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.
■ 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
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
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 6th 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 7th period.
The last main shell is 7. But the last electron falls in the f subshell of the main shell 5. That is., 5f
• f-block elements Actinoids belong to the 7th period. Their last electrons are filled in the 5f subshell
We have completed our present discussion about Subshell electronic configuration and properties of various blocks. In the next section, we will see Mol concept.
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
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 6th 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 7th period.
The last main shell is 7. But the last electron falls in the f subshell of the main shell 5. That is., 5f
So we can write:
• f-block elements Lanthanoids belong to the 6th period. Their last electrons are filled in the 4f subshell • f-block elements Actinoids belong to the 7th period. Their last electrons are filled in the 5f subshell
■ 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
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 3rd 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 3rd main shell + 12
• Thus we get: Number of electrons in the p subshell of the 3rd 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 3rd 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
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 3rd 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 3rd main shell + 12
• Thus we get: Number of electrons in the p subshell of the 3rd 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 3rd 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.
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.
We have completed our present discussion about Subshell electronic configuration and properties of various blocks. In the next section, we will see Mol concept.
No comments:
Post a Comment