## Chemistry FBISE SSC-1 Keybook Chapter 3

Q. What is periodic table? Also state periodic law?

Ans. Periodic table:

A table showing systematic arrangement of elements is called periodic table.

Periodic law:

If the elements are arranged in the order of their increasing atomic numbers, their properties are repeated in a periodic manner.

Q. Write a note on periods?

• The horizontal rows of the periodic table are called periods.
• There are seven periods in the periodic table.
• First three periods are called short periods.
• Remaining periods are called long periods.
• From left to right properties of the elements change gradually.
• Down the group, properties repeat in a periodic manner.

Q. Write a note on groups?

• The vertical columns in the periodic table are called groups.
• Similar valence shell electronic configuration elements are placed in the same group.
• Group A elements are called normal or representative elements.
• Group B elements are called transition elements.
• Some groups have names, for example:

IA alkali metals

IIA alkaline earth metals

VIIA halogens

VIIIA noble gases

In which period and group the following elements are present in the periodic table? (use periodic table)

 Element Period Group a. Mg 3 IIA b. Ne 2 VIIIA c. Si 3 IVA d. B 2 IIIA

Example 3.1:

Identify the group and period of , , on the basis of electronic configuration.

(a) = 1s2 2s2 , 2p6 3s2 3p1

K L M

Valence shell is M, n = 3 so

Period = 3rd

Group = 2 + 1 = 3 = III A

(b) = 1s2 2s2 , 2p1

K L

Valence shell is L, n = 2 so

Period = 2nd

Group = 2 + 1 = 3 = III A

(c) = 1s2 2s2 , 2p6 3s2

K L M

Valence shell is M, n = 3 so

Period = 3rd

Group = 2 = II A

Identify the group and period of the following elements on the basis of electronic configurations.

(a) = 1s2 2s2 , 2p6 3s2 3p2

K L M

Valence shell is M, n = 3 so

Period = 3rd

Group = 2 +2= 4 = IV A

(b) = 1s2 2s2 , 2p6 3s2 3p4

K L M

Valence shell is M, n = 3 so

Period = 3rd

Group = 2 +4 = 6 = VI A

(c) = 1s2 2s2 , 2p5 K L Valence shell is M, n = 2 so

Period = 2nd

Group = 2 +5 = 7 = VII A

(d) = 1s2 2s2 , 2p6 3s2, 3p6

K L M

Valence shell is M, n = 3 so

Period = 3rd

Group = 2 +6 = 8 = VIII A

Example. 3.2

Electronic configuration of atoms of some elements are given below. Classify them in groups and periods.

 Electronic configuration Group Period a. 1s2 2s2 IIA 2 b. 1s2 2s2 2p3 VA 2 c. 1s2 2s2 2p5 VIIA 2 d. 1s2 2s2 2p6 3s2 IIA 3 e. 1s2 2s2 2p6 3s2 3p5 VIIA 3 f. 1s2 2s2 2p6 3s2 3p3 VA 3

Electronic configuration of atoms of some elements are given below. Place them into groups and periods.

P= 1s2 2s2 2p2 Q= 1s2 2s2 3p1

R= 1s2 S= 1s2 2s2

T= 1s2 2s1 W= 1s2 2s2 2p6

X= 1s2 2s2 2p6 3s2 3p2 Y= 1s2 2s2 2p6 3s2 3p6

Z= 1s2 2s2 2p1

Solution:

 IA VIIIA IIA IIIA IVA VA VIA VIIA R T S Z P W Q X Y

Q. What are s-block and p-block elements?

s-Block:

Elements contain their valence electrons in s sub-shell is called s-block elements.

p-block

Elements contain their valence electrons in p sub shell is called p-block elements.

Example 3.3

Write the valence shell electronic configuration of the following elements from their position in the periodic table.

(a) Phosphorus (b) Neon

Solution:

1. Period number of phosphorus is 3, so n=3, valence shell is M. The group number is 5 so valence electron will be present in 3s and 3p sub-shells.

Two electrons will fill 3s sub-shell and remaining 3p sub-shell. Thus, valence shell electronic configuration = 3s2 3p3

1. Period number of Neon (Ne) is 2 so n= 2, valence shell is L. The group number is 8.

So valence electron will be present in 2s and 2p sub-shells.

Two electron will fill 2s and remaining six 2p sub-shell. Thus, valence shell electronic configuration = 2s2 2p6

Example 3.4

Find out the position of the following element in the periodic table from the electronic configuration:

1. Nitrogen (atomic number : 7)

= 1s2 2s2 , 2p3

K L

Period number = 2

Group number = 3+2=5 = V A

Nitrogen is present in the 2nd period of Group V-A

1. Oxygen (atomic number : 8)

= 1s2 2s2 , 2p4

K L

Period number = 2

Group number = 2+4=6 = VIA

Oxygen is present in the 2nd period of Group VIA

1. Obtain the valence shell configuration of Al and S from their position in the periodic table.
1. Aluminum (Al):

Period number of Al is 3

So n = 3 valence shell is M

The group number is 3

So valence electron will be present in 3s and 3p sub-shell.

Two electron will fill 3s sub-shell and remaining one 3p sub-shell. This valence shell electronic configuration – 3s2, 3p1

1. Sulphur (S):

Period number of S is 3

So n = 3 valence shell is M.

The group number is 6

So valence electron will be present in 3s and 3p.

Two electron will fill 3s sub-shell and remaining four 3p sub shell. Thus valence shell electronic configuration = 3s2 3p4

2. Find out the position of Ne (at No. 10) and Cl (at No. 17) in the periodic table.

(a) = 1s2 2s2 , 2p6

K L

Period number = 2

Group number = 2+6 = 8 = VIII A

Ne is present in the 2nd period of Group VIIIA

(b) = 1s2 2s2 , 2p6 3s2 3p5

K L M

Period number = 3

Group number = 2 + 5 = 7 = VII A

Cl is present in the 3rd period of Group VII A

Q. Why elements of same group have same chemical properties? (Periodicity of properties)

Ans. Chemical properties depend on the valence shell electronic configuration. All the elements of same group have similar valence shell electronic configuration they possess similar chemical properties.

Q. Why element show gradation in physical properties in a group? (Periodicity of properties)

Ans. Physical properties depend on the size of atoms. Sizes of atom change gradually from top to bottom in a group. Therefore elements show gradation in physical properties in a group.

Q. Why chemical and physical properties show variation in the same manner in a period (periodicity of properties)

Ans. The number of electrons in the valence shell increase gradually from left to right, so chemical and physical properties show variation in the same manner in a period.

Q. What is shielding effect? Also write its trends?

Ans. Shielding effect:

The reduction in force of attraction between nucleus and the valence electrons by the electrons present in the inner sub-shells is called shielding effect.

Trends:

In Group: From top to bottom shielding effect increases

In Period: From left to right shielding effect remain constant.

Q. Which atom has greater shielding effect Be or Mg? (Example 3.5)

Ans. Mg have 10 inner shell electrons and Be have 2. So Mg atom will have greater shielding effect due to greater number of inner shell electrons.

Q. Which atom has greater shielding effect Li or Be

Ans. Li have 2 inner shell electrons and Be have also 2. So Li or Be have same shielding effect due to same number of inner shell electrons.

Q. Which atom has greater shielding effect C or Si? (Example 3.5)

Ans. C have 2 inner shell electrons and Si have 10. So Si atom will have greater shielding effect due to greater number of inner shell electrons.

Choose the element whose atoms you expect to have smaller shielding effect.

(a) F or Cl:

F have 2 inner shell electrons and Cl have 10. So F atom will have smaller shielding effect due to smaller number of inner shell electrons.

(b) Li or Na:

Li have 2 inner shell electrons and Na have 10. So Li atom will have smaller shielding effect due to smaller number of inner shell electrons.

(c) B or Al:

B have 2 inner shell electrons and Al have 10, so B atom will have smaller shielding effect due to smaller number of inner shell electrons.

Q. What is atomic size? Also write its trends

Ans. Atomic size:

The size of an atom is the average distance between the nucleus of an atom and the outer electronic shell.

Trends:

In Group: From top to bottom atomic size increases.

In Period: From left to right atomic size decreases

Q. Why atomic radius decreases in any period as we move across the period?

Ans. Atomic radius decreases because from left to right in a period one electron is added. Due to which positive charge on electron increases by 1. Attractive force of nucleus increases for electron. Hence atomic radius decreases.

Q. Which atom has greater atomic size, Li or B?

Ans. The valence shell electronic configuration of Li (2s1) and B (2s2). There is no change in the shell number n, but atomic number increases from 3 to 4. Due to this force of the nucleus for the valence shell electron increases. Therefore, atomic radius of B decreases. So atomic size of Li is greater than B.

Q. Which atom has greater atomic size, Li or Na?

Ans. Li have 2 shells and Na have 3. Atomic size is determined by the size of its valence shell. So Na has grater atomic size.

Example 3.6

Choose the element whose atom you expect to have larger atomic radius in each of the following pairs.

(a) Mg or Al:

As we move from left to right atomic radius decreases, therefore, Mg has larger atomic radius.

(b) C or Si:

As we move from top to bottom in a group atomic size increases, therefore, Si has larger atomic radius.

Using the periodic table choose the element whose atomic radius is smaller.

(a) O or S

As we move from top to bottom in a group atomic radius increases. Therefore, O has smaller atomic radius.

(b) O or F

As we move from left to right in a period atomic radius decreases. Therefore, F has smaller atomic radius.

Q. What is Ionization energy? Also write trends.

Ans. Ionization energy:

Ionization energy is defined as the minimum amount of energy required to remove the outermost electron from an isolated gaseous atom.

High value of ionization energy means stronger attraction between the nucleus and the outermost electron.

Trends:

In Group: As we move from top to bottom in a group, Ionization energy decreases.

In Period: As we move from left to right in a period ionization energy increases.

Q. Which atom has higher ionization energy, Li or Be?

Ans. Ionization energy increases from left to right in a period. Therefore, B has higher ionization energy.

Example 3.7

Choose the element whose ionization energy is smaller.

a. B, C

Ans. Ionization energy increases from left to right in a period. Therefore, B has smaller ionization energy.

b. N, P

Ionization energy decreases from top to bottom in a group. Therefore, P has smaller ionization energy.

Which atom has smaller ionization energy?

(a) B or N

Ionization energy increases from left to right in a period. Therefore, B has smaller ionization energy.

(b) Be or Mg

Ionization energy decreases from top to bottom in a group. Therefore, Mg has smaller ionization energy.

(c) C or S

Ionization energy decreases from top to bottom in a group. Therefore Si has smaller ionization energy.

Q. What is electron affinity? Also write trends

Ans. Electron affinity:

The amount of energy released when an electron adds up in the valence shell of an isolated atom to form a uninegative gaseous ion.

Trend:

In Group: Electron affinity decreases from top to bottom in a group.

In Period: Electron affinity generally increases from left to right in a period.

Q. What is electro negativity? Also write its trend

Ans. Electro-negativity:

Electro-negativity is the ability of an atom to attract the electrons towards itself in a chemical bond.

The American chemist Linus Pauling devised a method for calculating relative electro-negativity of elements.

Trend:

In Group: Generally decreases from top to bottom in a group.

In Period: Generally increases from left to right in a period.

Review Questions

Q.1 Encircle the correct answer:

1. Number of periods in the periodic table are:
1. 8
2. 7
3. 16
4. 5
1. Which of the following groups contain alkaline earth metals?
1. IA
2. IIA
3. VIIA
4. VIIIA
1. Which of the following elements belongs to VIIIA?
1. Na
2. Mg
3. Br
4. Xe
1. Main group elements are arranged in _____ Groups
1. 6
2. 7
3. 8
4. 10
1. Period number of is:
1. 1
2. 2
3. 3
4. 4
1. Valence shell electronic configuration of an element M (atomic no. 14) is
1. 2s22p1
2. 2s22p2
3. 2s22p3
4. 4s1
1. Which of the following elements you expect to have greater shielding effect?
1. Li
2. Na
3. K
4. Rb
1. As you move from right to left across a period. which of the following do not increase:
1. electron affinity
2. ionization energy
3. nuclear charge
4. shielding effect
1. All the elements of Group IIA are less reactive than alkali ‘petals. This is because these elements have:
1. low ionization energies
2. relatively greater atomic sizes
3. similar electronic configuration
4. decreased nuclear charge

 i. B ii. B iii.D iv. C v. C vi. B vii. D viii. D ix. B

Q.2 Give short answers:

i) Write the valence shell electronic configuration of an element present in the 3rd period and group III A

Ans. The element present in 3rd period and group is

= 1s2 2s2 2p6 3s2 3p1

Valence shell electronic configuration is 3s2 3p1

(ii) Write two ways in which isotopes of an element differ.

Ans. Isotopes of an element differ by

(a) Number of neutrons

(b) Atomic mass

(iii) Which atom has higher shielding effect, Li or Na?

Ans. Na have 10 inner shell electron and Li have 2. So Na atoms will have greater shielding effect due to greater number of inner shell electrons.

(iv) Explain why, Na has higher ionization energy than K?

Ans. Na have 3 shells and K have 4. The size of Na is less than K. Ionization energy decreases from top to bottom. Therefore Na has higher ionization energy.

(v) Alkali metals belong to s-block in the periodic table why?

Ans. Alkali metals belong to s-block in the periodic table because the valence shell electrons are present in s-sub shells.

Q.3 Arrange the elements in each of the following groups in order to increasing ionization energy

(a) Li, Na, K (b) Cl, Br. I

Ans. The ionization energy decreases from top to bottom therefore

Li Na K

Cl Br I

Q.4 Arrange the elements in each of the following in order of decreasing shielding effect.

(a) Li Na K

(b) Cl Br I

(c) Cl Br

Q.5 Specify which of the following elements you would expect to have the greater electron affinity?

S, P, Cl

Ans. Electron affinity increases from left to right in a period so electron affinity of Cl is greater than S and P.

Q.6 Electronic configuration of some elements are given below, group the elements in pairs that would represent similar chemical properties.

A = 1s2 2s2

B = 1s2 2s2 2p6

C = 1s2 2s2 2p3

D = 1s2

E = 1s2 2s2 2p6 3s2 3p3

F = 1s2 2s1

G = 1s2 2s2 2p6 3s1

H = 1s2 2s2 2p6 3s2

Ans. Elements having similar electronic configuration in their outer shells have similar chemical properties so

F=G, A=H, C=E, D=B

Q.7 Arrange the elements in group and periods in Q. No. 6

 IA VIIIA IIA IIIA IVA VA VIA VIIA D F A C B G H F

Q.8 For normal elements, the number of valence electrons of an element is equal to the group number. Find the group number of the following elements.

, , ,

Ans: = 1s2 2s2 2p6 3s2 3p1

Group number = 2 + 1 = III A

= 1s2 2s2 2p6 3s2 3p4

Group Number = 2 + 4 = VI A

= 1s2 2s2 2p6 3s2 3p4 4s1

Group Number = 1 = I A

= 1s2 2s2 2p4

Group Number = 2 + 4 = VI A

Q.9 Write the valence shell electronic configuration for the following groups

(a) Alkali metals

= ns1

Alkaline earth metals

= ns2

Halogens

= ns2 np5

Noble gases

= ns2 np6

Q.10 Write electron dot symbols for an atom of the following elements

(a) Be (b) K (c) N (d) I

(a)

(b)

(c)

(d)

Q.11 Write the valence shell electronic configuration of the atoms of the following elements.

(a) An element present in period 3 of Group VA

Ans: 3s2 3p3

(b) An element present in period 2 of Group VI A

Ans: 2s2 2p4

Q.12 Copy and complete the following table

 Atomic number Mass number No. of Protons No. of Neutrons No. of electrons 11 23 11 12 11 14 29 14 15 14 22 47 22 25 22 13 27 13 14 13

Q.13 Imagine you are standing on the top of Neon-20 nucleus. How many kinds of sub-atomic particles you would see looking down into the nucleus and those you would see looking out from the nucleus?

Ans. Number of protons = 10

Number of neutrons = 20-10 = 10

Q.14 Chlorine is a reactive element used to disinfect swimming pools. It is made up of two isotopes Cl-35 and Cl-37. Because Cl-35 is more than Cl-37, the atomic mass of chlorine 35.5 amu is closer to 35 than 37. Write electronic configuration of each isotope of chlorine. A write symbol for these isotopes (atomic number for chlorine is 17)

Ans. Cl-35

= 1s2 2s2 2p6 3s2 3p5

Cl-37

= 1s2 2s2 2p6 3s2 3p5

Symbol:

Cl-35

Cl-37

Q.15 In which block, group and period in the periodic table where would you place each of the following elements with the following electronic configuration?

Ans.

 Block Group Period a. 1s2 2s2 s IA 2 b. 1s2 2s2 2p5 p VIIA 2 c. 1s2 2s2 2p6 3s2 s IIA 3 d. 1s2 s VIIIA 1

ThinkTank

1. What types of elements have the highest ionization energies and what types of elements have the lowest ionization energies?

Ans. Noble gases have the highest ionization energies due to complete outermost shells.

Alkali metals have the lowest ionization energies due to greater size.

1. Two atoms have electronic configuration 1s22s22p6 and 1s22s22p63s1. The ionization energy of one is 20801KJ/mole and that of the other is 496KJ/mole. Match each ionization energy with one of the given electronic configuration. Give reason for your choice.

Ans. First atom 1s2 2s22p6 has ionization energy 20801KJ/mole because it is noble gas electronic configuration. Noble gases have high ionization energies. Second atom 1s22s22p63s1 has ionization energy of 496KJ/mole because it is alkali metal electronic configuration, with Group IA. Alkali Metals have low ionization energies.

1. Use the second member of each group IA, IIA and VIIA to show that the number of valence electrons of an atom of the element is the same as its group number.

 Group Element Valence Shell Configuration Valence Electrons IA Li 2s1 1 IIA Mg 3s2 2 VIIA Cl 3s23p5 2+5=7

1. Letter A, B, C, D, E, F indicates elements in the following figure:
 C A B D E F

a. Which elements are in the same periods?

Ans. Elements A, B and D, E are in the same periods.

b. Write valence shell electronic configuration of element D.

Ans. Element D lies in Group IIA and 4th period, so electronic configuration is 4s2

c. Which elements are metals?

Ans. Elements A and D are metals

d. Which element can lose two electrons?

Ans. Element D can lose two electrons, because there are two electrons in its outermost shell

e. In which group, E is present?

Ans. Element E is present in Group VA

f. Which of the element is halogen?

Ans. Element F is halogen because group VIIA elements are known as halogens.

g. Which element will form dipositve cation?

Ans. The element D will form dipositive cation because it is in group IIA and can lose two electrons.

h. Write electronic configuration of element E.

Ans. The element E is in group VA and 4th period, so electronic configuration is 4s24p3

i. Which two elements can form ionic bond?

Ans. Elements A, F and D, C can form ionic bond.

j. Can element C form C2 molecule?

Ans. Yes, element C can form C2 molecule.

k. Which element can form covalent bonds?

Ans. Element C and F can form covalent bonds.

l. Is element F a metal or non-metal?

Ans. The element F is a non-metal.

1. Electronic configurations of four elements are given below:
1. 1s22s1
2. 1s22s22p5
3. 1s22s22p63s2
4. 1s2

Which of these elements is

1. An alkali metal

Ans. Element “a” is alkali metal because it is in group IA. Group IA elements are known as alkali metals.

1. An alkaline earth metal

Ans. Element “c” is alkaline earth metal because it is in group IIA. Group IIA elements are known alkaline earth metals.

1. A noble gas

Ans. Element “d” is noble gas because it is in group VIIIA. Group VIIIA elements are known as noble gases.

1. A halogen

Ans. Element “b” is halogen because it is in Group VIIA. Group VIIA elements are known as halogens

1. In what region of the periodic table you will find elements with relatively
1. high ionization energies

Ans. Group VIIIA have highest ionization energies. Group VIIIA elements are known as noble gases. Noble gases have the highest ionization energies due to complete outer most shells.

1. low ionization energies

Ans. Group IA have low ionization energies. Group IA elements are known as alkali metals. They have low ionization energies due to greater size.

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