MCQ Questions for Class 11 Medical Chemistry Chemical Bonding And Molecular Structure Quiz 5

Consider the structure pictured above, including the accompanying ligand.
Which of the following CORRECTLY identifies the pictured compound by name, and identifies the bonding involved in its structure?

0%
The compound is ammonium dichromate. It contains both ionic and covalent bonds.
0%
The compound in diammonium chromate. It contains both ionic and covalent bonds.
0%
The compound is ammonium dichromate. It contains only ionic bonds.
0%
The compound is ammonium chromate. It contains ionic bond.

Explanation

From the given image, the compound is ammonium dichromate with the chemical formula

${ \left( { NH }_{ 4 } \right) }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }$ . Which have

$2$

${ NH }_{ 4 }^{ + }$ ions and a

${ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }$ ions.

Both the ions are bonded through ionic bond

${ \left( { NH }_{ 4 } \right) }_{ 2 }\pm { Cr }_{ 2 }{ O }_{ 7 }^{ 2- }$ Ionic bond. and the bonds between

$N-H$ and

$Cr=O$ in Ammonium ion and

${Cr_2O_7}^{-2}$ respectively, are covalent. Hence, the given compound in Ammonium dichromate with both ionic and covalent bonds.

Option A is correct.

In C-C bond is

$C_2H_6$ undergoes heterolytic fission, the hybridisation of two resulting carbon atoms is/are :

0%
$sp^2$ both
0%
$sp^3$ both
0%
$sp^2,sp^3$
0%
$sp,sp^2$

Explanation

${ H }_{ 3 }C-{ CH }_{ 3 }\quad \underrightarrow { heterolytic } \quad \underbrace { { CH }_{ 3 }^{ + } } \quad \underbrace { { CH }_{ 3 }^{ - } }$

${ SP }^{ 2 }$

${ SP }^{ 3 }$

The shapes of

$IF_3$ and

$IF_7$ are respectively:

0%
square pyramidal and pentagonal bipyramidal
0%
octahedral and pyramidal
0%
trigonal bipyramidal and square antiprismatic
0%
distorted square planar and distorted octahedral

Explanation

As refer to IMAGE

$01$

$I\rightarrow 5s^25p^5$

at ground state.

For

$IF_3$ ;

$I$ at excited state.

Geometry is Trigonal bipyramidal as refer to IMAGE

$02$

As refer to IMAGE

$03$

For

$IF_7$ ,

$I$ at excited state is

$sp^3d^3$

Geometry

$=$ Pentagonal bipyramidal

Shape

$=$ Pentagonal bipyramidal as refer to IMAGE

$04$

The total number(s) of stable conformers with non-zero dipole moment for the following compound is (are):

0%
1
0%
2
0%
3
0%
none of these

Explanation

$\bf{Hint-}$ Two or more structures that are categorized as conformers differ only in terms of the angle about one or more sigma bonds.

$\bf{Explanation-}$

$\bullet$ A dipole moment arises in any system in which there is a separation of charge. Dipole moments occur due to the difference in electronegativity between two chemically bonded atoms.

$\bullet$ All the eclipsed and fully eclipsed conformations are unstable because of dipole-dipole interactions. so, we have to consider only gauche conformation. as staggered confirmation will have zero dipole moment.

$\bullet$ The three conformers with non zero dipole are shown as follows:

$\bullet$ In conformation shown in

$Image-2$ has two

$Cl$ opposite to each other but due to the presence of two

$Br$ groups, the dipole moment becomes non-zero.

$\bullet$ In confirmation shown in

$Image-3$ , there is a greater dipole due to the presence of two

$Cl$ and two

$Br$ which does not cancel out.

$\bullet$ In conformation shown in

$Image-4$ , it has two

$Br$ opposite to each other so the dipole due to two

$Br$ cancels our but there is a dipole due to two

$Cl$ atoms present in the compound.
So, Total stable conformers with non-zero dipole moment are

$3$ .

$\bf{Conclusion-}$ Hence, the correct option is

$C$ .

There is change in the type of hybridisation when:

0%
$NH_3$ combines with $H^+$
0%
$AlH_3$ combines with $H{^-}$
0%
$NH_3$ forms $NH_2{^-}$
0%
$SiF_4$ forms $SiF_4^{2-}$

Explanation

Change in hybridization will take place only when there is change in no. of electrons around the central atom in a molecule.

$\Rightarrow$ Combination of

${ H }^{ + }$ to N of

$N{ H }_{ 3 }$ will bring no change in hybridization as

${ H }^{ + }$ has no electrons.

$\Rightarrow$ Combination of

${ H }^{ - }$ to Al of

$Al{ H }_{ 3 }$ will bring a change in hybridization as there will be addition of

$2$ electrons around Al by

${ H }^{ - }$

$\Rightarrow N{ H }_{ 2 }^{ - }$ formation from

$N{ H }_{ 3 }$ will take place when

${ H }^{ + }$ leaves

$N{ H }_{ 3 }$ molecule. This will not lead to removal of electrons from N atom. There will be no change in type of hybridization.

$\Rightarrow$ Formation of

$Si{ F }_{ 4 }^{ 2- }$ from

$Si{ F }_{ 4 }$ will lead to addition of one lone pair on Si atom.

There will be a change in type of hybridization.

The hybridisation and geometry of

$BrF_3$ molecules are:

0%
$sp^3d$ and $T$ shaped
0%
$sp^2d^2$ and tetragonal
0%
$sp^3d$ and bent
0%
none of these

Explanation

Steric no. of

$BrF_3=\cfrac{1}{2}[7+3]=5 \rightarrow sp^3d$

Molecular geometry is trigonal bipyramidal.

Shape is

$T-$ shaped (Refer IMAGE

$01$ )

$Br$ at ground state (Refer IMAGE

$02$ )

$Br- 4s^{2}4p^{5}$

$Br$ at excited state (Refer IMAGE

$03$ )

$Br- sp^{3}d$ with

$2lp$

What is the angle between any two bonds in methane molecule?

0%
$105^o54'$
0%
$109^o28'$
0%
$119^o28'$
0%
$190^o28'$

Explanation

The angle between any two bonds to the central carbon atom in the methane molecule is

$109^o28'$ . The molecule of methane has a tetrahedral geometry.
1316795_625830_ans_bf6db116459d4bf19413e3a1bcd411e3.png

Which of the following structure of a molecule is expected to have three bond pairs and one loan pair of electrons

$?$

0%
Octahedral
0%
Trigonal planar
0%
Pyramidal
0%
Tetrahedral

Explanation

Pyramidal structure of a molecule is expected to have three bond pairs and one loan pair of electrons. An example is

$ammonia$ molecule

$(NH_3)$ . The electron pair geometry is tetrahedral and molecular geometry is pyramidal.

The number of lone pair(s) of electrons on the central atom in

$[BrF_4]^-$ ,

$XeF_6$ and

$[SbCl_6]^{3-}$ are respectively:

0%
$2, 1$ and $1$
0%
$2, 1$ and $0$
0%
$2, 0$ and $1$
0%
$1, 0$ and $0$

Explanation

Hence, option

$A$ is the answer.

The number of

$C-C$ sigma bonds in the compound is:

0%
$16$
0%
$17$
0%
$18$
0%
$11$

Explanation

A triple bond consists of 1 sigma and two pi bonds.

A double bond consists of 1 sigma and one pi bonds.

A single bond is always only sigma bond.

So in the given compound we can count

$16$ sigma bonds.

Hence option A is correct.

Which has a maximum repulsive interaction?

0%
$bp - bp$
0%
$lp - lp$
0%
$lp - bp$
0%
Equal

Explanation

Repulsive interactions in decreasing order

$= lp - lp > lp - bp > bp - bp$ .

Which of the following statements are true with respect to electronic displacement in a covalent bond?
I. Inductive effect operates through

$\pi$ -bond.
II. Resonance effect operates through

$\sigma$ -bond.
III. Inductive effect operates through

$\sigma$ -bond.
IV. Resonance effect operates through

$\pi$ -bond.
V. Resonance and inductive effects operate through

$\sigma$ -bond.

0%
III and IV
0%
I and II
0%
II and IV
0%
I and V

Explanation

The statements III and IV are true with respect to electronic displacement in a covalent bond.
III. Inductive effect operates through

$\displaystyle \sigma-$ bond.
Inductive effect is permanent but weak effect involving displacement of

$\displaystyle \sigma-$ electrons towards more electronegative species.
It can be transmitted through a chain of C atoms.

IV. Resonance effect operates through

$\displaystyle \pi-$ bond.

The following example illustrates resonance.

$\displaystyle CH_2=CH-CH=CH_2 \leftrightarrow CH_2^+-CH=CH-CH_2^- \leftrightarrow CH_2^--CH=CH-CH_2^+$

Considering x-axis as the internuclear axis, which out of the following will not form a sigma bond.

0%
$1s$ and $1s$
0%
$1s$ and $2{p}_{x}$
0%
$2{p}_{y}$ and $2{p}_{y}$
0%
$1s$ and $2s$

Explanation

${ 2p }_{ y }$ and

${ 2p }_{ y }$ will not form a

$\sigma$ bond because on taking x-axis as internuclear overlap sideway (lateral overlapping) resulting in the formation of a

$\pi$ bond.

Bond polarity of diatomic molecule is because of:

0%
difference in electron affinities of two atoms
0%
difference in electronegativities of two atoms
0%
difference in ionisation potentials
0%
all of the above

The table shown below gives the bond dissociation energies

$(E_{diss})$ for single covalent bonds of carbon (

$C$ ) atoms with the element

$P, Q, R$ and

$S$ . Which element has the smallest atoms?

Bond	$E_{diss}(kJmol^{-1})$
$C - P$	$240$
$C - Q$	$328$
$C - R$	$276$
$C - S$	$485$

0%
$P$
0%
$S$
0%
$R$
0%
$Q$

Explanation

The element

$'S'$ has the smallest atoms. It corresponds to strongest

$\displaystyle C-S$ bond and maximum bond dissociation energy (485 kJ/mol).
Smaller is the bond length, stronger is the bond and higher is the bond dissociation energy.

$TeCl_4$ , the central atom tellurium involves:

0%
$sp^3$ hybridisation
0%
$sp^3d$ hybridisation
0%
$dsp^2$ hybridisation
0%
$sp^3d^2$ hybridisation

Explanation

Number of hybrid orbitals

$=\dfrac{1}{2}$ (no. of electrons in valence shell of atom + no. of monovalent atoms - charge no cation + charge on the anion

Number of hybride orbitals

$=\dfrac{1}{2}(6+4+0+0)=5$

Hence,

$TeCl_4$ shows

$sp^3d$ hybridisation.

Therefore, option B is correct.

What fraction of charge is present on

$I$ in covalently bonded

$\overset { \delta + }{ H } -\overset { \delta - }{ I }$ if the dipole moment of

$HI$ is

$0.38\ D$ and the bond length is

$\ 1.61\ \mathring { A }$ ?

0%
$5$ %
0%
$16$ %
0%
$33$ %
0%
$79$ %

Explanation

$\delta ,charge=\cfrac { Dipole\quad moment }{ d } =\cfrac { 0.38\times { 10 }^{ -18 }esu-cm }{ 1.61\times { 10 }^{ -8 } } =2.4\times { 10 }^{ -11 }esu$
% of charge

$=\cfrac { 2.4\times { 10 }^{ -11 }esu }{ 4.8\times { 10 }^{ -10 }esu } \times 100=5$ %

Which molecule has minimum dipole moment?

0%
$H_{2}O$
0%
$AgI$
0%
$PbSO_{4}$
0%
$HBr$

Which carbonyl compound has maximum dipole moment?

Explanation

Double bond oxygen can undergo resonance so that

$3-$ membered ring species has

$2\pi -{ e }^{ - }$ and has aromatic character, as a result of this, this species exist in

$'B'$ form and hence has more dipole because of more charge separation.

If the element

$A$ has three electrons in an outer shell and

$B$ has six electrons in an outer shell, then the formula of the compound formed by these element will be:

0%
$A_3B_2$
0%
$A_2B_3$
0%
$AB_3$
0%
$A_6B_3$

Explanation

Since

$'Al'$

$(2,8,3)$ has three valance electron to loose, euhen it looses all

$3{ e }^{ - }s$ and then it becomes

${ Al }^{ 3+ }$ and oxygen has valence electron

$2$ less than nobel gas configuration. So,

$3'O'$ atom got

$6{ e }^{ - }$ from

$2'Al'$ atom result in

${ Al }_{ 2 }{ O }_{ 3 }$ or

${ A }_{ 2 }{ B }_{ 3 }$ .

In which one of the following the central atom is

$sp^{3}$ hydridised?

0%
$NH_{4}^{+}$
0%
$BF_{3}$
0%
$SF_{6}$
0%
$PCl_{5}$
0%
$XeF_{4}$

${ PO }_{ 4 }^{ 3- }$ , the formal charge on each oxygen atom and the

$P-O$ bond order respectively are:

0%
$-0.75,0.6$
0%
$-0.75,1.0$
0%
$-0.75,1.25$
0%
$-3,1.25$

Explanation

${ PO }_{ 4 }^{ 3- }$ , the formal charge on each oxygen atom and the

$P-O$ bond order are

$-0.75,1.25$ respectively.

$\left[O-\underset{O}{\underset{||}{\overset{O}{\overset{|}{P}}}}-O\right]^{3-} \leftrightarrow \left[O-\underset{P}{\underset{|}{\overset{O}{\overset{|}{P}}}}=O\right]^{3-}$

$\left[O-\underset{O}{\underset{|}{\overset{O}{\overset{||}{P}}}}-O\right]^{3-} \leftrightarrow \left[O=\underset{P}{\underset{|}{\overset{O}{\overset{|}{P}}}}-O\right]^{3-}$

Bond order

$=\cfrac { \text { Number of bonds} }{ \text {Number of Resonating structures} } =\cfrac { 5 }{ 4 } =1.25$

In a given resonance structure, the O atom that forms double bond has formal charge of 0 and the remaining 3 O atoms have formal charge of -1 each.
In the resonance hybrid, a total of -3 charge is distributed over 4 O atoms. Thus the formal charge of each O atom is

$\dfrac {-3}{4} = -0.75$ .

Note:
To calculate the formal charge, the following formula is used.
Formal Charge

$=$ [Number of valence electrons on atom]

$–$ [non-bonded electrons

$+$ number of bonds]

For O atom that forms double bond with P atom,
Formal Charge

$=$ 6]

$–$ [4

$+$ 2]

$=$ 0.

For O atom that forms single bond with P atom,
Formal Charge

$=$ 6]

$–$ [6

$+$ 1]

$=-$ 1.

Which of the following has maximum number of lone pairs associated with

$Xe ?$

0%
$XeO_3$
0%
$XeF_4$
0%
$XeF_6$
0%
$XeF_2$

Explanation

$XeF_2$ has the maximum number of lone pairs of electrons.

Number of lone pair of electrons for

$XeF_2$ is

$3.$

The dipole moment of the chlorobenzene is

$1.5D$ . The dipole moment of given compound is:

0%
$0D$
0%
$1.5D$
0%
$3.0D$
0%
$2.25D$

The total number of stable conformers with non-zero dipole moment for following compound is (are) [Eclipsed forms are considered to be highly unstable]

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Which of the following compounds have zero dipole moment?

0%
${ CH }_{ 3 }Cl$
0%
${ CH }_{ 2 }{ Cl }_{ 2 }$
0%
${ BF }_{ 3 }$
0%

Explanation

$BF_3$ has a

$sp^2$ hybridisation and thus has a symmetric trigonal planar structure with a net zero dipole moment.

Match the given chemical reactions with the products listed in the box and select the correct option from the given codes.
(i)

$CaO$ (ii)

$Mg(OH)_{2}$ (iii)

$NO_{2}$ (iv)

$Cu$ (v)

$MgCl_{2}$ (vi)

$O_{2}$ (vii)

$Fe_{2}O_{3}$ (viii)

$CaCO_{3}$ .
P. Mixing magnesium oxide with water
Q. Brown deposit on iron nail kept in copper sulphate solution
R. Carbon dioxide is passed through lime water
S. Absorption of ultraviolet radiations by ozone.

0%
$P - (iv), Q - (vii), R - (i), S - (vi)$
0%
$P - (iv), Q - (ii), R - (i), S - (iii)$
0%
$P - (ii), Q - (iii), R - (viii), S - (vii)$
0%
$P - (ii), Q - (iv), R - (viii), S - (vi)$

Explanation

$P - (ii) : \underset {Magnesium\ oxide}{MgO} + \underset {Water}{H_{2}O}\rightarrow \underset {Magnesium\ hydroxide}{Mg(OH)_{2}}$

$Q - (iv) : \underset {Iron}{Fe} + \underset {Copper\ sulphate}{CuSO_{4}} \rightarrow \underset {Ferrous\ sulphate}{FeSO_{4}} + \underset {Copper (Brown)}{Cu}$

$R - (viii) : \underset {Carbon\ dioxide}{CO_{2}} + \underset {Lime\ water}{Ca(OH)_{2}} \rightarrow \underset {Calcium\ Carbonate}{CaCO_{3}} + \underset {Water}{H_{2}O}$

$S - (vi) : \underset {Ozone}{O_{3}}\xrightarrow {UV\ radiations} \underset {Oxygen}{O_{2}} + [O]$ .

Dipole-dipole interactions

$d\pi-p\pi$ bond present in:

0%
$CO_3^{2-}$
0%
$PO_4^{3-}$
0%
$N{ O }_3^-$
0%
$N{ O }_2^-$

Explanation

1. $p\pi-d\pi$ bonding includes d-orbitals, if you observe the options only P belongs to 3rd period and has d-orbitals (though unoccupied in ground state)

2. Other options $C$ and $N$ belong to 2nd period and do not possess d orbitals.

Which of the following ions is mostly unlikely to exist?

0%
${ Li }^{ - }$
0%
${ Be }^{ 2+ }$
0%
${ B }^{ - }$
0%
${ F }^{ - }$

Explanation

${ Li }^{ \left( - \right) }$ is unlikely to exist, as it is an alkaline earth metal. Electron accepting capability is very low.

Which of the following would have permanent dipole moment?

0%
$SiF_4$
0%
$SF_4$
0%
$XeF_4$
0%
$BF_3$

Explanation

$\bf{Hint-}$ Dipole moments occur when there is a separation of charge.

$\bf{Explanation-}$

$\bullet$ In

$SiF_4$ , the dipole moments cancel each other. Hence it does not have a permanent dipole moment.

$\bullet$ The hybridization of a

$SF_4$ molecule is

$sp^3d$ , thus a see-saw structure (bent because of

$lp-lp$ repulsion) and thus having a net dipole moment.

$\bullet$

$XeF_4$ has

$4 \ S-F$ bonds and

$2$ lone pair on

$Xe$ so overall structure becomes symmetric and all the dipole cancels each other, so it is non-polar.

$\bullet$ In

$BF_3$ , the dipole moments cancel each other. Hence it does not have a permanent dipole moment.

$\bf{Conclusion-}$ As there is only one compound that has finite dipole moment that is

$SF_4$ . So, Option

$B$ is correct.

Observe the given figure carefully and filled in the blanks by choosing an appropriate option.

0%
Crystallisation, Iron, Chemical reversible, Iron oxide
0%
Sublimation, Zinc, Physical irreversible, Zinc oxide
0%
Melting, Copper, Physical reversible, Copper oxide
0%
Crystallisation, Iron, Chemical irreversible, Copper

In which pair of molecules is the permanent dipole in molecule I greater than that in molecule II?

0%
0%
$CH_{3}CH_{2}Cl, \ \ CH_{3}CHCl_{2}$
0%
0%

Which one of the following species is diamagnetic in nature?

0%
$He^+_2$
0%
$H_2$
0%
$H^+_2$
0%
${ H }^-_2$

Explanation

As all the electrons are paired in

$H_{2}$ molecular orbitals. So

$H_{2}$ is diamagnetic.
1171508_770076_ans_95fb42da614e4b08a147fbbf99f96c61.png

How many ways a bond between two atoms can cleave?

0%
$One$
0%
$Five$
0%
$Three$
0%
$Four$

Which of the following has highest dipole moment?

0%
${CH}_{3}Cl$
0%
${CH}_{3}F$
0%
${CH}_{3}Br$
0%
${CH}_{3}I$

Explanation

Reason:-

$CH_3F$ has highest dipole moment due to its electronegativity but also due to its smallest size it is unstable &

$CH_3Cl$ outlooks

$CH_3F$ . Therefore,

$CH_3Cl$ has the highest dipole moment.

Which of the following compound has non zero dipole moment?

0%
$PCl_{3}$
0%
$PCl_{2}F_{3}$
0%
$PCl_{3}F_{2}$
0%
$PClF_{4}$

Explanation

$PCl_{3}$ dipole moment is non zero as lone pair being more electronegative tends to attract

$e^{-}$ density towards its ey. Hence has dipole moment.

$PCl_{2}F_{3}\rightarrow$ The two

$F$ at axial positions cancel out each others dipole effect and

$2Cl$ and

$1F$ at equitorial.

$\because Cl$ has less electronegativity than

$F$ . So resultant dipole of

$Cl\sim F$ .

$\therefore$ cancelling each others dipole effects resulting in zero dipole moment.

$PCl_{3}F_{2}$ , The two

$F$ at axial positions cancel each others dipole effect and at equitorial the

$3Cl$ bonds cancels out each others dipole effect resulting in zero dipole moment.

$PClF_{4}$ , The two

$F$ at axial positions cancel out each others effect and at equitorial the the resultant electron attracting tendency of

$2F>1Cl$ thereby having a non-zero dipole.

So, according to me,

$PCl_{3}$ and

$PCl_{4}$ have non-zero dipole.

Which one of the following species has plane triangular shape :

0%
${ N }_{ 3 }^{ - }$
0%
${ NO}^{-}_{ 3 }$
0%
${ NO }_{ 2 }^{ - }$
0%
${ CO }_{ 2 }$

Among these compounds the correct order of

$C-N$ bond length is :

0%
$IV > I > II > III$
0%
$III > I > II > IV$
0%
$III > II > I > IV$
0%
$III > I > IV > II$

Explanation

As a result of resonance there will be no formation of the double bond between

$C-N$ . Therefore, bond length will increase from

$(iv)$ to

$(i)$ then to

$(ii)$ & then to

$(iii)$ .

Which of the following compounds shows strongest hydrogen bonding?

0%
$HF$
0%
$H_2O$
0%
$NH_3$
0%
$CH_3OH$

Explanation

Among the

$F$ ,

$O$ and

$N$ , Fluorine has the higher value of electronegativity, this is the reason that fluorine can form the strongest hydrogen bond with the

$H$ atom of another molecule.

Just take an idea from the value of energy of hydrogen bond that is formed with

$HF$ ,

$H_2O$ and

$NH_3$

$HF$	$10kcal/mol$
$H_2O$	$7Kcal/mol$
$NH_3$	$2Kcal/mol$

means to break the hydrogen of one mole

$HF$ molecules,

$10Kcal$ energy is required.

So the correct option is [A].

According to molecular orbital theory, which of the following will not exist?

0%
$H_2^+$
0%
$Be_2$
0%
$B_2$
0%
$C_2$

Explanation

According to the molecular orbital theory elements which have bond order negative or

$0$ will not exist.

Formula to calculate bond order

$=\dfrac { Number\ of \ electrons\ in\ BMO -\quad Number\ of\ electrons\ in\ ABMO }{ 2 }$

where,

$BMO=$ Bonding molecular orbital

$ABMO=$ Antibonding molecular orbital

Molecular electronic configuation of

$Be_2=\sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2$

$Be_2$ has

$4$ electrons in

$BMO$ and

$4$ in

$ABMO$ so the bond order will come

$0$ .

So, the correct option is

$[B].$

The synthetic steroid ethynylestardiol (1) is a compound used in the birth control pill. How many

$sp^3$ hybridised carbon atoms are present in compound (1) ?

0%
8
0%
9
0%
10
0%
11
0%
12

Explanation

$sp^3$ hybridised carbon are those carbon having

$4$ single bond with

$4$ different groups.

(Refer to Image)

There are

$12$

$sp^3$ hybridised carbon atoms.

1187305_988743_ans_e765a35c5e8847c78c1c6437424b7b1d.JPG

The synthetic steroid ethynylestardiol (1) is a compuond used in the birth control pill. How many

$sp^2$ hybridised carbon atoms are present in compound (1)?

0%
4
0%
5
0%
6
0%
7
0%
8

Explanation

$sp^2$ hybridised carbon are those carbon which form double bond

(Refer to Image)

There re

$6$

$sp^2$ hybridised carbon.

1187329_988744_ans_79eeb44c7c0e4b9082911636a71609dd.JPG

$1,3,5,7-$ Octatetraene contains

$\underline { X }$

$\sigma -$ bonds and

$\underline { Y }$

$\pi-$ bonds.

$X$ and

$Y$ are:

0%
$23,\ 4$
0%
$17,\ 4$
0%
$18,\ 5$
0%
$33,\ 2$

Explanation

A $\sigma$ bond is a type of covalent bond. It is formed by the head-on overlap of two atomic orbitals. These combinations of overlapping orbitals will be as $s-s, s-p_{z}\ or \ p_{z} - p_{z}$ .

Also, a $\pi$ bond is also a covalent bond. It is formed by the side-to-side overlap of two atomic orbitals. The atomic orbital may have combinations as $p_{x} - p_{x} or p_{y} - p_{y}$

Thus as in the figure, total $\sigma$ bonds will be 17 and $\pi$ bonds will be 4.

Thus option B is correct.

1499135_928720_ans_05db1e82a5d64515aa27c327fe1cbcd9.jpg

Arrange the following molecules in the increasing order of

$\alpha$ to

$\pi$ bond ratio?

0%
(2)<(3)<(4)<(1)
0%
(2)<(4)<(3)<(1)
0%
(3)<(2)<(1)<(4)
0%
(2)<(3)<(1)<(4)

Which of the following is an isoelectronic pair?

0%
$ICl_{2}$ , $ClO_{2}$
0%
$BrO_{2}{^-}$ , $BrF_{2}^{^+}$
0%
$ClO_{2}$ , $BrF$
0%
$CN^{-}$ , $O_{3}$

Explanation

$Br{ O }_{ 2 }^{ - } \Rightarrow$ no of electrons =

$1 \times35+2 \times8+1 \times1$

$=52$

$Br{F}_{2}^{+} \Rightarrow$

$1 \times35+2 \times 9+1 \left( -1 \right)$

$=52$

$Br{ O }_{ 2 }^{ - }$ ,

$Br{ F }_{ 2 }^{ + }$ isoelectronic pair

The synthetic steroid ethynylestardiol (1) is a compuond used in the birth control pill.
How many

$sp$ hybridised carbon atoms are present in compound (1) ?

0%
2
0%
4
0%
6
0%
8
0%
10

Explanation

$sp$ hybridized carbon are those who forms triple bond.

(Refer to Image)

These are two carbons which are

$sp$ hybridized.

1187334_988745_ans_f231c81421434221b9a50496c218c87a.JPG

In which of the following molecule/ion all the bonds are not equal?

0%
$XeF_4$
0%
$BF_4{^-}$
0%
$C_2H_4$
0%
$SiF_4$

Explanation

Explanation:

$XeF_4$ has $sp^{3}d^{2}$ hybridisation and square planar in shape.

Since the shape is square planar and all the bonds formed in this molecule are with the same element, thus the bonds are equal.

$BF_4^{-}$ has $sp^{3}$ hybridization and tetrahedral shape.

Since the shape is tetrahedral and all the bonds formed in this molecule are with the same element, thus the bonds are equal.

$SiF_4$ has $sp^{3}$ hybridization and tetrahedral shape.

Since the shape is tetrahedral and all the bonds formed in this molecule are with the same element, thus the bonds are equal.

But in $C_2H _4$ has $sp^{2}$ hybridisation and planar shape with bond angle $120^{\circ}$ .

Since the type of bonds in this molecule are of two types i.e., $C=C$ and $C-H$ . Hence in this molecule, all the bonds are not equal.

Final Answer:

Hence, the correct answer is option $C.$

A diatomic molecule has a dipole moment of 1.2 D. If the bond legth is

$1.0\times 10^{-8}$ cm, what fraction of charge does exist each atom?

0%
0.1
0%
0.2
0%
0.25
0%
0.3

Explanation

$\mu = 1.2D$ =

$1.2 \times 3.336 \times {10}^{-30} Cm$

$\mu = q \times l$

$q = \dfrac{1.2 \times 3.336 \times {10}^{-30}}{{10}^{-10}}$

$q = 4 \times {10}^{-20}\space C$

Fraction of charge =

$\dfrac{4 \times {10}^{-20}}{1.6 \times {10}^{-19}}$ =

$0.25$

Which of the following statements is true about hydrogen bonding?

0%
$Cl$ and $N$ have comparable electronegativities yet there is no $H-$ bonding in $HCI$ because size of $Cl$ is large
0%
Intermolecular $H-$ bonding results in decrease in m.p. and b.p
0%
Ice has maximum density at $0^o$ C due to $H-$ bonding
0%
$KHCl_2(HCl_2{^-})$ exists but $KHF_2(HF_2{^-})$ does not exist due to lack of $H-$ bonding in $HCl$

Explanation

There are some limitations to form hydrogen bond for a molecule-

1. Atom must be of

$2nd$ period means the size of the atom must be small.

Reason= As we go down in group size of the atom is increase means covalent character also increases.

2. Electronegativity of the atom must be

$3$ or more than

$3$ .

Chlorine:- It follows the

$2nd$ condition but the size of

$chlorine$ is large that's why some covalent characters also present in the

$Chlorine$ .

This is the reason

$HCl$ has not any hydrogen bonding.

So the correct option is [A].

Dipole moments of given compound will be:

0%
$(A) = 6.87 D, (B) = 4.11 D$
0%
$(A) = 4.11 D, (B) = 6.87 D$
0%
$(A) = 4.11 D, (B) = 4.11 D$
0%
$(A) = 6.87 D, (B) = 6.87 D$

CBSE Questions for Class 11 Medical Chemistry Chemical Bonding And Molecular Structure Quiz 5 - 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 5 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

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