MCQ Questions for Class 11 Medical Chemistry Chemical Bonding And Molecular Structure Quiz 7

The dipole moment of is

$1.5\ D$ . The dipole moment of?

0%
$0\ D$
0%
$1.5\ D$
0%
$2.86\ D$
0%
$2.25\ D$

The molecular orbital shown below can be described respectively as ?

0%
$\sigma, \sigma^-$
0%
$\pi, \pi^-$
0%
$\pi^-, \pi$
0%
$\pi^-, \sigma^-$

$16$ g of sulphur completely reacts with

$38$ g of

$F_2$ to form

$SF_x$ . The value of

$x$ is:

(Atomic mass of

$F = 19$ )

0%
$1$
0%
$2$
0%
$4$
0%
$6$

Explanation

$16g$ of

$S$ reacts with

$38g$ of

$F_2$

The ratio of moles reacted is

$\cfrac {16}{32}:\cfrac {38}{19}$

$\Rightarrow \cfrac {1}{2}:2$

$\Rightarrow 1:4$

So one

$S$ atom and

$4$ atoms forms

$SF_4$

So,

$x=4$ .

In the given structure of compound, the correct various bond moments direction involving atoms are shown as:

0%
$Br\overset{\longleftarrow}{-}N\overset{\longleftarrow}{-}CH_2\overset{\longrightarrow}{-}SiH_2\overset{\longleftarrow}{-}H_2\overset{\longrightarrow}{-}O\overset{\longleftarrow}{-}CH_3$
0%
$Br\overset{\longleftarrow}{-}N\overset{\longleftarrow}{-}CH_2\overset{\longleftarrow}{-}SiH_2\overset{\longleftarrow}{-}CH_2\overset{\longrightarrow}{-}O\overset{\longleftarrow}{-}CH_3$
0%
$Br\overset{\longrightarrow}{-}N\overset{\longleftarrow}{-}CH_2\overset{\longleftarrow}{-}SiH_2\overset{\longrightarrow}{-}CH_2\overset{\longrightarrow}{-}O\overset{\longleftarrow}{-}CH_3$
0%
$Br\overset{\longleftarrow}{-}N\overset{\longrightarrow}{-}CH_2\overset{\longrightarrow}{-}SiH_2\overset{\longleftarrow}{-}CH_2\overset{\longrightarrow}{-}O\overset{\longrightarrow}{-}CH_3$

Explanation

Bond moments direction depends on the electronegativity of the elements.

The electronegativity of the given elements is-

$O>N>Br>C>Si$

option C is correct.

On treatment of the following compound with a strong acid, the most susceptible site for bond cleavage is:

0%
$C1-O2$
0%
$O2-C3$
0%
$C4-O5$
0%
$O5-C6$

Explanation

On treatment of the following compound with a strong acid, the most susceptible site for bond cleavage is

${ O }_{ 5 }-{ C }_{ 6 }$ .

1197651_1121871_ans_0eb3cb3d82064f5c870b330304f87a0d.jpg

Which of the following molecule is theoretically not possible?

0%
$SF_4$
0%
$OF_2$
0%
$OF_4$
0%
$O_2F_2$

Explanation

$O$ belongs to period 2 having 2p valence shell and does not have vacant d-orbitals. Belonging to group 6, it already has 6 electrons in its valence shell and can accommodate only 2 more electrons to complete its octet. In the formation of

$OF_4$ the number of electrons in its outermost shell is 10. Since, it does not have vacant d-orbitals to accommodate the extra electrons, formation of

$OF_4$ is theoretically not possible.

Which combination of atomic orbital is correct for

$\sigma$ -bond formation?

0%
$s - P_z$
0%
$s - P_x$
0%
$s - P_y$
0%
All are correct

What is the nature of chemical bonding between

$Cs$ and

$F$ ?

0%
lonic
0%
Covalent
0%
Coordinate
0%
Metallic

Explanation

The nature of the bond between calcium and fluorine is electrovalent ionic.

As calcium is a metal,

$F$ is non-metal.

High electronegativity difference.

Charge transfer from fluorine to calcium.

Hence, the correct option is A.

$3B_2H_6 + 6NH_3\xrightarrow [ low{\,}T ]{ slow{\,} heating }3X\xrightarrow{Heat}2Y + 12H_2$ . In the anionic part of the intermediate

$X$ covalency of the central atom is:

0%
$2$
0%
$4$
0%
$6$
0%
$8$

Explanation

When

$1$ mole of

$B_2H_6$ reacts with

$2$ moles of

$NH_3$ by slow heating, an addition compound

$B_2H_6.2NH_3$ is formed which on further heating at

$200^oC$ forms borazine

$(B_3N_3H_6)$ .

(Refer to Image)

Covalency of amionic part=

$4$

1189450_1124339_ans_a4d45767bdcf4139a7ce33c099a3132f.JPG

$NH_3$ (excess)

$+ 3Cl_2\rightarrow A+N_2\uparrow$ ; the bond present in compound

$A$ is?

0%
Ionic, Covalent and dative
0%
Ionic and covalent
0%
Dative and ionic
0%
Dative only

What is correct increasing order of bond lengths of bond indicated as

$I, II, III,$ and

$IV$ in the following compounds?

0%
$I < II < III < IV$
0%
$II < III < IV < I$
0%
$IV < II < III < I$
0%
$IV < I < II < III$

Explanation

Generally, the triple bond is of short length than double and single bonds. Double bonds are shorter than single bonds.

So,

$IV < I$

In between

$II$ &

$III, III$ is shorter because it is near to triple bond. The triple bond effect bond

$III$ and it is shorter than

$II$ .

$III$ and

$II$ are longer than

$IV$ &

$I$ because, they are single bonds. So order is,

$IV < I < II < III$

1174504_1126343_ans_d8be184ec79e45c784b1724afc0e010e.png

Lone pair of electrons are electrons in orbitals?

0%
Which does not undergo hybridization
0%
Which undergoes hybridization
0%
Which undergoes hybridisation and participate in Bond formation
0%
Which undergo hybridization but not participate in Bond formation

Correct order of dipole moment is:

0%
$CH_3F > CH_3Cl > CH_3Br > CH_3I$
0%
$CH_3F > CH_3Br > CH_3Cl > CH_3I$
0%
$CH_3Cl > CH_3F > CH_3Br > CH_3I$
0%
$CH_3F > CH_3I > CH_3Cl > CH_3Br$

Explanation

The decreasing order of electronegativity among halogens is :

$F>Cl>Br>I$ . So due to highest electronegativity of

$F$ ,

$F$ atom attracts more attracts more electrons towards himself as a result more partial charge separation occurs in

${ CH }_{ 3 }F$ .

Therefore highest dipole moment is observed in

${ CH }_{ 3 }F$ .

${ \overset { \delta + }{ CH } }_{ 3 }\overset { \mapsto }{ - } \overset { \delta - }{ F } >{ \overset { \delta + }{ CH } }_{ 3 }\overset { \mapsto }{ - } \overset { \delta - }{ Cl } >{ \overset { \delta + }{ CH } }_{ 3 }\overset { \mapsto }{ - } \overset { \delta - }{ Br } >{ \overset { \delta + }{ CH } }_{ 3 }\overset { \mapsto }{ - } \overset { \delta - }{ I }$

Among the following molecular orbitals, how many have at least one nodal plane ?

$\sigma ^*1s,\sigma 2s,\sigma ^{ * }2p_{ z },\pi 2p_{ x },\pi ^{ * }2p_{ y },\sigma 2p_{ z }$

0%
$4$
0%
$5$
0%
$2$
0%
$6$

The increasing order of dipole moment
A.

${ SO }_{ 2 }(X),$

$B.$

$\quad { H }_{ 2 }O(Y),$

$C.$

$\quad { CO }_{ 2 }(Z)$

$is$ [

$X>Y>Z$ ]

0%
True
0%
False

In the given, lewis structure of

${ S }_{ 2 }{ O }_{ 3 }^{ 2- }$ formal charge present on sulphur atoms

$1$ and

$2$ respectively are:

0%
$zero, \ +1$
0%
$+1,+1$
0%
$+2,+2$
0%
$zero, \ zero$

Explanation

Formal charge is expressed as (a lewis dot structure)

$=$ (total no. of valence electron in free atom)

$-$ (total no. of non bonding electron)

$-\dfrac { 1 }{ 2 }$ (total no. of bonding electron)

$=V-L-\dfrac { 1 }{ 2 } \times S$

$L=$ lone pair electrons,

$S=$ bonding electrons

For Sulphur atom

$(1)$

$FC=6-4-\dfrac { 1 }{ 2 } \times 4\Rightarrow 6-4-2\Rightarrow 0$

For Sulphur atom

$(2)$

$FC=6-0-\dfrac { 1 }{ 2 } \times 12\Rightarrow 6-6=0$

Ans is

$(D)$ .

1 mole of

$H_3PO_4$ having

0%
Two $\pi$ - bond
0%
six mole of $\sigma$ -bond
0%
Sixteen lone pair of electron
0%
seven $\sigma$ -bond

Which bond angle

$\theta$ would result the maximum dipole moment for the triatonuc molecules

$XY_2$ as shown below?

0%
$90^o$
0%
$120^o$
0%
$150^o$
0%
$180^o$

Explanation

The dipole moment of

$2$ dipoles inclined at an angle

$\theta$ is given by,

$\mu =\sqrt { x^{ 2 }+{ y }^{ 2 }+2xycos\theta } ,\quad \quad cos90=0$

$\therefore$ The dipole moment is maximum for

$\theta ={ 90 }^{ 0 }$

$P_4O_6$ , the number of

$P-O$ bonds and number of lone pairs on

$P$ are:

0%
12 and 4
0%
8 and 8
0%
12 and 16
0%
12 and 12

For which of the above molecules dipole moment

$\mu \neq 0$ ?

0%
Only (c)
0%
(c) and (d)
0%
only (a)
0%
(a) and (b)

Which of the following has dipole moment ?

0%
$H_2S$
0%
$CO_2$
0%
$CCl_4$
0%
$BF_3$

Which of the following has the weakest carbon-chloride bond?

0%
0%
0%
0%
$CH_2=CH-CI$

Find out the species with zero dipole moment
(i)

$trans-{CH}_3{CH}=CHCI$
(ii)

${CS}_{2}$

0%
only one
0%
i and ii
0%
ii only
0%
none of these

The inter nuclear distance in

$H_2$ and

$Cl_2$ molecules are

$74$ and

$198$ pm. respectively. The bond length of

$HCl$ may be:

0%
$136 \,pm$
0%
$272 \,pm$
0%
$135.919 \,pm$
0%
$271.919 \,pm$

Explanation

$r_{H_2} = \dfrac{74}{2} = 37 \,pm$

$r_{Cl_2} = \dfrac{198}{2} = 99 \,pm$

bond length

$= 37 + 99 = 136 \,pm$

Option A is the correct answer.

Non - zero dipole moment is shown by :

0%
$CCl_4$
0%
$CS_2$
0%
$H_2O$
0%
$p-Cl-C_6H_4-Cl$

Which of the following has zero value of dipole moment:

0%
Benzene
0%
Naphthalene
0%
p-dichlorobenzene
0%
all the these

Which of following has been arranged in order of decreasing dipole moment?

0%
$CH_{3}CI> CH_{3}F> CH_{3}Br> CH_{3}I$
0%
$CH_{3}F> CH_{3}CI> CH_{3}Br> CH_{3}I$
0%
$CH_{3}CI> CH_{3}Br> CH_{3}I> CH_{3}F$
0%
$CH_{3}F> CH_{3}CI> CH_{3}I> CH_{3}Br$

Most acidic among the following is

Explanation

Hint: Electron withdrawing groups increases the acidity of the compound while electron-donating group decreases it.

Correct Option:

$D$

Explanation of Correct Option:

$-{NO}_{2}$ group at ortho and para position withdraws electrons of the

$O-H$ bond towards itself by the stronger

$-R$ and

$-I$ effect while the

$-{NO}_{2}$ group at m-position withdraw electrons of the

$O-H$ bond by the weaker

$-I$ effect.

Meta cresol (Option

$C$ ) will be the least acidic as there is the presence of a methyl group that is an electron-donating group.

Thus, ortho and para nitrophenols are more acidic than meta nitrophenol.

Ortho nitrophenol is a little less acidic than p-nitrophenol due to intramolecular H-bonding which makes the loss of a proton a little more difficult.
p - Nitrophenol is a stronger acid than o - nitrophenol.

p - Nitrophenol is a stronger acid than o - nitrophenol.

$\therefore$ The order of acidity of nitrophenol is:

$\bf{p- nitrophenol> o-nitrophenol> m-nitrophenol>Phenol>m-Cresol}$

As m-Cresol has electron releasing methyl group which is least acidic.

Hence Correct answer is Option "D"

With a change in hybridization of the carbon bearing the charge, the stability of a carbanion decreases in the order

0%
$sp < { sp }^{ 2 } < { sp }^{ 3 }$
0%
$sp < { sp }^{ 3 } < { sp }^{ 2 }$
0%
${ sp }^{ 3 } < { sp }^{ 2 } < sp$
0%
${ sp }^{ 2 } < sp < { sp }^{ 3 }$

Explanation

Stability of carbanions increases with increase in

$s-$ character of hybrid orbital of carbon bearing charge hence correct order is:

$sp^{3} < sp^{2} < sp$

Which of the following molecules(s) have dipole moment?
I)

$Trans-pent-2-ene$
II)

$cis-hex-3-ene$
III)

$2,2-Dimethyl propane$
IV)

$2,2,3,3-tertramethyl butane$

0%
I,II
0%
II,III
0%
I,III
0%
I,IV

The number of antibonding electron paris in

${O}_{2}^{2-}$ molecular ion on the basis of molecular orbital theory are (At. no

$O=8$ )

0%
$2$
0%
$3$
0%
$4$
0%
$5$

Which of the following is the decreasing order of their dipole moments?

0%
${NH}_{3}> {BF}_{3}> {NF}_{3}$
0%
${NH}_{3}> {NF}_{3}> {BF}_{3}$
0%
${BF}_{3}> {NH}_{3}> {NF}_{3}$
0%
${BF}_{3}> {NF}_{3}> {NH}_{3}$

Explanation

$NH_3$ all bond moments assists each other and so net moment adds up.

$NF_3$ , moment of low pairs is opposite to that of

$N-F.$ So, net dipole moment reduces.

$BF_3$ , the molecule being planer, the moments cancel each other. Thus net moment is

$zero.$

Hence, the order,

$NH_3>NF_3>BF_3$

See the figure.

1132514_1138750_ans_578c6cde4352469188cb33e5fa90362a.png

Formal charge on nitrogen and oxygen in

$NO^-_3$ ion are respectively:

0%
$0, \dfrac{-1}{3}$
0%
$+1, \dfrac{-1}{3}$
0%
$+1, \dfrac{-2}{3}$
0%
$0, \dfrac{-2}{3}$

Explanation

Formula change is basically a change given to an atom, on assuming that the electrons in a chemical bond are shared equally between the atoms, neglecting theirrelative,

$EN.$

$FC=V-N-\cfrac{B}{2}$

$(V=\text{no. of valance electron}\\N=\text{No. of non bonding electrons}\\B=\text{No. of electrons in covalent bond})$

$FC$ on

$N$ in

$NO^-_3=0$

$FC$ on

$O$ in

$NO_3^-=\cfrac{-1}{3}$

The compoumd with no dipole moment is:

0%
${CH}_{3}Cl$
0%
$C{Cl}_{4}$
0%
methylene chloride ( ${CH}_{2}{Cl}_{2}$ )
0%
chloroform ( $CH{Cl}_{3}$ )

Explanation

$CCl_4$ has a net dipole moment of

$zero$ . Even though each of four

$C-Cl$ bonds are polar, all the bond moments cancel due to tetrahedral structure of

$CCl_4.$

WHich of the following substance has dipole moment more than zero?

0%
Water
0%
Methane
0%
Carbon dioxide
0%
Nitrogen

Explanation

Dipole moment: It is the measurement of bond polarity.

Denoted by

$[\mu=e.d]$

$e=4.8\times 10^{-10}e.s.u$

$d=$ distance in centimeter

Unit of

$\mu=$ debye (D)

(i)Lone pair moment due to lone pair

(ii)Bond pair moment due to bond direction from low electronegative to high E.N

So,considering molecules

$H_2O,CH_4,CO_2,N_2$

Refer to image,

(iv)

$N_2$

$N\equiv N$

$[\mu=0]$

Answer is water has dipole moment more than zero.

1080921_1139868_ans_c93f2d8d0e1c485cad2ca9fe330743f6.PNG

Effective overlapping will be shown by :

0%
0%
0%
0%
All of these

The dipole moment of m-dichlorobenzene is 1.5 D. The value of dipole moment of o-dichlorobenzene is

0%
0.02 D
0%
1.5 D
0%
2.6 D
0%
2.12 D

Molecule AB has a bond length of 1.617 A and a dipole moment of 0.38 D. The fractional charge on each atom (absolute magnitude) is :
(

${ e }_{ 0 }=\quad 4.802\quad \times \quad { 10 }^{ -10 }$ esu)

0%
$0$
0%
$0.05$
0%
$0.5$
0%
$1.0$

Explanation

Dipole moment

$(\mu )= q\times d$

$\mu = 0.38D$ (given) =

$0.38\times 10^{-18}$ esu cm

$q=\frac{\mu }{d}=\frac{0.38\times 10^{-18}}{1.617\times 10^{-8}}$

fractional charge =

$\frac{particle\, charge}{Total\, charge}$

$=\frac{0.38\times 10^{-18}}{1.617\times 10^{-8}\times 4.802\times 10^{-10}}$

$= 0.048$

Which of the following involves transfer of five electrons?

0%
$MnO_4^ - \to M{n^{2 + }}$
0%
$CrO_4^{2 - } \to C{r^{3 + }}$
0%
$MnO_4^{2 - } \to Mn{O_2}$
0%
$C{r_2}O_7^{2 - } \to 2C{r^{3 + }}$

Explanation

$x + 4( - 2) = - 1$

$x - 8 = - 1$

$x = + 7$

1381125_1255302_ans_4a6ff40bf35c4b4686ba563a0f74e34c.jpg

The correct order of increasing bond angles in the following triatomic species is

0%
$NO_2^+ < NO_2 < NO_2^-$
0%
$NO_2^+ < NO_2^- < NO_2$
0%
$NO_2^- < NO_2^+ < NO_2$
0%
$NO_2^- < NO_2 < NO_2^+$

Which of the following statements is true about hybridization ?

0%
Hybrid orbitals frequently undergo linear overlaps making sigma bonds.
0%
Hybrid orbitals frequently undergo lateral overlaps maing $\pi$ -bonds. In other words, there are several compounds in which $\pi$ -bonds are formed using hybrid orbitals.
0%
Hybrid orbitals are molecular orbitals.
0%
A hybrid orbital bigger in size makes shorter bond.

Explanation

$(A)$

Linear overlaps of bybridabitak from sigma bonds

$True$

$(B)$

HyBrid orbitals Do Not Form

$\pi$ -bonds.

$\pi$ -bonds are formed by arbybridized orbitals of the molecule by lateral overlap.

Statement is

$False$

$(C)$

Hybrid are not same as molecules orbital hence

$False$ statement

While of same atom, molecular orbitals are formed by mixing orbitals of different atoms

$False$

$(D)$

Only first Statement is

$True$

1364817_1178386_ans_c3ecf24b34804b6281759722f712588b.png

Which one of the following has the strongest

$O-O$ bond?

0%
$O_{2}^{+}$
0%
$O_{2}$
0%
$O_{2}^{-}$
0%
$O_{2}^{2-}$

Explanation

$O^+_2$ has

$15$ electrons whose bond order is

$2.5$

(Refer to Image) Stronger bond has higher bond order.

Hence,

$O^+_2$ has strongest

$O-O$ bond.

1141578_1154576_ans_dfec3097083b4d4fbfc97b0004ad35bc.PNG

In 2sp hybridization, 2s-orbital can be mixed with

0%
Only $2p_x$
0%
Only $2p_y$
0%
Only $2p_z$
0%
Any one of $2p_x,2p_y$ and $2p_z$

Explanation

$S$ orbital is spherically symmetric and hence can be mixed with

$p$ orbital from any axis

So, answer should be

$d$

1364935_1178415_ans_97706cfe9181462b89cbaed0ee093e16.png

One of the resonating structure of

$SO_{4}^{-2}$ is as shown.
Which set formal charge on oxygen and bond order is correct?

0%
$-0.5$ and $1.5$
0%
$1.5$ and $3$
0%
$2$ and $3$
0%
$1.5$ and $1.5$

Explanation

All resonating structures of

$SO_4^{2-}$ are

(Refer to Image)

Formal charge on (1)=

$\cfrac {-1+0-1+0-1+0}{6}=-0.5$

Bond order=

$\cfrac {1+2+1+2+1+2}{6}= \cfrac {3}{2}=1.5$

1134615_1154766_ans_1c21bf02a13d4b1597bdc0624239f1f5.png

If the dipole moment of

$C-Cl$ bond is

$1.2$

$D$ , then the resulting value of dipole moment in:

0%
$1.2$ and $0.6$ respectively
0%
$0.6$ and $1.2$ respectively
0%
Less than $1.2$ $D$ and more than $1.2$ $D$ respectively
0%
More than $1.2$ $D$ in both

Explanation

net dipole moment's magnitude

$=\sqrt{ (1.2)^{2} + (1.2)^{2} +2 (1.2)^{2} \cos \theta }$

$=1.2\ D$

net dipole moment's magnitude

$=\sqrt{(1.2)^{2} + (1.2)^{2} +2 (1.2)^{2} \cos \theta }$

$= (\sqrt{3})(1.2)\ D$

1364968_1178317_ans_46b293c361a84e26bd3162f4f9451746.png

Which of the following has least dipole moment:

0%
$NF_3$
0%
$CO_2$
0%
$SO_2$
0%
$NH_3$

Explanation

$CO_2$ is a linear molecule, so our dipoles are symmetrical; the dipoles are equal in magnitude but point in opposite directions.

$CO_2$ has linear shape like shown in the image.

Due to this shape the dipoles of two oxygen atoms cancel each other and that is why $CO_2$ has

$0$ dipole moment.

The bond angle and hybridization in ether

$(CH_3OCH_3)$ is ______.

0%
$106.51^0 , sp^{2}$
0%
$110^0$ , $sp^3$
0%
$109.28^0, sp^3$
0%
None of these.

Choose the correct options:

0%
In $O_2F_2$ , the $O-F$ bond length is lesser than expected value.
0%
In trislyl amine $(SiH_3)_3$ N, the bond length is lesser than expected value
0%
The bond angle in $OF_2$ is lesser than in $Cl_2O$
0%
The Be atom in $BeCl_2(s)$ is $sp^3$ hybridised.

The dipole moment of chlorobenzene is

$1.73 D$ . The dipole moment of

$p-dichlorobenzene$ is excepted to be:

0%
$3.46 D$
0%
$0.00 D$
0%
$1.73 D$
0%
$1.00 D$

Explanation

Dipole - moment of chlorobenzene is

$1.73$ D. The dipole moment of a molecule is determined by the electronegativity of elements in the molecule and the symmetry of the molecule. Because, p-dichlorobenzene i.e,

$1,4-$ dichlorobenzene is symmetrical, the molecule will have a zero dipole moment.

Which of the following has the highest dipole moment?

0%
$AsH_3$
0%
$SbH_3$
0%
$PH_3$
0%
$NH_3$

Explanation

As we go down a group, Electronegativity of elements decreases.

So, the difference between electronegativities of elements of a group and Hydogen decreases.

So, the dipole moment of trihydrides of a group decreases as we go down a group.

So,

$\mathrm{NH_3}$ has the highest dipole moment among trihydrides of Nitrogen Family.

Hence, Option "D" is the correct answer.

CBSE Questions for Class 11 Medical Chemistry Chemical Bonding And Molecular Structure Quiz 7 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Chemical Bonding And Molecular Structure Quiz 7 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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