Chemical Bonding and Molecular Structure Test

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Chemical Bonding and Molecular Structure Test - 5

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Weekly Quiz Competition

Question 1
4 / -1

According to VSEPR theory

A
The shape of the molecule depends upon the bonded electron pairs

B
Pair of electrons attract each other in valence shells

C
The pairs of electrons tend to occupy such positions that minimise repulsions

D
The pairs of electrons tend to occupy such positions that mimimise distances from each other

Solution

The pairs of electrons tend to occupy such positions that place them farthest from each other and minimise repulsions.
Question 2
4 / -1

In a bonded molecule, the order of repulsion between the bonded and non-bonded electrons is

A
lone pair - lone pair > bond pair - bond pair > lone pair - bond pair

B
bond pair-bond pair > lone pair - lone pair > lone pair - bond pair

C
lone pair - lone pair > lone pair - bond pair > bond pair - bond pair

D
bond pair- bond pair > lone pair - bond pair > lone pair - lone pair

Solution

Lone pair-lone pair lp - lp repulsions are considered stronger than lone pair-bonding pair lp - bp repulsions, which in turn are considered stronger than bonding pairbonding pair bp - bp repulsions, This is because the lone pair of electrons are attracted towards single nuclei only whereas bond pair electrons are attracted towards two nuclei. Thus the strength of the lone pair-lone pair repulsion is supposed to be more than bond pair-bond pair or bond pair-lone pair repulsion.
Question 3
4 / -1

The shape of water molecule which should be tetrahedral has a bent or distorted tetrahedral shape with a bond angle 104.5°. What could be the reason for this?

A
Ip-lp repulsion is more than Ip-bp repulsion

B
Ip-bp repulsion is more than Ip-lp repulsion

C
Ip-lp repulsionis equal to Ip-bp repulsion

D
Presence of lone pair does not affect the bond angle

Solution

Due to presence of lone pairs, the repulsion is more which changes the bond angle to 104.5°.
Question 4
4 / -1

Which of the following shapes of SF₄ is more stable and why?

A
(i), due to 3 Ip-bp repulsions at 90°

B
(ii), due to 2 Ip-bp repulsions

C
Both are equally stable due to 2 Ip-bp repulsions

D
Both are unstable since SF₄ has tetrahedral shape

Solution

In structure (ii), Ip-bp repulsions are minimum.
Question 5
4 / -1

The most stable shape of ClF₃ is shown by

A
(i) only

B
(i) and (iii)

C
(ii) only

D
(iii) only

Solution

In (i), the Ip are atequatorial position so there are less Ip - bp repulsions as compared to other positions.
Hence, T-shape is moststable.
Question 6
4 / -1

In which of the following molecules the central atomdoes not retain any lone pair of electrons?

A
NO₂

B
NH₃

C
BF₃

D
H₂O

Solution

In all other molecules, N and O retain lone pair
of electrons.
Question 7
4 / -1

Few examples of the compounds formed by chemical bonding are given below. Mark the incorrect example.

A
A molecule with central atom devoid of octet - BF₃

B
A molecule with linear shape - CO₂

C
A non-polar covalent compound between two different atoms - CH₄

D
A molecule which is V-shaped with a bond angle 104.5⁰ - NH₃

Solution

The correct example is H₂O. NH₃ is pyramidal and bond angle is 104.5⁰.
Question 8
4 / -1

Which of the following statements is correct regarding the structure of PCI₅?

A
Three P—Cl bonds lie in one plane and two P—Cl bonds lie above and below the equatorial plane

B
Five P—Cl bonds lie in the same plane

C
The bond angle in all P—Cl bonds is 90°

D
The bond length of all P—Cl bonds is same

Solution

In PCI₅, three axial and two equatorial bonds are present.
Question 9
4 / -1

Match the molecules given in column I with their shapes given in column II and mark the appropriate choice

A
(A)→(iv),(B)→(ii),(C)→(iii),(D)→(i)

B
(A)→(iv),(B)→(i),(C)→(ii),(D)→(iii)

C
(A)→(iii),(B)→(i),(C)→(ii),(D)→(iv)

D
(A)→(ii),(B)→(iii),(C)→(i),(D)→(iv)

Solution

(A) SF₆− Octahedral
(B) SiCl₄− Tetrahedral
(C) AsF₅− Trigonal bipyramidal
(D) BCl₃− Trigonal planar
Question 10
4 / -1

What is common between the following molecules
SO₃, CO^2-₃, No^-₃

A
All have linear shape

B
All have trigonal planar shape

C
All have tetrahedral shape

D
All have trigonal pyramidal shape

Solution

All the moleculeshave trigonal planar structure.
Question 11
4 / -1

Which molecule is depicted by the given ball and stick models?

A
(i) BeCl₂, (ii)CH₄

B
(i) BF₃, (ii) PCl₅

C
(i) BF4, (ii) CH₄

D
(i) BeCl₂, (ii) PCl₅

Solution

BF₃ - Trigonal planar with bond angle 120^º.
PCI₅ - Trigonal bipyramidal with bond angles 120^º and 90^º.
Question 12
4 / -1

Which of the following does not show octahedral geometry?

A
SF_`6

B
IF₅

C
SiF₆^2-

D
SF₄

Solution

SF₄ has trigonal bipyramidal geometry. SF₆, IF₅, and SiF₆^2- have octahedral geometry.
Question 13
4 / -1

Given below is the table showing shapes of some molecules having lone pairs of electrons. Fill up the blanks left in it.

A
a

B
b

C
c

D
d
Question 14
4 / -1

The BCI₃ is a planar molecule whereas NCI₃ is pyramidal, because

A
B-Cl bond is more polar than N-Cl bond

B
N-Cl bond is more covalent than B-Cl bond

C
Nitrogen atom is smaller than boron atoms

D
BCI₃ has no lone pair but NCI₃has a lone pairofelectrons

Solution

No lone pair ofelectrons is available in BCI₃
One lone pair of electrons is available on N atom, occupies a corner in the tetrahedral arrangement. Therefore, NCI₃ appears pyramidal in shape.
Question 15
4 / -1

CF₄, SF₄ and XeF₄ contain the following electronicstructure on their central atoms. Which one iscorrect option?

A
1, 2 and 3 lone pairs of electrons respectively

B
0,1 and 2 lone pairs of electrons respectively

C
1,1 and 1 lone pairs of electrons respectively

D
No lone pairs ofelectrons on any molecule

Solution

CF₄−sp³ , tetrahedral but no unpaired electron.
SF₄−sp³d, trigonal bipyramidal with one unpaired electron.

XeF₄−sp³d², square planar, two lone pairs of electrons.