MCQ Questions for Class 11 Medical Chemistry Chemical Bonding And Molecular Structure Quiz 3

In which of the following pairs, indicated bond is of greater strength?

0%
I
0%
II
0%
Both are equal
0%
None of the above

Explanation

Chlorine is smaller than bromine. So there will be more attraction to the nucleus and less electron shielding, so the bond length will be less and

$C-Cl$ is stronger than

$C-Br$ .

In which of the above compounds, the indicated bond is of greater strength?

0%
I
0%
II
0%
Both are equal
0%
Cannot determine

Explanation

$sp^2$ carbon atom is more electronegative than

$sp^3$ carbon atom. The electronegativity difference between

$sp^2$ carbon atom and bromine atom is less than the electronegativity difference between

$sp^3$ carbon atom and bromine atom.
Hence, the extent of sharing of electron pair in the bond between

$sp^2$ carbon atom and bromine atom is more than the extent of sharing of electron pair in the bond between

$sp^3$ carbon atom and bromine atom.
So, the bond between

$sp^2$ carbon atom and bromine atom is stronger than the bond between

$sp^2$ carbon atom and bromine atom.

The maximum covalency is equal to :

0%
the number of unpaired p-electrons
0%
the number of paired d-electrons
0%
the number of unpaired s and p-electrons
0%
total number of unpaired electron in ground state or in excited state

Which of the following molecules will not have a dipole moment?

0%
$CH_3Cl$
0%
trans $CHCl = CHCl$
0%
$CH_2Cl_2$
0%
$CCl_2 = CH_2$

Explanation

In trans-

$1,2$ -dichloroethane, the dipole vectors of two

$C-Cl$ bonds are equal in magnitude and opposite in direction. Hence, they cancel each other.
Similarly the dipole vectors of two

$C-H$ bonds are equal in magnitude and opposite in direction. Hence, they cancel each other.
Thus the molecule has no net dipole moment.

The total number of lone-pair of electrons in melamine is:

0%
$3$ lone pairs from the triazine moeity
0%
$3$ lone pairs from the $-NH_2$ moeity
0%
$6$ lone pairs from all the $N$
0%
None of these

Explanation

Each

$N$ uses

$3$ bonds and hence has

$1$ lone pair. Hence total of

$6$ lone pairs.

S in

$\displaystyle SF_{6}$ is ______________ hybridized.

0%
$\displaystyle sp^{3}d^{3}$
0%
$\displaystyle sp^{3}d^{4}$
0%
$\displaystyle sp^{3}d^{2}$
0%
None of these

Explanation

$SF_6$ , sulphur has six valence electrons and there are six fluorine atoms that are monovalent. So, by applying the hybridization formula answer comes for 6 hybrid orbitals, which is

$sp^3d^2$ hybridisation.
1297346_147061_ans_c3d52ff7feeb4135abfc9fc081c2de60.PNG

The total spin of two electrons forming a covalent bond is :

0%
0
0%
$\pm 1/2$
0%
1
0%
$\pm 3/2$

Explanation

The total spin of two electrons forming a covalent bond is zero. When two electrons form a covalent bond, they have opposite spins. One electron has

$+\frac {1} {2}$ spin and the other electron has

$-\frac {1} {2}$ spin. Thus, sum of the spins of two electrons is zero.

$+\frac {1} {2}-\frac {1} {2}=0$ .

Which of the following is/are covalent binary compound(s)?

0%
$BF_3$
0%
$KBr$
0%
$CaI_2$
0%
$AlF_3$

Explanation

A binary compound is a compound composed of only two elements i.e., a binary compound is made from exactly two different elements. There are two types of binary compounds.
1) Ionic binary compound: Binary compound which is formed by the combination of a metal atom and non metal atom is called ionic binary compound, e.g., KBr, NaCl,

$CaI_2$ ,

$AlF_3$ ,

$MgBr_2$ .
2) Covalent binary compound: Binary compound which is formed by the combination of two non metal atoms is called binary covalent compound. In binary covalent compound sharing of electrons takes place, e.g., BF,

$N_2O$ ,

$P_2O_5$ ,

$N_2H_4$ ,

$CH_4$ and

$H_2O$ .
A) Binary compound formed by B and F is $BF_3$ in which both boron(B) and florine(F) are non metals, hence, it is covalent binary compound.
B) Binary compound formed by K and Br is KBr in which potassium(K) is metal and bromine(Br) is non metal. Hence, its is ionic binary compound.

$\displaystyle N_{2}$ molecule contains ________ o and _______

$\displaystyle \pi-$ bonds.

0%
one , four
0%
three , two
0%
one , two
0%
none

Explanation

Sigma bond always between S-S and Px-Px overlap. While pi bond is formed between Py-Py and Pz-Pz overlap.

So according to above diagram it is clear that there two pi bonds and one sigma bond in

$N_2$ .

Which one would better represent bonding in the molecule

$HNO$ ?
I :

$H - \ddot{N} = \ddot{O}:$ and II :

$H - \ddot{O} = \underset{\cdot\cdot}{\ddot{N}}$

0%
$I > II$
0%
$II > I$
0%
$I = II$
0%
None of these

Explanation

$I$ is better as octets of

$N$ and

$O$ are complete and duplet of

$H$ is also complete.

True structure is predicted by :

0%
valence bond approach
0%
Sidgwick approach
0%
hybrid orbital formation
0%
None of the above

The nitrogen atom has an atomic number

$7$ and oxygen has an atomic number

$8$

Calculate the total number of electrons in nitrate ion.

0%
$8$
0%
$16$
0%
$32$
0%
$24$

Explanation

No.of electrons in

$NO^-_3$ = (Electrons in

$N$ )+(

$3\times$ electrons in

$O$ )+[

$1$ (due to negative charge)]

$7$ +

$3\times 8+1=32$ .

So, the correct answer is 32

The formal charge on the O atoms in the ion

$[:\ddot{O} = N = \ddot{O}:]^+$ is:

0%
$-2$
0%
$-1$
0%
$0$
0%
$+1$

Explanation

Formal charge

$=$ Valence electrons-non bonding electrons+no of bonds

$=6-(4+2)=0$

Consider the above Lewis Dot structure:

The formal charges on

$S$ and

$O$ are respectively:

0%
$0 , 0$
0%
$+2 , 0$
0%
$0 , -2$
0%
$+1 , -1$

Explanation

Charge=valence electrons-[non-bonded electrons+no.of bonds]

$S:6-[2+4]=0$

$O:6-[4+2]=0$

Which of these molecules have non-bonding electron pairs on the central atom?

$I : SF_4;\, \, \, \, \, II : ICl_3;\, \, \, \, \, III : SO_2$

0%
II only
0%
I and II only
0%
I and III only
0%
I, II and III

Number of lone pairs around phosphorus in

$PCl_5, PCl_4^+$ and

${PCl_6}^-$ are respectively

0%
$0, 1, 2$
0%
$0, 0, 0$
0%
$1, 2, 3$
0%
$0, 2, 1$

Explanation

$P$ atom has

$5$ valence electrons,

$PCl_{ 5 }$ , all

$5$ electrons involed in bonding.

$PCl_{ 4 }^{ + }$ :

$+$ sign signifies loss of

$1$ electron, rest

$4$ involved in bonding.

$PCl_{ 6 }^{ - }$ :

$-$ sign signifies gain of

$1$ electron, all

$6$ involved in bonding.

By the axial overlapping of two p-orbitals of the same energy, a strong sigma bond is formed.

0%
True
0%
False
0%

Ambiguous
0%
None of these

Explanation

Sigma bond is formed by the linear or head to head or end on overlapping of orbitals. Sigma bonds are the strong bonds due to maximum overlapping of orbitals. Electron density is maximum around the bond axis. Therefore the axial overlapping of two p-orbitals of the same energy results in a strong sigma bond.

Hence, the correct option is

$\text{A}$

Number of bond pairs (X) and lone pairs (Y) around the central atom in the

${ I }_3^-$ ion are

0%
$X-2, Y-2$
0%
$X-2, Y-3$
0%
$X-3, Y-2$
0%
$X-4, Y-3$

Explanation

Explanation:

Molecular geometry of

$I^{-}_{3}$ is linear. In this there are three Iodine atoms, one of the atoms has a negative charge which further gives 3 lone pairs of electrons and 2 bond pairs

How is the structure of triiodide ion (I3-) possible? - Chemistry Stack Exchange

From the structure it is clear that there are

$3$ lone pairs and

$2$ bond pairs. Option B is correct answer

The maximum covalency is equal to the number of:

0%
paired p-electrons
0%
unpaired s-electrons
0%
unpaired s-and p-electrons
0%
s-and p-electrons in the valence shell

Which contains both polar and non-polar covalent bonds?

0%
$N{ H }_{ 4 }Cl$
0%
$HCN$
0%
${ H }_{ 2 }{ O }_{ 2 }$
0%
$C{ H }_{ 4 }$

Explanation

${ H }_{ 2 }{ O }_{ 2 }$ contains both polar and non-polar covalent bonds. In the structure of

${ H }_{ 2 }{ O }_{ 2 }$ , there are 3 sigma bonds,2 between

$O$ and

$H$ and one bond is between 2

$O$ atoms.The two bonds between

$O$ and

$H$ are polar while that between two

$O$ atoms are non-polar.

Calculate the steric numbers for iodine in

$(IF_4^-)$ and for bromine in

$(BrO_4^-)$ .

0%
$6$ and $4$
0%
$5$ and $6$
0%
$4$ and $6$
0%
$3$ and $4$

Explanation

The central

$I^-$ has eight valence electrons
Each

$F$ atom has seven valence electrons of its own and needs to one of the electrons from

$I^-$ to attain noble-gas configuration

$(s^2p^6)$ . Thus, four, of the

$I^-$ valence electrons take part in covalent bonds, leaving the remaining four to form two lone pairs. Thus,

$S_N =$ 4 (bonded atoms) + 2 (lone pairs)

$= 6$
The central

$Br^-$ has also eight electrons. In

$BrO_4^-$ , each oxygen atom needs to share two of electrons from

$Br^-$ to attain noble-gas configurations thus all the electrons are used in bonding and no lone pair exists.
Thus,

$S_N =$ 4 (bonded pair) + 0 (lone pairs)

$=4$ .

The number of lone pairs of electron on

$Xe$ in

$XeO{F}_{4}$ is:

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

Explanation

The number of lone pairs of electron on $Xe$ in $XeO{F}_{4}$ is 1. $Xe$ in $XeO{F}_{4}$ has ${ sp }^{ 3 }{d}^{2}$ -hybridization having one lone pair on $Xe$ -atom and is square pyramidal in shape.

The ratio of lone pair and bond pair electrons on central

$I$ atom in

${I}_{3}^{-}$ :

0%
$1$
0%
$1.5$
0%
$0.6$
0%
$0.5$

Explanation

The ratio of lone pair and bond pair electrons on central $I$ atom in ${I}_{3}^{-}$ is $1.5$ .

Thus, No. of lone pair of electrons $=3$

No. of bond pair of electrons $=2$

The ratio of lone pair and bond pair electrons $= 3/2 = 1.5$

Number of bond pair and lone pair of electrons in

${OF}_{2}$ are:

0%
$2,6$
0%
$2,8$
0%
$2,10$
0%
$1,9$

Explanation

Number of bond pair and lone pair of electrons in

$OF_2$ are

$2,8$ .

In which, central atom has one lone pair of electron?

0%
${Cl}_{2}$
0%
${NH}_{3}$
0%
$P{Cl}_{3}$
0%
$Xe{F}_{6}$

Explanation

$Cl_2$ molecule central atom contains three lone pair of electrons.

$NH_3$ molecule central atom nitrogen contains one lone pair of electrons.

$PCl_3$ central atom phosphorous contains one lone pair of electron.

$XeF_6$ central atom Xenon contains one lone pair of electron.

So answers are B,C,D.

Which statements are correct?

0%
A pi bond is weaker than sigma bond
0%
A sigma bond is weaker than pi bond
0%
A double bond is stronger than single bond
0%
A covalent bond is stronger than $H$ -bond

Explanation

Option

$(A),\ (C)$ and

$(D)$ are correct.

$(A)$ : A pi bond is weaker than sigma bond.A sigma bond is the end-to-end overlap of the bonding orbitals.whereas, a pi bond is the side-to-side overlap of (usually) two p-orbitals. This places the electron density above and below the plane of the nuclei with a node in the middle. The nodal region reduces the degree to which the orbitals overlap and therefore a pi bond is less strong than a sigma bond between the same two elements.

$(B)$ : A sigma bond is stronger than

$\pi$ bond.

$(C)$ : A(=) bond is stronger than a single bond. It is because there are more electrons to be shared by the two atoms, so the nuclei are more attracted. The bonds to be longer because there are more electrons and the electrons would repel each other more.

$(D)$ : A covalent bond is stronger than

$H$ -bond. The difference in strength is due to the fact that hydrogen bonding is an intermolecular force due to dipole formation in molecules. Covalent bonding is due to the sharing of valence electrons between atoms (intramolecular attraction).

In the cyanide ion the formal negative charge is on:

0%
$C$
0%
$N$
0%
both $C$ and $N$
0%
resonate between $C$ and $N$

Explanation

In the cyanide ion the formal negative charge resonates between $C$ and $N$ . Cyanide ion has resonating structure, $-\overset { \_ }{ C } \equiv N\longleftrightarrow \overset { \_ }{ N } \underrightarrow { = } C$

Select the correct statements.

0%
van der Waals' radii is always larger than the covalent radii
0%
The bond length of a particular bond depends on the state of hybridization of the involved atoms
0%
When %s - character increases, then bond length increases
0%
All are incorrect

Select the correct statement about chemical bond.

0%
It is attraction between atoms
0%
It is caused by the electrostatic force of attraction
0%
The strength of chemical bonds varies considerably
0%
All of the above

What is the value of

$1D$ in

$SI$ units ?

0%
$3.336\, \times 10^{-30}\, cm$
0%
$33.36\, \times\, 10^{-30}\, cm$
0%
$333.6\, \times\, 10^{-30}\, cm$
0%
None of these

Explanation

The debye is a CGS unit of electric dipole moment.

It is defined as

$1 \times 10^{−18}$ statcoulomb-centimetre and

$3.33564\times 10^{−30} C.m$ .

Which of the following statements is/are not correct?

0%
$\pi$ - bond always exists with sigma -bond according to Valence Bond Theory
0%
$\pi$ - bond can exist independently according to Valence Bond Theory
0%
$\sigma$ - bond is weaker than pi-bond
0%
$\pi$ - bond is less reactive than sigma-bond

Explanation

(A)

$\pi-$ bond always exists with sigma-bond according to Valence Bond Theory.

(B)

$\pi-$ bond cannot exist independently.

$\pi$ bond are usually weaker than sigma bonds. Quantum mechanics says this is because the orbital paths are parallel so there is much less overlap between the p-orbitals.

$\pi$ bonds happen when two atomic orbitals are in contact through two areas of overlap.

(D) Unsaturated compounds (containing pi bonds due to the existence of a double bond) are more reactive than saturated compounds (containing only sigma bonds) because the electron density at the internuclear axis is zero in a

$\pi$ bond ( in contrast to the sigma bond).

$\pi$ bonds involve sideways overlap while sigma bond involves a head-on or axial overlap.

Shape of the molecules is decided by:

0%
$\sigma$ -bond
0%
$\pi$ -bond
0%
both $\sigma$ and $\pi$ bonds
0%
neither $\sigma$ nor $\pi$ bond

Explanation

The reason that

$\sigma$ bonds ( and lone pairs) determine the geometry is that they form the basic skeleton of the molecule,

$\sigma$ bonds are formed by head-on overlap of atomic orbitals, meaning that they are oriented a long the imaginery axis connecting two atomic nuclei, and hence concentrate electron density in the region directly between the nuclei,

$\pi$ bonds, on the other hand, are essentially orthogonal to the

$\sigma$ bond skeleton, and are substantially weaker. Moreover,

$\pi$ bonds do not exist in isolation, meaning any

$\pi$ bond between two given atoms is always formed secondarily to the

$\sigma$ bond between said atoms. As such,

$\pi$ bonds do not alter the basic idealized geometry of a molecule as dictated by

$\sigma$ bonding.

$XeF_{2},\, XeF_{4}$ , and

$XeF_{6}$ , the number of lone pairs on

$Xe$ , is__________.

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

Explanation

$XeF_2$ has a linear shape. Its hybridization is

$sp^3d$ which means it should have a trigonal bipyramidal shape. Since it has

$3$ lone pairs, they occupy the equatorial triangle.

$XeF_4$ has

$sp^3d^2$ hybridization. It has a square planar shape and

$2$ lone pair of electrons.

$XeF_6$ has one lone pair of electrons. The

$6$ out of

$8$ electrons of Xenon are shared by

$6$ atoms of fluorine.

Hence option D is the right answer.

The correct order of a dipole moment is :

0%
$CH_4 < NF_3 < NH_3 < H_2O$
0%
$NF < CH_4 < NH_3 < H_2O$
0%
$NH_3 < NF_3 < CH_4 < H_2O$
0%
$H_2O < NH_3 < NF_3 < CH_4$

Explanation

The correct order of dipole moment is

$CH_4 < NF_3 < NH_3 < H_2O$ .
methane is zero dipole moment as it is tetrahedral molecule in which

$C-H$ bond dipoles cancel each other.
In

$NF_3$ , the bond dipole of the lone pair and the resultant dipole of three

$N-F$ bonds are in opposite direction wheres in ammonia, they are in same direction. Hence, the dipole moment of

$NF_3$ is smaller than that of methane.
The dipole moment of water is greater than the dipole moment of ammonia as oxygen is more electronegative than nitrogen.

Select the correct order of dipole moment.

0%
$BF_3\, <\, H_2O\, <\, H_2S$
0%
$H_2O\, >\, BF_3\, >\, H_2S$
0%
$H_2S\, <\, H_2O\, <\, BF_3$
0%
$BF_3\, <\, H_2S\, <\, H_2O$

Explanation

$BF_{ 3 }$ is symmetric and therefore, its dipole moment is zero.

In case of

${ H }_{ 2 }O$ ,

$O$ is more electronegative and hence, it is more polar. So,the correct order is as follows:

$BF_3\, <\, H_2S\, <\, H_2O$

A lone pair of electrons in an atom implies :

0%
a pair of valence electrons
0%
a pair of electrons
0%
a pair of electrons involved in bonding
0%
a pair of electrons not involved in bonding

Which of the following species has the maximum number of lone pair of electrons on the central atom?

0%
${ClO_{3}}^{-}$
0%
$XeF_{4}$
0%
$SF_{4}$
0%
${I_{3}}^{-}$

Explanation

Explanation:

$ClO_3^-$ has central $Cl$ as $sp^3$ hybridized and $1$ lone pair present on the central $Cl$ .
It is trigonal pyramidal in shape.

$Structure \space of \space XeF_4$
$XeF_4$ has central $4 \ Xe$ as $sp^3d^2$ hybridisation and there are $2$ lone pair present on the central $Xe$ .
It has square planar shape.

$Structure of SF_4$
$SF_4$ has central $S$ as $sp^3d$ hybridization and there is $1$ lone pair present on the central $S$ .
It has a distorted tetrahedral shape.

$Structure \space of \space I_{3}^{-}$
$I_3^-$ has central $I$ having $sp^3d$ hybridization and there are $3$ lone pair present on the central $I$ .
It has linear shape.
So, $I_3^-$ has maximum lone pair present on the central atom.

Correct Option : $D$ .

Which of the following species has four lone pairs of electrons in its outer shell ?

0%
$I$
0%
$O^-$
0%
$Cl^-$
0%
$He$

Explanation

Consider the electronic configuration of each of these options:-

$I$ -

$[Kr] 4d^{10} 5s^2 5p^5 \rightarrow$ 3 lone pairs.

$O^-$ -

$1s^2 2s^2 2p^5 \rightarrow$ 3 lone pairs.

$Cl^-$ -

$1s^2 2s^2 2p^6 3s^2 3p^6\rightarrow$ 4 lone pairs.

$He$ -

$1s^2\rightarrow$ 1 lone pair.

Hence option C is the right answer.

Arrange the following types of interactions in order of increasing stability (covalent, van der Waals' force, hydrogen bonding).

0%
Hydrogen bonding < covalent < van der Waals' force
0%
Covalent < hydrogen bonding < van der Waals' force
0%
Hydrogen < Van der Waals' force < covalent bonding
0%
Van der Waals' force < hydrogen bonding < covalent

Which of the following phenomenon will occur when two atoms of same spin will react?

0%
Bonding will not occur.
0%
Orbital overlap will not occur.
0%
Both (a) and (b)
0%
None of these.

How many bonding pairs and lone pairs surround the central atom in the

${ I }_3^-$ ion ?

0%
$2, 2$
0%
$2, 3$
0%
$3 ,2$
0%
$4 ,3$

Explanation

$H=\dfrac {1 }{2 }\left[ V+M-C+A \right]$
where,
H= Number of orbitals involved in hybridization.
V=Valence electrons of the central atom.
M- Number of monovalent atoms linked to a central atom.
C= Charge of a cation.
A= Charge of an anion.
Consider the hybridization and shape of the options:-
A)

${ I }_{3}^-$

$H = \dfrac { 1}{2 } \left[ 7+2+1\right] =5$ .

$\Rightarrow sp^3d$ hybridized state with 3 lone pairs.
Linear in shape.
It has 2 bond pairs.

For the type of interactions : (I) covalent bond, (II) van der Waals' forces, and (III) hydrogen bonding, which represents the correct order of increasing stability?

0%
$(I) < (III) < (II)$
0%
$(II) < (III) < (I)$
0%
$(III) < (II) < (I)$
0%
$(II) < (I) < (III)$

Explanation

A hydrogen bond is the electrostatic attraction between polar groups that occurs when a hydrogen

$(H)$ atom bound to a highly electronegative atom such as nitrogen

$(N)$ , oxygen

$(O)$ or fluorine

$(F)$ experiences attraction to some other nearby highly electronegative atom. These hydrogen-bond attractions can occur between molecules (intermolecular) or within different parts of a single molecule (intramolecular). The hydrogen bond (5 to 30 kJ/mole) is stronger than a van der Waals interaction, but weaker than covalent or ionic bonds.

$OF_{2}$ , the number of bond pairs and lone pairs of electrons is, respectively :

0%
$2, 6$
0%
$2, 8$
0%
$2, 10$
0%
$2, 9$

Explanation

$H=\dfrac {1 }{2 }\left[ V+M-C+A \right]$
where,
H= Number of orbitals involved in hybridization.
V=Valence electrons of central atom.
M- Number of monovalent atoms linked to central atom.
C= Charge of cation.
A= Charge of anion.
Now consider

$OF_2$

$H = \dfrac { 1}{2 } \left[6+2\right] =4$ .

$\Rightarrow sp^3$ hybridized state with 2 lone pairs on central atom.
Bent shape.
Hence it has 2 bond pairs and

$2+3+3 = 8$ lone pairs.
Hence option B is the right answer.

State True or False.

Dipole moment is completely based on ionic nature of bond.

0%
True
0%
False

Which of the following compounds have the same no. of lone pairs with their central atom:

0%
$XeF_{5}^{-}$
0%
$BrF_{3}$
0%
$XeF_{2}$
0%
Triple methylene

Explanation

Hence,

$XeF_{5}^{-}$ and

$BrF_{3}$ both have 2 lone pairs.

1155067_256976_ans_5701c5900fdb4164b93dfc0ffdbee0b2.JPG

The net dipole in the water molecules is the resultant of its bond dipoles.

0%
True
0%
False

Nodal planes of

$\pi$ -bonds (s) in

$CH_2\, =\, C\, =\, C\, =\, CH_2$ are located in:

0%
all are in molecular plane
0%
perpendicular plane which contains $C- C$ $\sigma$ -bond
0%
one in molecular plane and two in a plane perpendicular to molecular plane which contains $C -C$ $\sigma$ -bonds
0%
two in molecular plane and one in a plane perpendicular to molecular plane which bisects $C- C$ $\sigma$ -bonds at right angle

Explanation

Nodal planes of

$\pi$ - bonds (s) in

${ CH }_{ 2 }=C=C={ CH }_{ 2 }$
The pi bond is formed perpendicular to the plane in which one lobe of electron density is above the plane and one lobe of electron density is below the plane.
The nodal plane of the pi bond is in between the two lobes of electron density of the pi bond.
So 2 in a molecular plane and 1 in the plane perpendicular to the molecular lane in which has

$C-C$ bond.

How many bonded electron pairs are present in

$IF_7$ molecule?

0%
$6$
0%
$7$
0%
$5$
0%
$8$

Explanation

$IF_7$ molecule contains seven bonds in its structure. So

$7$ bonded electron pairs are present in

$IF_7$ molecule.

So answer is B.

Which is not characteristic of

$\pi$ -bond?

0%
$\pi$ - bond is formed when a sigma bond already formed
0%
$\pi$ - bond are formed from hybrid orbitals
0%
$\pi$ - bond may be formed by the overlapping of p-orbitals
0%
$\pi$ - bond results from lateral overlap of atomic orbitals

Explanation

$\pi$ - bonds are formed from unhybridized orbitals.
Thus in ethylene, unhybridized

$2p$ orbital of one

$C$ atom overlaps with unhybridized

$2p$ orbital of second

$C$ atom to form

$\pi$ - bond.

Consider the following molecules/ions:
(1)

$SF_4$ (2)

$CF_4$ (3)

$XeF_4$ (4)

${BF_4}^-$
Which pair doesnot contain lone pair of electrons around its central atom?

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

Explanation

(1) In

$SF_4$ , the central

$S$ atom has

$6$ valence electrons, out of which

$4$ are involved in the formation of

$4\ S-F$ bonds. The remaining

$2$ electrons are present as a lone pair.
(2) In

$CF_4$ , the carbon atom has

$4$ valence electrons, all of which are involved in the formation of bonds with four

$F$ atoms. Hence, no lone pairs are present.

(3) In

$XeF_4$ , the

$Xe$ atom has

$8$ valence electrons out of which

$4$ are involved in the formation of bonds with four

$F$ atoms. Hence, two lone pairs are present.
(4) In

$BF_4^-$ , the

$B$ atom has

$4$ valence electrons all of which are involved in the formation of bonds with four

$F$ atoms. Hence, no lone pairs are present.

CBSE Questions for Class 11 Medical Chemistry Chemical Bonding And Molecular Structure Quiz 3 - 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 3 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

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