Chemical Bonding and Molecular Structure Test 36

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Chemical Bonding and Molecular Structure Test 36

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Weekly Quiz Competition

Question 1
1 / -0

Match list I (Complex ions) with list II (number of unpaired electrons) and select the correct answers:
List I List II
A $$[CrF_6]^{4-}$$ 1 One
B $$[MnF_6]^{4-}$$ 2 two
C $$[Fe(CN)_6]^{4-}$$ 3 zero
D $$[Mn(CN)_6]^{4-}$$ 4 four
5 five

A
$$A-4, B-1, C-2, D-5$$

B
$$A-2, B-5, C-3, D-1$$

C
$$A-4, B-5, C-3, D-1$$

D
$$A-2, B-1, C-3, D-5$$

Solution

(A) The complex ion $$[CrF_6]^{4-}$$ contains four unpaired electrons.
(B) The complex ion $$[MnF_6]^{4-}$$ contains five unpaired electrons.
(C) The complex ion $$[Fe(CN)_6]^{4-}$$ contains zero unpaired electrons.
(D) The complex ion $$[Mn(CN)_6]^{4-}$$ contains one unpaired electrons.
Question 2
1 / -0

Assertion: In dilute benzene solutions, equimolar addition of $$R_3N$$ and $$HCl$$ produce a substance with a dipole moment. In the same solvent, equimolar addition of $$R_3N$$ and $$SO_3$$ produce a substance having an almost identical dipole moment.
Reason:($$ N-S$$) bond is a more polar bond.

A
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

B
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

C
Assertion is correct but Reason is incorrect

D
Assertion is incorrect but Reason is correct

E
Both Assertion and Reason are incorrect

Solution

In dilute benzene solutions, equimolar addition of $$R_3N$$ and $$HCl$$ produce a substance with a dipole moment. In the same solvent, equimolar addition of $$R_3N$$ and $$SO_3$$ produce a substance having an almost identical dipole moment.
This is because, both $$HCl$$ and $$SO_3$$ are Lewis acids and can react with the amine base to form polar substances which undergo ionic dissociation in a solvent sufficiently more polar than benzene.
Moreover, ($$N-S$$) bond is a more polar bond.
Question 3
1 / -0

Consider the given lewis dot structure:
The formal charges on $$B$$ and $$F$$ are respectively.

A
$$-1 , 0$$

B
$$-1 , -1$$

C
$$0 , -1$$

D
$$-1 , +1$$

Solution

Formal Charge = [valence electrons on atom] - [non-bonded electrons + number of bonds].
$$B=3-[0+4]=-1$$
$$F=7-[1+6]=0$$
Question 4
1 / -0

Directions For Questions

In order to explain the shape and geometry of molecules, the valence bond theory was supplemented by the concept of hybridization. This is a hypothetical concept and has been introduced by Pauling and Slater. According to this concept, any number of atomic orbitals of an atom which differ in energy slightly may mix with each other to form new orbitals called hybrid orbitals. The process of mixing or amalgamation of atomic orbitals of nearly same energy to produce a set of entirely new orbitals of equivalent energy is known as hybridization.

...view full instructions

In $$XeF_2$$, $$XeF_4$$ and $$XeF_6$$ the number of lone pairs on Xe is respectively__________.

A
2, 3, 1

B
1, 2, 3

C
4, 1, 2

D
3, 2, 1

Solution

$$XeF_2$$ has 2 bond pairs and 3 lone pairs.
$$XeF_4$$ has 4 bond pairs and 2 lone pairs.
$$XeF_6$$ has 6 bond pairs and 1 lone pairs.
Question 5
1 / -0

For the type of interactions:$$(I)$$ Covalent bond, $$(II)$$ van der waal's forces, $$(III)$$ Hydrogen bonding, $$(IV)$$ Dipole-dipole interaction, which represents the correct order of increasing stability?

A
$$(I)< (III)< (II)< (IV)$$

B
$$(II)< (III)< (IV)< (I)$$

C
$$(II)< (IV)< (III)< (I)$$

D
$$(IV)< (II)< (III)< (I)$$

Solution
Question 6
1 / -0

In which of the following molecules/species, all following characteristics are found?
(a) Tetrahedral hybridisation.
(b) Hybridisation can be considered to have taken place with the help of empty orbital(s).
(c) All bond lengths are identical, i.e., all $$A - B$$ bond lengths are identical.

A
$$B_4H_6$$

B
$$Al_2Cl_6$$

C
$$BeCl_2\, (g)\, $$

D
$${BF_4}^-$$

Solution

In $${BF_4}^-$$ ion, all following characteristics are found.
(a) Tetrahedral hybridisation which is $$sp^3$$ hybridization due to the presence of $$4$$ bond pairs of electrons and zero lone pairs of electrons..
(b) Hybridisation can be considered to have taken place with the help of empty orbital. $$B$$ contains $$3$$ valence electrons, whereas $$4$$ orbitals participate in the hybridization. Thus, one empty orbital participates in the hybridization.
(c) All bond lengths are identical, i.e., all $$A - B$$ bond lengths are identical as there are four identical $$B-F$$ single bonds.
Question 7
1 / -0

The formal charges of $$N_{(1)}, N_{(2)}$$ and $$O$$ atoms in $$: \overset {. .}{N_{(1)}} = N_{(2)} = \overset {. .}{O}:$$ are respectively

A
$$+1, -1, 0$$

B
$$-1, +1, 0$$

C
$$+1, +1, 0$$

D
$$-1, -1, 0$$

Solution

You can write this two ways:
$$:N:::N:O::: $$
That has a (+) on the middle nitrogen and a $$(-)$$ on the oxygen
Or you can write it as
$$::N::N::O:: $$
That has a $$(+)$$ on the middle nitrogen and a $$(-)$$ on the left nitrogen. The first way is preferred because the oxygen (the more electronegative element) gets the negative charge.
Question 8
1 / -0

Which of the following best describes the term dipole moment?

A
When electrons are unevenly distributed within a molecule

B
When several atoms in a compound are electronegative

C
A non polar molecule

D
Having more than four dipoles

Solution

When electrons are unevenly distributed within a molecule, the dipole moment occur there.
The electric dipole moment is a measure of the separation of positive and negative electrical charges within a system, that is, a measure of the system's overall polarity. The electric field strength of the dipole is proportional to the magnitude of dipole moment.
Question 9
1 / -0

What is a non-bonding electron domain?

A
A lone pair of electrons

B
An area of electron density

C
A double bond

D
The area of electron density in an anion

Solution
Question 10
1 / -0

Calculate the formal charges on oxygen atoms 1, 2, 3 respectively.

A
0, -1, +1

B
0, +1, -1

C
1, 0, -1

D
1, 0, +1

Solution

The formula of formal charge, $$FC = V - N - \dfrac B2$$

where V the number of valence electrons of the neutral atom in isolation (in its ground state); N is the number of non-bonding valence electrons on this atom in the molecule, and B is the total number of electrons shared in bonds with other atoms in the molecule.

The atoms have been numbered as 1, 2, and 3. The formal charge on:
The central O atom marked 1:
$$= 6 - 2 - \dfrac 12 (6) = +1 $$

The end O atom marked 2:
$$= 6 - 4 - \dfrac 12 (4) = 0$$

The end O atom marked 3:
$$= 6 - 6 - \dfrac 12 (2) = -1$$

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Re-Attempt Weekly Quiz Competition

	List I		List II
A	$$[CrF_6]^{4-}$$	1	One
B	$$[MnF_6]^{4-}$$	2	two
C	$$[Fe(CN)_6]^{4-}$$	3	zero
D	$$[Mn(CN)_6]^{4-}$$	4	four
		5	five

Chemical Bonding and Molecular Structure Test 36

Result

Chemical Bonding and Molecular Structure Test 36

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Question 1

$$A-4, B-1, C-2, D-5$$

$$A-2, B-5, C-3, D-1$$

$$A-4, B-5, C-3, D-1$$

$$A-2, B-1, C-3, D-5$$

Solution

Question 2

Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Assertion is correct but Reason is incorrect

Assertion is incorrect but Reason is correct

Both Assertion and Reason are incorrect

Solution

Question 3

$$-1 , 0$$

$$-1 , -1$$

$$0 , -1$$

$$-1 , +1$$

Solution

Question 4

2, 3, 1

1, 2, 3

4, 1, 2

3, 2, 1

Solution

Question 5

$$(I)< (III)< (II)< (IV)$$

$$(II)< (III)< (IV)< (I)$$

$$(II)< (IV)< (III)< (I)$$

$$(IV)< (II)< (III)< (I)$$

Solution

Question 6

$$B_4H_6$$

$$Al_2Cl_6$$

$$BeCl_2\, (g)\, $$

$${BF_4}^-$$

Solution

Question 7

$$+1, -1, 0$$

$$-1, +1, 0$$

$$+1, +1, 0$$

$$-1, -1, 0$$

Solution

Question 8

When electrons are unevenly distributed within a molecule

When several atoms in a compound are electronegative

A non polar molecule

Having more than four dipoles

Solution

Question 9

A lone pair of electrons

An area of electron density

A double bond

The area of electron density in an anion

Solution

Question 10

0, -1, +1

0, +1, -1

1, 0, -1

1, 0, +1

Solution

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