Covalent bonding Questions in English

(C) The nitrogen molecule $(N_2)$ consists of two nitrogen atoms connected by a triple bond $(N \equiv N)$.
$A$ triple bond consists of one sigma $(\sigma)$ bond and two pi $(\pi)$ bonds.
Each bond represents a shared pair of electrons,meaning $3 \times 2 = 6$ electrons are involved in the bonding process.

(B) The bond dissociation energy of halogens follows the order: $Cl_2 > Br_2 > F_2 > I_2$.
In $F_2$,the interelectronic repulsion between the lone pairs of the small $F$ atoms is very high,which weakens the $F-F$ bond.
Therefore,$Cl_2$ has the highest bond energy among the given halogens.

(A) At room temperature,$BeCl_2$ exists as a dimer $(Be_2Cl_4)$ or polymer in the solid state.
It has residual bonding capacity because the $Be$ atom is electron-deficient (incomplete octet) and can accept electron pairs from other molecules to form coordinate bonds,thereby attaining further stability.

(D) $CCl_4$ is a covalent compound and does not ionize to provide $Cl^-$ ions in the solution.
Since $AgNO_3$ reacts with free $Cl^-$ ions to form a white precipitate of $AgCl$,no reaction occurs between $CCl_4$ and $AgNO_3$.
Therefore,the correct answer is $D$.

(B) Lewis acid is defined as an electron-pair acceptor.
$BCl_3$ has an incomplete octet on the central Boron atom (only $6$ electrons),making it an electron-deficient species and a strong Lewis acid.
$PCl_3$ has a lone pair on the Phosphorus atom,which can be donated,making it a Lewis base.
$NCl_3$ also has a lone pair on the Nitrogen atom,making it a Lewis base.
Therefore,only $BCl_3$ acts as a Lewis acid.

(B) The bond energy of halogens is influenced by the small size of the fluorine atom,which leads to significant inter-electronic repulsion between the lone pairs of the two fluorine atoms.
This makes the $F-F$ bond weaker than the $Cl-Cl$ bond.
The general order of bond dissociation energy for halogens is $Cl_2 > Br_2 > F_2 > I_2$.
Therefore,$Cl_2$ has the highest bond energy.

(A) An electron-deficient molecule is one in which the central atom has fewer than $8$ electrons in its valence shell,or it lacks sufficient electrons to form conventional covalent bonds.
In $B_2H_6$ (diborane),each boron atom is bonded to four hydrogen atoms,but there are only $12$ valence electrons available for the entire molecule,which is insufficient to form standard two-center-two-electron $(2c-2e)$ bonds for all linkages.
It contains two $3$-center-$2$-electron $(3c-2e)$ bonds,also known as banana bonds,making it an electron-deficient molecule.
In contrast,$C_2H_6$,$PH_3$,and $SiH_4$ are electron-precise molecules where the octet rule is satisfied for the central atoms.

(D) In $PH_6^+$,the phosphorus atom carries a positive charge,which increases its electronegativity and electron-attracting tendency.
This results in a higher degree of ionic character in the $P-H$ bond compared to the other neutral phosphine species.
Therefore,$PH_6^+$ has the least covalent $P-H$ bond.

(C) The bond energy depends on the size of the atoms and the extent of orbital overlap. $O-O$ bond has the lowest bond energy due to high inter-electronic repulsion between the lone pairs of small oxygen atoms. As we move down the group,the atomic size increases,which decreases the bond energy. However,the $S-S$ bond is stronger than $Se-Se$ and $Te-Te$ due to better orbital overlap compared to the larger atoms. Therefore,among the given options,the $S-S$ bond has the highest bond energy.

(B) In an $HCl$ molecule,the electronegativity difference between $H$ $(2.1)$ and $Cl$ $(3.0)$ is approximately $0.9$.
Because the electronegativity difference is greater than $0$ but less than $1.7$,the bond formed is a polar covalent bond,where the shared pair of electrons is shifted towards the more electronegative chlorine atom.

(B) The bond length decreases as the bond order increases.
The bond lengths are:
$C-C$ (single bond) = $1.54 \ \mathring{A}$
$C=C$ (double bond) = $1.34 \ \mathring{A}$
$C\equiv C$ (triple bond) = $1.20 \ \mathring{A}$
Therefore,the correct order of increasing bond length is $C\equiv C < C=C < C-C$.

(D) Anhydrous $AlCl_3$ acts as a Lewis acid because it is an electron-deficient compound.
In the Friedel-Crafts reaction,it reacts with the alkyl halide to generate a carbocation (electrophile).
The reaction is: $CH_3Cl + AlCl_3 \to CH_3^+ + AlCl_4^-$

(C) Covalent compounds dissolve in non-polar solvents,while ionic compounds dissolve in polar solvents.
Naphthalene $(C_{10}H_8)$ is a non-polar covalent compound,and kerosene is a non-polar solvent.
According to the principle of 'like dissolves like',non-polar solutes dissolve in non-polar solvents. Therefore,naphthalene dissolves easily in kerosene.

(D) . In graphite,each carbon atom is $sp^2$ hybridized and is covalently bonded to three other carbon atoms in the same layer to form a hexagonal network. The layers are held together by weak Vander Waal's forces,but the atoms within the layers are joined by covalent bonding.

(C) $CCl_4$ is a covalent compound.
It does not ionize in water to produce $Cl^-$ ions.
Since $AgNO_3$ reacts with $Cl^-$ ions to form a white precipitate of $AgCl$,the absence of $Cl^-$ ions means no precipitate is formed.
$CCl_4 + AgNO_3 \to \text{No reaction}$

(C) In $BF_{3}$,the boron atom is $sp^{2}$ hybridized and has an empty $p$-orbital.
Fluorine atoms have lone pairs in their $2p$-orbitals.
There is a back-bonding effect where fluorine donates electron density into the empty $p$-orbital of boron,creating a $p\pi-p\pi$ bond.
This back-bonding gives the $B-F$ bond partial double bond character,making it shorter and stronger than a typical single bond.
Therefore,the statement that this interaction cannot explain the shortness and strength of the bond is incorrect.

(B) In $BCl_3$,the boron atom has an empty $p$-orbital and chlorine has lone pairs of electrons. This leads to $p\pi - p\pi$ back bonding,which reduces the electron deficiency of the boron atom and stabilizes the monomeric form.
In $BH_3$,there is no such back bonding because hydrogen does not have lone pairs. To complete its octet and achieve stability,$BH_3$ undergoes dimerization to form $B_2H_6$ (diborane),which involves $3c-2e^-$ (three-center two-electron) bonds.

(B) In diamond,each carbon atom is $sp^3$ hybridized and is bonded to four other carbon atoms through strong covalent bonds,forming a rigid three-dimensional network structure.

(B) The bond dissociation energy of the $N - N$ bond is higher than that of $C - C$,$O - O$,and $F - F$ bonds due to the small size of nitrogen atoms and the presence of lone pair-lone pair repulsions in the other cases,although $N - N$ single bond is relatively weak compared to $N \equiv N$ triple bond,it is stronger than the $O - O$ and $F - F$ single bonds.

(C) Aluminum chloride $(AlCl_3)$ has an incomplete octet around the central aluminum atom.
It has only $6$ electrons in its valence shell.
Therefore,it acts as an electron-pair acceptor.
According to the Lewis concept,a substance that accepts an electron pair is a Lewis acid.

(B) Calcium carbide $(CaC_2)$ is an ionic carbide containing the acetylide ion,$[C \equiv C]^{2-}$.
In the acetylide ion,the two carbon atoms are bonded by a triple bond.
$A$ triple bond consists of $1$ sigma $(\sigma)$ bond and $2$ pi $(\pi)$ bonds.
Therefore,the correct answer is $1$ sigma and $2$ pi bonds.

(B) single covalent bond involves the sharing of $2$ electrons.
Therefore,a double bond involves the sharing of $2 \times 2 = 4$ electrons.

(A) The bond in $H-Cl$ is a polar covalent bond,formed by the sharing of electrons between $H$ and $Cl$ atoms.

(A) The carbon suboxide molecule,$C_3O_2$,consists of a linear chain of three carbon atoms with two oxygen atoms attached to the terminal carbons.
Each carbon atom must satisfy the octet rule and form four bonds.
The structure is $O=C=C=C=O$,where each carbon atom is $sp$ hybridized.

(C) The maximum covalent character is found between identical atoms.
This is because the electronegativity difference between two identical atoms is $0$,resulting in a purely non-polar covalent bond.

(D) Covalent compounds are generally non-polar in nature. According to the principle of $ \text{'like dissolves like'} $,non-polar solutes dissolve in non-polar solvents.

(D) In a bond between two identical atoms,the difference in electronegativity is zero. Therefore,the shared pair of electrons is distributed equally between them,resulting in a non-polar covalent bond.

(C) According to the valence bond theory,a covalent bond is formed by the overlapping of atomic orbitals containing electrons with opposite spins.
If the electrons have the same spin,the orbitals will not overlap effectively,and consequently,bonding will not be possible.

(A) covalent bond is formed by the sharing of electrons between atoms of non-metals.
$H_2$ is a homonuclear diatomic molecule where two hydrogen atoms share a pair of electrons to form a covalent bond.
$CaS$,$KCl$,and $Na_2S$ are ionic compounds formed by the electrostatic attraction between metal cations and non-metal anions.

(C) The bond energy is directly proportional to the bond order (number of bonds between atoms).
$N_2$ has a triple bond $(N \equiv N)$,which has a bond order of $3$.
$O_2$ has a double bond $(O=O)$,which has a bond order of $2$.
$H_2$ has a single bond $(H-H)$,which has a bond order of $1$.
$CH_4$ involves $C-H$ single bonds.
Since $N \equiv N$ has the highest bond order,it requires the highest energy to dissociate.

(C) The bond length of a molecule $AB$ is given by the sum of the covalent radii of atoms $A$ and $B$,i.e.,$d_{AB} = r_A + r_B$.
For $H_2$,the internuclear distance is $74 \, pm$,so the covalent radius of $H$ is $r_H = \frac{74}{2} = 37 \, pm$.
For $Cl_2$,the internuclear distance is $198 \, pm$,so the covalent radius of $Cl$ is $r_{Cl} = \frac{198}{2} = 99 \, pm$.
Therefore,the bond length of $HCl$ is $d_{HCl} = r_H + r_{Cl} = 37 + 99 = 136 \, pm$.

(B) 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.
In $BF_3$,the central Boron atom is bonded to $3$ Fluorine atoms.
Each $B-F$ bond consists of $2$ shared electrons,resulting in a total of $3 \times 2 = 6$ electrons in the valence shell of Boron.
Since $6 < 8$,the central Boron atom in $BF_3$ does not follow the octet rule (it is an electron-deficient molecule).

(A) The bond strength is directly proportional to the bond order and the $s$-character of the hybrid orbitals involved in the bond formation.
For $C-C$ (alkane),the hybridization is $sp^3$ ($25\% \ s$-character).
For $C=C$ (alkene),the hybridization is $sp^2$ ($33.3\% \ s$-character).
For $C \equiv C$ (alkyne),the hybridization is $sp$ ($50\% \ s$-character).
As the $s$-character increases,the bond becomes shorter and stronger.
Therefore,the correct order of bond strength is $C-C < C=C < C \equiv C$.

(D) In $BF_3$,the central boron atom is bonded to three fluorine atoms by single covalent bonds.
This results in a total of $6$ valence electrons around the boron atom.
Since the octet rule requires $8$ electrons for stability,$BF_3$ is an electron-deficient compound that does not follow the octet rule.

(C) $AlCl_3$ is an electron-deficient compound with an incomplete octet of $Al$ atom.
It acts as a Lewis acid because it can accept a lone pair of electrons.
In many organic reactions,such as Friedel-Crafts reactions,it is used as an acid catalyst.

(D) Lewis acid is defined as an electron-pair acceptor.
In $(CH_3)_3B$,the central boron atom has only $6$ electrons in its valence shell,meaning it has an incomplete octet.
Therefore,it can accept a lone pair of electrons to complete its octet,acting as a Lewis acid.
In contrast,$(CH_3)_2O$,$(CH_3)_3P$,and $(CH_3)_3N$ all have central atoms with complete octets and at least one lone pair of electrons,which allows them to act as Lewis bases (electron-pair donors).

(B) $BF_3$ is a Lewis acid due to the incomplete octet of Boron.
In $BF_3$,the Boron atom has a vacant $2p$-orbital,and the Fluorine atom has lone pairs in its $2p$-orbital.
This allows for $p\pi - p\pi$ back-bonding from the Fluorine to the Boron,which increases the bond order and strengthens the $B-F$ bond.
In $CF_4$,the Carbon atom has a complete octet and no vacant orbitals,so no such back-bonding is possible.
Therefore,the $B-F$ bond in $BF_3$ is stronger than the $C-F$ bond in $CF_4$ due to this back-bonding interaction.

(C) In $SF_6$,the central sulfur atom $(S)$ undergoes excitation to form six covalent bonds with fluorine atoms.
Ground state electronic configuration of $S$ is $[Ne] 3s^2 3p^4 3d^0$.
To form $6$ bonds,$S$ promotes electrons from $3s$ and $3p$ orbitals to the empty $3d$ orbitals.
This results in the $2^{nd}$ excited state configuration: $3s^1 3p^3 3d^2$.
Since it has $6$ unpaired electrons,its covalency is $6$.
The oxidation number of $S$ in $SF_6$ is calculated as: $x + 6(-1) = 0 \implies x = +6$.
Thus,$S$ is in an excited state and its oxidation number is $+6$.

(B) The molecules are $B_2F_4$ and $B_2Cl_4$. The bond length $y$ (in $B_2Cl_4$) is greater than the bond length $x$ (in $B_2F_4$).
Fluorine is more electronegative than chlorine. In $B_2F_4$, the $B-F$ bond is highly polar, which increases the $s$-character in the $B-B$ bond due to Bent's rule (electronegative atoms prefer orbitals with more $p$-character, leaving more $s$-character for the $B-B$ bond). Higher $s$-character leads to a shorter bond length.
Additionally, the back-bonding from $F$ to $B$ is more effective than from $Cl$ to $B$, which also influences the bond order and length. Experimental values confirm $x \approx 172 \ pm$ and $y \approx 175 \ pm$.
Therefore, $y > x$.

(A) Carbon belongs to the second period and has a maximum covalency of $4$ because it lacks $d$-orbitals in its valence shell.
Therefore,it cannot expand its octet to form a hexacoordinated species like $[CCl_6]^{2-}$.
Silicon,Germanium,and Tin belong to the third,fourth,and fifth periods respectively,and they possess vacant $d$-orbitals,allowing them to form stable octahedral complexes like $[SiF_6]^{2-}$,$[GeF_6]^{2-}$,and $[SnCl_6]^{2-}$.

(C) In $BCl_3$,the large size of chlorine atoms creates steric hindrance,preventing the formation of a bridged dimer structure. Additionally,$p\pi - p\pi$ back bonding stabilizes the monomeric $BCl_3$ molecule.
In $BH_3$,the small size of hydrogen atoms allows them to easily fit between boron atoms to form the bridged structure $(B_2H_6)$,which is stabilized by $3c-2e$ (three-center two-electron) bonds.

(D) Aluminium bromide $(AlBr_3)$ exists as a dimer,$Al_2Br_6$,in the solid and vapour states to complete the octet of the aluminium atom.
In this dimeric structure,two $AlBr_3$ units are linked by two bridging bromine atoms,where each aluminium atom achieves a tetrahedral geometry.
Thus,the structure is best represented as a dimer.

(B) The molecular formula of phosphorus $(V)$ oxide is $P_4O_{10}$.
In its structure,each phosphorus atom is bonded to four oxygen atoms.
There are $6$ $P-O-P$ linkages,which involve $12$ $P-O$ single bonds.
Additionally,each of the $4$ phosphorus atoms forms one terminal $P=O$ double bond.
Therefore,the structure contains $12$ $P-O$ single bonds and $4$ $P=O$ double bonds.

(A) When two hydrogen atoms approach each other to form a bond,the attractive forces between the nuclei and electrons dominate,leading to a decrease in the potential energy of the system.
This release of energy results in the formation of a stable $H_2$ molecule.

(D) Lewis acid is defined as an electron pair acceptor.
In $AlCl_3$,the central aluminum atom has only $6$ electrons in its valence shell,which is an incomplete octet.
Therefore,$AlCl_3$ can accept an electron pair to complete its octet,making it a Lewis acid.

(B) The formation of a $p \pi-d \pi$ bond requires one atom to have a lone pair of electrons in a $p$-orbital and the other atom to have a vacant $d$-orbital.
Oxygen has a lone pair in its $p$-orbital. For $p \pi-d \pi$ bonding,the other element must possess vacant $d$-orbitals in its valence shell.
Carbon $(2s^2 2p^2)$,Nitrogen $(2s^2 2p^3)$,and Boron $(2s^2 2p^1)$ belong to the second period and lack $d$-orbitals in their valence shell.
Phosphorus $(3s^2 3p^3 3d^0)$ belongs to the third period and possesses vacant $3d$-orbitals,allowing it to participate in $p \pi-d \pi$ bonding with oxygen (e.g.,in $PO_4^{3-}$ or $P=O$ bonds).

(D) The octet rule states that atoms tend to form bonds such that they have $8$ electrons in their valence shell.
$BH_3$ is an electron-deficient molecule with $6$ valence electrons around $B$.
$ZnCl_2$ is a covalent compound where $Zn$ does not complete its octet.
$NO_2$ is an odd-electron molecule with $7$ valence electrons on the nitrogen atom.
$SnCl_4$ has a central $Sn$ atom bonded to $4$ $Cl$ atoms,resulting in $8$ electrons in the valence shell of $Sn$,thus following the octet rule.

(A) When two hydrogen atoms approach each other,the potential energy of the system decreases due to attractive forces between the nucleus of one atom and the electron of the other.
As they get closer,the potential energy reaches a minimum value at the equilibrium bond distance.
If they get even closer,the repulsive forces between the nuclei and between the electrons dominate,causing the potential energy to increase sharply.
The plot in option $A$ correctly depicts this behavior,showing a minimum potential energy at the equilibrium distance.

(A) Bond energy depends on the size of the atoms involved in the bond. As the atomic size increases down the group,the bond length increases and the bond strength (bond energy) decreases. The order of atomic size is $C < Si < Ge < Sn$. Therefore,the order of bond energy is $C-C > Si-Si > Ge-Ge > Sn-Sn$. Thus,the $C-C$ bond has the highest bond energy.