Mix Examples-Chemical Bonding Questions in English

(D) An ionic bond is formed by the complete transfer of electrons between atoms,typically between a metal and a non-metal.
In $NaCl$,$CsCl$,and $LiF$,the bonds are primarily ionic due to the large electronegativity difference between the constituent atoms.
In $H_2O$,the bond between Hydrogen and Oxygen is formed by the sharing of electrons,which characterizes a covalent bond.
Therefore,the ionic bond is absent in $H_2O$.

(B) $CO$ and $CN^{-}$ are isoelectronic species.
Total electrons in $CO = 6 + 8 = 14$.
Total electrons in $CN^{-} = 6 + 7 + 1 = 14$.
Since both have $14$ electrons,they are isoelectronic.

(D) Isoelectronic species are those that have the same number of electrons.
$I: CH_3^+ = 6 + 3 - 1 = 8 \ e^-$
$II: H_3O^+ = 3(1) + 8 - 1 = 10 \ e^-$
$III: NH_3 = 7 + 3(1) = 10 \ e^-$
$IV: CH_3^- = 6 + 3(1) + 1 = 10 \ e^-$
Thus,$II, III,$ and $IV$ are isoelectronic as each contains $10 \ e^-$. The correct option is $D$.

(B) Isoelectronic species are those that have the same number of electrons.
$I. CH_{3}^{+}$: $6 + 3(1) - 1 = 8$ electrons.
$II. H_{3}O^{+}$: $3(1) + 8 - 1 = 10$ electrons.
$III. NH_{3}$: $7 + 3(1) = 10$ electrons.
$IV. CH_{3}^{-}$: $6 + 3(1) + 1 = 10$ electrons.
Thus,$II, III,$ and $IV$ have $10$ electrons each and are isoelectronic.

(A) Isosteres are molecules or ions that have the same number of atoms and the same number of electrons.
For $NO_2^-$: Number of atoms = $3$. Total electrons = $7 + (8 \times 2) + 1 = 24$.
For $O_3$: Number of atoms = $3$. Total electrons = $8 \times 3 = 24$.
Since both have $3$ atoms and $24$ electrons,they are isosteres.
Therefore,the correct option is $A$.

(B) The octet rule states that atoms tend to gain,lose,or share electrons to achieve a stable configuration of $8$ valence electrons.
In $CO_2$,$H_2O$,and $O_2$,all atoms (except $H$ in $H_2O$) achieve an octet.
In $NO$ (Nitric oxide),the nitrogen atom has $5$ valence electrons and the oxygen atom has $6$ valence electrons,totaling $11$ valence electrons.
Since the total number of valence electrons is odd,it is impossible for all atoms to satisfy the octet rule,making $NO$ an odd-electron molecule.

(A) $CaH_2$ is an ionic hydride where $Ca^{2+}$ forms ionic bonds with $H^-$ ions.
In $CCl_4$,carbon shares its valence electrons with four chlorine atoms to form $4$ covalent bonds.

(C) $NH_4Cl$ (Ammonium chloride) contains both ionic and covalent bonds.
The ammonium ion $(NH_4^+)$ is formed by covalent bonds between the nitrogen atom and four hydrogen atoms.
The bond between the ammonium cation $(NH_4^+)$ and the chloride anion $(Cl^-)$ is ionic.

(B) The $N_2O$ molecule (nitrous oxide) exhibits resonance. The most stable Lewis structure involves a triple bond between the two nitrogen atoms and a single bond between the terminal nitrogen and oxygen,represented as $:\,N \equiv N^{+} - \mathop O\limits_{..}^{..} :^-$. This structure satisfies the octet rule for all atoms and accounts for the formal charges.

(D) The octet rule states that atoms tend to gain,lose,or share electrons to achieve a stable configuration of $8$ electrons in their valence shell.
$CO$ (Carbon monoxide) has a total of $10$ valence electrons ($4$ from $C$ and $6$ from $O$). In the structure $:C \equiv O:$,the carbon atom has only $6$ electrons in its valence shell,which is an incomplete octet.
$PCl_5$ (Phosphorus pentachloride) has $5$ valence electrons from $P$ and $5$ from $5$ $Cl$ atoms,resulting in $10$ electrons around the central $P$ atom,which is an expanded octet.
Therefore,both $CO$ and $PCl_5$ do not obey the octet rule.

(A) In the structure of sulphuric acid $(H_2SO_4)$,the central sulphur atom is bonded to two hydroxyl groups $(-OH)$ via single covalent bonds.
Additionally,it is bonded to two oxygen atoms via double covalent bonds.
However,in the Lewis structure representation,these double bonds are often described as two covalent bonds and two coordinate (dative) bonds,where the sulphur atom donates a lone pair to each of the two terminal oxygen atoms.
Thus,the molecule contains both covalent and coordinate bonds.

(D) In the structure $H-C=C(O)-OH$,the carbon atom involved in the double bond with oxygen and the single bond with the other carbon and the hydroxyl group has a total of $5$ bonds ($1$ with $H$,$2$ with $C$,and $2$ with $O$). Since carbon can only form a maximum of $4$ covalent bonds,this representation is incorrect.

(C) The bond length is inversely proportional to the bond order.
Bond orders for the given species are:
$O_2$: Bond order = $2.0$
$O_3$: Bond order = $1.5$
$H_2O_2$: Bond order = $1.0$
Since bond length increases as bond order decreases,the order of bond length is $H_2O_2$ $(1.48 \ \mathring{A})$ > $O_3$ $(1.28 \ \mathring{A})$ > $O_2$ $(1.21 \ \mathring{A})$.
Thus,the correct option is $C$.

(D) $BCl_3$ has a trigonal planar shape due to $sp^2$ hybridization.
$PCl_3$ has a pyramidal shape due to $sp^3$ hybridization with one lone pair.
$ICl_3$ has a $T$-shaped geometry due to $sp^3d$ hybridization with two lone pairs.
Since the shapes are different,the correct answer is that all the above options are incorrect.

(B) The bond length depends on the bond order. As the bond order increases,the bond length decreases.
For carbon-carbon bonds,the typical lengths are:
Single bond $(C-C)$: $\approx 1.54 \ \mathring{A}$
Double bond $(C=C)$: $\approx 1.34 \ \mathring{A}$
Triple bond $(C\equiv C)$: $\approx 1.20 \ \mathring{A}$
However,in the context of $CO_2$ $(O=C=O)$,the $C=O$ bond length is approximately $1.15 \ \mathring{A}$. The question asks for the general trend of single,double,and triple bond lengths involving carbon,which are approximately $1.54 \ \mathring{A}, 1.34 \ \mathring{A}$,and $1.20 \ \mathring{A}$ respectively. Given the provided options,the sequence $1.22, 1.15, 1.10 \ \mathring{A}$ represents the decreasing order of bond lengths as bond order increases.

(B) The bond angle of $109^o 28'$ is characteristic of a tetrahedral geometry,which occurs in species with $sp^3$ hybridization and no lone pairs on the central atom.
In $(NH_4)^+$,the nitrogen atom is $sp^3$ hybridized with four bonding pairs and zero lone pairs,resulting in a tetrahedral geometry with a bond angle of $109^o 28'$.
In $(BF_4)^-$,the boron atom is $sp^3$ hybridized with four bonding pairs and zero lone pairs,resulting in a tetrahedral geometry with a bond angle of $109^o 28'$.
Therefore,both $(NH_4)^+$ and $(BF_4)^-$ exhibit this bond angle.

(D) The carbonate ion $(CO_3^{2-})$ exhibits the following properties:
$1$. The $C$ atom is $sp^2$ hybridized,resulting in a trigonal planar geometry.
$2$. Due to resonance,all three $C-O$ bonds are equivalent in length,having a bond order of $1.33$.
$3$. The structure is stabilized by resonance,where the negative charge is delocalized over the three oxygen atoms.
Therefore,all the given characteristics are correct.

(C) Isoelectronic species are those that have the same number of electrons.
$A$: $PO_4^{3-} (15+8 \times 4+3 = 50)$,$SO_4^{2-} (16+8 \times 4+2 = 50)$,$ClO_4^- (17+8 \times 4+1 = 50)$. All have $50$ electrons.
$B$: $CN^- (6+7+1 = 14)$,$N_2 (7+7 = 14)$,$C_2^{2-} (6+6+2 = 14)$. All have $14$ electrons.
$C$: $SO_3^{2-} (16+8 \times 3+2 = 42)$,$CO_3^{2-} (6+8 \times 3+2 = 32)$,$NO_3^- (7+8 \times 3+1 = 32)$. These are not isoelectronic.
$D$: $BO_3^{3-} (5+8 \times 3+3 = 32)$,$CO_3^{2-} (6+8 \times 3+2 = 32)$,$NO_3^- (7+8 \times 3+1 = 32)$. All have $32$ electrons.
Therefore,the set that does not contain isoelectronic species is $C$.

(D) To determine the isoelectronic pair,we calculate the total number of valence electrons for each species.
For $ClO_2^-$: $7 (Cl) + 2 \times 6 (O) + 1 (\text{charge}) = 20$ valence electrons.
For $ClF_2^+$: $7 (Cl) + 2 \times 7 (F) - 1 (\text{charge}) = 20$ valence electrons.
Since both $ClO_2^-$ and $ClF_2^+$ have $20$ valence electrons,they form an isoelectronic pair.
Therefore,the correct option is $D$.

(B) To determine the bond angle,we look at the hybridization and the number of lone pairs:
$1$. $C_2H_2$ ($sp$ hybridization,linear): $180^{\circ}$
$2$. $CH_4$ ($sp^3$ hybridization,tetrahedral): $109.5^{\circ}$
$3$. $NH_3$ ($sp^3$ hybridization,trigonal pyramidal,$1$ lone pair): $107^{\circ}$
$4$. $H_2O$ ($sp^3$ hybridization,bent,$2$ lone pairs): $104.5^{\circ}$
Thus,the correct decreasing order is $C_2H_2 > CH_4 > NH_3 > H_2O$.

(A) lone pair of electrons is defined as a pair of valence electrons that do not participate in chemical bonding between atoms.

(D) All the given molecules,$BCl_3$,$BBr_3$,and $BF_3$,possess a trigonal planar geometry due to $sp^2$ hybridization of the central boron atom.
In a perfect trigonal planar geometry,the bond angle between the atoms is $120^o$.
Since all three molecules have the same hybridization and geometry,the bond angle is identical for all of them.

(A) The chemical formula of acetic acid is $CH_3COOH$.
Total valence electrons in $CH_3COOH$ are: $4(C) + 3(H) + 6(O) + 6(O) + 1(H) = 24$ electrons.
In the Lewis structure of $CH_3COOH$:
- There are $8$ single bonds ($C-H$ $\times$ $3$,$C-C$ $\times$ $1$,$C-O$ $\times$ $1$,$O-H$ $\times$ $1$,$C=O$ $\times$ $1$ double bond).
- Total shared electrons = $8 \text{ bonds} \times 2 = 16$ shared electrons.
- Unshared electrons (lone pairs) are on oxygen atoms: The carbonyl oxygen has $2$ lone pairs ($4$ electrons) and the hydroxyl oxygen has $2$ lone pairs ($4$ electrons).
- Total unshared electrons = $4 + 4 = 8$ electrons.
Therefore,there are $16$ shared and $8$ unshared electrons.

(C) The bond lengths are as follows:
$1$. In $O_2$, the bond order is $2$, resulting in a bond length of approximately $121 \text{ pm}$.
$2$. In $O_3$, resonance results in a bond order of $1.5$, leading to a bond length of approximately $128 \text{ pm}$.
$3$. In $H_2O_2$, the $O-O$ bond is a single bond (bond order $1$), resulting in a bond length of approximately $148 \text{ pm}$.
Therefore, the increasing order of $O-O$ bond length is $O_2 < O_3 < H_2O_2$.

(A) The bond length is inversely proportional to the bond order.
$1$. For $CO$,the bond order is $3$.
$2$. For $CO_2$ $(O=C=O)$,the bond order is $2$.
$3$. For $CO_3^{2-}$,the resonance hybrid results in a bond order of $1.33$.
Since the bond order follows the order $CO > CO_2 > CO_3^{2-}$,the bond length follows the reverse order: $CO_3^{2-} > CO_2 > CO$.

(C) When two atoms approach each other to form a molecule,both attractive forces (between the nucleus of one atom and the electrons of the other) and repulsive forces (between the nuclei of both atoms and between the electrons of both atoms) come into play. $A$ stable bond is formed when the net attractive force balances the repulsive force at a specific internuclear distance.

(B) The correct answer is $(B)$.
In the ammonium ion $(NH_4^+)$,the nitrogen atom is covalently bonded to three hydrogen atoms.
The fourth hydrogen ion $(H^+)$ is attached to the nitrogen atom via a coordinate (dative) bond,where the lone pair on nitrogen is shared with the $H^+$ ion.
Since the entire species carries a positive charge and exists as an ion,the interaction between the $NH_4^+$ cation and any counter-anion (like $Cl^-$ in $NH_4Cl$) is electrovalent (ionic) in nature.
Thus,$NH_4Cl$ contains all three types of bonds.

(A) The formation of a chemical bond is an exothermic process,which means energy is released during the process.
Therefore,the potential energy of the system decreases,leading to a more stable state.

(D) chemical bond is formed due to the balance between the forces of attraction and repulsion between atoms at a specific internuclear distance.
Thus,it involves both $(a)$ and $(b)$.

(C) $A.$ The polarity increases in the order $HI < HBr < HCl < HF$ due to electronegativity differences. $HF$ is more polar than $HBr$,so option $A$ is incorrect.
$B.$ The statement that absolutely pure water does not contain any ions is false. Pure water undergoes auto-ionization $(2H_2O \rightleftharpoons H_3O^+ + OH^-)$,resulting in a small concentration of ions.
$C.$ Chemical bond formation occurs when the forces of attraction between atoms or ions overcome the forces of repulsion,leading to a stable,lower-energy state. This is true.
$D.$ In covalency,electrons are shared between atoms,not transferred. The transference of electrons is characteristic of ionic bond formation. Thus,option $D$ is incorrect.

(D) Blue vitriol is $CuSO_4 \cdot 5H_2O$.
In this compound,the $Cu^{2+}$ and $SO_4^{2-}$ ions are held together by ionic bonds.
The water molecules are coordinated to the $Cu^{2+}$ ion via coordinate covalent bonds.
Furthermore,the fifth water molecule is held in the crystal lattice by hydrogen bonds.
Therefore,it contains all the mentioned types of bonds.

(A) The structure of ammonium chloride $(NH_4Cl)$ consists of an ammonium ion $(NH_4^+)$ and a chloride ion $(Cl^-)$.
$1$. The bond between $NH_4^+$ and $Cl^-$ is an ionic bond $(1)$.
$2$. Inside the $NH_4^+$ ion,there are three $N-H$ covalent bonds $(3)$.
$3$. The fourth $N-H$ bond is formed by the donation of a lone pair from nitrogen to the hydrogen ion $(H^+)$,which is a coordinate (dative) bond $(1)$.
Therefore,the number of ionic,covalent,and coordinate bonds are $1, 3$,and $1$ respectively.

(C) The chemical species $O_2^{2-}$ represents the peroxide ion.
In this ion,each oxygen atom has an oxidation state of $-1$,and they are linked by a single covalent bond: $^-O-O^-$.
In contrast,$O^{2-}$ is the oxide ion,and $O_2^-$ is the superoxide ion.

(C) The bond length depends on the hybridization of the carbon atoms involved in the bond.
Ethane $(I)$ has $sp^3$ hybridized carbons, resulting in a $C-C$ single bond length of approximately $154 \ pm$.
Ethene $(II)$ has $sp^2$ hybridized carbons, resulting in a $C=C$ double bond length of approximately $133 \ pm$.
Acetylene $(III)$ has $sp$ hybridized carbons, resulting in a $C \equiv C$ triple bond length of approximately $120 \ pm$.
Benzene $(IV)$ has $sp^2$ hybridized carbons with resonance, resulting in a $C-C$ bond length of approximately $139 \ pm$ (intermediate between single and double bonds).
Therefore, the order of bond length is: Ethane $(154 \ pm) >$ Benzene $(139 \ pm) >$ Ethene $(133 \ pm) >$ Acetylene $(120 \ pm)$.
This corresponds to the order: $I > IV > II > III$.

(D) The structure of the $SO_3$ trimer $(S_3O_9)$ consists of a cyclic ring of alternating $S$ and $O$ atoms,where each $S$ atom is also bonded to two terminal oxygen atoms via double bonds.
In this structure,there are no direct $S-S$ bonds.
Therefore,the number of $S-S$ bonds is $0$.

(C) In $CuSO_4 \cdot 5H_2O$,there is an electrovalent (ionic) bond between $Cu^{2+}$ and $SO_4^{2-}$ ions.
Covalent bonds exist within the $SO_4^{2-}$ ion (between $S$ and $O$ atoms) and within the $H_2O$ molecules (between $H$ and $O$ atoms).
Coordinate covalent bonds exist between the $Cu^{2+}$ ion and four $H_2O$ molecules,forming the complex $[Cu(H_2O)_4]^{2+}$.

(A) Lewis acid is defined as an electron pair acceptor.
$SiCl_4$ has vacant $d-$orbitals on the $Si$ atom,allowing it to accept electron pairs.
$SO_3$ has an electron-deficient $S$ atom due to the electronegative oxygen atoms.
$Zn^{2+}$ is a metal cation that can accept electron pairs to form coordinate bonds.
$CO$ (carbon monoxide) has a lone pair on the carbon atom,which it can donate,making it a Lewis base rather than a Lewis acid.
Therefore,the correct answer is $A$.

(D) Lewis acid is defined as an electron pair acceptor.
$BF_3$ has an incomplete octet on the boron atom,making it an electron-deficient species and a Lewis acid.
$FeCl_3$ has an empty $d$-orbital and an incomplete octet,acting as a Lewis acid.
$SiF_4$ has vacant $d$-orbitals on the silicon atom,allowing it to accept electron pairs,thus acting as a Lewis acid.
$C_2H_4$ (ethene) has a $\pi$-bond but does not have an empty orbital to accept an electron pair in the standard sense of a Lewis acid.
Therefore,$C_2H_4$ is not a Lewis acid.

(A) According to the Lewis acid-base theory,a Lewis acid is an electron pair acceptor.
In the reaction $SnCl_2 + 2Cl^{-} \to SnCl_4$,the $SnCl_2$ molecule accepts a pair of electrons from each $Cl^{-}$ ion to form the complex ion $[SnCl_4]^{2-}$.
Therefore,$SnCl_2$ acts as a Lewis acid.

(C) $BF_3$ acts as a catalyst in various industrial processes because it is a strong Lewis acid.
According to the Lewis acid-base theory,a Lewis acid is an electron-pair acceptor.
In $BF_3$,the boron atom has an incomplete octet (only $6$ electrons in its valence shell),which makes it highly electron-deficient and capable of accepting a lone pair of electrons from other molecules.

(C) Lewis acid is defined as an electron pair acceptor.
$SnCl_4$,$FeCl_3$,and $BF_3$ have vacant orbitals or incomplete octets,allowing them to accept electron pairs.
$KCl$ is an ionic compound that dissociates into $K^+$ and $Cl^-$ ions in solution and does not act as an electron pair acceptor.
Therefore,the correct answer is $(C)$.

(A) Lewis acid is defined as an electron pair acceptor.
In the reaction $BCl_3 + PH_3 \to BCl_3:PH_3$,the boron atom in $BCl_3$ has an incomplete octet (only $6$ valence electrons) and accepts a lone pair of electrons from the phosphorus atom in $PH_3$.
Therefore,$BCl_3$ acts as the Lewis acid and $PH_3$ acts as the Lewis base.

(A) Lewis acid is a substance that can accept a pair of electrons.
In $ClF_3$,the central chlorine atom has vacant $d$-orbitals and can expand its octet to accept electron pairs.
$H_2O$ and $NH_3$ act as Lewis bases because they possess lone pairs of electrons that they can donate.
Therefore,$ClF_3$ is the correct Lewis acid.

(D) The average bond energies for the given bonds are approximately as follows:
$S = O$: $\approx 125 \ kcal/mol$
$C \equiv C$: $\approx 200 \ kcal/mol$
$C \equiv N$: $\approx 213 \ kcal/mol$
$N \equiv N$: $\approx 226 \ kcal/mol$
Comparing these values,the $N \equiv N$ bond has the highest average bond energy.

(D) The Born-Haber cycle is a thermodynamic cycle used to relate the lattice energy of an ionic crystal to other thermodynamic properties. It is primarily used to calculate $Lattice \ energy$,but it can also be used to determine $Electron \ affinity$,$Ionization \ energy$,and $Crystal \ energy$ by applying Hess's Law. Therefore,all of these properties can be determined using the cycle.

(A) In $OF_4$,the oxidation state of oxygen would be $+4$.
Oxygen is the second most electronegative element and typically exhibits oxidation states of $-2, -1, +1$,and $+2$.
Since oxygen cannot expand its octet due to the absence of $d$-orbitals,it cannot achieve a $+4$ oxidation state.
Therefore,$OF_4$ is not a theoretically possible molecule.

(D) The cyclobutadienyl anion $({C_4}{H_4})^{2-}$ consists of a four-membered ring with two double bonds and two negative charges on the carbons.
Each double bond contributes $2$ $\pi$ electrons,totaling $4$ $\pi$ electrons.
The two negative charges each represent a lone pair of electrons in a $p$-orbital,which also participate in the $\pi$ system.
Therefore,total $\pi$ electrons $= 4 \text{ (from double bonds)} + 2 \times 2 \text{ (from negative charges)} = 8$ $\pi$ electrons.
Thus,the correct option is $D$.

(A) The bond length is inversely proportional to the bond order.
$I.$ $C_2H_4$ $(CH_2=CH_2)$: Bond order = $2$, Bond length $\approx 134 \ pm$
$II.$ $C_2H_2$ $(CH \equiv CH)$: Bond order = $3$, Bond length $\approx 120 \ pm$
$III.$ $C_6H_6$ (Benzene): Bond order = $1.5$, Bond length $\approx 139 \ pm$
$IV.$ $C_2H_6$ $(CH_3-CH_3)$: Bond order = $1$, Bond length $\approx 154 \ pm$
Comparing the bond orders: $C_2H_6 (1) < C_6H_6 (1.5) < C_2H_4 (2) < C_2H_2 (3)$.
Therefore, the decreasing order of bond length is: $IV (154 \ pm) > III (139 \ pm) > I (134 \ pm) > II (120 \ pm)$.

(C) Isoelectronic species are those that have the same number of electrons.
For $CN^{-}$: $C(6) + N(7) + 1 = 14$ electrons.
For $CO$: $C(6) + O(8) = 14$ electrons.
Since both have $14$ electrons,they are isoelectronic.
Additionally,they are isosteres because they have the same number of atoms and the same number of electrons.