Dipole moment Questions in English

(B) molecule has a zero dipole moment if its net dipole moment is zero. This occurs in highly symmetric molecules where the bond dipoles cancel each other out.
$1$. $CCl_4$: Tetrahedral geometry,all $C-Cl$ bond dipoles cancel. $\mu = 0$.
$2$. $BF_3$: Trigonal planar geometry,all $B-F$ bond dipoles cancel. $\mu = 0$.
$3$. $CHCl_3$: Asymmetric,$\mu \neq 0$.
$4$. $CS_2$: Linear geometry,$S=C=S$ bond dipoles cancel. $\mu = 0$.
$5$. $NH_3$: Pyramidal geometry with a lone pair,$\mu \neq 0$.
$6$. $1,4$-dichlorobenzene: Para-substituted benzene,the two $C-Cl$ bond dipoles are equal and opposite. $\mu = 0$.
$7$. $CO_2$: Linear geometry,$O=C=O$ bond dipoles cancel. $\mu = 0$.
Thus,the molecules with zero dipole moment are $CCl_4, BF_3, CO_2, CS_2, 1,4$-dichlorobenzene.

(A) The dipole moment $(\mu)$ is defined as the product of the magnitude of the charge $(q)$ and the distance $(d)$ between the charges: $\mu = q \times d$.
The unit of dipole moment is Debye $(D)$.
One Debye is defined as $1 \ D = 10^{-18} \ esu \ cm$.
To convert this to $SI$ units (coulomb meter,$C \ m$):
$1 \ D = 10^{-18} \ esu \ cm \times (3.33564 \times 10^{-10} \ C / 1 \ esu) \times (10^{-2} \ m / 1 \ cm)$
$1 \ D \approx 3.33 \times 10^{-30} \ C \ m$.

(A) The dipole moments of the given molecules are as follows:
$A$. $HBr$: The dipole moment is $0.79 \ D$ $(III)$.
$B$. $H_2S$: The dipole moment is $0.95 \ D$ $(IV)$.
$C$. $NH_3$: The dipole moment is $1.47 \ D$ $(V)$.
$D$. $CHCl_3$: The dipole moment is $1.04 \ D$ $(I)$.
Therefore,the correct match is $A-III, B-IV, C-V, D-I$.

(A) The compounds are:
$(i)$ Toluene $(C_6H_5CH_3)$
(ii) $m$-Dichlorobenzene
(iii) $o$-Dichlorobenzene
(iv) $p$-Dichlorobenzene
$1$. For $p$-dichlorobenzene (iv),the two $C-Cl$ bond dipoles are equal and opposite,so they cancel each other out,resulting in a dipole moment of $0$.
$2$. For toluene $(i)$,the dipole moment is small due to the weak electron-donating effect of the $-CH_3$ group.
$3$. For $m$-dichlorobenzene (ii),the angle between the two $C-Cl$ bonds is $120^{\circ}$. The resultant dipole moment is $\sqrt{\mu^2 + \mu^2 + 2\mu^2 \cos(120^{\circ})} = \mu$.
$4$. For $o$-dichlorobenzene (iii),the angle between the two $C-Cl$ bonds is $60^{\circ}$. The resultant dipole moment is $\sqrt{\mu^2 + \mu^2 + 2\mu^2 \cos(60^{\circ})} = \sqrt{3}\mu$.
Comparing these,the order is $iv (0) < i < ii < iii$.
Thus,the correct option is $A$.

(B) The dipole moments of the given compounds are as follows:
$(a)$ For $HBr$,$\mu = 0.82 \ D$.
$(b)$ For acetone $(CH_3COCH_3)$,the electron-releasing methyl groups increase the electron density towards the oxygen atom,resulting in a net dipole moment of $\mu_{net} = 2.69 \ D$.
$(c)$ For $H_2S$,$\mu_{net} = 0.97 \ D$.
$(d)$ For $COCl_2$,$\mu_{net} = 1.17 \ D$.
Comparing these values,acetone $(CH_3COCH_3)$ has the highest dipole moment.

(C) $BF_3$ has a zero dipole moment due to its symmetrical trigonal planar geometry.
In $NH_3$,$N$ is more electronegative than $H$. The bond dipoles of the three $N-H$ bonds and the lone pair on the nitrogen atom are in the same direction,which reinforces each other,resulting in a high net dipole moment.
In $NF_3$,$F$ is more electronegative than $N$. The bond dipoles of the three $N-F$ bonds are in the opposite direction to the dipole moment of the lone pair on the nitrogen atom,which partially cancels out the net dipole moment.
Therefore,the dipole moment of $NH_3$ is greater than that of $NF_3$,and $BF_3$ has the lowest (zero) dipole moment.
The correct order is $NH_3 > NF_3 > BF_3$.

(A) $1$. $CO_2$ is linear and $BF_3$ is trigonal planar. Due to their symmetrical geometry,the net dipole moment is zero. This statement is correct.
$2$. In $NH_3$,the dipole moments of $N-H$ bonds and the lone pair are in the same direction,leading to a higher net dipole moment. In $NF_3$,the dipole moments of $N-F$ bonds are in the opposite direction to the lone pair moment,resulting in a smaller net dipole moment. Thus,the dipole moment of $NF_3$ is lower than that of $NH_3$. This statement is incorrect.
$3$. The dipole moment $\mu = q \times d$. Although the bond length of $HI$ is greater than $HCl$,the electronegativity difference in $HCl$ is much larger than in $HI$,making the charge separation $q$ significantly higher in $HCl$. Thus,$HI$ has a lower dipole moment than $HCl$. This statement is correct.

(C) From the definition of dipole moment,$\mu = \delta \times d$,where $\delta$ is the magnitude of the electric charge and $d$ is the bond length.
Therefore,$\delta = \frac{\mu}{d}$.
The ratio of the fraction of electric charge for $HCl$ and $HI$ is given by:
$\frac{\delta_{HCl}}{\delta_{HI}} = \frac{\mu_{HCl}}{d_{HCl}} \times \frac{d_{HI}}{\mu_{HI}}$
Substituting the given values:
$\frac{\delta_{HCl}}{\delta_{HI}} = \frac{1.03 \times 1.6}{1.3 \times 0.38} = \frac{1.648}{0.494} \approx 3.33: 1$.
Thus,the ratio is approximately $3.3: 1$.

(D) Given: Observed dipole moment $= 1.03 \ \text{D}$.
Bond length of $HCl$ molecule,$d = 1.275 \ \text{Å} = 1.275 \times 10^{-8} \ \text{cm}$.
Charge of electron,$e = 4.8 \times 10^{-10} \ \text{esu}$.
Theoretical dipole moment $= e \times d = (4.8 \times 10^{-10} \ \text{esu}) \times (1.275 \times 10^{-8} \ \text{cm}) = 6.12 \times 10^{-18} \ \text{esu} \cdot \text{cm} = 6.12 \ \text{D}$.
Percentage ionic character $= (\text{Observed dipole moment} / \text{Theoretical dipole moment}) \times 100$.
Percentage ionic character $= (1.03 / 6.12) \times 100 = 16.83 \%$.

(C) The dipole moment of a purely ionic bond is calculated as $\mu_{\text{theoretical}} = q \times d$.
Given: bond length $d = 1.25 \ \mathring{A} = 1.25 \times 10^{-8} \ cm$.
The charge of an electron $q = 4.8 \times 10^{-10} \ esu$.
$\mu_{\text{theoretical}} = 1.25 \times 10^{-8} \ cm \times 4.8 \times 10^{-10} \ esu = 6.0 \times 10^{-18} \ esu \ cm = 6.0 \ D$.
Given experimental dipole moment $\mu_{\text{experimental}} = 1.0 \ D$.
Percent ionic character $= \frac{\mu_{\text{experimental}}}{\mu_{\text{theoretical}}} \times 100 = \frac{1.0}{6.0} \times 100 = 16.66 \%$.

(A) In the $CO$ molecule,the electronegativity difference suggests a dipole moment,but the actual dipole moment is very small $(0.11 \ D)$.
This is because the $\sigma$-electron drift from $C$ to $O$ (due to the lone pair on $C$) is almost completely nullified by the back-donation of $\pi$-electrons from $O$ to $C$ through $\pi$-bonding.
Thus,the net dipole moment is nearly zero,making it practically non-polar.

(A) molecule has a permanent dipole moment if it is polar,meaning its net dipole moment $(\mu_{net})$ is not equal to zero.
$I$: $CH_{2}Cl_{2}$ (dichloromethane) is a non-symmetrical tetrahedral molecule. The bond dipoles of $C-H$ and $C-Cl$ bonds do not cancel each other out,resulting in a non-zero net dipole moment $(\mu_{net} \neq 0)$.
$II$: $trans-1,2-dichloroethene$ is a symmetrical molecule where the bond dipoles cancel each other out,resulting in $\mu_{net} = 0$.
$III$: $CCl_{4}$ (carbon tetrachloride) is a highly symmetrical tetrahedral molecule where all four $C-Cl$ bond dipoles cancel each other out,resulting in $\mu_{net} = 0$.
$IV$: $1,2-dibromoethyne$ is a linear,symmetrical molecule where the bond dipoles cancel each other out,resulting in $\mu_{net} = 0$.
Therefore,the compound with a permanent dipole moment is $I$.

(C) The dipole moment of a molecule is the vector sum of the individual bond dipoles. For chlorobenzene,the dipole moment is $1.5 \ D$. Let the dipole moment of a $C-Cl$ bond be $\mu$.
For $o-$dichlorobenzene,the two $C-Cl$ bonds are at an angle of $60^{\circ}$ to each other.
The resultant dipole moment is given by $\mu_{net} = \sqrt{\mu^2 + \mu^2 + 2\mu^2 \cos(60^{\circ})}$.
Since $\cos(60^{\circ}) = 0.5$,we have $\mu_{net} = \sqrt{2\mu^2 + 2\mu^2(0.5)} = \sqrt{3\mu^2} = \mu \sqrt{3}$.
Given $\mu = 1.5 \ D$,the dipole moment of $o-$dichlorobenzene is $1.5 \times 1.732 \approx 2.54 \ D$.

(C) The dipole moments of the given molecules are as follows:
$H_2S = 0.95 \ D$
$H_2O = 1.85 \ D$
$NF_3 = 0.23 \ D$
$NH_3 = 1.47 \ D$
$CHCl_3 = 1.04 \ D$
Comparing these values,$NF_3$ has the lowest dipole moment.
Therefore,the molecule $(X)$ is $NF_3$.
In $NF_3$,the central atom is Nitrogen $(N)$.
Nitrogen has $5$ valence electrons. It forms $3$ single bonds with $3$ Fluorine atoms,leaving $1$ lone pair on the Nitrogen atom.
Thus,the number of lone pairs on the central atom is $1$.