Covalent bonding Questions in English

(A) The strength of a covalent bond depends on the electronegativity difference between the bonded atoms.
Greater electronegativity difference leads to higher ionic character,which increases the bond strength.
The electronegativity values are: $F = 4.0$,$Cl = 3.0$,$Br = 2.8$.
The electronegativity differences are:
$Br-F$: $|4.0 - 2.8| = 1.2$
$F-F$: $|4.0 - 4.0| = 0$
$Cl-F$: $|4.0 - 3.0| = 1.0$
$Br-Cl$: $|3.0 - 2.8| = 0.2$
Since $Br-F$ has the largest electronegativity difference,it is the strongest bond.

(C) In the $N_2$ molecule,the two nitrogen atoms are connected by a triple bond.
Each nitrogen atom contributes $3$ electrons to the bond formation.
Therefore,the total number of electrons involved in the bond formation is $3 + 3 = 6$ electrons.

(C) The covalency of an atom is the number of electrons it shares with other atoms when forming chemical bonds.
In a covalent bond,electrons are shared between atoms to achieve a stable electronic configuration.
For example,in $CH_4$,carbon shares its four valence electrons with four electrons from four different $H$ atoms,forming $4$ covalent bonds.
Thus,the covalency of carbon is $4$.

(A) In graphite,each carbon atom is $sp^{2}$-hybridised and is bonded to three other carbon atoms by covalent bonds,forming a hexagonal layer structure.

(D) When two similar non-metallic atoms form a bond,they share their valence electrons equally because their electronegativities are identical.
This type of bond is known as a non-polar covalent bond.
Therefore,the electron pair is equally shared between the two atoms.

(A) covalent bond is formed when the electronegativity difference between two atoms is $1.7$ or less.
If the electronegativity difference is $1.7$,the bond is considered to have $50\%$ ionic character and $50\%$ covalent character.
Therefore,the correct condition for covalent bond formation is an electronegativity difference $\leq 1.7$.

(B) Similar atoms have identical electronegativity,so they share electrons equally to form a covalent bond.

(B) Water is a polar solvent,while covalent compounds are generally non-polar. According to the principle of "like dissolves like",non-polar substances do not dissolve in polar solvents. Therefore,covalent compounds are usually insoluble in water.

(C) The correct answer is $(C)$.
$BCl_3$ is an electron-deficient compound because the central Boron atom has only $6$ electrons in its valence shell after forming three covalent bonds with Chlorine atoms,which is less than the stable octet of $8$ electrons.

(C) To complete an octet,silicon requires $4$ more electrons.
Since the energy required to remove $4$ electrons is very high and gaining $4$ electrons is also energetically unfavorable,silicon prefers to share its $4$ valence electrons with other atoms to form covalent bonds.

(A) When two hydrogen atoms approach each other to form a $H_2$ molecule,the attractive forces between the nuclei and electrons of the atoms dominate over the repulsive forces.
As the atoms get closer,the potential energy of the system decreases,reaching a minimum at the equilibrium bond distance.
This decrease in potential energy results in the release of energy,making the $H_2$ molecule more stable than the two isolated hydrogen atoms.

(C) In $NaCl$ and $MgCl_2$,the bonding is ionic due to the transfer of electrons between metal and non-metal atoms.
Brass is an alloy of copper and zinc,which exhibits metallic bonding.
$H_2O$ is formed by the sharing of electrons between oxygen and hydrogen atoms,which is a characteristic of covalent bonding.
Therefore,the correct option is $C$.

(A) Diamond is organized in a giant lattice structure with strong covalent bonds between carbon atoms.
Each carbon atom forms $4$ covalent bonds with other carbon atoms.
Therefore,it has a rigid structure and cannot conduct electricity due to the lack of free electrons.

(A) covalent bond is formed by the sharing of electrons between atoms,typically between non-metals.
$H_2$ consists of two hydrogen atoms sharing a pair of electrons,making it a covalent molecule.
$CaO$,$KCl$,and $Na_2S$ are ionic compounds formed by the electrostatic attraction between metal cations and non-metal anions.
Therefore,the correct option is $A$.

(A) covalent bond is formed by the sharing of electrons between atoms.
This occurs most readily when the atoms involved have similar electronegativities,as neither atom is strong enough to completely remove an electron from the other.
For example,in the $H_2$ molecule,two $H$ atoms with identical electronegativity share electrons to form a covalent bond.
Similarly,in $CH_4$,$C$ and $H$ have relatively similar electronegativities,leading to the formation of $C-H$ covalent bonds.

(D) Bonds between identical non-metal atoms are purely covalent because both atoms have the same electronegativity.
Therefore,the bonded atoms exert an equal pull on the shared pair of electrons,resulting in an equal sharing of the electron pair.
Examples include $F_2$,$O_2$,and $N_2$.

(A) Germanium $(Ge)$ exhibits a giant covalent network structure,similar to diamond.
$A$. Germanium: Covalent bonding.
$B$. Sodium chloride $(NaCl)$: Ionic bonding.
$C$. Solid neon: Van der Waals forces (London dispersion forces).
$D$. Copper $(Cu)$: Metallic bonding.
Therefore,the correct answer is $A$.

(A) The $HCl$ molecule is formed by the sharing of electrons between the hydrogen atom and the chlorine atom.
Since both atoms share a pair of electrons to complete their valence shells,the bond formed is a covalent bond.

(B) The bond length of a molecule $AB$ can be estimated using the covalent radii of atoms $A$ and $B$ as $d_{AB} = r_A + r_B$.
For $H_2$ molecule,the bond length is $74 \, pm$,so the covalent radius of $H$ is $r_H = \frac{74}{2} = 37 \, pm$.
For $Cl_2$ molecule,the bond length is $198 \, pm$,so the covalent radius of $Cl$ is $r_{Cl} = \frac{198}{2} = 99 \, pm$.
Therefore,the bond length of $HCl$ is $r_H + r_{Cl} = 37 + 99 = 136 \, pm$.

(C) Covalent compounds consist of molecules held together by weak intermolecular forces.
Because of these weak forces,they generally possess low boiling points and low melting points.
They are typically poor conductors of electricity and heat,and they are generally insoluble in polar solvents like water.

(B) $ (b) $ In a water molecule $(H_2O)$,the oxygen atom is bonded to two hydrogen atoms by sharing electron pairs. This sharing of electrons results in the formation of a covalent bond. The structure is represented as:
$ H-O-H $

(B) covalent bond,also called a molecular bond,is a chemical bond that involves the sharing of electron pairs between atoms.
These electron pairs are known as shared pairs or bonding pairs.
The stable balance of attractive and repulsive forces between atoms,when they share electrons,is known as covalent bonding.

(B) covalent bond is formed by the sharing of electrons between atoms.
It is a directional bond because it depends on the orientation of the atomic orbitals involved in overlapping.
Therefore,the statement that the bond is non-directional is incorrect.
Ionic bonds are typically considered non-directional.

(D) Bond dissociation energy is inversely proportional to the bond length,which in turn depends on the atomic size of the bonded atoms.
Smaller atoms form shorter,stronger bonds,resulting in higher bond dissociation energy.
Comparing the given values: $485 \ kJ \ mol^{-1} > 328 \ kJ \ mol^{-1} > 276 \ kJ \ mol^{-1} > 240 \ kJ \ mol^{-1}$.
Since element $D$ has the highest bond dissociation energy $(485 \ kJ \ mol^{-1})$,it must have the smallest atomic size among the given elements.

(B) triple bond is formed when an atom needs $3$ electrons to complete its octet,typically seen in group $15$ elements like nitrogen $(N_2)$.
For $N$ $(Z=7)$,the electronic configuration is $1s^2, 2s^2, 2p^3$.
Thus,the correct configuration is $1s^2, 2s^2, 2p^3$.

(A) 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 $PCl_5$,the central phosphorus atom is bonded to $5$ chlorine atoms,resulting in $10$ electrons in its valence shell (an expanded octet).
Therefore,$PCl_5$ does not follow the octet rule.

(B) The electronic configuration of nitrogen $(N)$ is $1s^2 2s^2 2p^3$.
To complete its octet,each nitrogen atom requires $3$ more electrons.
Therefore,in the $N_2$ molecule,each nitrogen atom shares $3$ electrons with the other to form a triple covalent bond $(N \equiv N)$.

(B) The correct answer is $(B)$.
Electronic configuration of $_{16}S$ is $1s^2, 2s^2, 2p^6, 3s^2, 3p^4$.
Sulfur has $6$ valence electrons in its outermost shell.
To complete its octet,sulfur requires $2$ more electrons.
It shares $2$ electrons with two hydrogen atoms,resulting in the formation of $2$ covalent bonds.

(B) The electronic configuration of nitrogen $(Z=7)$ is $1s^2, 2s^2, 2p^3$.
Nitrogen has $5$ valence electrons.
In an ammonia $(NH_3)$ molecule,nitrogen shares $3$ of its valence electrons with $3$ hydrogen atoms to form $3$ covalent bonds.
After sharing,nitrogen gains $3$ electrons from the hydrogen atoms,completing its octet.
Therefore,the total number of electrons in the valence shell of nitrogen in $NH_3$ is $5 + 3 = 8$.

(C) The bond joining two hydrogen atoms in a hydrogen gas molecule $(H_2)$ is a classic $Covalent \ Bond$.
Since both hydrogen atoms are identical,neither can completely transfer an electron to the other to form an ionic bond.
Instead,each hydrogen atom shares its single electron with the other,resulting in the formation of a shared electron pair,which constitutes a covalent bond.

(C) The strength of a covalent bond is directly proportional to the bond order and inversely proportional to the bond length.
Among the given options,the $C - N$ bond can exist as a triple bond $(C \equiv N)$ in compounds like hydrogen cyanide $(HCN)$,which has a significantly higher bond energy compared to single bonds like $C - C$,$C - H$,or $C - O$.
Therefore,the $C \equiv N$ bond is the strongest.

(C) The major binding force in diamond,silicon,and quartz is the covalent bond.
These substances form giant covalent network structures (network solids) where atoms are linked by strong covalent bonds.

(D) multiple covalent bond (double or triple bond) exists when two atoms share more than one pair of electrons.
In $H_2$,there is a single bond $(H-H)$.
In $N_2$,there is a triple bond $(N \equiv N)$.
In $C_2H_4$ (ethene),there is a double bond between the carbon atoms $(CH_2=CH_2)$.
Therefore,both $N_2$ and $C_2H_4$ contain multiple covalent bonds.

(D) covalent bond involves the sharing of electrons between atoms.
$A$ covalent bond can be polar or non-polar depending on the electronegativity difference between the bonded atoms.
Therefore,both statements $(a)$ and $(b)$ are correct.

(B) The bond between $P$ and $Cl$ in $PCl_5$ is primarily covalent due to the electronegativity difference between $P$ $(2.19)$ and $Cl$ $(3.16)$.
Since there is a significant difference in electronegativity,the bond exhibits partial ionic character.
Therefore,the $P-Cl$ bond is covalent with some ionic character.

(A) When two atoms share electrons to form a bond,it is called a covalent bond.
In the compound $AB$,since the atoms are different ($A$ and $B$),they will have different electronegativities.
The sharing of electrons between two different atoms results in a polar covalent bond due to the electronegativity difference between $A$ and $B$.

(B) Calcium carbide $(CaC_2)$ consists of $Ca^{2+}$ and acetylide ions $(C_2^{2-})$.
The acetylide ion has a triple bond between the two carbon atoms,which is represented as $[C \equiv C]^{2-}$.
$A$ triple bond consists of $1 \ \sigma$ bond and $2 \ \pi$ bonds.
Therefore,there is $1 \ \sigma$ and $2 \ \pi$ bond between the two carbon atoms.

(C) In a double bond,two pairs of electrons are shared between two atoms,which corresponds to a total of $4$ electrons.
For example,in ethene $(H_2C=CH_2)$,each carbon atom shares two electrons with the other carbon atom to form the double bond.

(C) The bond strength depends on the bond order.
$C\equiv C$ has a bond order of $3$,$C=C$ has a bond order of $2$,and $C-C$ has a bond order of $1$.
As the bond order increases,the bond strength increases.
Therefore,$C\equiv C$ is the strongest bond.

(D) In trisilylamine,the nitrogen atom is $sp^2$ hybridized and has a lone pair of electrons in a $p$-orbital. Silicon has vacant $3d$-orbitals. The lone pair on nitrogen is donated into the vacant $d$-orbital of silicon,resulting in $p\pi - d\pi$ back bonding.

(D) The structure of $P_4O_{10}$ consists of a $P_4$ tetrahedron where each edge is bridged by an oxygen atom ($P-O-P$ bonds).
There are $6$ such $P-O-P$ bridges,contributing $12$ sigma bonds.
Additionally,each of the $4$ phosphorus atoms is bonded to a terminal oxygen atom via a double bond $(P=O)$.
Each $P=O$ bond consists of $1$ sigma bond and $1$ pi bond.
Thus,there are $4$ sigma bonds from the $P=O$ groups.
Total number of sigma bonds = $12$ (from $P-O-P$ bridges) + $4$ (from $P=O$ bonds) = $16$ sigma bonds.

(A) The correct electronic formula for the $Cl_{2}$ molecule is $:\,\mathop {Cl}\limits_{..}^{..} \,:\,\mathop {Cl}\limits_{..}^{..} \,:$ (often represented as $:\,\mathop {Cl}\limits_{..}^{..} \, - \,\mathop {Cl}\limits_{..}^{..} \,:$).
Each chlorine atom has $7$ valence electrons.
To achieve a stable octet,each chlorine atom shares $1$ electron with the other,forming a single covalent bond.
Thus,each chlorine atom is surrounded by $3$ lone pairs and $1$ shared pair of electrons.

(B) $Double \ covalent \ bond$ is formed when two pairs of electrons are shared between the atoms rather than just one pair.

(A) The diborane $(B_2H_6)$ molecule contains two types of $B-H$ bonds:
$(i)$ $B-H_t$ (terminal): These are normal two-centre two-electron $(2c-2e)$ covalent bonds.
(ii) $B-H_b$ (bridging): These are three-centre two-electron $(3c-2e)$ bonds,often called banana bonds.
Therefore,the two types of bonds present are covalent and three-centre bonds.

(C) The bond dissociation energy depends on the bond order and the strength of the bond.
$N_2$ contains a triple bond $(N \equiv N)$,which is significantly stronger than the single bond in $H_2$ $(436 \ kJ/mol)$,the $C-H$ bond in $CH_4$ $(413 \ kJ/mol)$,or the double bond in $O_2$ $(498 \ kJ/mol)$.
The bond dissociation energy of $N_2$ is approximately $945 \ kJ/mol$.
Therefore,the $N \equiv N$ bond requires the largest amount of energy to dissociate.

(A) The Lewis structure of the cyanide ion $(CN^-)$ consists of a triple bond between carbon and nitrogen $(C \equiv N)$.
In this structure,the carbon atom has $4$ valence electrons and the nitrogen atom has $5$ valence electrons.
For the $CN^-$ ion,the formal charge on an atom is calculated as: $\text{Formal Charge} = \text{Valence electrons} - \text{Non-bonding electrons} - \frac{1}{2} \times \text{Bonding electrons}$.
For Carbon: $4 - 2 - \frac{1}{2}(6) = 4 - 2 - 3 = -1$.
For Nitrogen: $5 - 2 - \frac{1}{2}(6) = 5 - 2 - 3 = 0$.
Wait,re-evaluating: The structure is $[:C \equiv N:]^-$. Carbon has $4$ valence electrons,$2$ non-bonding,and $6$ bonding. Formal charge on $C = 4 - 2 - 3 = -1$. Nitrogen has $5$ valence electrons,$2$ non-bonding,and $6$ bonding. Formal charge on $N = 5 - 2 - 3 = 0$.
Therefore,the formal negative charge is on the carbon atom $(C)$.

(C) When a chemical bond forms between two atoms,the potential energy of the resulting molecule is lower than the sum of the potential energies of the separate atoms. This is because bond formation is an exothermic process,where energy is released to achieve a more stable state.

(A) The strength of a covalent bond is inversely proportional to the bond length.
Among the given interhalogen compounds and the fluorine molecule,the bond length increases as the size of the halogen atom increases $(F < Cl < Br < I)$.
Therefore,the bond length order is $F-F < Cl-F < Br-F < I-F$.
Since the $F-F$ bond has the shortest bond length,it is the strongest bond among the options provided.

(A) The structure of white phosphorus $(P_4)$ consists of four phosphorus atoms arranged at the corners of a regular tetrahedron.
Each phosphorus atom is bonded to the other three phosphorus atoms by single covalent bonds.
There are a total of $6$ $P-P$ single bonds in the $P_4$ molecule.
Since each single bond represents one bonding electron pair,there are $6$ bonding electron pairs in white phosphorus.