Chemical Bonding and Molecular Structure Test -5

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Chemical Bonding and Molecular Structure Test -5

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Weekly Quiz Competition

Question 1
1 / -0

During the formation of a chemical bond

A
energy of the system does not change

B
energy decreases

C
electron-electron repulsion becomes more than the nucleus-electron attraction

D
energy increases

Solution

A chemical bond is a lasting attraction between atoms that enables the formation of chemical compounds. The bond may result from the electrostatic force of attraction between atoms with opposite charges, or through the sharing of electrons as in the covalent bonds.

When a bond forms, electrons are attracted to the space between nuclei where the electrostatic force of attraction is greater. As the electrons fall to a position of lower potential energy, the total mechanical energy of the molecular system decreases. Part of the mechanical energy of the unbound atoms is lost when they form the molecular system with a lower total mechanical energy.

Lower the energy higher the stability of bond.
Question 2
1 / -0

Among the following the maximum covalent character is shown by the compound

A
AlCl₃

B
SnCl₂

C
FeCl₂

D
MgCl₂
Solution
Few ionic bonds have partial covalent characteristics which were first discussed by a scientist, Kazimierz Fajans in 1923. He gave the rule to predict whether a chemical bond will be covalent or ionic which is now commonly known as Fajan’s rule. The rule can be stated as:
- Size of the ion: Smaller the size of cation, larger the size of anion, greater is the covalent character of the ionic bond.
- Charge of Cation: Greater the charge of cation, greater is the covalent character of the ionic bond.
- Electronic configuration- For cations with same charge and size, the one, with (n-1)dⁿns⁰ which is found in transition elements have greater covalent character than the cation with ns²np⁶ electronic configuration, which is commonly found in alkali or alkaline earth metals.
Keeping these rules in mind, we can say that AlCl₃ has high covalent character.
Question 3
1 / -0

Which one of the following is paramagnetic?

A
N₂

B
NO

C
O₃

D
CO

Solution

The number of electrons in Nitrogen are 7 and in oxygen are 8. Hence the number of electrons that would be present in the molecular orbitals in NO are 7 + 8 = 15. As the number of electrons are odd, all the electrons in NO molecule cannot be paired. Hence, a single electron would be present in a π*2p orbital. Therefore NO is an odd electron species and the gas is hence paramagnetic due to the presence of unpaired electron.
Question 4
1 / -0

Hybridization of C₂and C₃of H₃C − CH = C = CH − CH₃are

A
Sp, Sp

B
Sp, Sp³

C
Sp², Sp²

D
Sp², Sp

Solution

C₂ is sp² while C₃ is sp (special case of allenes)
Question 5
1 / -0

A species $$H_2O^+$$ pyramidal in shape. The number of lone pairs of electrons is:

A
0

B
1

C
2

D
3

Solution

$${ H }_{ 2 }{ O }^{ + }$$
Since 'O' has 6 electrons in valence shell where 2 electron bonded with hydrogen and one electron is donated due to +ve charges. one electron is as non-bonding electron and thus there is one lone pair of electrons.
Question 6
1 / -0

The number of $$\sigma$$ and $$\pi$$ bonds in $$C_2H_2$$ is:

A
0 and 4

B
2 and 2

C
3 and 2

D
4 and 2

Solution

Acetylene form cylindrical $$\pi$$- electron cloud where nodal plane at one $$\pi$$- bond occupied by antinode of other $$\pi$$-bond.
It has 3 $$\sigma$$-bond and 2 $$\pi$$ bond.
Question 7
1 / -0

Which of the following would not have a dipole moment?
(i) $$CCl_4$$ (iii) $$1, 2$$-dichloroacetylene
(ii) trans-$$1, 2$$-dichloroethene (iv) $$CH_2Cl_2$$

A
(i) and (iv)

B
(i), (ii) and (iv)

C
(i), (ii) and (iii)

D
(ii) and (iii)

Solution

In carbon tetrachloride, the resulting dipole moment of three $$C-Cl$$ bonds is equal in magnitude and opposite in direction to the dipole moment of the fourth $$C-Cl$$ bond. Hence the molecule has no net dipole moment.
In trans-$$1,2$$-dichloroethane, the dipole vectors of two $$C-Cl$$ bonds are equal in magnitude and opposite in direction. Hence, they cancel each other.
Similarly the dipole vectors of two $$C-H$$ bonds are equal in magnitude and opposite in direction. Hence, they cancel each other.
Thus the molecule has no net dipole moment.
In $$1, 2$$-dichloroacetylene, the dipole moment vectors of two $$C-Cl $$ bonds are equal in magnitude and opposite in direction. Hence, they cancel each other. Hence, the molecule has no net dipole moment.
In dichloromethane, the resultant vector of dipole moments of two $$C-Cl $$ bonds is not equal to the resultant vector of the dipole moments of two $$C-H$$ bonds. Hence, the molecule has net dipole moment.
Question 8
1 / -0

The heat of combustion of carbon to $$CO_2$$ is –393.5 kJ/mol. The heat released upon formation of 35.2 g of $$CO_2$$ from carbon and oxygen gas is:

A
-315kJ

B
+315kJ

C
-630Kj

D
+3.15kJ

Solution

$$\Delta H_f$$ = $$-393.5kJ/mol$$

Therefore heat released on formation of44 gm or 1 mole of $$CO_2$$ = $$-393.5kJ/mol$$

Heat released on formation of 35.2 gm of $$CO_2$$ =$$\dfrac{-393.5kJ/mol}{44g}$$$$\times35.2g$$ =$$ -315 kJ/mol$$

Hence,option A is correct.
Question 9
1 / -0

Directions For Questions

Between ionic and covalent bonds, there are a large majority of bonds, in which the bonding electrons are shared unequally between two atoms but are not completely transferred. Such bonds are said to be polar covalent bonds and the bond polarity is due to the difference in electronegativity, the ability of an atom in a molecule to attract the shared electrons in a covalent bond.
The measure of net polarity is a quantity called the dipole moment, $$\mu$$ which is defined as the magnitude of the charge Q at either end of the molecular dipole times the distance $$r$$ between the charges: $$\mu = Q \times r$$. Molecular polarities give rise to some of the forces that occur between molecules and these forces are of several different types including dipole-dipole forces, London dispersion forces, and hydrogen bonds. All these intermolecular forces are electrical in origin and result from the mutual attraction of unlike charges or the mutual repulsion of like charges.

...view full instructions

In which of the following mixtures, the London dispersion force acts as a major intermolecular force of attraction?

A
Sodium chloride and water

B
Cyclohexane and carbon tetrachloride

C
Water and ethyl alcohol

D
Benzene and acetone

Solution

Cyclohexane $$(C_6H_{12})$$ and carbon tetrachloride $$(CCl_4)$$ both are non-polar compounds so there exists London dispersion forces between their constituent molecules.
Question 10
1 / -0

In which of the following pairs, indicated bond is of greater strength?

A
I

B
II

C
Both are equal

D
None of the above

Solution

Chlorine is smaller than bromine. So there will be more attraction to the nucleus and less electron shielding, so the bond length will be less and $$C-Cl$$ is stronger than $$C-Br$$.
Question 11
1 / -0

In which of the above compounds, the indicated bond is of greater strength?

A
I

B
II

C
Both are equal

D
Cannot determine

Solution

$$sp^2$$ carbon atom is more electronegative than $$sp^3$$ carbon atom. The electronegativity difference between $$sp^2$$ carbon atom and bromine atom is less than the electronegativity difference between $$sp^3$$ carbon atom and bromine atom.
Hence, the extent of sharing of electron pair in the bond between $$sp^2$$ carbon atom and bromine atom is more than the extent of sharing of electron pair in the bond between $$sp^3$$ carbon atom and bromine atom.
So, the bond between $$sp^2$$ carbon atom and bromine atom is stronger than the bond between $$sp^2$$ carbon atom and bromine atom.
Question 12
1 / -0

Which of the following molecules will not have a dipole moment?

A
$$CH_3Cl$$

B
trans $$CHCl = CHCl$$

C
$$CH_2Cl_2$$

D
$$CCl_2 = CH_2$$

Solution

In trans-$$1,2$$-dichloroethane, the dipole vectors of two $$C-Cl$$ bonds are equal in magnitude and opposite in direction. Hence, they cancel each other.
Similarly the dipole vectors of two $$C-H$$ bonds are equal in magnitude and opposite in direction. Hence, they cancel each other.
Thus the molecule has no net dipole moment.
Question 13
1 / -0

Among the following molecules, the correct order of $$C - C$$ bond length is:

A
$$C_2 H_6 > C_2 H_4 > C_6 H_6 > C_2 H_2$$

B
$$C_2 H_6 > C_6 H_6 > C_2 H_4 > C_2 H_2 \;(C_6 H_6$$ is benzene)

C
$$C_2 H_4 > C_2 H_6 > C_2 H_2 > C_6 H_6$$

D
$$C_2 H_6 > C_2 H_4 > C_2 H_2 > C_6 H_6$$

Solution

Ethane contains carbon-carbon single bond.
Benzene is aromatic in nature. It has resonance. All the carbon-carbon bonds in benzene have partial double bond character.
Ethylene contains carbon-carbon double bond.
Acetylene contains carbon-carbon triple bond.
The order of bond length is, single bond>partial double bond> double bond > triple bond.
Hence, the correct order of carbon-carbon bond length is $$C_2 H_6 > C_6 H_6 > C_2 H_4 > C_2 H_2$$.
Question 14
1 / -0

The total number of lone-pair of electrons in melamine is:

A
$$3$$ lone pairs from the triazine moeity

B
$$3$$ lone pairs from the $$-NH_2$$ moeity

C
$$6$$ lone pairs from all the $$N$$

D
None of these

Solution

Each $$N$$ uses $$3$$ bonds and hence has $$1$$ lone pair. Hence total of $$6$$ lone pairs.

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Chemical Bonding and Molecular Structure Test -5

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Question 1

energy of the system does not change

energy decreases

electron-electron repulsion becomes more than the nucleus-electron attraction

energy increases

Solution

Question 2

AlCl3

SnCl2

FeCl2

MgCl2

Solution

Question 3

N2

NO

O3

CO

Solution

Question 4

Sp, Sp

Sp, Sp3

Sp2, Sp2

Sp2, Sp

Solution

Question 5

0

1

2

3

Solution

Question 6

0 and 4

2 and 2

3 and 2

4 and 2

Solution

Question 7

(i) and (iv)

(i), (ii) and (iv)

(i), (ii) and (iii)

(ii) and (iii)

Solution

Question 8

-315kJ

+315kJ

-630Kj

+3.15kJ

Solution

Question 9

Sodium chloride and water

Cyclohexane and carbon tetrachloride

Water and ethyl alcohol

Benzene and acetone

Solution

Question 10

I

II

Both are equal

None of the above

Solution

Question 11

I

II

Both are equal

Cannot determine

Solution

Question 12

$$CH_3Cl$$

trans $$CHCl = CHCl$$

$$CH_2Cl_2$$

$$CCl_2 = CH_2$$

Solution

AlCl₃

SnCl₂

FeCl₂

MgCl₂

N₂

O₃

Sp, Sp³

Sp², Sp²

Sp², Sp