Chemical Bonding and Molecular Structure Test -2

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Chemical Bonding and Molecular Structure Test -2

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Weekly Quiz Competition

Question 1
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Octet of electrons, represents a particularly stable electronic arrangement.. Atoms achieve the stable octet when they are linked by chemical bonds. This rule is associated with one of the following theories

A
Molecular Orbital (MO) Theory

B
Valence Shell Electron Pair Repulsion VSEPR Theory

C
Lewis approach

D
Valence Bond (VB) Theory

Solution

Lewis developed electronic theory of valence or theory of chemical bonding to explain the formation of chemical bond between the two atoms.

According to the electronic theory of valence, every atom tries to attain octet configuration (presence of eight electrons) in its valence shell by losing or gaining or by sharing of electrons. This is known as the "Octet Rule".

The electrostatic forces of attraction that holds the two oppositely charged ions together are known as "electrovalent bond".
Question 2
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Inertness of noble gas was observed to be due to their electronic configurations: Choose the most appropriate

A
Noble gases form ions to complete their outermost orbitals

B
Total number of electrons in the outermost shell is duplets

C
Outermost orbitals of the noble gases are fully filled.

D
Atomic mass is low so they are stable

Solution

The six noble gases (Helium Neon Argon Krypton Xenon Radon) are found in group 18 of the periodic table. These elements were considered to be inert gases until the 1960's, because their oxidation number of 0 prevents the noble gases from forming compounds readily. All noble gases have the maximum number of electrons possible in their outer shell (2 for Helium, 8 for all others), making them stable.
Question 3
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Stable outer octet of electrons is achieved in chlorine atom during the formation of NaCl by:

A
The formation of a hybrid orbital

B
The loss of an electron

C
The formation of a coordinate bond

D
The gain of an electron

Solution

Na lose its one electron forming Na⁺ while Cl gains the electron formimg Cl^-
Question 4
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The bond formed, as a result of the electrostatic attraction between the positive and negative ions was termed as

A
coordinate bond

B
covalent bond

C
single covalency

D
electrovalent bond

Solution

Chemical bond formed between two atoms due to transfer of electron(s) from one atom to the other. atom is called "Ionic bond" or "electrovalent bond".

in electrovalent bond one atom loses an electron to form a positive ion and the other atom gains the electron to form a negative ion. The resulting ions are held together by electrostatic attraction
Question 5
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The number of dots around the Lewis symbols for the elements represent

A
electrovalency

B
coordinate valency

C
the number of valence electrons of the element

D
group valence of the element

Solution

Gilbert N. Lewis is widely known for his use of simple symbolic representations of elements that show valence electrons as dots.

The Lewis electron-dot symbols focus on the electrons in the highest principal energy level in the atom, the valence electrons.

After all, these are the electrons that participate in chemical reactions. Lewis electron-dot symbols work well for the representative elements.
Question 6
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The formation of the Cl₂molecule can be understood in terms of the sharing of a pair of electrons between the two chlorine atoms, each chlorine atom contributing one electron to the shared pair. Choose the most appropriate name of the bond that is formed

A
double bond

B
ionic bond

C
multiple bond

D
a single covalent bond

Solution

Two chlorine atoms will each share one electron to get a full outer shell and form a stable Cl₂ molecule.

By sharing the two electrons where the shells touch each chlorine atom can count 8 electrons in its outer shell. These full outer shells with their shared electrons are now stable and the Cl₂ molecule will not react further with other chlorine atoms. One pair of shared electrons form a single covalent bond.
Question 7
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The molecular formula of the compound formed from B and C will be

A
BC₂

B
BC₃

C
B₂C

D
BC

Solution

BC₃
Question 8
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788 kJ of energy is required to separate one mole of solid NaCl into one mole of Na⁺ (g) and one mole of Cl⁻ (g) to an infinite distance. This process explains

A
ionization enthalpy

B
Electron affinity

C
lattice enthalpy

D
electron gain enthalpy

Solution

Lattice enthalpy is simply the change in Enthalpy associated with the formation of one mole of an ionic compound from its oppositely charged ions in their standard states under standard conditions.
Question 9
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A qualitative measure of the stability of an ionic compound is provided by

A
lattice enthalpy

B
Electron affinity

C
electron gain enthalpy

D
ionization enthalpy

Solution

Lattice energy is an estimate of the bond strength in ionic compounds. It is defined as the heat of formation for ions of opposite charge in the gas phase to combine into an ionic solid.

The stability of ionic bond is directly propotional to lattice energy.
Question 10
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bond lengths are lower in elements having

A
crystal structure

B
double bond

C
single bond

D
triple bond

Solution

The bond length depends on the strength of the bond. The stronger the bond is, the shorter it will be. The triple bonds are the strongest and hence the shortest. Then comes double bonds which are of intermediate strength between the triple and single bonds. And finally the single bonds are weaker than the other two. This way, Triple bonds are the shortest. Then comes double bonds. Finally, single bonds are the longest among the three.

The order of bond lengths is given as, Triple bond < Double bond < Single bond
Question 11
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Which has a maximum repulsive interaction?

A
$$bp-bp$$

B
$$lp-lp$$

C
$$lp-bp$$

D
Equal

Solution

A bonding pair in a $$ \sigma $$ , on average lies further from the atom that a non - bonding pair, which is closer to the +ve nucleus.
The force of repulsion between two negative charge is proportional to $$ \cfrac { 1 }{ { r }^{ 2 } } $$.
where $$ { r }^{ 2 }$$ is the separation. Since $$ { r }^{ 2 }$$ is less for non-bonding pairs, the repulsion is greater. (since they are inversely related)

The repulsive interaction order is-
$$l.p.-l.p. > l.p.-b.p. > b.p-b.p.$$
Question 12
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For the given bond pairs, decide which is the more polar?

$$Si - O$$ and $$P - P $$

A
$$ Si - O > P - P $$

B
$$ Si - O < P - P $$

C
$$ Si - O = P - P $$

D
None of these

Solution

$$Si$$ and $$O$$ have different electronegativity while it is same in $$P-P$$. Hence, $$Si-O$$ is a more polar bond.
Question 13
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Which of the following compounds exhibits H-bonding?
$$CH_3CH_2OH$$ and $$CH_3OCH_3$$

A
$$CH_3CH_2OH$$

B
$$CH_3OCH_3$$

C
Both

D
None

Solution

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. The presence of $$OH$$ group in ethanol leads to hydrogen bonding.
Question 14
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The nitrogen atom has an atomic number $$7$$ and oxygen has an atomic number $$8$$
Calculate the total number of electrons in nitrate ion.

A
$$8$$

B
$$16$$

C
$$32$$

D
$$24$$

Solution

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
Question 15
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Does the molecule $$OCS$$ have a higher or lower dipole moment than $$CS_2$$?

A
Higher dipole moment than of $$CS_2$$

B
Lower dipole moment than of $$OCS$$

C
Equal dipole moment than of $$OCS$$

D
None of these

Solution

$$\displaystyle CS_2$$ or $$\displaystyle S=C=S$$ is a symmetrical, linear molecule. The individual bond dipole moments are equal in magnitude and opposite in direction. It has zero dipole moment as the individual bond dipoles cancel each other.
On the other hand $$\displaystyle OCS$$ is unsymmetrical molecule with non zero dipole moment. Hence, the dipole moment of $$\displaystyle OCS$$ is higher than the dipole moment of $$\displaystyle CS_2$$
Question 16
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Arrange $$B - F$$ and $$B - Cl$$ in the order of polarity.

A
$$ B - F $$ > $$ B - Cl $$

B
$$ B - F $$ < $$ B - Cl $$

C
$$ B - F $$ = $$ B - Cl $$

D
None of these

Solution

$$B-F$$ bond is more polar because of high electronegativity of $$F$$ than $$Cl$$.
Question 17
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Which of these molecules have non-bonding electron pairs on the central atom?
$$I : SF_4;\, \, \, \, \, II : ICl_3;\, \, \, \, \, III : SO_2$$

A
II only

B
I and II only

C
I and III only

D
I, II and III

Solution
Question 18
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Directions For Questions

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Number of lone pairs around phosphorus in $$PCl_5, PCl_4^+$$ and $${PCl_6}^-$$ are respectively

A
$$0, 1, 2$$

B
$$0, 0, 0$$

C
$$1, 2, 3$$

D
$$0, 2, 1$$

Solution

$$P$$ 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.
Question 19
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The formal charge on the O atoms in the ion $$[:\ddot{O} = N = \ddot{O}:]^+$$ is:

A
$$-2$$

B
$$-1$$

C
$$0$$

D
$$+1$$

Solution

Formal charge$$=$$Valence electrons-non bonding electrons+no of bonds$$ =6-(4+2)=0$$
Question 20
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Consider the above Lewis Dot structure:

The formal charges on $$S$$ and $$O$$ are respectively:

A
$$0 , 0$$

B
$$+2 , 0$$

C
$$0 , -2$$

D
$$+1 , -1$$

Solution

Charge=valence electrons-[non-bonded electrons+no.of bonds]
$$S:6-[2+4]=0$$
$$O:6-[4+2]=0$$

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Chemical Bonding and Molecular Structure Test -2

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Question 1

Molecular Orbital (MO) Theory

Valence Shell Electron Pair Repulsion VSEPR Theory

Lewis approach

Valence Bond (VB) Theory

Solution

Question 2

Noble gases form ions to complete their outermost orbitals

Total number of electrons in the outermost shell is duplets

Outermost orbitals of the noble gases are fully filled.

Atomic mass is low so they are stable

Solution

Question 3

The formation of a hybrid orbital

The loss of an electron

The formation of a coordinate bond

The gain of an electron

Solution

Question 4

coordinate bond

covalent bond

single covalency

electrovalent bond

Solution

Question 5

electrovalency

coordinate valency

the number of valence electrons of the element

group valence of the element

Solution

Question 6

double bond

ionic bond

multiple bond

a single covalent bond

Solution

Question 7

BC2

BC3

B2C

BC

Solution

Question 8

ionization enthalpy

Electron affinity

lattice enthalpy

electron gain enthalpy

Solution

Question 9

lattice enthalpy

Electron affinity

electron gain enthalpy

ionization enthalpy

Solution

Question 10

crystal structure

double bond

single bond

triple bond

Solution

Question 11

$$bp-bp$$

$$lp-lp$$

$$lp-bp$$

Equal

Solution

Question 12

$$ Si - O > P - P $$

$$ Si - O < P - P $$

$$ Si - O = P - P $$

None of these

Solution

BC₂

BC₃

B₂C