Chemical Bonding and Molecular Structure Test 0

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Chemical Bonding and Molecular Structure Test 0

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Weekly Quiz Competition

Question 1
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Arrange the following types of interactions in order of increasing stability (covalent, van der Waals' force, hydrogen bonding).

A
Hydrogen bonding < covalent < van der Waals' force

B
Covalent < hydrogen bonding < van der Waals' force

C
Hydrogen < Van der Waals' force < covalent bonding

D
Van der Waals' force < hydrogen bonding < covalent

Solution

The hydrogen bond (5 to 30 kJ/mole) is stronger than a van der Waals interaction, but weaker than covalent or ionic bonds. Stability order of bond is given by Van der Waals' force < hydrogen bonding < covalent
Question 2
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Among the following bonds, which has the most polar character ?

A
$$C - O$$

B
$$C - Br$$

C
$$C - F$$

D
$$C - S$$

Solution

The bond with the polar character is the one in which there is a high electronegativity difference between the atoms.
We know that Flourine is the most electronegative element. Hence, in this case $$C-F$$ bond will be the most polar of them all. Hence option C is the right answer.
Question 3
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Directions For Questions

Different types of bonds are formed in the chemical compounds. These bonds have diferent strengths and bond energies associated with them. These bonds are formed in different environments associated with atoms and the compounds in which they are present.

...view full instructions

Shape of the molecules is decided by:

A
$$\sigma$$-bond

B
$$\pi$$-bond

C
both $$\sigma$$ and $$\pi$$ bonds

D
neither $$\sigma$$ nor $$\pi$$ bond

Solution

The reason that $$\sigma $$ bonds ( and lone pairs) determine the geometry is that they form the basic skeleton of the molecule, $$\sigma $$ bonds are formed by head-on overlap of atomic orbitals, meaning that they are oriented a long the imaginery axis connecting two atomic nuclei, and hence concentrate electron density in the region directly between the nuclei, $$\pi $$ bonds, on the other hand, are essentially orthogonal to the $$\sigma $$ bond skeleton, and are substantially weaker. Moreover, $$\pi$$ bonds do not exist in isolation, meaning any $$\pi$$ bond between two given atoms is always formed secondarily to the $$\sigma $$ bond between said atoms. As such, $$\pi$$ bonds do not alter the basic idealized geometry of a molecule as dictated by $$\sigma $$ bonding.
Question 4
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H-bonding is not present in:

A
glycerin.

B
water.

C
hydrogen Sulphide.

D
hydrogen Fluoride.

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.
Hydrogen bonding is not present in $$H_2S$$.
Question 5
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$$\pi$$-Bonding occurs in each of the following except :

A
$$CO_{2}$$

B
$$C_{2}H_{4}$$

C
$$CN^{-}$$

D
$$CH_{4}$$

Solution

$$\pi$$-Bonding occurs in $$CO_{2}$$, $$C_{2}H_{4}$$ and $$CN^{-}$$.
However, it does not occur in $$CH_{4}$$ as it is a saturated molecule having all the four sigma bonds.
Question 6
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In $$XeF_{2},\, XeF_{4}$$, and $$XeF_{6}$$, the number of lone pairs on $$Xe$$, is__________.

A
$$2, 3, 1$$

B
$$1, 2, 3$$

C
$$4, 1, 2$$

D
$$3, 2, 1$$

Solution

$$XeF_2 $$ has a linear shape. Its hybridization is $$sp^3d$$ which means it should have a trigonal bipyramidal shape. Since it has $$3$$ lone pairs, they occupy the equatorial triangle.

$$XeF_4$$ has $$sp^3d^2$$ hybridization. It has a square planar shape and $$2$$ lone pair of electrons.

$$XeF_6$$ has one lone pair of electrons. The $$6$$ out of $$8$$ electrons of Xenon are shared by $$6$$ atoms of fluorine.

Hence option D is the right answer.
Question 7
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Which bond is more polar ?

A
$$Cl - Cl$$

B
$$N - F$$

C
$$C - F$$

D
$$O - F$$
Solution
Explanation:
The polarity of a bond depends on its electronegativity difference between the two elements of the bond. More the difference the bond will be more polar.
- · The electro negativity of $$Cl,N,F,C,O$$ are $$3.5,3,4,2.5,3.5$$.
- · The electro negativity difference in $$N-F$$ bond is $$4.0-3.0=1$$.
- · The electro negativity difference in $$C-F$$ bond is $$4.0-2.5=1.5$$.
- · The electro negativity difference in $$O-F$$ bond is $$4.0-3.5=0.5$$.
- · The electronegativity difference in $$Cl-Cl$$ bond is $$3.5-3.5=0$$. This bond is nonpolar.
- · Among the following bonds $$C-F$$ bond has the highest electronegativity difference so, it is more polar.
Correct Option : $$C$$
Question 8
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A lone pair of electrons in an atom implies :

A
a pair of valence electrons

B
a pair of electrons

C
a pair of electrons involved in bonding

D
a pair of electrons not involved in bonding

Solution

A lone pair of electrons is a pair of electrons that do not take part in bonding. Hence option D is the right answer.
Question 9
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Which of the following are arranged in the decreasing order of dipole moment ?

A
$$CH_{3}Cl,\, CH_{3}Br,\, CH_{3}F$$

B
$$CH_{3}Cl,\, CH_{3}F,\, CH_{3}Br$$

C
$$CH_{3}Br,\, CH_{3}Cl,\, CH_{3}F$$

D
$$CH_{3}Br,\, CH_{3}F,\, CH_{3}Cl$$

Solution

The bond polarity will be in the order $$F>Cl>Br$$.
$$F$$ is a very small ion, which reples the bonds, decreasing the bond angle, order of polarity:
$$CH_{3}Cl,\, CH_{3}F,\, CH_{3}Br$$
Question 10
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Which of the following gives the correct arrangement of compounds based on their bond strength ?

A
$$HF > HCl > HBr > HI$$

B
$$HI > HBr > HCl > HF$$

C
$$HF > HBr > HCl > HI$$

D
$$HCl > HF > HBr > HI$$

Solution

Electronegativity plays a role, but there is an even bigger effect. As you go down a group, the shell number (n) increases, meaning that the outer valence electrons (the ones to make that bond with hydrogen) are further and further away from the nucleus.

The bigger the element, the less "tightly" it can hold onto the hydrogen.

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Chemical Bonding and Molecular Structure Test 0

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Question 1

Hydrogen bonding < covalent < van der Waals' force

Covalent < hydrogen bonding < van der Waals' force

Hydrogen < Van der Waals' force < covalent bonding

Van der Waals' force < hydrogen bonding < covalent

Solution

Question 2

$$C - O$$

$$C - Br$$

$$C - F$$

$$C - S$$

Solution

Question 3

$$\sigma$$-bond

$$\pi$$-bond

both $$\sigma$$ and $$\pi$$ bonds

neither $$\sigma$$ nor $$\pi$$ bond

Solution

Question 4

glycerin.

water.

hydrogen Sulphide.

hydrogen Fluoride.

Solution

Question 5

$$CO_{2}$$

$$C_{2}H_{4}$$

$$CN^{-}$$

$$CH_{4}$$

Solution

Question 6

$$2, 3, 1$$

$$1, 2, 3$$

$$4, 1, 2$$

$$3, 2, 1$$

Solution

Question 7

$$Cl - Cl$$

$$N - F$$

$$C - F$$

$$O - F$$

Solution

Question 8

a pair of valence electrons

a pair of electrons

a pair of electrons involved in bonding

a pair of electrons not involved in bonding

Solution

Question 9

$$CH_{3}Cl,\, CH_{3}Br,\, CH_{3}F$$

$$CH_{3}Cl,\, CH_{3}F,\, CH_{3}Br$$

$$CH_{3}Br,\, CH_{3}Cl,\, CH_{3}F$$

$$CH_{3}Br,\, CH_{3}F,\, CH_{3}Cl$$

Solution

Question 10

$$HF > HCl > HBr > HI$$

$$HI > HBr > HCl > HF$$

$$HF > HBr > HCl > HI$$

$$HCl > HF > HBr > HI$$

Solution

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