Chemical Bonding and Molecular Structure Test 25

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Chemical Bonding and Molecular Structure Test 25

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Weekly Quiz Competition

Question 1
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Which of the following molecule/ion has a zero dipole moment?

A
$$ ClF_3 $$

B
$$ ICl_2^- $$

C
$$ SF_4 $$

D
None of these

Solution

Iodine is $$sp^{3}d$$ hybridised in the complex $$ICl_{2}^{-}$$. Here two chlorine atoms occupy the axial positions. So the compound has a linear structure. Hence dipole moment of one $$I-Cl$$ bond gets cancelled by the other $$I-Cl$$ bond. So the ion has zero dipole moment. Option B is correct.
Question 2
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Linear combination of two hybridized orbitals, belonging to two atoms and each having one electron, leads to:

A
Sigma-bond

B
Double bound

C
Coordinate covalent bond

D
Pi-bond

Solution

Linear combination of two hybridized orbitals, belonging to two atoms and each having one electron, leads to $$ \sigma$$ - bond formation.
Question 3
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The bond between two identical non metal atoms has a pair of electron

A
unequally shared betweem the two

B
transferred fully from one atom to another

C
with identical spins

D
equally shared between them

Solution

The bond between two identical non-metal atoms has a pair of electrons equally shared between them.
they form a covalent bond which is totally non polar because of the similar electronegativity.
Question 4
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Directions For Questions

The electronic configuration of three elements $$A, B, C$$ and $$D$$ are given below. Answer the question on the basis of these configurations.
$$ A) \quad 1s^2 \quad 2s^2 \quad 2p^6 \quad -\ Ne $$
$$ B) \quad 1s^2 \quad 2s^2 \quad 2p^6 \quad 3s^2 \quad 3p^3 \quad -\ P $$
$$ C) \quad 1s^2 \quad 2s^2 \quad 2p^6 \quad 3s^2 \quad 3p^5 \quad -\ Cl $$
$$ D) \quad 1s^2 \quad 2s^2 \quad 2p^6 \quad 3s^2 \quad 3p^4 \quad -\ S $$

...view full instructions

The bond between $$B$$ and $$C$$ will be

A
Ionic

B
Covalent

C
Hydrogen

D
Cordiante
Solution
Correct Answer: Option B
Explanation of Correct Option:
- A covalent bond is formed by sharing of electrons.
- The elements in which the last electron enters the $$p$$−subshell belong to the $$p$$ block.
- The electronegativity difference between $$p$$− block elements is very less. Therefore, the bond is formed by sharing of electrons and is covalent in nature.
Explanation of Incorrect Options:
An ionic bond is formed by the complete exchange of electrons between the atoms of metal and non-metal, which results in the formation of cations and anions when the bond is broken.
A hydrogen bond is formed by a hydrogen atom and an electronegative atom and is weaker than ionic and covalent bonds.
A coordinate bond is a type of covalent bond in which the electrons are shared by only one atom.
Final Step: Correct answer - (B) Covalent.
Question 5
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On hybridization of one s- and one p-orbital we get :

A
Two mutually perpendicular orbitals

B
Two orbitals at $$ 180^0 $$

C
Four orbitals directed tetrahedrally

D
Three orbitals in a plane

Solution
Question 6
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Carbon tetrachloride has no net dipole moment because of :

A
Its planar structure

B
Its regular tetrahedral structure

C
Similar sizes of carbon and chloride atoms

D
Similar electron afinities of carbon and chloride

Solution

In carbon tetrachloride, there is a regular tetrahedral shape. All the dipole moment vectors cancel each other so that the resultant dipole moment comes out to be zero.
Question 7
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The molecule which has zero dipole moment is :

A
$$ CH_2Cl_2 $$

B
$$ BF_3 $$

C
$$NF_3 $$

D
$$ ClO_3 $$

Solution

$$ BF_3$$ is a triangular planar molecule where the resultant of three dipole moment vectors is equal to zero.
Question 8
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Among the following species, identify the isostructural pairs:

$$ NF_3,\ NO^-_3,\ H_3O^+ ,\ N_3H,\ BF_3 $$

A
$$ [NF_3,\ NO^-_3]$$ and $$[ BF_3, \ H_3O^+] $$

B
$$[ NF_3,\ N_3H]$$ and $$[ NO^-_3,\ BF_3 ]$$

C
$$ [NF_3 ,\ H_3O^+]$$ and $$[ NO^-_3, \ BF_3] $$

D
$$ [NF_3,\ H_3O^+ ] $$ and $$ [N_3H,\ BF_3 ] $$

Solution

$$ NF_ 3 $$ and $$ H_3O^+ $$ are pyaramidal structures with three bond pairs and two lone paird due to $$ sp^3 $$ hybridization.
Question 9
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The number and type of bonds in $$ C^{2-}_2 $$ in $$ CaC_2 $$ are:

A
Two $$ \sigma $$ bond and one $$ \pi $$ bond

B
Two $$ \sigma $$ bond and two $$ \pi $$ bond

C
One $$ \sigma $$ bond and two $$ \pi $$ bond

D
One $$ \sigma $$ bond and one $$ \pi $$ bond

Solution

$$ Ca^{+2} [ C \equiv C]^{-2} $$
A single bond between two atoms is always considered as sigma bond.
A double bond between two atoms is always considered as one sigma and one pi bond.
A triple bond between two atoms is always considered as one sigma bond and two pi bonds.
Question 10
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The correct order of increasing C-O bond length in $$ CO, CO^{2-}_3, CO_2 $$ is :

A
$$ CO^{2-}_3 < CO_2 < CO $$

B
$$ CO_2 < CO^{2-}_3 < CO $$

C
$$ CO < CO^{2-}_3 < CO_2 $$

D
$$ CO < CO_2< CO^{2-}_3 $$

Solution

Bond orders:
$$ CO^{2-}_3 : 1.33 $$

$$ CO_2 : 2 $$

$$ CO :3 $$
The higher the bond order, the lesser the bond length.

So, the order of bond length of C-O is:
$$ CO < CO_2< CO^{2-}_3 $$