Chapter 9.5 - Properties of s-block and p-block elements

In the previous section, we saw how to determine the group number of an element from it's S.E.C. We also saw which all groups fall under each block.  In this section we will see the properties of s-block and p-block elements.

Properties of s-block elements 

■ Position: The s-block consists of the groups I and II of the periodic table
■ Common names:
• Group I elements have the common name Alkali metals 
• Group II elements have the common name Alkaline earth metals (Details here)
■ Acceptance or Donation of electrons during chemical reactions:
Consider the elements Li, Na and K of Group I. Let us write their S.E.C:
1. Li - 1s²2s¹ 
• The outer most main shell is 2.
• This main shell has one electron, which is in 2s subshell
• During chemical reactions, this one electron is donated. This is because, donating one electron is easier than accepting 7 electrons to attain octet 
2. Na - 1s²2s²2p⁶3s¹
• The outer most main shell is 3.
• This main shell has one electron, which is in 3s subshell
• During chemical reactions, this one electron is donated. This is because, donating one electron is easier than accepting 7 electrons to attain octet
3. K - 1s²2s²2p⁶3s²3p⁶4s¹
• The outer most main shell is 4.
• This main shell has one electron, which is in 4s subshell
• During chemical reactions, this one electron is donated. This is because, donating one electron is easier than accepting 7 electrons to attain octet

Consider the elements Be, Mg and Ca of Group II. Let us write their S.E.C:  
1. Be - 1s²2s²
• The outer most main shell is 2.
• This main shell has two electrons, which is in 2s subshell
• During chemical reactions, these two electrons are donated. This is because, donating two electrons is easier than accepting 6 electrons to attain octet 
2. Mg - 1s²2s²2p⁶3s²
• The outer most main shell is 3.
• This main shell has two electrons, which is in 3s subshell
• During chemical reactions, these two electrons are donated. This is because, donating two electrons is easier than accepting 6 electrons to attain octet 
3. Ca - 1s²2s²2p⁶3s²3p⁶4s²
• The outer most main shell is 4.
• This main shell has two electrons, which is in 4s subshell
• During chemical reactions, these two electrons are donated. This is because, donating two electrons is easier than accepting 6 electrons to attain octet 

■ Oxidation state
• All the elements in group I always donate 1 electron. So they always show an oxidation state of +1
• All the elements in group II always donate 2 electrons. So they always show an oxidation state of +2 
• We have seen that some elements show fixed oxidation states which ever be the reaction. While some others show variable oxidation states. (We have seen details about oxidation number or oxidation state here) 
    ♦ Groups I and II elements always show definite fixed oxidation states. 
■ Atomic radius
• We have learned about periodic trends in an earlier chapter. Details here. We have seen that the atomic radius decreases as we move from left to right in a period. 
• The s-block elements are at the left most end of the various periods. So we can say that, the s-block elements have high atomic radius
• Also, when we move from top to bottom in any group in the s-block, the atomic radius increases
■ Ionization energy
• We have learned about periodic trends in an earlier chapter. Details here. We have seen that the ionization energy increases as we move from left to right in a period. 
• The s-block elements are at the left most end of the various periods. So we can say that, the s-block elements have low ionization energy
• That means lower energies are sufficient to remove electrons from the s-block elements
• Also these elements form mostly ionic compounds
■ Electronegativity
• We have learned about periodic trends in an earlier chapter. Details here. We have seen that the electronegativity increases as we move from left to right in a period. 
• The s-block elements are at the left most end of the various periods. So we can say that, the s-block elements have low electronegativity
■ Metallic nature
• We have learned about periodic trends in an earlier chapter. Details here. We have seen that the metallic nature decreases as we move from left to right in a period. 
• The s-block elements are at the left most end of the various periods. So we can say that, the s-block elements have high metallic nature
■ Reactivity
• The Group I elements will be the first element in respective periods. 
• These first elements have the greatest reactivity in the respective periods
• Also, the reactivity increases as we move from top to bottom in the groups
■ Oxides and hydroxides
• Magnesium oxide (MgO) is an example of an oxide of an s-block element. It is formed when magnesium ribbon burns in air. 
    ♦ Other examples of oxides from this s-block are: Calcium oxide (CaO) and Barium oxide (BaO) 
• Sodium hydroxide (NaOH) and Potassium hydroxide (KOH) are examples of hydroxides formed from this block. They are alkaline in nature.
• Metallic oxides give alkalies and non-metallic oxides give acids. We have seen the details in an earlier chapter here.
    ♦ Above we have seen that s-block elements are metallic in nature. So we can say that the oxides and hydroxides formed from this block are alkaline in nature.

Properties of p-block elements 

■ Position: The p-block consists of the groups 13 to 18 of the periodic table
■ S.E.C:
• We know that, the last electron of each element in the p-block will be filled in the p subshell. That means, there will always be a p subshell in the last main shell of any p-block element
• Now, p subshell will be filled only after completing the s subshell. So, the last main shell will contain a s subshell also.
• Thus, The general form of the last main shell configuration is: ns²np^{(1 to 6)}.
• Where n is the last main shell. The superscript for s will always be 2
• The superscript for p varies from 1 to 6. This is shown below:
The superscript for p for all elements in group 13 will be 1 (Add 12 to 1, and we get the group no.13) 
The superscript for p for all elements in group 14 will be 2 (Add 12 to 2, and we get the group no.14) 
The superscript for p for all elements in group 15 will be 3 (Add 12 to 3, and we get the group no.15) 
The superscript for p for all elements in group 16 will be 4 (Add 12 to 4, and we get the group no.16) 
The superscript for p for all elements in group 17 will be 5 (Add 12 to 5, and we get the group no.17) 
The superscript for p for all elements in group 18 will be 6 (Add 12 to 6, and we get the group no.18)
An example:
• ₃₄Se    - 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁴  OR [Ar]3d¹⁰4s²4p⁴.
    ♦ Add 12 to the subscript 4 of the 4p subshell. We get 16, which is indeed the group number of Selenium

■ Common names:
• Group 13 elements have the common name Boron family
• Group 14 elements have the common name Carbon family
• Group 15 elements have the common name Nitrogen family
• Group 16 elements have the common name Oxygen family
• Group 17 elements have the common name Halogens
• Group 18 elements have the common name Noble gases

■ Oxidation state
• Some of the elements in the p-block shows variable oxidation states. But in general, we can write the values as follows:
• +3 oxidation state for group 13 elements 
    ♦ donating 3 electrons in the outermost main shell
• +4 oxidation state for group 14 elements 
    ♦ sharing 4 pairs of electrons in the outermost main shell. But all the 4 electrons will move away because of low electronegativity
• -3 oxidation state for group 15 elements 
    ♦ there are 5 electrons in the outermost main shell. Will accept 3 electrons to attain octet 
• -2 oxidation state for group 16 elements 
    ♦ there are 6 electrons in the outermost main shell. Will accept 2 electrons to attain octet  
• -1 oxidation state for group 17 elements 
    ♦ there are 7 electrons in the outermost main shell. Will accept 1 electron to attain octet 
• 0 oxidation state for group 18 elements 
    ♦ Already have an octet configuration
• The above values should be used only as general guide lines. Accurate values should be carefully calculated  for each compound.

■ Atomic radius
• We have learned about periodic trends in an earlier chapter. Details here. We have seen that the atomic radius decreases as we move from left to right in a period.
• So, as we move from left to right in any period in the p-block, the atomic radius decreases
• Also, when we move from top to bottom in any group in the p-block, the atomic radius increases
■ Ionization energy
• We have learned about periodic trends in an earlier chapter. Details here. We have seen that the ionization energy increases as we move from left to right in a period. 
• So, as we move from left to right in any period in the p-block, the ionization energy increases
• Also, when we move from top to bottom in any group in the p-block, the ionisation energy decreases
■ Electronegativity
• We have learned about periodic trends in an earlier chapter. Details here. We have seen that the electronegativity increases as we move from left to right in a period. 
• So, as we move from left to right in any period in the p-block, the electronegativity increases
• Also, when we move from top to bottom in any group in the p-block, the electronegativity decreases

■ Metallic nature
• We have learned about periodic trends in an earlier chapter. Details here. We have seen that the metallic nature decreases as we move from left to right in a period. 
• So, as we move from left to right in any period in the p-block, the metallic nature decreases
• Also, when we move from top to bottom in any group in the p-block, the metallic nature increases
• Among the p-block elements, both metals and non-metals are present

■ Reactivity
• Consider the Group 17 elements. 
• They have the smallest atomic size among the p-block elements
• They have the highest electronegativity among the p-block elements
• They need only one more electron to attain octet
• So they have the highest reactivity among the p-block elements
• These first elements have the greatest reactivity in the respective periods
• Also, the reactivity increases as we move from top to bottom in the groups
■ The Group 18 elements needs special mention
• They have eight electrons in the outermost shell
• Their S.E.C ends with ns²np⁶  
• They already have octet. So they does not show any reactivity.
• All the elements in this group are gases
• They are mono atomic because they do not need to combine with other atoms for stability

---

In the next section, we will see the properties of d-block and f-block elements. 

PREVIOUS      CONTENTS       NEXT

                        Copyright©2017 High school Chemistry lessons. blogspot.in - All Rights Reserved

Chapter 9.4 - Group number from Subshell Electronic Configuration

In the previous section, we saw how to determine the block name and period number of any element using the S.E.C. In this section we will see the method to determine the group number.

Arrangement of elements into Groups

The rule for determining the group number of a given element will depend on the block in which that element resides. That means:
• The elements in the s-block have their own unique rule to determine group numbers
• The elements in the p-block have their own unique rule to determine group numbers
• The elements in the d-block have their own unique rule to determine group numbers
■ However there is basic step which  is common for all blocks. It will be come clear when we write the rule for finding the group for any element.

Rule 9.3
Given any element. We want to know the group in which that element resides in the periodic table. For that, use the following steps:
■ Basic step for any element:
• Apply rule 9.1 and find the block in which the element resides in the periodic table
    ♦ Once we complete the above basic step, we can choose the appropriate one from A, B or C given below:
A. Rule for s-block
1. From the S.E.C, find the outer most Main shell
2. In the outer most main shell, find the s subshell
3. In this s subshell, find the number of electrons
4. This number is the 'group number' of that element

B. Rule for p-block
1. From the S.E.C, find the outer most Main shell
2. In the outer most main shell, find the s and p subshells
3. Find the total number of electrons in those s and p subshells
4. Add '10' to this sum
5. The final sum obtained by adding '10' is the 'group number' of that element
OR
1. From the S.E.C, find the outer most Main shell
2. In the outer most main shell, find the p subshell
3. Find the number of electrons in that p subshell
4. Add '12' to this number
5. The sum obtained by adding '12' is the 'group number' of that element

C. Rule for d-block
1. From the S.E.C, find the outer most Main shell
• In the outer most main shell, find the s subshell
• Write down the number of electrons in this s subshell
2. From the S.E.C, find the second last Main shell
• In the second last main shell, find the d subshell
• Write down the number of electrons in this d subshell
3. Add the number of electrons in (1) and (2)
4. The sum thus obtained is the 'group number' of that element

---

Based on the above rules we can determine the group number of any given element. Consider the fig.9.2 that we saw in the previous section. For convenience, it is shown again below: 

Fig.9.2

[Above fig.9.2 is taken from wikipedia. See details here]
■ We get the following information:
• Groups 1 and 2 belong to the s-block  
• Groups 3 to 12 belong to the d-block
• Groups 13 to 18 belong to the f-block
■ We know that, the f-block elements consist of Lanthanides and Actinids. 
• They are kept away from the main body of the periodic table. If we accommodate them, they will also get group numbers. 
• But then,  the periodic table will become very large. We saw those details in a previous chapter here.

---

Now we will see some solved examples
Solved example 9.3
The atomic number of an element is 16. Find the following:
(i) Subshell configuration (S.E.C)  (ii) Block (iii) Period  (iv) Group
Solution:
(i) 1s²2s²2p⁶3s²3p⁴  OR [Ne]3s²3p⁴.
(ii) Block: Apply rule 9.1:
• The last electron is filled in the 3p subshell. So the element resides in the p-block
(iii) Period: Apply rule 9.2:
• The highest main shell number in the S.E.C is 3. So the element resides in the third period  
(iv) Group: Apply rule 9.3 that we saw above.
We have already seen that the element belongs to the p-block. So apply Rule.9.3(B)
1. The outermost main shell is 3
2. The p subshell in this main shell is 3p
3. The number of electrons in this subshell is 4
4. Adding 12 to this number, we get: 4+12 = 16
5. So the group number is 16

Solved example 9.4
(a) Which are the subshells in which electrons are filled in the s-block elements of the third period?
(b) Write the subshell electronic configuration (S.E.C) of the last element of this period
Solution:
Part (a):
1. We have two clues:
a. The elements belong to the s-block.
• So the last electron will be filled in the s subshell
b. The elements belong to the third period.
• So the highest 'main shell number' in the S.E.C will be 3
2. Combining the above two clues, we get:
• The last term in the S.E.C is 3s
3. If the last term is 3s, the following subshells would be already filled:
1s, 2s, and 2p
4. So the subshells present are:
1s, 2s, 2p and 3s
Part (b)
1. The clue is:
• The element is the last one of the third period
2. So we can write:
• The element belongs to the last group 18.
• So it belongs to the p-block
• So the last electron will be filled in the 3p subshell
3. Now we take Rule 9.3 (B)
• The group number is 18. It is obtained by adding 12 to the number of electrons in the 3p subshell. 
• Thus, the number of electrons in the 3p subshell is 18-12 = 6
4. So the last term in the S.EC is 3p⁶.
• So the terms before that are: 1s²,2s², 2p⁶ , and 3s²
• Thus the S.E.C is: 1s²2s²2p⁶3s²3p⁶. 

Solved example 9.5

When the last electron of an atom was filled in the 3d subshell, the S.E.C was recorded as 3d⁸. Answer the following questions related to this atom

(i) Write the complete S.E.C

(ii) What is the atomic number, block name, period number and group number of this atom?
Solution:
1. When the last electron is filled, 3d⁸ is obtained. 
2. The filling of 3d can begin only after completely filling up 4s 
3. That means 4s is completely filled up. This gives us 4s².
4. If 4s is completely filled, all the subshells preceding it should be already filled up. Let us write those preceding subshells in order:
1s, 2s, 2p, 3s and 3p 
5. The above subshells are completely filled up. So the S.E.C is:
1s²2s²2p⁶3s²3p⁶4s²3d⁸  OR [Ar]4s²3d⁸

Solved example 9.6

The S.E.C of 5 elements are given below. Pick out the wrong ones from among them
(i) 1s²2s²2p⁷ (ii)1s²2s²2p² (iii)1s²2s²2p⁵3s¹ (iv)1s²2s²2p⁶3s²3p⁶4s¹3d² (v)1s²2s²2p⁶3s²3p⁶4s²3d² 
Solution:
■ (i), (iii) and (iv) are wrong. 
Explanation:
• Consider (i) 1s²2s²2p⁷: The p subshell can carry a maximum of 6 electrons only
• Consider (iii)1s²2s²2p⁵3s¹: The 3s can begin only after completely filling the 2p with 6 electrons
• Consider (iv)1s²2s²2p⁶3s²3p⁶4s¹3d²: The 3d can begin only after completely filling the 4s with 2 electrons

---

So we saw how to find the 'Group number' of any given element. We can now proceed to learn the properties of the elements of each block. In the next section, we will see the properties of s-block and p-block elements. 

PREVIOUS      CONTENTS       NEXT

                        Copyright©2017 High school Chemistry lessons. blogspot.in - All Rights Reserved

Chapter 9.3 - Arrangement of Elements into Blocks

In the previous section, we saw the electronic configuration of elements upto the thirtieth element. That is., up to zinc. In this section we will see some solved examples. Later we will also see some applications of writing the electronic configuration using subshells.

Solved example 9.1
In the main shell 4, Which subshell has the least energy and which subshell has the maximum energy?
Solution:
1. In the main shell 4, there will be 4 subshells. The 4 subshells in general are s, p, d and f. 
2. In the main shell 4, they are written as 4s, 4p, 4d and 4f. We are asked which of these four has the least and maximum energies
3. Consider fig.9.1(c). For convenience, it is shown again below: 

4. We follow the path shown by the arrow. Among the four subshells 4s, 4p, 4d and 4f, the first one to reach is 4s. So it has the least energy among the four.
5. The last one to reach is 4f. So it has the maximum energy among the four
■ This is also clear from the graphical representation given here. Among the four subshells, 4s occupies the bottom most position and 4f occupies the top most position.

Solved example 9.2
The subshell wise electronic configuration (S.E.C) of an atom is 1s²2s²2p⁶3s². Find the answers to the following:
(i) How many shells are present in this atom?
(ii) Which are the subshells of each shells?
(iii) What is the total number of electrons in the atom?
(iv) What is it's atomic number?
(v) How can this subshell electronic configuration be written in a short form?
Solution:
(i) We have S.E.C as 1s²2s²2p⁶3s².
• The numbers just before the subshell names give us the names of the main shells (main shells are some times simply called as 'shells')
• In the given S.E.C, the maximum number is '3'
• So it means that shells upto 3 (which is same as M) are present in the atom
• If M is present, K and L will also be present. So 3 shells are present in the atom
(ii) In the shell 1, the s subshell is present
• In shell 2, s and p subshells are present
• In shell 3, s subshell is present
(iii) Total number of electrons is the sum of all superscripts
This is equal to 2 + 2 + 6 + 2 = 12
(iv) The number of electrons = number of protons = 12
• So atomic number = 12
(v) The given S.E.C is 1s²2s²2p⁶3s².  
• Detach the last 3s². We get: 1s²2s²2p⁶.
• This is the S.E.C of Neon (Ne)
• So the short form is: [Ne]3s².

---

Now we will see some of the applications of subshell wise electronic configuration (S.E.C) of elements.

Arrangement of elements into Blocks

The following fig.9.2 shows that, all the elements in the periodic table are classified into four blocks. They are: s-block, p-block, d-block and f-block.

Fig.9.2

[Above fig.9.2 is taken from wikipedia. See details here] 
• We want to know the basis of such a classification. Let us analyse: 
Following are some elements taken at random from the periodic table. Some details are also written about them:
■ Element: ₃Li
1. Atomic number: 3
2. S.E.C: 1s²2s¹
3. Subshell to which the last electron is added: s
4. Block: s-block
■ Element: ₁₂Mg
1. Atomic number: 12
2. S.E.C: 1s²2s²2p⁶3s²
3. Subshell to which the last electron is added: s
4. Block: s-block
■ Element: ₇N
1. Atomic number: 7
2. S.E.C: 1s²2s²2p³
3. Subshell to which the last electron is added: p
4. Block: p-block
■ Element: ₂₁Sc
1. Atomic number: 21
2. S.E.C: 1s²2s²2p⁶3s²3p⁶4s²3d¹
3. Subshell to which the last electron is added: d
4. Block: d-block

---

• We can see that items (3) and (4) are the same for all the elements. 

• That means, the following two items are same for all the elements:
    ♦ Subshell to which the last electron is added
    ♦ The block in which the element resides in the periodic table. 
• As these two items are same for every element, we can write it as a rule. This rule can be written in steps.
Rule 9.1:
■ Given any element. We want to know the block in which that element resides in the periodic table. For that, use the following steps:
1. Write the S.E.C. 
2. Find the subshell to which the last electron is added
3. Name of that subshell is the required name of the block

---

Some examples:

■ ₄Be
1. S.E.C → 1s²2s² 
2. Subshell to which the last electron is added → s
3. Required name of the block → s
• So beryllium resides in the s-block
■ ₂₆Fe
1. S.E.C → 1s²2s²2p⁶3s²3p⁶4s²3d⁶
2. Subshell to which the last electron is added → d [∵ based on fig.9.1(c), 4s is completely filled up before filling of 3d begins]
3. Required name of the block → d
• So iron resides in the d-block
■ ₁₈Ar
1. S.E.C → 1s²2s²2p⁶3s²3p⁶
2. Subshell to which the last electron is added → p
3. Required name of the block → p
• So argon resides in the p-block

---

Arrangement of elements into Periods

Following are some elements taken at random from the periodic table. Some details are also written about them:
■ Element: ₄Be
1. Atomic number: 4
2. S.E.C: 1s²2s²
3. Highest shell number in the S.E.C: 2
4. Period number: 2
■ Element: ₆C
1. Atomic number: 6
2. S.E.C: 1s²2s²2p²
3. Highest shell number in the S.E.C: 2
4. Period number: 2
■ Element: ₁₁Na
1. Atomic number: 11
2. S.E.C: 1s²2s²2p⁶3s¹
3. Highest shell number in the S.E.C: 3
4. Period number: 3
■ Element: ₁₉K
1. Atomic number: 19
2. S.E.C: 1s²2s²2p⁶3s²3p⁶4s¹
3. Highest shell number in the S.E.C: 4
4. Period number: 4

---

• We can see that items (3) and (4) are the same for all the elements. 

• That means, the following two items are same for all the elements:
    ♦ Highest shell number in the S.E.C
    ♦ Period number of the element. 
• As these two items are same for every element, we can write it as a rule. This rule can be written in steps.
Rule 9.2:
■ Given any element. We want to know the period in which that element resides in the periodic table (that is, the period number of the element). For that, use the following steps:
1. Write the S.E.C. 
2. Find the highest shell number in the S.E.C
3. That shell number is the required period number

---

So we saw how to find the 'Block name' and 'Period number' from the S.E.C. In the next section, we will see how to find the 'Group number' from the S.E.C. 

PREVIOUS      CONTENTS       NEXT

                        Copyright©2017 High school Chemistry lessons. blogspot.in - All Rights Reserved