contains 14 electrons ($7 \times 2 = 14$), not 16 as stated. Therefore, the statements that Xe has 5 electron pairs, the $\mathrm{O}-\mathrm{Xe}-\mathrm{O}$ bond angle is $180^{\circ}$, and Xe has 16 valence electrons are all incorrect.

Molecular Orbital Theory states that the bond order of a species is determined by the formula $\frac{N_b - N_a}{2}$, where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons, while paramagnetic character arises from the presence of one or more unpaired electrons. Both $O_2^{+}$ and $N_2^{-}$ contain 15 total electrons, which follow an electronic configuration that places the 15th electron \in an antibonding orbital, such as $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2, \pi^* 2p_x^1$. This distribution results \in $N_b = 10$ and $N_a = 5$, yielding a bond order of $\frac{10 - 5}{2} = 2.5$ for both species, and the single electron \in the $\pi^*$ orbital confirms they are both paramagnetic.

The octet rule requires atoms to surround themselves with eight valence electrons to achieve stability, yet many molecules exhibit hypervalency where the central atom exceeds this count. Among the specified substances, $\mathrm{SO}_2, \mathrm{SO}_3, \mathrm{SF}_4,$ and $\mathrm{SF}_6$ all possess expanded octets, meaning four of these compounds violate the rule rather than three, which makes the first statement incorrect.

Regarding the lone pair distribution, central atoms with exactly one lone pair are found in $\mathrm{NH}_3, \mathrm{BrF}_5,$ and $\mathrm{SF}_4$, while molecules like $\mathrm{H}_2\mathrm{O}, \mathrm{ClF}_3,$ and $\mathrm{XeF}_4$ contain two lone pairs on their central atoms. Because only the specific set containing $\mathrm{NH}_3, \mathrm{BrF}_5,$ and $\mathrm{SF}_4$ consists entirely of molecules with one lone pair, the assertion that there is exactly one such set is valid.

Structural similarity in Lewis dot representations generally depends on species being isoelectronic and having analogous atomic connectivity. Pairs such as $\mathrm{O}_2^{2-}$ and $\mathrm{F}_2$, $\mathrm{CN}^{-}$ and $\mathrm{CO}$, as well as $\mathrm{NH}_3$ and $\mathrm{H}_3\mathrm{O}^+$ qualify as having similar structures because they contain the same number of valence electrons. Conversely, $\mathrm{SO}_3^{2-}$ and $\mathrm{CO}_3^{2-}$ lack this structural similarity due to the inherent differences between the central atoms involved in their bonding. Similarly, the pair $\mathrm{MnO}_4^{2-}$ and $\mathrm{CrO}_4^{2-}$ fails to share a common Lewis structure because $\mathrm{MnO}_4^{2-}$ possesses an unpaired electron, whereas $\mathrm{CrO}_4^{2-}$ contains no such unpaired electrons in its configuration.

According to VSEPR theory, the molecular shape of xenon compounds is determined by the arrangement of bonding and lone electron pairs surrounding the central atom, which dictates their hybridization and geometry. $XeO_3$ possesses three bonding pairs and one lone pair resulting in a trigonal pyramidal shape like $NH_3$, while $XeF_2$ contains two bonding pairs and three lone pairs, creating a linear geometry similar to $[I_3]^-$. Furthermore, $XeO_2F_2$ forms a see-saw structure with four bonding pairs and one lone pair, analogous to $SF_4$, whereas $XeOF_4$ adopts a square pyramidal geometry with five bonding pairs and one lone pair, matching $BrF_5$.

The dipole moment of $NH_3$ is approximately $1.47$ D because the nitrogen-hydrogen bond dipoles and the lone pair dipole reinforce each other, whereas in $NF_3$ the highly electronegative fluorine atoms pull electron density away from nitrogen, opposing the lone pair dipole and resulting in a lower net dipole moment, making statement A incorrect. In contrast, $BeH_2$ adopts a linear geometry where the two $Be-H$ bond dipoles cancel exactly, resulting in a net dipole moment of zero, confirming statement B is correct. Furthermore, both $O_2^{2-}$ and $F_2$ are isoelectronic with 18 electrons, resulting in a bond order of $\frac{10-8}{2} = 1$ according to molecular orbital theory, which validates statement C.

The formal charge on the central oxygen atom in ozone is calculated as $6 - (2 + 3) = +1$, where 6 is the valence count, 2 represents the lone pair electrons, and 3 represents the bonding electrons, which contradicts statement D. Finally, $NO_2$ contains 17 valence electrons, forming an odd-electron radical where the central nitrogen atom cannot complete its octet, rendering statement E incorrect.

Bond length is governed by the bond multiplicity and the covalent radii of the bonded atoms, where higher bond order reduces length and smaller atomic radii lead to shorter bonds. The $\mathrm{C}-\mathrm{H}$ bond (A) is the shortest due to the minimal atomic radius of hydrogen, followed by the $\mathrm{C} \equiv \mathrm{N}$ bond (D) due to its triple bond order, the $\mathrm{C}=\mathrm{O}$ bond (C) as a double bond, and finally the $\mathrm{C}-\mathrm{O}$ bond (B) as a single bond. Consequently, the increasing order of bond lengths is $\mathrm{A} < \mathrm{D} < \mathrm{C} < \mathrm{B}$.

Stability of resonance structures depends upon the completeness of octets, the number of covalent bonds, and the placement of formal charges. Structures exhibiting incomplete octets, like charges on adjacent atoms, or positive charges on highly electronegative elements are inherently higher in energy and thus less stable. Option D is identified as the least stable because it represents an arrangement that violates these electronic requirements most severely, often by placing positive charge on an electronegative atom or featuring adjacent like-charged atoms, which creates strong repulsive interactions.

The total number of lone pairs in each molecule is determined by summing the non-bonding electron pairs on all constituent atoms. $\mathrm{HNO}_3$ contains 7 lone pairs, $\mathrm{H}_2\mathrm{SO}_4$ contains 8 lone pairs, $\mathrm{O}_3$ has 6 lone pairs, and $\mathrm{NF}_3$ possesses 10 lone pairs, comprised of 1 on the nitrogen atom and 9 on the fluorine atoms, identifying $\mathrm{NF}_3$ as molecule $\mathrm{X}$. Due to the high electronegativity of fluorine in $\mathrm{NF}_3$, the bonding electron pairs are displaced toward the fluorine atoms, which reduces the repulsion between bonding pairs and allows the lone pair on the central nitrogen atom to compress the $\mathrm{F-N-F}$ bond angle to 102 degrees from the ideal tetrahedral angle of 109.5 degrees.

The molecular geometry of a species is determined by the hybridization of the central atom and the presence of lone pairs, while the octet rule dictates that atoms typically seek to be surrounded by eight electrons in their valence shell. In Statement I, $NH_4^+$ and $[NiCl_4]^{2-}$ possess tetrahedral geometry due to $sp^3$ hybridization. However, $SF_4$ and $SeF_4$ exhibit see-saw geometry ($sp^3d$ with one lone pair), $XeF_4$ has square planar geometry ($sp^3d^2$ with two lone pairs), $[PtCl_4]^{2-}$ is square planar because $5d$ metal complexes adopt this geometry regardless of the ligand, and $[Ni(CN)_4]^{2-}$ is square planar ($dsp^2$ with strong field ligands). Since only two species are tetrahedral, Statement I is false.

In Statement II, the central atoms in $NO_2$, $BeH_2$, $BF_3$, and $AlCl_3$ are electron-deficient, possessing fewer than eight electrons in their valence shells. Specifically, $NO_2$ has 7 electrons around the nitrogen atom, $BeH_2$ has 4 electrons around beryllium, $BF_3$ has 6 electrons around boron, and $AlCl_3$ has 6 electrons around aluminum. Thus, all molecules listed have incomplete octets, making Statement II true.

The bond dissociation enthalpy of fluorine is anomalously low compared to chlorine and bromine due to significant inter-electronic repulsion between the lone pairs of the small fluorine atoms, resulting in the correct order $Cl_2 > Br_2 > F_2 > I_2$. According to Fajans' rule, the covalent character of metal halides increases with the oxidation state of the metal ion, meaning the species with the higher charge density should be more covalent. Consequently, while the trends for $SnCl_4 > SnCl_2$ and $PbCl_4 > PbCl_2$ are correct, the assertion $UF_4 > UF_6$ is incorrect because the higher oxidation state of uranium in $UF_6$ confers it greater covalent character than in $UF_4$.

Bond angles in molecules such as $\mathrm{F}_2\mathrm{O}$, $\mathrm{H}_2\mathrm{O}$, and $\mathrm{Cl}_2\mathrm{O}$ are determined by the interplay of lone pair repulsions and steric factors associated with the size of the peripheral atoms, resulting in experimentally measured angles of 103 degrees, 104.5 degrees, and 111 degrees, respectively. This sequence confirms that the bond angle increases with the size of the peripheral atom, making the trend $\mathrm{F}_2\mathrm{O} < \mathrm{H}_2\mathrm{O} < \mathrm{Cl}_2\mathrm{O}$ accurate. The second statement is incorrect because compounds like $\mathrm{SiF}_4$ are covalent, rather than ionic, a characteristic that generally holds for the fluorides of the group 14 elements listed.

$BF_4^-$, $SiF_4$, and $XeF_4$ exhibit high-symmetry tetrahedral or square planar geometries in which all central atom-fluorine bonds are equivalent, while $SF_4$ possesses a see-saw geometry with distinct axial and equatorial bond lengths, meaning only one species has unequal bond lengths. The bond order is defined by the formula $BO = \frac{N_b - N_a}{2}$, yielding values of $2.5$ for $O_2^+$, $1.5$ for $O_2^-$, and $1$ for both $O_2^{2-}$ and $F_2$. Since $O_2^+$ possesses the highest bond order, both provided statements are incorrect.

Lewis acidity in $EF_3$ type molecules arises from an incomplete octet, characteristic of group 13 elements which have three valence electrons and form three single bonds, resulting in $sp^2$ hybridization. Conversely, Lewis basicity in $EF_3$ compounds is characteristic of group 15 elements, which possess five valence electrons and form three single bonds while retaining one lone pair of electrons, leading to $sp^3$ hybridization.

The formal charge of an atom in a molecule is calculated using the formula $FC = V - L - \frac{1}{2}B$, where $V$ represents the number of valence electrons, $L$ represents the number of non-bonding electrons, and $B$ represents the number of bonding electrons. In the Lewis structure of nitric acid ($HNO_3$), the oxygen atom bonded to hydrogen has a formal charge of $6 - 4 - 2 = 0$, the central nitrogen atom forming four bonds has a formal charge of $5 - 0 - 4 = +1$, the double-bonded oxygen atom has a formal charge of $6 - 4 - 2 = 0$, and the single-bonded oxygen atom carries a formal charge of $6 - 6 - 1 = -1$. This results in a sequence of $0, +1, 0, -1$ for the corresponding atoms in the molecular representation.

The covalent bond length between two atoms A and B is defined as the distance separating their nuclei, which is simply the sum of their covalent radii, expressed as $r_A + r_B$. Since the total length of the AB molecule encompasses the full diameters of both bonded atoms, this value is determined by doubling that sum, resulting in $2(r_A + r_B)$.

VSEPR theory explains the geometry of these ions by minimizing the repulsion between bonding pairs and lone pairs around the central bromine atom. For the cation $\mathrm{BrF}_2^{\oplus}$, the central bromine atom is surrounded by four electron pairs-two bonding and two lone-resulting in $sp^3$ hybridization and a bent molecular shape. Conversely, the anion $\mathrm{BrF}_4^{\ominus}$ involves $sp^3d^2$ hybridization where the central bromine atom coordinates with four bonding pairs and maintains two lone pairs, forcing the fluorine atoms into a square planar arrangement to achieve the most stable electronic configuration.

Bond length is inversely proportional to bond order, meaning that species with a higher bond order have shorter bond lengths. Based on bond orders of 2.5 for $\mathrm{O}^{+}_{2}$, 2 for $\mathrm{O}_{2}$, 1.5 for $\mathrm{O}^{-}_{2}$, and 1 for $\mathrm{O}^{2-}_{2}$, the bond lengths must increase as the bond order decreases, confirming the sequence $\mathrm{O}^{+}_{2} < \mathrm{O}_{2} < \mathrm{O}^{-}_{2} < \mathrm{O}^{2-}_{2}$ as a true statement.

When examining the number of unpaired electrons, $\mathrm{O}_{2}$ contains two, $\mathrm{O}^{+}_{2}$ and $\mathrm{O}^{-}_{2}$ each possess one, and $\mathrm{O}^{2-}_{2}$ has zero. Since both $\mathrm{O}^{+}_{2}$ and $\mathrm{O}^{-}_{2}$ have the same number of unpaired electrons, the claimed strict inequality $\mathrm{O}_{2} > \mathrm{O}^{+}_{2} > \mathrm{O}^{-}_{2} > \mathrm{O}^{2-}_{2}$ is inaccurate because it incorrectly suggests that $\mathrm{O}^{+}_{2}$ has more unpaired electrons than $\mathrm{O}^{-}_{2}$, which makes the second statement false.

The dipole moment of a molecule is determined by the vector sum of its bond dipoles and lone pair dipoles, and in $NF_3$, the dipole moment associated with the lone pair on the nitrogen atom acts in direct opposition to the resultant vector of the three N-F bond dipoles, which leads to substantial cancellation and results in the lowest dipole moment of $0.23 \, \text{D}$ among the listed species. Since the central nitrogen atom has five valence electrons and engages three of them in bonding with fluorine atoms, the remaining two valence electrons constitute exactly one lone pair.

According to VSEPR theory, the molecular shape and hybridization of a central atom are determined by its steric number, calculated as the sum of sigma bonds and lone pairs. In the cases of $IF_3$, $IF_5$, and $IF_7$, the central Iodine atom exhibits steric numbers of 5, 6, and 7, which correspond to $sp^3d$, $sp^3d^2$, and $sp^3d^3$ hybridization and result in T-shaped, square pyramidal, and pentagonal bipyramidal geometries, respectively. Conversely, the $ClO_4{ }^{-}$ ion features four oxygen atoms bonded to the central Chlorine with zero lone pairs, leading to $sp^3$ hybridization and a tetrahedral geometry rather than the incorrect square planar configuration.