Chemical Bonding and Molecular Structure Test -8

Result

Chemical Bonding and Molecular Structure Test -8

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Polarity in a molecule and hence the dipole moment depends primarily on electronegativity of the constituent atoms and shape of a molecule. Which of the following has the highest dipole moment?

A
HI

B
SO₂

C
H₂O

D
CO₂

Solution

Because of presence of lp on head of O atom in H₂O therefore it has max dipole moment.
Question 2
1 / -0

The types of hybrid orbitals of nitrogen in NO₂⁺, NO⁻₃ and NH⁺₄ respectively are expected to be

A
sp², sp³and sp.

B
sp,sp³ and sp²

C
sp, sp² and sp³

D
sp², sp and sp³
Solution
- In NO₂⁺
  Number of electron pairs =2
  Number of bond pairs = 2
  Number of lone pair = 0
  So, the species is linear with sp hybridisation.
- In NO₃⁻
  Number of electron pairs =3
  Number of bond pairs = 3
  Number of lone pair = 0
  So, the species is trigonal planar with sp2 hybridisation
- In NH₄⁺
  Number of electron pairs =4
  Number of bond pairs = 4
  Number of lone pair = 0
  So, the species is tetrahedral with sp³ hybridisation.
Question 3
1 / -0

Isostructural species are those which have the same shape and hybridisation. Among the given species identify the isostructural pairs.

A
[BCl₃and BrCl₃]

B
[BF⁻₄ and NH⁺₄]

C
[NH₃ and NO⁻₃]

D
[NF₃and BF₃]

Solution

BF₄^- and NH₄⁺ both have sp³ hybridisation.
Question 4
1 / -0

Species having same bond order are :

A
O⁻₂ and N⁻₂

B
F⁺₂ and N⁺₂

C
F⁺₂ and O⁻₂

D
N₂ and N⁻₂

Solution

bond order of F₂⁺ and O₂^- is same i.e 1.5
Question 5
1 / -0

Which of the following statements are not correct?

A
NaCl being an ionic compound is a good conductor of electricity in the solid state.

B
In canonical structures there is no difference in the arrangement of atoms.

C
VSEPR Theory can explain the square planar geometry of XeF₄.

D
Hybrid orbitals form stronger bonds than pure orbitals.

Solution

NaCl is not good conductor in solid state. It conducts electricity only in molten state or in aqueous solution.
Question 6
1 / -0

Rank the bonds in the set C=O, C-O, C≡O in order of decreasing bond length

A
C=O > C≡O > C-O

B
C-O > C≡O > C=O

C
C≡O < C-O < C=O

D
C-O > C=O > C≡O

Solution

The length of the bond is determined by the number of bonded electrons (the bond order). The higher the bond order, the stronger the pull between the two atoms and the shorter the bond length. Generally, the length of the bond between two atoms is approximately the sum of the covalent radii of the two atoms. CO has triple bond so has minimum bond length.
Question 7
1 / -0

Rank the bonds in the set C=O, C-O, C≡O in order of decreasing bond strength

A
C=O < C≡O > C-O

B
C≡O > C-O < C=O

C
C-O < C≡O > C=O

D
C≡O > C=O > C-O

Solution

The more bonds between atoms, the shorter the bond is. for example a triple bond will be shorter than a double bond. Shorter=stronger, so the shorter the bond is the stronger it is (higher dissociation energy). Bond order is basically the number of bonds between atoms, so a larger number bond order=more bonds=shorter bonds=stronger bonds.
Question 8
1 / -0

Quartz is very hard and melts at1 550^∘C. Reason is:

A
Quartz (Si0₂) has silicon-oxygen covalent bonds that extend throughout the sample

B
separate Si0₂ molecules exist that are held together by pi bonds that extend in three dimensions throughout the sample

C
Quartz (Si0₂) has silicon-oxygen electrovalent bonds that extend throughout the sample

D
separate Si0₂ molecules exist they are held together by coordinate bonds that extend throughout the sample

Solution

It's because quartz have network of covalent bonds which means that they are covalently bonded in a network solid. The many covalent bonds must be broken which is why they have very high melting points.
Question 9
1 / -0

Chemical bond implies ___________.

A
repulsion

B
attraction

C
attraction and repulsion balanced at a particular distance

D
attraction and repulsion

Solution

Chemical bond is an attraction between atoms that allows the formation of chemical substances. The bond will be formed where attraction forces due to opposite charges were balanced by repulsion forces due to similar charges.
Question 10
1 / -0

$$Cl+Cl\rightarrow Cl_{2}$$, this is an example for __________.

A
Endothermic reaction

B
Exothermic reaction

C
Either exothermic or endothermic

D
Neither exothermic nor endothermic

Solution

$$Cl^{\cdot} + Cl^{\cdot} \longrightarrow Cl : Cl $$
As two Chlorine atoms come together their electrostatic energy decreases. According to the law of conservation of energy, this energy will be transformed into heat energy.
So, $$Cl + Cl \longrightarrow Cl_{2}$$ is exothermic reaction.
Question 11
1 / -0

The number of electron pairs present in the valency shell of central atom in $$SF_{6}$$ molecule are:

A
$$4$$

B
$$6$$

C
$$8$$

D
$$7$$

Solution

$$1$$ electron pair for each bond and there are $$6\: S-F$$ bonds in the given molecule.
Question 12
1 / -0

Molecule having maximum number of covalent bonds is:

A
$$NH_{4}OH$$

B
$$NH_{4}Cl$$

C
$$CO(NH_{2})_{2}$$

D
$$CH_{3}OH$$

Solution

Urea has maximum covalent bonds - $$8$$.
Question 13
1 / -0

Which of the following statement is correct?

A
Energy difference should be more between orbitals which undergo hybridization.

B
Number of hybrid orbitals formed should be same as the number of atomic orbitals involved in hybridization.

C
Hybrid orbitals arrange around the center of the atom unsymmetrically.

D
Hybrid orbitals can form $$\pi$$ bonds.

Solution

Number of hybrid orbitals formed is equal to number of atomic orbitals involved in hybridisation. $$\pi$$ bonds will be formed always by pure orbitals (p or d orbitals). Hybrid orbitals arrange around the centre of atom gymmetrically depending on number of bond pairs and lone pairs.
Question 14
1 / -0

The molecule having maximum number of sigma bonds in it is:

A
Ethane

B
Ethene

C
Ethyne

D
Sulphur hexafluoride

Solution

Ethane, $${C}_{ 2 }{ H}_{6 }$$ has seven sigma bonds.
Ethene $${C}_{ 2 }{H}_{4 }$$ has five sigma bonds.
Ethyne $${C}_{ 2 }{H}_{2 } $$ has three sigma bonds.
sulphur hexafluoride $$S{F }_{6 }$$ has six sigma bonds.
Question 15
1 / -0

The maximum number of H-bonds formed by a water molecule is:

A
$$2$$

B
$$8$$

C
$$1$$

D
$$4$$

Solution

A hydrogen atom can form only one H-bond and oxygen can form two H-bonds with two hydrogen atoms. In $$H_2O$$, there are two hydrogens and one oxygen. So, the maximum number of H-bonds formed by water molecules is $$4$$.
Question 16
1 / -0

Valence bond theory of Pauling and Slater accounts for the following characteristic of covalent bond:

A
directional

B
ionic

C
strength

D
hybrid

Solution

Valence bond theory was proposed by Heitler and London to explain the shapes of covalent molecules, their bond angles and bond lengths.It was extended by Pauling and Slater to explain the directional nature of covalent bonds.
Question 17
1 / -0

The electronegative elements between which the H-bond is formed principally in aqueous ammonia are:

A
$$O, O$$

B
$$N,N$$

C
$$O, N$$

D
Both $$O,O$$ & $$N,N$$

Solution

In aqueous ammonia, hydrogen atoms are attached to the oxygen and nitrogen (small size & high electronegative atoms). So, the hydrogen of one molecule forms H-bond with high electronegative atom of other molecule. So, a hydrogen bond is formed between oxygen and nitrogen.
Question 18
1 / -0

Molecule having maximum number of lone pairs of electrons on central atom is:

A
$$PH_{3}$$

B
$$H_{2}S$$

C
$$CH_{4}$$

D
$$BrF_{5}$$

Solution

$$P$$ belongs to group $$15$$ and has $$5$$ electrons in valence shell. $$3$$ are involved in bonding with the $$3$$ $$H$$ atoms.
So, lone pair $$= 1$$.

$$S$$ belongs to group $$16$$ and has $$6$$ electrons in valence shell. $$2$$ are involved in bonding with the $$2$$ $$H$$ atoms.
So, lone pairs $$= 2$$.

$$C$$ belongs to group $$14$$ and has $$4$$ electrons in valence shell. $$4$$ are involved in bonding with the $$4$$ $$H$$ atoms.
So, lone pair $$= 0$$.

$$Br$$ belongs to group $$17$$ and has $$7$$ electrons in valence shell. $$5$$ are involved in bonding with the $$5$$ $$F$$ atoms.
So, lone pair $$= 1$$.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Chemical Bonding and Molecular Structure Test -8

Result

Chemical Bonding and Molecular Structure Test -8

Score

-

Rank

-

SHARING IS CARING

Question 1

HI

SO2

H2O

CO2

Solution

Question 2

sp2, sp3and sp.

sp,sp3 and sp2

sp, sp2 and sp3

sp2, sp and sp3

Solution

Question 3

[BCl3and BrCl3]

[BF−4 and NH+4]

[NH3 and NO−3]

[NF3and BF3]

Solution

Question 4

O−2 and N−2

F+2 and N+2

F+2 and O−2

N2 and N−2

Solution

Question 5

NaCl being an ionic compound is a good conductor of electricity in the solid state.

In canonical structures there is no difference in the arrangement of atoms.

VSEPR Theory can explain the square planar geometry of XeF4.

Hybrid orbitals form stronger bonds than pure orbitals.

Solution

Question 6

C=O > C≡O > C-O

C-O > C≡O > C=O

C≡O < C-O < C=O

C-O > C=O > C≡O

Solution

Question 7

C=O < C≡O > C-O

C≡O > C-O < C=O

C-O < C≡O > C=O

C≡O > C=O > C-O

Solution

Question 8

Quartz (Si02) has silicon-oxygen covalent bonds that extend throughout the sample

separate Si02 molecules exist that are held together by pi bonds that extend in three dimensions throughout the sample

Quartz (Si02) has silicon-oxygen electrovalent bonds that extend throughout the sample

separate Si02 molecules exist they are held together by coordinate bonds that extend throughout the sample

Solution

Question 9

repulsion

attraction

attraction and repulsion balanced at a particular distance

attraction and repulsion

Solution

Question 10

Endothermic reaction

Exothermic reaction

Either exothermic or endothermic

Neither exothermic nor endothermic

Solution

Question 11

$$4$$

$$6$$

$$8$$

$$7$$

Solution

Question 12

$$NH_{4}OH$$

$$NH_{4}Cl$$

$$CO(NH_{2})_{2}$$

$$CH_{3}OH$$

Solution

SO₂

H₂O

CO₂

sp², sp³and sp.

sp,sp³ and sp²

sp, sp² and sp³

sp², sp and sp³

[BCl₃and BrCl₃]

[BF⁻₄ and NH⁺₄]

[NH₃ and NO⁻₃]

[NF₃and BF₃]

O⁻₂ and N⁻₂

F⁺₂ and N⁺₂

F⁺₂ and O⁻₂

N₂ and N⁻₂

VSEPR Theory can explain the square planar geometry of XeF₄.

Quartz (Si0₂) has silicon-oxygen covalent bonds that extend throughout the sample

separate Si0₂ molecules exist that are held together by pi bonds that extend in three dimensions throughout the sample

Quartz (Si0₂) has silicon-oxygen electrovalent bonds that extend throughout the sample

separate Si0₂ molecules exist they are held together by coordinate bonds that extend throughout the sample