Dipole moment Questions in English

(B) $CO_2$ has a linear geometry with the two $C=O$ bond dipoles pointing in opposite directions,which cancel each other out. Therefore,the net dipole moment of $CO_2$ is $0 \ D$.

(D) $BF_3$ has a trigonal planar geometry,$CCl_4$ has a tetrahedral geometry,and $BeCl_2$ has a linear geometry. In all these molecules,the bond dipoles cancel each other out due to their symmetrical shapes,resulting in a net dipole moment of $0 \ D$.

(D) The dipole moment $\mu$ is defined as the product of the magnitude of the charge $(q)$ and the distance of separation $(d)$ between the charges, i.e., $\mu = q \times d$.
In the hydrogen halides $(HF, HCl, HBr, HI)$, the electronegativity of the halogen decreases as we move down the group $(F > Cl > Br > I)$.
Consequently, the electronegativity difference between hydrogen and the halogen is highest for $HF$, leading to the largest partial charges $(q)$.
Although the bond length $(d)$ increases from $HF$ to $HI$, the effect of the large electronegativity difference in $HF$ dominates, making $HF$ the most polar molecule with the largest dipole moment.

(A) The unequal sharing of a bonded pair of electrons between two atoms with different electronegativities in a molecule results in the formation of a $Dipole$.
This occurs because the more electronegative atom attracts the shared electron pair more strongly,creating partial negative $(\delta^-)$ and partial positive $(\delta^+)$ charges,which leads to the development of a dipole moment.

(C) The dipole moment of a molecule depends on the polarity of its bonds and its molecular geometry.
$Water$ $(H_2O)$,$Ethanol$ $(C_2H_5OH)$,and $Ether$ $(C_2H_5OC_2H_5)$ are polar molecules due to the presence of electronegative oxygen atoms and their bent or asymmetric structures.
$Ethane$ $(C_2H_6)$ is a non-polar hydrocarbon with a symmetrical structure where the bond dipoles cancel each other out.
Therefore,$Ethane$ shows the least dipole character.

(C) molecule shows a dipole moment if it has a net non-zero dipole moment due to its geometry and the electronegativity difference between atoms.
$CH_4$ (Methane) and $CCl_4$ (Carbon tetrachloride) have a tetrahedral geometry,which results in the cancellation of individual bond dipoles,leading to a net dipole moment of $0$.
$CO_2$ (Carbon dioxide) is a linear molecule where the two $C=O$ bond dipoles cancel each other out,resulting in a net dipole moment of $0$.
$CHCl_3$ (Chloroform) has a tetrahedral geometry,but because the atoms attached to the central carbon are not identical ($3$ $Cl$ atoms and $1$ $H$ atom),the bond dipoles do not cancel out,resulting in a net dipole moment. Thus,the correct option is $(C)$.

(A) In $H_2O$,the $O-H$ bond dipoles are equal in magnitude,but they are oriented at an angle of $104.5^{\circ}$ to each other.
Because of the bent geometry,the vector sum of the bond dipoles is not zero.
Therefore,$H_2O$ possesses a net dipole moment.
In contrast,$BF_3$,$C_6H_6$,and $CCl_4$ have symmetric structures (trigonal planar,planar hexagonal,and tetrahedral,respectively) where the individual bond dipoles cancel each other out,resulting in a net dipole moment of zero.

(A) The resultant dipole moment $\mu_R$ of two individual bond dipoles $\mu$ inclined at an angle $\theta$ is given by the formula: $\mu_R = \sqrt{\mu^2 + \mu^2 + 2 \mu^2 \cos \theta} = \sqrt{2 \mu^2 (1 + \cos \theta)}$.
As the bond angle $\theta$ increases from $0^o$ to $180^o$,the value of $\cos \theta$ decreases from $1$ to $-1$.
Consequently,the term $(1 + \cos \theta)$ decreases as $\theta$ increases.
Therefore,the dipole moment $\mu_R$ is maximum when $\theta$ is minimum,which is $90^o$ among the given options.

(C) The dipole moment of a molecule depends on its geometry and the polarity of its bonds.
$BF_3$ has a trigonal planar geometry,$SiF_4$ has a tetrahedral geometry,and $XeF_4$ has a square planar geometry. Due to their high symmetry,the bond dipoles in these molecules cancel each other out,resulting in a net dipole moment of zero.
$SF_4$ has a see-saw geometry due to the presence of one lone pair on the sulfur atom. This lack of symmetry prevents the bond dipoles from canceling out,resulting in a permanent dipole moment.

(B) $CCl_4$ has no net dipole moment because of its regular tetrahedral structure.
In a regular tetrahedral geometry,the four $C-Cl$ bond dipoles are oriented at an angle of $109.5^{\circ}$ to each other,which results in the cancellation of individual bond moments,leading to a net dipole moment of zero.

(B) The dipole moment of a molecule depends on the polarity of its bonds and its molecular geometry.
$CH_4$ and $CCl_4$ have a tetrahedral geometry and are non-polar,so their net dipole moment is $0 \ D$.
Between $CHCl_3$ and $CHI_3$,the electronegativity difference between $C$ and $Cl$ is greater than that between $C$ and $I$.
Therefore,the $C-Cl$ bond is more polar than the $C-I$ bond.
Consequently,$CHCl_3$ has a larger net dipole moment compared to $CHI_3$.

(D) The dipole moment of a molecule is non-zero if it is polar.
$CCl_4$,$C_6H_6$,and $BF_3$ are symmetric molecules with a net dipole moment of $0$.
$HF$ is a polar molecule due to the significant difference in electronegativity between $H$ and $F$,resulting in a net dipole moment.

(C) $BCl_3$ has a trigonal planar geometry where the three $B-Cl$ bond dipoles cancel each other out due to symmetry,resulting in a net dipole moment of $0$.

(D) The dipole moment of a molecule is the vector sum of the individual bond dipoles.
In $NH_3$ and $H_2O$,the molecules have a net dipole moment due to their bent or pyramidal geometry and the presence of lone pairs.
In $cis-1,2-dichloroethene$,the two $C-Cl$ bonds are on the same side,leading to a non-zero resultant dipole moment.
In $trans-1,2-dichloroethene$,the two $C-Cl$ bonds are in opposite directions,and the two $C-H$ bonds are also in opposite directions. The bond dipoles cancel each other out,resulting in a net dipole moment of zero.

(D) The polarity of a molecule is determined by its dipole moment.
$CCl_4$ is a non-polar molecule due to its symmetrical tetrahedral geometry.
$CHCl_3$ has a dipole moment of $1.04 \ D$.
$CH_3Cl$ has a dipole moment of $1.86 \ D$.
$CH_3OH$ has a dipole moment of $1.70 \ D$ in the gas phase,but due to the presence of the highly electronegative oxygen atom and the ability to form hydrogen bonds,it exhibits significant polarity.
Comparing the values,$CH_3Cl$ has a higher dipole moment than $CH_3OH$ in the gas phase. However,in the context of general chemistry problems,$CH_3OH$ is often considered highly polar due to the hydroxyl group. Given the options,$CH_3Cl$ is the most polar molecule based on dipole moment values.

(C) The dipole moment of a molecule depends on its geometry and the polarity of its bonds.
In $trans$-but-$2$-ene,the two methyl groups are on opposite sides of the double bond.
Due to the symmetry of the molecule,the bond dipoles of the $C-CH_3$ bonds cancel each other out.
Therefore,the net dipole moment of $trans$-but-$2$-ene is zero.

(A) The dipole moment of a molecule depends on its geometry and the polarity of its bonds.
$CCl_4$ has a tetrahedral geometry where the four $C-Cl$ bonds are arranged symmetrically.
The individual bond dipoles cancel each other out due to the symmetric tetrahedral structure,resulting in a net dipole moment of $0 \ D$.
In contrast,$CH_3Cl$,$CH_3F$,and $CHCl_3$ are asymmetric molecules,which leads to a non-zero net dipole moment.

(C) molecule possesses a permanent dipole moment if it has a non-zero net dipole moment due to its geometry and electronegativity difference.
$H_2S$ has a bent geometry and a net dipole moment.
$SO_2$ has a bent geometry and a net dipole moment.
$CS_2$ has a linear geometry $(S=C=S)$,where the two $C=S$ bond dipoles are equal in magnitude and opposite in direction,canceling each other out,resulting in a net dipole moment of $0$.
$SO_3$ is a planar trigonal molecule with a net dipole moment of $0$,but among the given options,$CS_2$ is a classic example of a linear non-polar molecule.

(B) $CH_4$ (Methane) has a regular tetrahedral geometry.
Due to the symmetrical arrangement of the four $C-H$ bonds,the individual bond dipoles cancel each other out,resulting in a net dipole moment of zero.

(B) The dipole moment of a molecule is zero if it has a symmetrical geometry where the bond dipoles cancel each other out.
$BF_3$ is trigonal planar,$CCl_4$ is tetrahedral,and $CH_4$ is tetrahedral; all these have a net dipole moment of $0 \ D$.
$NH_3$ has a trigonal pyramidal geometry with a lone pair on the nitrogen atom,which results in a net dipole moment of approximately $1.47 \ D$.
Therefore,$NH_3$ does not show zero dipole moment.

(B) The charge of an electron is $q = 1.6 \times 10^{-19} \ C$.
The observed dipole moment of $HBr$ is $\mu_{obs} = 1.6 \times 10^{-30} \ C \ m$.
The interatomic spacing (bond length) is $d = 1 \ \mathring{A} = 1 \times 10^{-10} \ m$.
The theoretical dipole moment for $100\%$ ionic character is $\mu_{theo} = q \times d = (1.6 \times 10^{-19} \ C) \times (1 \times 10^{-10} \ m) = 1.6 \times 10^{-29} \ C \ m$.
The $\%$ ionic character is calculated as: $\frac{\mu_{obs}}{\mu_{theo}} \times 100 = \frac{1.6 \times 10^{-30}}{1.6 \times 10^{-29}} \times 100 = 0.1 \times 100 = 10\%$.

(B) molecule is non-polar if its net dipole moment is zero.
Carbon tetrachloride $(CCl_4)$ has a tetrahedral geometry where the four $C-Cl$ bond dipoles cancel each other out due to symmetry,resulting in a net dipole moment of $0$.
Therefore,$CCl_4$ is a non-polar solvent.
In contrast,$Dimethyl \text{ } sulphoxide$,$Ammonia$,and $Ethyl \text{ } alcohol$ are all polar molecules due to the presence of lone pairs or electronegativity differences that do not cancel out.

(A) $CCl_4$ (Carbon tetrachloride) has a regular tetrahedral geometry.
Due to the symmetry of the molecule,the individual bond dipoles of the four $C-Cl$ bonds cancel each other out completely.
Therefore,the net dipole moment of $CCl_4$ is $0 \ D$.

(B) molecule has a zero dipole moment if it is non-polar,meaning the vector sum of its individual bond dipoles is zero.
$BF_3$ has a trigonal planar geometry with three $B-F$ bonds at $120^{\circ}$ to each other.
The resultant dipole moment of two $B-F$ bonds is equal and opposite to the third $B-F$ bond,resulting in a net dipole moment of $0 \ D$.
$H_2O$,$NH_3$,and $HBr$ are polar molecules with non-zero dipole moments due to their bent,pyramidal,and linear asymmetric structures respectively.

(B) $BF_3$ (Boron trifluoride) has a trigonal planar geometry where the three $B-F$ bond dipoles cancel each other out,resulting in a net dipole moment of $0 \ D$.

(B) $N_2$ is a neutral,non-polar molecule with a very strong triple bond $(N \equiv N)$ and no dipole moment,making it chemically inert under normal conditions.
In contrast,$CN^{-}$ is an anion with a negative charge and a dipole moment,which makes it a strong nucleophile and highly reactive compared to $N_2$.

(C) Given: Ionic charge $(q) = 4.8 \times 10^{-10} \ e.s.u.$
Inter-ionic distance $(d) = 1 \ \mathring{A} = 10^{-8} \ cm$.
We know that dipole moment $(\mu) = q \times d$.
$\mu = (4.8 \times 10^{-10} \ e.s.u.) \times (10^{-8} \ cm) = 4.8 \times 10^{-18} \ e.s.u. \ cm$.
Since $1 \ debye = 10^{-18} \ e.s.u. \ cm$,the dipole moment is $4.8 \ debye$.

(A) compound is polar if it has a net dipole moment,which arises due to the difference in electronegativity between bonded atoms.
$HCl$ is a polar molecule because the electronegativity difference between $H$ $(2.1)$ and $Cl$ $(3.0)$ is significant,leading to a permanent dipole moment.
While $H_2Se$ and $HI$ are also polar,$HCl$ is the most commonly cited example of a simple polar covalent molecule in this context.
$CH_4$ is non-polar due to its symmetric tetrahedral geometry.

(A) The dipole moment of a molecule depends on its geometry and the polarity of its bonds.
$CO_2$ has a linear structure $(O=C=O)$,where the bond dipoles of the two $C=O$ bonds are equal in magnitude and opposite in direction,canceling each other out.
Therefore,the net dipole moment of $CO_2$ is $0$.
$SO_3$ is trigonal planar and also has a net dipole moment of $0$.
However,in standard multiple-choice contexts,$CO_2$ is the most common example of a linear molecule with zero dipole moment.
Note: Both $A$ and $B$ have zero dipole moment,but $CO_2$ is the primary answer.

(C) The correct answer is $(C)$.
$SF_6$ has an octahedral geometry which is highly symmetrical.
In this structure,the individual bond dipoles cancel each other out,resulting in a net dipole moment of zero.
Therefore,$SF_6$ is non-polar.

(A) molecule is polar if there is a difference in electronegativity between the bonded atoms,resulting in a net dipole moment.
In the given set,$HCl$,$HF$,and $HBr$ consist of two different atoms with different electronegativities,making them polar.
$H_2$ consists of two identical atoms with zero electronegativity difference,making it a non-polar molecule.

(D) molecule exhibits a dipole moment if it is polar and has a non-zero net dipole moment.
$1, 4-$dichlorobenzene is a symmetric molecule where the dipole moments of the two $C-Cl$ bonds cancel each other out,resulting in a net dipole moment of $0$.
$cis-1, 2-$dichloroethene is polar because the two $Cl$ atoms are on the same side,leading to a non-zero net dipole moment.
$trans-1, 2-$dichloro$-2-$pentene is an unsymmetrical molecule because the groups attached to the double-bonded carbons are different (one side has $H$ and $Cl$,the other has $CH_3$ and $Cl$). Due to this lack of symmetry,the dipole moments do not cancel out,resulting in a non-zero dipole moment.
Therefore,both $(b)$ and $(c)$ exhibit a dipole moment.

(A) The percentage of ionic character is calculated using the formula:
$\% \text{ of ionic character} = \frac{\text{Experimental dipole moment}}{\text{Expected dipole moment}} \times 100$
Substituting the given values:
$\% \text{ of ionic character} = \frac{1.03}{6.12} \times 100 = 16.83\% \approx 17\%$

(D) $BF_3$ has a trigonal planar geometry ($sp^2$ hybridization) where the bond dipoles cancel each other out,resulting in a net dipole moment of $0$.
In $NF_3$,the nitrogen atom has one lone pair of electrons,leading to a trigonal pyramidal geometry ($sp^3$ hybridization).
The presence of the lone pair and the pyramidal shape prevent the bond dipoles from canceling,making $NF_3$ a polar molecule.

(C) $H_2O$ is a polar molecule because it has a bent geometry and a net dipole moment due to the electronegativity difference between oxygen and hydrogen atoms. In contrast,$CO_2$,$CCl_4$,and $CH_4$ are non-polar due to their symmetric structures which cause the individual bond dipoles to cancel each other out.

(B) When there is a difference in electronegativity between two bonded atoms,the shared electron pair is attracted more towards the more electronegative atom.
As a result,the electron pair does not remain in the centre,leading to the formation of a polar bond.

(D) liquid is deflected by a non-uniform electrostatic field if it possesses a permanent dipole moment (i.e.,it is polar).
$A$. Water $(H_2O)$ is polar.
$B$. Chloroform $(CHCl_3)$ is polar.
$C$. Nitrobenzene $(C_6H_5NO_2)$ is polar.
$D$. Hexane $(C_6H_{14})$ is a non-polar hydrocarbon with a symmetrical structure and zero net dipole moment.
Therefore,it is not deflected by an electrostatic field.

(C) molecule is considered non-polar if the net dipole moment is zero.
In the case of $Cl_2$,the bond is formed between two identical atoms with the same electronegativity.
Therefore,the shared pair of electrons is attracted equally by both atoms,resulting in a net dipole moment of $0$.

(C) The polarity of a covalent bond depends on the difference in electronegativity between the bonded atoms.
As we move down the halogen group $(F > Cl > Br > I)$,the electronegativity decreases.
Therefore,the electronegativity difference between $H$ and the halogen is greatest for $HCl$ compared to $HBr$ and $HI$.
$H_2$ is a non-polar molecule because the electronegativity difference is zero.
Thus,the $H-Cl$ bond is the most polar among the given options.

(C) $CH_2Cl_2$ is a polar molecule.
The molecule has a tetrahedral geometry due to $4$ electron pairs around the central $C$ atom.
Although the geometry is tetrahedral,the presence of two different types of bonds ($C-H$ and $C-Cl$) with different electronegativities results in a non-zero net dipole moment.
Therefore,the bond dipoles do not cancel each other out,making the molecule polar.

(D) The $ICl$ molecule is polar because there is an electronegativity difference between $I$ $(2.66)$ and $Cl$ $(3.16)$.
Since the electronegativity of $Cl$ is greater than that of $I$,the shared pair of electrons is shifted towards the $Cl$ atom.
This results in a partial negative charge $(\delta-)$ on the $Cl$ atom and a partial positive charge $(\delta+)$ on the $I$ atom.
Therefore,the molecule is polar with the negative end on chlorine.

(A) compound is considered polar if there is a significant difference in electronegativity between the bonded atoms,resulting in a permanent dipole moment.
In the given options,all are polar,but $HF$ exhibits the highest degree of polarity due to the largest electronegativity difference between $H$ $(2.1)$ and $F$ $(4.0)$.
Therefore,$HF$ is the most polar compound among the choices.

(C) $SiF_4$ has a tetrahedral geometry where all four $Si-F$ bonds are identical and arranged symmetrically around the central $Si$ atom.
Due to this symmetric arrangement,the individual bond dipole moments cancel each other out,resulting in a net dipole moment of $0$.

(D) The dipole moment of a molecule depends on its geometry and the polarity of its bonds.
$BF_3$ has a trigonal planar geometry with three $B-F$ bonds oriented at $120^{\circ}$ to each other.
Due to the symmetric arrangement,the resultant dipole moment of the three $B-F$ bonds cancels out to zero.
$PH_3$,$CHCl_3$,and $NH_3$ all have non-zero dipole moments due to their pyramidal or asymmetric structures.
Therefore,$BF_3$ has the least dipole moment.

(A) $CO_2$ is a linear molecule $(O=C=O)$,so the individual bond dipoles cancel each other out,resulting in a net dipole moment of zero.
$H_2O$ has a bent structure $(H-O-H)$ due to the presence of lone pairs on the oxygen atom,which prevents the bond dipoles from canceling,resulting in a net dipole moment.

(B) The overall value of the dipole moment of a polar molecule depends on its geometry and shape,i.e.,the vectorial addition of the dipole moments of the constituent bonds.
Water $(H_2O)$ has an angular structure with a bond angle of $105^{\circ}$,which results in a net dipole moment.
However,$BeF_2$ is a linear molecule where the dipole moments of the two $Be-F$ bonds are equal in magnitude and opposite in direction,thus canceling each other out,resulting in a net dipole moment of zero.