Mix Examples-Chemical Bonding Questions in English

(A) The anhydride of sulphuric acid $(H_2SO_4)$ is sulphur trioxide $(SO_3)$.
In the structure of $SO_3$,there are three $S=O$ double bonds.
Each double bond consists of one $\sigma$-bond and one $\pi$-bond.
Therefore,the total number of $\sigma$-bonds is $3$ and the total number of $\pi$-bonds is $3$.

(A) Both $SnCl_2$ and $H_2O$ have a bent geometry.
In $SnCl_2$,the central atom $Sn$ has $2$ bond pairs and $1$ lone pair,resulting in $sp^2$ hybridization.
In $H_2O$,the central atom $O$ has $2$ bond pairs and $2$ lone pairs,resulting in $sp^3$ hybridization.
Since both have a bent geometry but different hybridization states,the correct pair is $SnCl_2$ and $H_2O$.

(A) In the resonance structure of ozone $(O_3)$:
$1$. The central oxygen atom has $1$ lone pair.
$2$. The terminal oxygen atom with the double bond has $2$ lone pairs.
$3$. The terminal oxygen atom with the single bond has $3$ lone pairs.
Total number of lone pairs $= 1 + 2 + 3 = 6$.
Total number of bond pairs (counting each bond as one pair) $= 3$ (one double bond counts as $2$ bond pairs and one single bond counts as $1$ bond pair).
Ratio of lone pairs to bond pairs $= 6:3 = 2:1$.

(A) Isostructural species have the same shape and hybridization.
$PCl_5$ and $BrF_5$ are not isostructural.
$PCl_5$: The central atom $P$ has $5$ bond pairs and $0$ lone pairs,resulting in $sp^3d$ hybridization and a trigonal bipyramidal shape.
$BrF_5$: The central atom $Br$ has $5$ bond pairs and $1$ lone pair,resulting in $sp^3d^2$ hybridization and a square pyramidal shape.
Other pairs like $(CH_4, SiCl_4)$,$(CO_3^{2-}, NO_3^{-})$,and $(AlF_6^{3-}, SF_6)$ are isostructural as they have the same hybridization and geometry.

(D) The formal charge $(FC)$ is calculated using the formula: $FC = V - L - \frac{1}{2} B$,where $V$ is the number of valence electrons,$L$ is the number of lone pair electrons,and $B$ is the number of bonding electrons.
For the $CO_2$ molecule with the structure $\ddot{O}=C=\ddot{O}$:
For the $O$ atom: $V = 6$,$L = 4$,$B = 4$. Thus,$FC = 6 - 4 - \frac{1}{2}(4) = 0$.
For the $C$ atom: $V = 4$,$L = 0$,$B = 8$. Thus,$FC = 4 - 0 - \frac{1}{2}(8) = 0$.
Therefore,the formal charges of $C$ and $O$ atoms are $0$ and $0$,respectively.

(B) The formal charge $(FC)$ is calculated as: $FC = (\text{Total valence electrons}) - (\text{Non-bonding electrons}) - \frac{1}{2}(\text{Bonding electrons})$.
For $N_{(1)}$ (terminal nitrogen with $4$ non-bonding electrons and $2$ bonding electrons from a double bond): $FC = 5 - 4 - \frac{1}{2}(4) = -1$.
For $N_{(2)}$ (central nitrogen with $0$ non-bonding electrons and $8$ bonding electrons from two double bonds): $FC = 5 - 0 - \frac{1}{2}(8) = +1$.
For $O$ (oxygen with $4$ non-bonding electrons and $4$ bonding electrons from a double bond): $FC = 6 - 4 - \frac{1}{2}(4) = 0$.
Thus,the formal charges are $-1, +1, 0$.

(A) To determine the number of lone pairs on the central atom,we use the formula: $\text{Lone pairs} = \frac{V - M}{2}$,where $V$ is the number of valence electrons of the central atom and $M$ is the number of monovalent atoms bonded to it.
$A. \ NH_3$: Central atom $N$ has $5$ valence electrons. It is bonded to $3$ $H$ atoms. $\text{Lone pairs} = \frac{5-3}{2} = 1$ (Option $5$).
$B. \ H_2O$: Central atom $O$ has $6$ valence electrons. It is bonded to $2$ $H$ atoms. $\text{Lone pairs} = \frac{6-2}{2} = 2$ (Option $1$).
$C. \ XeF_2$: Central atom $Xe$ has $8$ valence electrons. It is bonded to $2$ $F$ atoms. $\text{Lone pairs} = \frac{8-2}{2} = 3$ (Option $2$).
$D. \ CH_4$: Central atom $C$ has $4$ valence electrons. It is bonded to $4$ $H$ atoms. $\text{Lone pairs} = \frac{4-4}{2} = 0$ (Option $3$).
Therefore,the correct matching is $A-5, B-1, C-2, D-3$.

(A) To determine the correct set,we analyze the hybridization and geometry of each molecule:

Molecule	$bp + lp$	Hybridisation	Shape
$H_2O$	$2 + 2$	$sp^3$	angular
$BCl_3$	$3 + 0$	$sp^2$	trigonal planar
$NH_4^{+}$	$4 + 0$	$sp^3$	tetrahedral
$CH_4$	$4 + 0$	$sp^3$	tetrahedral

Comparing the options with the table:
$A$. $H_2O$ has $sp^3$ hybridization and an angular shape. This is correct.
$B$. $BCl_3$ has $sp^2$ hybridization and trigonal planar shape.
$C$. $NH_4^{+}$ has $sp^3$ hybridization and tetrahedral shape.
$D$. $CH_4$ has $sp^3$ hybridization and tetrahedral shape.
Therefore,the correct set is $A$.

(A) Option $A$ is correct because in $SF_6$,the sulfur atom forms $6$ covalent bonds with $6$ fluorine atoms,resulting in $12$ electrons in its valence shell,which violates the octet rule.
Option $B$ is incorrect because ionic reactions occur almost instantaneously due to the presence of free ions in solution.
Option $C$ is incorrect because $SnCl_2$ has a bent geometry due to the presence of one lone pair on the $Sn$ atom.
Option $D$ is incorrect because according to Fajan's rule,the ability to form ionic compounds decreases as the charge density increases. Thus,the correct order of ionic character is $Na^{+} > Mg^{2+} > Al^{3+}$.

(C) $NH_3$ has $sp^3$ hybridisation with a trigonal pyramidal shape.
$BeCl_2$ is linear and $SO_2$ is bent ($V$-shaped).
$SF_6$ has $sp^3d^2$ hybridisation,resulting in an octahedral geometry where all $F-S-F$ bond angles are $90^{\circ}$.
$CO_2$ is a linear molecule with a net dipole moment of zero.

(B) The bond length depends on the bond order (multiplicity) and the atomic radii of the bonded atoms.
Bond length is inversely proportional to bond order: $Triple \ bond < Double \ bond < Single \ bond$.
Comparing the given bonds:
$1. C \equiv C$ (Bond order $3$)
$2. C=O$ (Bond order $2$)
$3. C=N$ (Bond order $2$)
$4. N-O$ (Bond order $1$)
Since $C=O$ has a higher bond polarity and smaller atomic radii compared to $C=N$,the $C=O$ bond is shorter than $C=N$.
Thus,the order of bond length is $C \equiv C < C=O < C=N < N-O$.

(B) An ionic bond is formed by the electrostatic attraction between oppositely charged ions,while a covalent bond is formed by the sharing of electrons between atoms.
In $K_2SO_4$ (potassium sulfate),the compound consists of $K^+$ ions and $SO_4^{2-}$ polyatomic ions,which are held together by ionic bonds.
Within the sulfate ion $(SO_4^{2-})$,the sulfur atom is covalently bonded to the four oxygen atoms.
Therefore,$K_2SO_4$ contains both ionic bonds (between $K^+$ and $SO_4^{2-}$) and covalent bonds (within the $SO_4^{2-}$ ion).

(A) Let us analyze each pair:
$(A)$ Bond angle: The bond angles are $NO_2$ $(134^{\circ})$,$O_3$ $(116.8^{\circ})$,and $H_2O$ $(104.5^{\circ})$. Thus,the order $NO_2 > O_3 > H_2O$ is correct.
$(B)$ Dipole moment: The dipole moments are $H_2O$ $(1.85 \ D)$,$HF$ $(1.78 \ D)$,and $NH_3$ $(1.47 \ D)$. Thus,the order $H_2O > HF > NH_3$ is correct.
$(C)$ Bond length: $I_2$ has a single bond with a large atomic size,$F_2$ has a single bond with a smaller atomic size,and $N_2$ has a triple bond with a very small atomic size. Bond length order is $I_2 > F_2 > N_2$. This is correct.
Therefore,all three pairs $(A), (B),$ and $(C)$ are correctly matched.

(C) Statement $(A)$: $CO_2$ has a linear geometry,so the bond dipoles cancel each other,resulting in a net dipole moment of $0$. $SO_2$ has a bent geometry due to the presence of a lone pair on $S$,and $H_2O$ has a bent geometry due to two lone pairs on $O$,so both have a non-zero net dipole moment. Thus,Statement $(A)$ is correct.
Statement $(B)$: According to Fajan's rule,the polarizing power of a cation increases with its charge. $Sn^{4+}$ has a higher charge than $Sn^{2+}$,so $SnCl_4$ exhibits significant covalent character,whereas $SnCl_2$ is predominantly ionic. Thus,Statement $(B)$ is correct.

(A) The bond lengths for the given molecules are as follows:
$H_2 = 74 \ pm$
$F_2 = 144 \ pm$
$Cl_2 = 199 \ pm$
$I_2 = 267 \ pm$
Therefore, the order for $F_2, H_2, Cl_2, I_2$ is $144, 74, 199, 267 \ pm$.
Thus, the correct option is $A$.

(C) To determine the number of electrons in the valence shell,we calculate the total number of bonding and lone pairs around the central atom.
In $SCl_2$,the central atom is $S$ (Sulfur).
Sulfur has $6$ valence electrons.
It forms $2$ single bonds with $Cl$ atoms,using $2$ electrons.
Number of lone pairs on $S = \frac{6 - 2}{2} = 2$.
Total electron pairs around $S = 2 \text{ (bonding pairs)} + 2 \text{ (lone pairs)} = 4 \text{ pairs}$.
Total electrons in the valence shell $= 4 \times 2 = 8$ electrons.
Thus,$SCl_2$ follows the octet rule.

(C) Lewis acid is an electron-pair acceptor.
$(i)$ $NH_3$: Has a lone pair on $N$,acts as a Lewis base.
$(ii)$ $AlCl_3$: $Al$ has an incomplete octet ($6$ electrons),acts as a Lewis acid.
$(iii)$ $SnCl_4$: $Sn$ has vacant $d$-orbitals and can expand its octet,acts as a Lewis acid.
$(iv)$ $CO_2$: The $C$ atom is electron-deficient due to the high electronegativity of oxygen atoms,acts as a Lewis acid.
$(v)$ $Ag^{+}$: $A$ metal cation with vacant orbitals,acts as a Lewis acid.
$(vi)$ $HSO_4^{-}$: Can act as a Lewis base (due to lone pairs on $O$) or a Brønsted-Lowry acid,but is not typically classified as a Lewis acid in this context.
Therefore,$AlCl_3$,$SnCl_4$,$CO_2$,and $Ag^{+}$ act as Lewis acids.
The total count is $4$.

(B) The structure of $Al_2Cl_6$ is a dimer where each $Al$ atom is bonded to $4$ chlorine atoms (two terminal and two bridging).
$(I)$ Oxidation state $(n)$ of $Al$ in $Al_2Cl_6$: $2x + 6(-1) = 0 \implies 2x = 6 \implies x = +3$.
$(II)$ Coordination number $(CN)$ of $Al$: Each $Al$ atom is surrounded by $4$ chlorine atoms,so $CN = 4$.
$(III)$ Number of valence electrons $(N)$ around $Al$: Since each $Al$ atom is bonded to $4$ chlorine atoms by covalent bonds (including coordinate bonds),it shares $4$ electron pairs,resulting in $4 \times 2 = 8$ valence electrons around each $Al$ atom.
Therefore,the values are $3, 4, 8$. The correct option is $(B)$.

(B) The chemical formula of peroxodisulphuric acid (Marshall's acid) is $H_2S_2O_8$.
Its structure consists of two $SO_3$ groups linked by a peroxide linkage $(-O-O-)$.
In the structure:
- There are $2$ $S=O$ double bonds per sulfur atom,contributing $2$ $\pi$ bonds each,totaling $4$ $\pi$ bonds.
- Counting the $\sigma$ bonds: $4$ $S=O$ bonds,$2$ $S-OH$ bonds,$2$ $O-H$ bonds,$2$ $S-O$ bonds (to peroxide oxygen),and $1$ $O-O$ bond.
- Total $\sigma$ bonds = $4 + 2 + 2 + 2 + 1 = 11$.
- Total $\pi$ bonds = $4$.
Thus,the number of $\sigma$ and $\pi$ bonds are $11$ and $4$ respectively.

(A) The bond dissociation energies are determined by the strength of the bond,which is related to the overlap of orbitals and bond length.
The bond dissociation energies are approximately:
$H-H \approx 436 \ kJ/mol$
$C-H \approx 413 \ kJ/mol$
$C-C \approx 348 \ kJ/mol$
Therefore,the decreasing order is $H-H > C-H > C-C$.

(D) The molecular formula of $p$-nitrobenzonitrile is $C_{7}H_{4}N_{2}O_{2}$.
In this molecule,the benzene ring is planar ($sp^{2}$ hybridized carbons).
The nitro group $(-NO_{2})$ is attached to the benzene ring,and due to resonance,it tends to lie in the same plane as the benzene ring.
The cyano group $(-CN)$ is linear ($sp$ hybridized carbon) and is also in the same plane as the benzene ring.
Therefore,all $15$ atoms ($7$ carbons,$4$ hydrogens,$2$ nitrogens,and $2$ oxygens) lie in the same plane.

(A) The formula for formal charge is: $\text{Formal charge} = \text{Valence electrons} - (\frac{1}{2} \times \text{Bonding electrons} + \text{Non-bonding electrons})$.
For the structure $: \ddot{N}_1 = N_2 = \ddot{N}_3 :$
For $N_1$ (left nitrogen): $\text{Valence electrons} = 5$,$\text{Bonding electrons} = 4$,$\text{Non-bonding electrons} = 4$.
$\text{Formal charge} = 5 - (\frac{1}{2} \times 4 + 4) = 5 - (2 + 4) = -1$.
For $N_2$ (middle nitrogen): $\text{Valence electrons} = 5$,$\text{Bonding electrons} = 8$,$\text{Non-bonding electrons} = 0$.
$\text{Formal charge} = 5 - (\frac{1}{2} \times 8 + 0) = 5 - 4 = +1$.
For $N_3$ (right nitrogen): $\text{Valence electrons} = 5$,$\text{Bonding electrons} = 4$,$\text{Non-bonding electrons} = 4$.
$\text{Formal charge} = 5 - (\frac{1}{2} \times 4 + 4) = 5 - (2 + 4) = -1$.
Thus,the formal charges are $-1, +1, -1$.

(B) The bond length is inversely proportional to the bond order.
$1$. In $O_{2}$,the bond order is $2$.
$2$. In $O_{3}$,the bond order is $1.5$ due to resonance.
$3$. In $H_{2}O_{2}$,the $O-O$ bond is a single bond with a bond order of $1$.
Comparing the bond orders: $O_{2} (2) > O_{3} (1.5) > H_{2}O_{2} (1)$.
Therefore,the order of bond length is $H_{2}O_{2} > O_{3} > O_{2}$.

(B) Dipole moment: $NF_3 (0.24 \ D) < NH_3 (1.47 \ D)$. Statement $A$ is incorrect.
$(B)$ $BeH_2$ is $sp$ hybridized and linear,so its dipole moment is $0$. Statement $B$ is correct.
$(C)$ Bond order of $O_2^{2-}$ is $\frac{10-8}{2} = 1$. Bond order of $F_2$ is $\frac{8-6}{2} = 1$. Statement $C$ is correct.
$(D)$ In $O_3$,the central oxygen atom has a formal charge of $+1$. Statement $D$ is incorrect.
$(E)$ In $NO_2$,nitrogen has an odd number of electrons ($7$ valence electrons),so it does not follow the octet rule. Statement $E$ is incorrect.
Therefore,only $B$ and $C$ are correct.

(C) The formula for formal charge is: $\text{Formal charge} = \text{Valence electrons} - \text{Non-bonding electrons} - \frac{1}{2}(\text{Bonding electrons})$.
For atom $(1)$ (Oxygen): Valence $e^- = 6$,Non-bonding $e^- = 4$,Bonding $e^- = 4$. $\text{FC} = 6 - 4 - \frac{4}{2} = 0$.
For atom $(2)$ (Nitrogen): Valence $e^- = 5$,Non-bonding $e^- = 0$,Bonding $e^- = 8$. $\text{FC} = 5 - 0 - \frac{8}{2} = +1$.
For atom $(3)$ (Oxygen): Valence $e^- = 6$,Non-bonding $e^- = 4$,Bonding $e^- = 4$. $\text{FC} = 6 - 4 - \frac{4}{2} = 0$.
For atom $(4)$ (Oxygen): Valence $e^- = 6$,Non-bonding $e^- = 6$,Bonding $e^- = 2$. $\text{FC} = 6 - 6 - \frac{2}{2} = -1$.
Thus,the formal charges are $0, +1, 0, -1$.

(B) To determine if the species have similar Lewis dot structures,we analyze their valence electrons and geometry:
$A: SO_3^{2-}$ has $26$ valence electrons and a trigonal pyramidal geometry. $CO_3^{2-}$ has $24$ valence electrons and a trigonal planar geometry. These are not similar.
$B: O_2^{2-}$ has $14$ valence electrons $(:O-O:)$ and $F_2$ has $14$ valence electrons $(:F-F:)$. These are isoelectronic and have similar structures.
$C: CN^-$ has $10$ valence electrons $([:C \equiv N:]^-)$ and $CO$ has $10$ valence electrons $(:C \equiv O:)$. These are isoelectronic and have similar structures.
$D: NH_3$ has $8$ valence electrons and a trigonal pyramidal geometry. $H_3O^+$ has $8$ valence electrons and a trigonal pyramidal geometry. These are similar.
$E: MnO_4^{2-}$ and $CrO_4^{2-}$ are both $d^0$ transition metal oxoanions with a tetrahedral geometry. These are similar.
Therefore,only pair $A$ does not have similar structures.

(C) Statement $I$: The bond angles are $F_2O (103^\circ) < H_2O (104.5^\circ) < Cl_2O (111^\circ)$. This trend is correct because in $Cl_2O$,the large size of $Cl$ atoms leads to steric repulsion,increasing the bond angle.
Statement $II$: $SiF_4$ is a covalent molecule due to the small size and high electronegativity of $Si$. While $SnF_4$ and $PbF_4$ exhibit significant ionic character,classifying all three as ionic is incorrect. Therefore,Statement $II$ is false.

(D) The formula for formal charge is: $\text{Formal charge} = (\text{Valence electrons}) - (\text{Non-bonding electrons}) - \frac{1}{2}(\text{Bonding electrons})$.
For oxygen atom $2$ (terminal oxygen with double bond): Valence electrons = $6$,Non-bonding electrons = $4$,Bonding electrons = $4$. Formal charge = $6 - 4 - \frac{1}{2}(4) = 6 - 4 - 2 = 0$.
For oxygen atom $1$ (central oxygen): Valence electrons = $6$,Non-bonding electrons = $2$,Bonding electrons = $6$. Formal charge = $6 - 2 - \frac{1}{2}(6) = 6 - 2 - 3 = +1$.
For oxygen atom $3$ (terminal oxygen with single bond): Valence electrons = $6$,Non-bonding electrons = $6$,Bonding electrons = $2$. Formal charge = $6 - 6 - \frac{1}{2}(2) = 6 - 6 - 1 = -1$.
Thus,the formal charges on oxygen atoms $2, 1$ and $3$ are $0, +1$ and $-1$ respectively.

(A) The bonding in the given molecules is as follows:
$(A)$ $C_{2}H_{4}$ (Ethene): Structure is $CH_{2}=CH_{2}$. It contains $5\sigma$ bonds and $1\pi$ bond.
$(B)$ $C_{2}H_{2}$ (Ethyne): Structure is $CH \equiv CH$. It contains $3\sigma$ bonds and $2\pi$ bonds.
$(C)$ $CH_{4}$ (Methane): Structure is $CH_{4}$ with $sp^{3}$ hybridization. It contains $4\sigma$ bonds.
$(D)$ $NH_{3}$ (Ammonia): Structure is $NH_{3}$ with $sp^{3}$ hybridization. It contains $3\sigma$ bonds and $1$ lone pair on the nitrogen atom.
Therefore,the correct matching is: $A-IV, B-I, C-III, D-II$.