11th Class Chemistry Chapter six Short Questions with Answers Solved

Question: 1

The abnormality of bond length and bond strength in HI is less prominent than that of HCI. Give reason.

Answer: 1

1-22

Chlorine has higher electronegativity than iodine. So, the polarities of HCI and HI bonds are unequal. Therefore, abnormality of bond length and bond strength of HCI is more prominent than HI.

Question: 2

Ionization energy of the element is the binding energy of the nucleus and the electron. Why is it measured when the atom is in the gaseous state?

Answer: 2

2-22

The outermost electrons are attracted by the nucleus. If the atom is closer to some other atom of its own kind or of different kinds, then the outermost electrons are being attracted or repelled by other species as well. In that situation, it is not possible to guess the forces of attractions within the nucleus and the outermost electron.

Question: 3

Why the ionization energies decrease down the group although the nuclear charges increase?

Answer: 3

3-22

When we go down the group, the number of shells increase and shielding effects also increase. These two factors decrease the force of attraction between the nucleus and the outermost electrons and is a cause of decreasing ionization energy.

Question: 4

Why the atomic radii increase down the group?

Answer: 4

4-22

The number of shells increase along with the increasing shielding effect down the group. These two factors are dominant in increasing the sizes. no doubt, the nuclear charges are increased, but this is not a dominant factor.

Question: 5

Why the bond angles of H₂O and NH₃ are not 109.5° like that of CH₄, although O and N-atoms are sp³ —hybridized?

Answer: 5

5-22

Like CH₄, the molecules of H₂O and NH₃ are also AB₄ type molecules. Carbon, oxygen and nitrogen atoms undergo sp³ -hybridization. CH₄ is perfectly tetrahedral with the angle of 109.5°. In case of ammonia, there are three bond pairs and one lone pair. Lone pair-bond pair repulsion is greater than bond pair-bond pair repulsion. Due to this reason, angle reduces to 107.5° In case of H₂O, there are two lone pairs on oxygen. Due to this increased repulsion of two lone pairs, the angle further reduces to 104.5°

Question: 6

Answer: 6

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Question: 7

How shielding effect is thought to be one of the parameters for the variation of ionization potential in group and periods?

Answer: 7

7-22

Greater the number of the inner electons, greater the shielding effect and lesser the I.E. values. In a period, the shielding effect remains the same from left to the right. So the I.E. value are not affected by this parameter. No doubt, other parameters affect it.

Question: 8

Why positively charged ions are mostly smaller in sizes than their neutral atoms?

Answer: 8

8-22

By the removal of electrons the cloud of outermost orbitals becomes thin. In this way, the remaining electrons are accommodated in smaller space. The nuclei have better attractive forces for them, and so the size are decreased as compared to the neutral atoms.

Question: 9

Most of He elements of the periodic table attain the electronic configuration of inert gases during bond formation. Justify it.

Answer: 9

9-22

Inert gases are not reactive due to complete octet except He. Most of He s-and p-block elements may attain eight electrons in the outermost orbitals. They do so either by losing, gaining or sharing the electrons.

Question: 10

The melting points, boiling points, heat of vaporization and heat of sublimation of electrovalent compounds are higher as compared to those of covalent compounds. Why?

Answer: 10

10-22

Electrovalent or ionic compounds have high melting and boiling points due to the close packing of oppositely charged ions. The positively charged ions are surrounded by negatively charged ions and vice versa. That is why, they have very high melting points, boiling points, heat of vaporizations and heats of sublimation.

Question: 11

Why CO₂ has linear structure?

Answer: 11

11-22

Carbon has four electros in the outermost orbitals. It makes two sigma and two pi-bonds with two oxygen atoms. It means that it has two double bonds. Caron has no lone pair in CO₂ is AB₂ type molecule having the linear structure.0=C=0.

Question: 12

Answer: 12

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Question: 13

Solid sodium chloride does not conduct electricity, but, when electric current is passed through molten sodium chloride or its aqueous solution, electrolysis takes place. Give reason.

Answer: 13

13-22

In solid, NaCI, the oppositely charged ions are fixed at their positions. So they do not conduct electricity in the solid state. In the molten state or solution state the ions are free to move towards the respective electrodes.

Question: 14

How the %age of ionic character of the polar bond can be determined?

Answer: 14

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Question: 15

The bent structure of H₂O show that it should have a dipole moment.

Answer: 15

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Question: 16

How do you justify that all the bonds between I-A and II-A with VI-A and VII-A are not equally ionic?

Answer: 16

16-22

The I.E. values of the I-A are less than II-A and the E.A. of VII-A are greater than VI-A. So, the bond between IA and VII-A should be ionic to a good extent. The bonds between II-A and VI-A should be poorly ionic. It means that all the above mentioned compounds are not equally ionic.

Question: 17

How does the electronegativity difference decide the nature of ionic bond?

Answer: 17

17-22

When the electronegativity difference between two bonded atoms is 1.7 or more than that, then the bond is said to be ionic, otherwise covalent. The % age of ionic character is more than 51% when the electronegativity difference is 1.7.

Question: 18

Dipole moment of CO₂ is zero but that of CO is 0.12 Debye. Why?

Answer: 18

18-22

CO₂ is linear molecule and the two dipoles cancel the effect of each other. In CO there is a single dipole directed from carbon to oxygen and it not cancelled.

Question: 19

Why MOT is superior to VBT?

Answer: 19

19-22

Molecular orbital theory tells us the reason for no bond between noble gases. It also tells us the number of bonds whether sigma or pi in N₂ and O₂. Moreover it tells about the paramagnetic and diamagnetic nature of the substance, but V.B.T. does not given such answers.

Question: 20

Why the molecule of BF₃ is triangular planar?

Answer: 20

20-22

'B' has three electrons in the outermost orbitals. It promotes the electron from 2s orbital to one of the 2p orbitals. Boron undergoes sp²' hybridization. Three sp² -hybridized orbitals lie in one plane and adjust themselves at angle of 120^°. There F atom make three sigma bonds which lie in one plane. So, the molecule of BF₃ is planar.

Question: 21

Why the electron affinities of II-A are less then those of I-A?

Answer: 21

21-22

The elements of II-A have fulfilled outermost s-orbitals, so electron has to be accommodated in the higher orbitals. Their electron affinities are positive. The elements of I-A can accommodate incoming electron in partially filled s-orbital.

Question: 22

Why the lone pairs of electrons on an atom occupy greater space?

Answer: 22

22-22

The lone of electrons of the central atom is only attracted by the nucleus of the central atom. It creates more forces of repulsion due to nearness of nucleus and occupies greater space as compared to the bond pair.

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11th Class Chapter six Chemistry Short Questions

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