Chapter 9.6 - Properties of d-block and f-block elements

In the previous section, we saw the properties of s-block and p-block elements.  In this section we will see the properties of d-block elements.

Properties of d-block elements 

■ Position: The d-block consists of the groups from 3 to 12 of the periodic table
■ Common names:
The elements of the groups from 3 to 12 are known as transition elements
■ Oxidation state
• Let us find the oxidation state of Fe in FeCl₂. 
We have seen the method of calculation here. From the table we have: oxidation state of Cl = -1
Let the oxidation state of Fe be 'x'. Then we get: x + 2 × -1 = 0 ⇒ x -2 = 0 ⇒ x = +2   
So the oxidation state of Fe in FeCl₂ is +2
• Let us find the oxidation state of Fe in FeCl₃. 
From the table we have: oxidation state of Cl = -1
Let the oxidation state of Fe be 'x'. Then we get: x + 3 × -1 = 0 ⇒ x -3 = 0 ⇒ x = +3   
So the oxidation state of Fe in FeCl₃ is +3
• So we see that Fe shows variable oxidation states. Let us analyse and find the reason for this:
1. The S.E.C of  Fe is: 1s²2s²2p⁶3s²3p⁶4s²3d⁶ . 
Fe loses two electrons and becomes Fe²⁺. The two electrons are lost from the s subshell in the outermost main shell. That is., the 4s subshell. 
So the S.E.C of  Fe²⁺ is: 1s²2s²2p⁶3s²3p⁶3d⁶ .
2. The S.E.C of  Fe is: 1s²2s²2p⁶3s²3p⁶4s²3d⁶ . 
Fe loses three electrons and becomes Fe³⁺. 
First, two electrons are lost from the s subshell in the outermost main shell. That is., the 4s subshell. 
Then one electron is lost from the d subshell in the penultimate main shell. That is., the 3d subshell. 
So the S.E.C of  Fe³⁺ is: 1s²2s²2p⁶3s²3p⁶3d⁵.
• When we consider the main shells alone, this phenomenon cannot be explained. 
• But the explanation becomes clear when we consider subshells also:
• The difference in energy between 3d subshell and 4s subshell is very small. So, under suitable conditions, the electron in 3d subshell also take part in chemical reactions. 
• This accounts for the variable oxidation states shown by transition elements  

Another example:
• Let us find the oxidation state of Mn in MnCl₂. 
We have seen the method of calculation here. From the table we have: oxidation state of Cl = -1
Let the oxidation state of Mn be 'x'. Then we get: x + 2 × -1 = 0 ⇒ x -2 = 0 ⇒ x = +2   
So the oxidation state of Mn in MnCl₂ is +2
• Let us find the oxidation state of Mn in MnO₂. 
From the table we have: oxidation state of O = -2
Let the oxidation state of Mn be 'x'. Then we get: x + 2 × -2 = 0 ⇒ x -4 = 0 ⇒ x = +4   
So the oxidation state of Mn in MnO₂ is +4
• Let us find the oxidation state of Mn in Mn₂O₃. 
From the table we have: oxidation state of O = -2
Let the oxidation state of Mn be 'x'. Then we get: 2 × x + 3 × -2 = 0 ⇒ 2x -6 = 0 ⇒ x = +3   
So the oxidation state of Mn in Mn₂O₃ is +3
• Let us find the oxidation state of Mn in Mn₂O₇. 
From the table we have: oxidation state of O = -2
Let the oxidation state of Mn be 'x'. Then we get: 2 × x + 7 × -2 = 0 ⇒ 2x -14 = 0 ⇒ x = +7   
So the oxidation state of Mn in Mn₂O₇ is +7

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1. The S.E.C of Mn is: 1s²2s²2p⁶3s²3p⁶4s²3d⁵.

Mn loses two electrons and becomes Mn²⁺. The two electrons are lost from the s subshell in the outermost main shell. That is., the 4s subshell. 
So the S.E.C of  Mn²⁺ is: 1s²2s²2p⁶3s²3p⁶3d⁵ . 
2. The S.E.C of  Mn is: 1s²2s²2p⁶3s²3p⁶4s²3d⁵ . 
Mn loses four electrons and becomes Mn⁴⁺. 
First, two electrons are lost from the s subshell in the outermost main shell. That is., the 4s subshell. 
Then two electrons are lost from the d subshell in the penultimate main shell. That is., the 3d subshell. 
So the S.E.C of  Mn⁴⁺ is: 1s²2s²2p⁶3s²3p⁶3d³.
3. The S.E.C of  Mn is: 1s²2s²2p⁶3s²3p⁶4s²3d⁶ . 
Mn loses three electrons and becomes Mn³⁺. 
First, two three electrons are lost from the s subshell in the outermost main shell. That is., the 4s subshell. 
Then one electron is lost from the d subshell in the penultimate main shell. That is., the 3d subshell. 
So the S.E.C of  Mn³⁺ is: 1s²2s²2p⁶3s²3p⁶3d⁴.
4. The S.E.C of  Mn is: 1s²2s²2p⁶3s²3p⁶4s²3d⁵ . 
Mn loses seven electrons and becomes Mn⁷⁺. 
First, two electrons are lost from the s subshell in the outermost main shell. That is., the 4s subshell. 
Then five electrons are lost from the d subshell in the penultimate main shell. That is., the 3d subshell. 
So the S.E.C of  Mn⁴⁺ is: 1s²2s²2p⁶3s²3p⁶.
• We see that electrons in the d subshell in the penultimate main shell also takes partin the reactions. 
• This is because, the difference in energy level between the following two is very low:
    ♦ s subshell in the outermost main shell
    ♦ d subshell in the penultimate main shell
• So transition elements in general show variable oxidation states
■ Metallic property
The transition elements are metals
■ Coloured compounds
Most of the compounds formed by transition elements are coloured. Some examples are given below:
• Copper sulphate is a compound of copper, which is a transition element. This compound is blue in colour  
• Cobalt nitrate is a compound of cobalt, which is a transition element. This compound is light pink in colour
• Potassium permanganate is a compound of manganese, which is a transition element. This compound is violet in colour    
• Ferrous sulphate is a compound of iron, which is a transition element. This compound is light green in colour  
• Ammonium dichromate is a compound of chromium, which is a transition element. This compound is orange in colour  
• Compounds of transition elements are used to give colour to glass, prepare some dyes etc.,
■ Uses of transition elements
• Copper is used for making electric cables
• Nickel is used for making coins
• Titanium is used aircraft parts

Properties of f-block elements 

■ Position: The f-block elements are those elements which are placed in two rows at the bottom of the periodic table. We have seen the reason for placing them separately in an earlier chapter. Details here.
The elements of the first row are called Lanthanoids and those in the second row are called Actinoids. They belong to the sixth and seventh periods respectively.
■ S.E.C:
• We know that, the last electron of each element in the f-block will be filled in the f subshell. 
Let us examine the S.E.C of two elements:
• First we will consider ₅₈Ce (Cerium). It is a Lanthanoid 
It has the S.E.C: [Xe]4f¹5d¹6s². 
From the S.E.C it is clear that, cerium falls in the 6^th period.
The last main shell is 6. But the last electron falls in the f subshell of the main shell 4. That is., 4f    
• Next we will consider ₉₀Th (Thorium). It is an Actinoid 
It has the S.E.C: [Rn]5f³6d¹7s². 
From the S.E.C it is clear that, thorium falls in the 7^th period.
The last main shell is 7. But the last electron falls in the f subshell of the main shell 5. That is., 5f

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So we can write:

• f-block elements Lanthanoids belong to the 6^th period. Their last electrons are filled in the 4f subshell 
• f-block elements Actinoids belong to the 7^th period. Their last electrons are filled in the 5f subshell 

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■ Characteristics and uses of f-block elements

• Most of the f-block elements show variable oxidation states like the d-block elements

• Most of the actinoids are radioactive and are artificial elements

• Uranium, thorium, plutonium etc., are used as fuels in nuclear reactors

• Many of them are used as catalysts in the petroleum industry

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Now we will see some solved examples

Solved example 9.7

The element X in group 17 has 3 shells. If so,

(i) Write the S.E.C of the element

(ii) Write the period number

(iii)What will be the chemical formula of the compound formed if the element X reacts with element Y of the third period which contains one electron in the p subshell?

Solution:
Part (i):
1. We have two clues:
• The element belongs to group 17
    ♦ From this it is clear that it belongs to the p-block
• It has 3 shells
    ♦ From this it is clear that it belongs to the 3^rd period
2. For any element in the p-block, the group number is obtained as follows:
• Number of electrons in the p subshell of the last main shell + 12
3. So we can write: 17 =  Number of electrons in the p subshell of the 3^rd main shell + 12
• Thus we get: Number of electrons in the p subshell of the 3^rd main shell = 5
4. For any element in the p-block, the last electron is filled in the p subshell of the last main shell
So we can write:
• The last term of the S.E.C is 3p⁵.
5. If the last term is 3p⁵, the preceding terms will be:
1s²2s²2p⁶3s².
• So the S.E.C is: 1s²2s²2p⁶3s²3p⁵. 
Part (ii):
The period number is 3
Part (iii):
1. First we will write the S.E.C of element Y
We have two clues:
• The element belongs to the third period
    ♦ From this it is clear that the suffix of the last term of the S.E.C is 3
• It has one electron in the p subshell
    ♦ The p subshell can hold 6 electrons. So, if there is only one electron, it will be the last electron
    ♦ So the last term is 3p¹
If the last term is 3p¹, the preceding terms will be:
1s²2s²2p⁶3s²
• So the S.E.C is: 1s²2s²2p⁶3s²3p¹. 
2. The last main shell has 3 electrons. So the valency of element Y is 3
From the S.E.C of element X, it's valency is 8 -7 = 1
3. We have seen how to write the chemical formula from valency. Details here.
4. We have to find the element with lower electronegativity between X and Y 
From the S.E.C of Y, the group number of Y is 13
(Number of electrons in the p subshell of the 3^rd main shell + 12 = 1 + 12 = 13)
So Y is in the group 13 and period 3. X is in group 17 and same period
Y is on the left of X in the same period
Thus Y is less electronegative than X
So Y is to be written first
5. So the required chemical formula is YX₃

Solved example 9.8

The element Cu with atomic number 29 undergoes chemical reaction to form an ion with oxidation state +2

(i) Write down the S.E.C of this ion

(ii) Can Cu show variable valency? Why?

(iii) Write down the chemical formula of one compound formed when Cu reacts with ₁₇Cl. 
Solution:
Part (i):
1. The S.E.C of Cu (copper) is 1s²2s²2p⁶3s²3p⁶4s¹3d¹⁰  OR [Ar]4s¹3d¹⁰.
2. There is only a small energy difference between 4s and 3d subshells. So if conditions are favourable,  electrons in the 3d subshell may also take part in chemical reactions
3. When two electrons are lost to form +2 oxidation state, it is clear that, one electron from the 3d is also lost in addition to the last electon in 4s
4. So the S.E.C of the ion is: 1s²2s²2p⁶3s²3p⁶3d⁹.
Part (ii)
Cu can show variable valency because, if conditions are favourable,  electrons in the 3d subshell may also take part in chemical reactions  
Part (iii):
1. The S.E.C of Chlorine is: 1s²2s²2p⁶3s²3p⁵.
It requires 1 more electron to complete octect. So it's valency is 1
2. Cu is ready to give two electrons. One from the 4s subshell and the other from the 3d subshell. So it's valency is 2
3. We have seen how to write the chemical formula from valency. Details here. 
4. Thus the chemical formula of the compound is CuCl₂. 
Note that Cu, which has a lower electronegativity is written first.

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We have completed our present discussion about Subshell electronic configuration and properties of various blocks. In the next section, we will see Mol concept. 

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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

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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.

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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

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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. 

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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.

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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

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• 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

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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

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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

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• 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

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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. 

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