In the previous section 3.8, we completed a discussion on periodic trends in electron gain enthalpy. In this section, we will see the periodic trends in electronegativity
1. In our previous classes, we have learned about covalent bonds (Details here)
• In this type of bonds, two atoms share ‘two electrons’ (a pair of electrons)
• Both the atoms have claim over both the electrons
2. There are two possibilities:
(i) Both the atoms are of the same type (eg: O-O, Cl-Cl)
♦ In this case, the ‘shared pair of electrons’ is equally attracted by the two atoms
(ii) The two atoms are of different types
♦ In this case, the ‘shared pair of electrons’ may get attracted towards one of the atoms
3. Some atoms have greater ability to attract the ‘shared pair of electrons’
■ A qualitative measure of the ability of an atom in a chemical compound to attract shared electrons to itself is called electronegativity
4. Properties like ionization enthalpy, electron gain enthalpy etc., can be measured. But electronegativity cannot be measured
• That is why we use the word ‘qualitative’
5. Let us see an example of ‘qualitative measure’
We will write it in steps:
(i) We want a color which is close to red
(ii) We examine different available colors
♦ Some colors are very reddish
♦ Some colors are moderately red
♦ Some colors have only a reddish tint
(iii) In this situation, we assume a convenient value of say ‘10’ for ‘perfect red’
♦ ‘Very reddish’ colors will be given values like 8 or 9
♦ ‘Moderate reddish’ colors will be given values like 4 or 5
♦ ‘Colors with only reddish tints’ will be given values like 2 or 3
(iv) In this way, we make a ‘qualitative measure’ of the color
6. Another example would be the ‘quality of work’ done by a mechanic when he repairs a car
♦ The owner of the car may give a ‘rating of 10’ if the work is ‘excellent’
♦ Rating may be ‘5’ or ‘6’ if the work ‘satisfactory’
♦ Rating may be ‘1’ or ‘2’ if the work is ‘poor’
7. In this way, electronegativity can be measured only in a qualitative way
■ Scientists have developed different scales for measuring electronegativity
• Some of them are:
♦ Pauling scale
♦ Mulliken-Jaffe scale
♦ Allred-Rochow scale
8. Pauling scale is the most widely used scale
• It was developed in 1922 by the American scientist Linus Pauling
9. In the Pauling scale, fluorine is considered to have the ‘greatest ability to attract electrons’
■ Pauling assigned an arbitrary value of ‘4’ to fluorine
• ‘Arbitrary’ means:
Based on random choice, rather than any reason or system
(The dictionary meaning can be seen here)
10. Approximate values for the electronegativity of a few elements are given in the tables 3.6 and 3.7 below:
• Note that, fluorine (which has the ‘greatest ability to attract electrons’) is given a value of ‘4’. So all other elements will be having values less than ‘4’
11. Electronegativity of a given element is not constant
• It depends on the other elements to which the ‘given element’ is bound
• We can start the discussion on it’s periodic trends
First we will see the trend along periods. We will write it in steps:
1. Imagine that, we are moving from left to right along a period
• We will see that, the electronegativity increases
2. For example, consider the 2nd period
• In this period,
♦ Li will have the smallest electronegativity
♦ F will have the largest electronegativity
3. Recall that, along the periods, ‘atomic radii’ also has a similar (but opposite) trend:
■ When we move from left to right, the radii decreases
• So naturally, we are inclined to make a comparison between two items:
♦ ‘Ability to attract electrons (electronegativity)’ increases towards the right
♦ ‘Atomic radii’ decreases towards the right
Are these two, related to each other?
• We can easily see the relation. It can be explained in 2 steps:
(i) The atomic radii decreases because, the outermost electrons are attracted more and more tightly towards the nucleus
(ii) In the same way, the ‘shared pair of electrons’ will also be attracted more and more tightly towards the nucleus
4. Again recall that, along the periods, the ‘ionization enthalpy’ also has a similar trend:
■ When we move from left to right, the ‘ionization enthalpy’ increases
• That is., it is more and more difficult to remove an electron
• So naturally, we are inclined to make a comparison between two items:
♦ ‘Ability to attract electrons (electronegativity)’ increases towards the right
♦ ‘Ionization enthalpy’ increases towards the right
Are these two, related to each other?
• We can easily see the relation. It can be explained in 2 steps:
(i) The ionization enthalpy increases because, the outermost electrons are attracted more and more tightly towards the nucleus
(ii) In the same way, the ‘shared pair of electrons’ will also be attracted more and more tightly towards the nucleus
5. Again recall that, along the periods, the ‘electron gain enthalpy’ also has a similar trend:
■ When we move from left to right, the ‘electron gain enthalpy’ becomes more and more negative
• That is., elements are more and more happier to gain an electron
• So naturally, we are inclined to make a comparison between two items:
♦ ‘Ability to attract electrons (electronegativity)’ increases towards the right
♦ ‘Electron gain enthalpy’ becomes more and more negative towards the right
Are these two, related to each other?
• We can easily see the relation. It can be explained in 2 steps:
(i) The electron gain enthalpy becomes more negative because, the elements are happier to add a new electron
♦ That is., the newly added electron will be held more tightly towards the nucleus
(ii) In the same way, the ‘shared pair of electrons’ will also be attracted more and more tightly towards the nucleus
6. So we have seen four items along a period:
(i) Atomic radii (ii) Ionization energy (iii) Electron gain enthalpy (iv) Electronegativity
• We saw that, all the four are closely related
• Since we compared their trends along periods (horizontal rows), we will use four horizontal arrows to show their increasing and decreasing trends
• This is shown in fig.3.14 below:
♦ This image is obtained from Wikimedia commons
♦ The link is given below:
♦ Link to image
1. Imagine that, we are moving from top to bottom along a period
• We will see that, the electronegativity decreases
2. For example, consider the 17th group
• In this period,
♦ F will have the largest electronegativity
♦ At will have the smallest electronegativity
3. Recall that, along the groups, ‘atomic radii’ also has a similar (but opposite) trend:
■ When we move from top to bottom, the radii increases
• So naturally, we are inclined to make a comparison between two items:
♦ ‘Ability to attract electrons (electronegativity)’ decreases towards the bottom
♦ ‘Atomic radii’ increases towards the bottom
Are these two, related to each other?
• We can easily see the relation. It can be explained in 2 steps:
(i) The atomic radii increases because, the outermost electrons are attracted less and less tightly towards the nucleus
♦ With the passage of each element, one more main-shell is added
♦ The repulsion from the inner electron increases
(ii) In the same way, the ‘shared pair of electrons’ will also be attracted less and less tightly towards the nucleus
4. Again recall that, along the groups, the ‘ionization enthalpy’ also has a similar trend:
■ When we move from top to bottom, the ‘ionization enthalpy’ decreases
• That is., it is more and more easy to remove an electron
• So naturally, we are inclined to make a comparison between two items:
♦ ‘Ability to attract electrons (electronegativity)’ decreases towards the bottom
♦ ‘Ionization enthalpy’ decreases towards the bottom
Are these two, related to each other?
• We can easily see the relation. It can be explained in 2 steps:
(i) The ionization enthalpy decreases because, the outermost electrons are attracted less and less tightly towards the nucleus
(ii) In the same way, the ‘shared pair of electrons’ will also be attracted less and less tightly towards the nucleus
5. Again recall that, along the periods, the ‘electron gain enthalpy’ also has a similar trend:
■ When we move from top to bottom, the ‘electron gain enthalpy’ becomes less and less negative
• That is., elements are less and less happier to gain an electron
• So naturally, we are inclined to make a comparison between two items:
♦ ‘Ability to attract electrons (electronegativity)’ decreases towards the bottom
♦ ‘Electron gain enthalpy’ becomes less and less negative towards the bottom
Are these two, related to each other?
• We can easily see the relation. It can be explained in 2 steps:
(i) The electron gain enthalpy becomes less negative because, the elements are less happier to add a new electron
♦ That is., the newly added electron will be held less tightly towards the nucleus
(ii) In the same way, the ‘shared pair of electrons’ will also be attracted less and less tightly towards the nucleus
6. So we have seen four items along a group:
(i) Atomic radii (ii) Ionization energy (iii) Electron gain enthalpy (iv) Electronegativity
• We saw that, all the four are closely related
• Since we compared their trends along groups (vertical columns), we will use four vertical arrows to show their increasing and decreasing trends
• This is shown in fig.3.15 below:
♦ This image is obtained from Wikimedia commons
♦ The link is given below:
♦ Link to image
Relation between electronegativities and 'non-metallic character of elements'
1. Non-metallic elements have a strong tendency to gain electrons
2. We have seen that ‘electronegativity’ indicates the tendency to gain electrons
• So elements having ‘greater electronegativity’ will be ‘more non-metallic' in nature
3. Now we will bring ‘periodic trends in electronegativity’ into this discussion:
(i) Consider the ‘horizontal arrow of electronegativity’ in fig.3.14 above
• We see that, it points towards the right
♦ That means, electronegativity increases towards the right
♦ That means, elements towards the right will be more non-metallic
(ii) Consider the ‘vertical arrow of electronegativity’ in fig.3.15 above
• We see that, it points upwards
♦ That means, electronegativity increases towards the top
♦ That means, elements towards the top will be more non-metallic
4. So we have two arrows
• They are shown in figs.3.16 (a) and (b) below
♦ Fig.a indicates that, the elements towards the right will be more non-metallic
♦ Fig.b indicates that, the elements towards the top will be more non-metallic
5. Combining the two information, we get:
■ Elements towards the top-right will be more non-metallic
• So we can draw an arrow towards the top-right corner of the periodic table
6. We can also write it in this way:
• A horizontal force acts towards the right as indicated by fig.3.16(a)
• A vertical force acts towards the top as indicated by fig.3.16(b)
■ The resultant of the two forces will obviously be a slanting force acting towards the top-right. This is shown in fig.3.16(c)
7. Now we know the periodic trend in non-metallic character
• We can show it in the periodic table. This is shown as the brown arrow in fig.3.17 below:
Relation between electronegativities and 'metallic character of elements'
1. Metallic elements have a strong tendency to lose electrons
2. We have seen that ‘electronegativity’ indicates the tendency to gain electrons
• So elements having ‘lesser electronegativity’ will be ‘more metallic' in nature
3. Now we will bring ‘periodic trends in electronegativity’ into this discussion:
(i) Consider the ‘horizontal arrow of electronegativity’ in fig.3.14 above
• We see that, it points towards the right
♦ That means, electronegativity increases towards the right
♦ That means, electronegativity decreases towards the left
♦ That means, elements towards the left will be more metallic
(ii) Consider the ‘vertical arrow of electronegativity’ in fig.3.15 above
• We see that, it points upwards
♦ That means, electronegativity increases towards the top
♦ That means, electronegativity decreases towards the bottom
♦ That means, elements towards the bottom will be more metallic
4. So we have two arrows
• They are shown in figs.3.18 (a) and (b) below
♦ Fig.a indicates that, the elements towards the left will be more metallic
♦ Fig.b indicates that, the elements towards the bottom will be more metallic
5. Combining the two information, we get:
■ Elements towards the bottom-left will be more metallic
• So we can draw an arrow towards the bottom-left corner of the periodic table
6. We can also write it in this way:
• A horizontal force acts towards the left as indicated by fig.3.18(a)
• A vertical force acts towards the bottom as indicated by fig.3.18(b)
■ The resultant of the two forces will obviously be a slanting force acting towards the bottom-left. This is shown in fig.3.18(c)
7. Now we know the periodic trend in metallic character
We can show it in the periodic table. This is shown as the magenta arrow in fig.3.17 above
Solved example 3.19
Considering the elements B, Al, Mg, and K, the correct order of their metallic character is :
(a) B > Al > Mg > K (b) Al > Mg > B > K
(c) Mg > Al > K > B (d) K > Mg > Al > B
Solution:
1. All the given four options are in the decreasing order
• That means:
♦ The most metallic element is written first
♦ The next most metallic element is written second
♦ so on . . .
2. We know that, the metallic character increases towards the bottom-left
• Out of the four given elements, the 'left-most' and 'bottom-most' is K
• So, out of the four given elements, K is the most metallic. It should be written first
3. Out of the four given elements, the 'right-most' and 'top-most' will be the least metallic
• We can see that, it is B
• So, out of the four given elements, B is the least metallic. It should be written last
4. Now, two elements remain: Al and Mg
• They belong to the same period
♦ So there is no question of 'being on top or bottom' of the other
• The element on the left will be more metallic
• So we get: Mg is more metallic. It should be written before Al
5. So the fourth option gives the correct order: K > Mg > Al > B
Solved example 3.20
Considering the elements B, C, N, F, and Si, the correct order of their non-metallic character is :
(a) B > C > Si > N > F (b) Si > C > B > N > F
(c) F > N > C > B > Si (d) F > N > C > Si > B
Solution:
1. All the given four options are in the decreasing order
• That means:
♦ The most non-metallic element is written first
♦ The next most non-metallic element is written second
♦ so on . . .
2. We know that, the non-metallic character increases towards the top-right
• Out of the five given elements, the 'right-most' and 'top-most' is F
• So, out of the five given elements, F is the most non-metallic. It should be written first
3. Out of the five given elements, the 'left-most' and 'bottom-most' will be the least non-metallic
♦ We see that, the left-most is B
♦ But the 'bottom-most' is Si
• So we have to pick the correct one from B and Si
• Which one is least non-metallic?
• To find the answer, we compare the electronegativities
♦ The electronegativity of B is 2.0
♦ The electronegativity of Si is 1.8
• B has greater electronegativity
♦ So B is more non-metallic than Si
♦ So the element with least non-metallic character is Si
♦ It should be written last
■ Note: To compare B and Si, we cannot seek the help of C. This is because, C is more non-metallic than both B and Si
4. Three elements remain: N, C and Si
• Out of them, the 'right-most' and 'top-most' is N
• So N should be written next to F
5. Two elements remain: C and Si
• They both belong to the same group
♦ So there is no question of being on left or right of the other
• The element on the top will be more non-metallic
• So C is more non-metallic than Si
• C must be written before Si
6. Thus the third option gives the correct order: F > N > C > Si > B
1. In our previous classes, we have learned about covalent bonds (Details here)
• In this type of bonds, two atoms share ‘two electrons’ (a pair of electrons)
• Both the atoms have claim over both the electrons
2. There are two possibilities:
(i) Both the atoms are of the same type (eg: O-O, Cl-Cl)
♦ In this case, the ‘shared pair of electrons’ is equally attracted by the two atoms
(ii) The two atoms are of different types
♦ In this case, the ‘shared pair of electrons’ may get attracted towards one of the atoms
3. Some atoms have greater ability to attract the ‘shared pair of electrons’
■ A qualitative measure of the ability of an atom in a chemical compound to attract shared electrons to itself is called electronegativity
4. Properties like ionization enthalpy, electron gain enthalpy etc., can be measured. But electronegativity cannot be measured
• That is why we use the word ‘qualitative’
5. Let us see an example of ‘qualitative measure’
We will write it in steps:
(i) We want a color which is close to red
(ii) We examine different available colors
♦ Some colors are very reddish
♦ Some colors are moderately red
♦ Some colors have only a reddish tint
(iii) In this situation, we assume a convenient value of say ‘10’ for ‘perfect red’
♦ ‘Very reddish’ colors will be given values like 8 or 9
♦ ‘Moderate reddish’ colors will be given values like 4 or 5
♦ ‘Colors with only reddish tints’ will be given values like 2 or 3
(iv) In this way, we make a ‘qualitative measure’ of the color
6. Another example would be the ‘quality of work’ done by a mechanic when he repairs a car
♦ The owner of the car may give a ‘rating of 10’ if the work is ‘excellent’
♦ Rating may be ‘5’ or ‘6’ if the work ‘satisfactory’
♦ Rating may be ‘1’ or ‘2’ if the work is ‘poor’
7. In this way, electronegativity can be measured only in a qualitative way
■ Scientists have developed different scales for measuring electronegativity
• Some of them are:
♦ Pauling scale
♦ Mulliken-Jaffe scale
♦ Allred-Rochow scale
8. Pauling scale is the most widely used scale
• It was developed in 1922 by the American scientist Linus Pauling
9. In the Pauling scale, fluorine is considered to have the ‘greatest ability to attract electrons’
■ Pauling assigned an arbitrary value of ‘4’ to fluorine
• ‘Arbitrary’ means:
Based on random choice, rather than any reason or system
(The dictionary meaning can be seen here)
10. Approximate values for the electronegativity of a few elements are given in the tables 3.6 and 3.7 below:
 |
| Table 3.6 |
 |
| Table 3.7 |
11. Electronegativity of a given element is not constant
• It depends on the other elements to which the ‘given element’ is bound
• Now we know the basics about electronegativity
First we will see the trend along periods. We will write it in steps:
1. Imagine that, we are moving from left to right along a period
• We will see that, the electronegativity increases
2. For example, consider the 2nd period
• In this period,
♦ Li will have the smallest electronegativity
♦ F will have the largest electronegativity
3. Recall that, along the periods, ‘atomic radii’ also has a similar (but opposite) trend:
■ When we move from left to right, the radii decreases
• So naturally, we are inclined to make a comparison between two items:
♦ ‘Ability to attract electrons (electronegativity)’ increases towards the right
♦ ‘Atomic radii’ decreases towards the right
Are these two, related to each other?
• We can easily see the relation. It can be explained in 2 steps:
(i) The atomic radii decreases because, the outermost electrons are attracted more and more tightly towards the nucleus
(ii) In the same way, the ‘shared pair of electrons’ will also be attracted more and more tightly towards the nucleus
4. Again recall that, along the periods, the ‘ionization enthalpy’ also has a similar trend:
■ When we move from left to right, the ‘ionization enthalpy’ increases
• That is., it is more and more difficult to remove an electron
• So naturally, we are inclined to make a comparison between two items:
♦ ‘Ability to attract electrons (electronegativity)’ increases towards the right
♦ ‘Ionization enthalpy’ increases towards the right
Are these two, related to each other?
• We can easily see the relation. It can be explained in 2 steps:
(i) The ionization enthalpy increases because, the outermost electrons are attracted more and more tightly towards the nucleus
(ii) In the same way, the ‘shared pair of electrons’ will also be attracted more and more tightly towards the nucleus
5. Again recall that, along the periods, the ‘electron gain enthalpy’ also has a similar trend:
■ When we move from left to right, the ‘electron gain enthalpy’ becomes more and more negative
• That is., elements are more and more happier to gain an electron
• So naturally, we are inclined to make a comparison between two items:
♦ ‘Ability to attract electrons (electronegativity)’ increases towards the right
♦ ‘Electron gain enthalpy’ becomes more and more negative towards the right
Are these two, related to each other?
• We can easily see the relation. It can be explained in 2 steps:
(i) The electron gain enthalpy becomes more negative because, the elements are happier to add a new electron
♦ That is., the newly added electron will be held more tightly towards the nucleus
(ii) In the same way, the ‘shared pair of electrons’ will also be attracted more and more tightly towards the nucleus
6. So we have seen four items along a period:
(i) Atomic radii (ii) Ionization energy (iii) Electron gain enthalpy (iv) Electronegativity
• We saw that, all the four are closely related
• Since we compared their trends along periods (horizontal rows), we will use four horizontal arrows to show their increasing and decreasing trends
• This is shown in fig.3.14 below:
♦ This image is obtained from Wikimedia commons
♦ The link is given below:
♦ Link to image
 |
| Fig.3.14 |
Next we will see the trend along groups. We will write it in steps:
• We will see that, the electronegativity decreases
2. For example, consider the 17th group
• In this period,
♦ F will have the largest electronegativity
♦ At will have the smallest electronegativity
3. Recall that, along the groups, ‘atomic radii’ also has a similar (but opposite) trend:
■ When we move from top to bottom, the radii increases
• So naturally, we are inclined to make a comparison between two items:
♦ ‘Ability to attract electrons (electronegativity)’ decreases towards the bottom
♦ ‘Atomic radii’ increases towards the bottom
Are these two, related to each other?
• We can easily see the relation. It can be explained in 2 steps:
(i) The atomic radii increases because, the outermost electrons are attracted less and less tightly towards the nucleus
♦ With the passage of each element, one more main-shell is added
♦ The repulsion from the inner electron increases
(ii) In the same way, the ‘shared pair of electrons’ will also be attracted less and less tightly towards the nucleus
4. Again recall that, along the groups, the ‘ionization enthalpy’ also has a similar trend:
■ When we move from top to bottom, the ‘ionization enthalpy’ decreases
• That is., it is more and more easy to remove an electron
• So naturally, we are inclined to make a comparison between two items:
♦ ‘Ability to attract electrons (electronegativity)’ decreases towards the bottom
♦ ‘Ionization enthalpy’ decreases towards the bottom
Are these two, related to each other?
• We can easily see the relation. It can be explained in 2 steps:
(i) The ionization enthalpy decreases because, the outermost electrons are attracted less and less tightly towards the nucleus
(ii) In the same way, the ‘shared pair of electrons’ will also be attracted less and less tightly towards the nucleus
5. Again recall that, along the periods, the ‘electron gain enthalpy’ also has a similar trend:
■ When we move from top to bottom, the ‘electron gain enthalpy’ becomes less and less negative
• That is., elements are less and less happier to gain an electron
• So naturally, we are inclined to make a comparison between two items:
♦ ‘Ability to attract electrons (electronegativity)’ decreases towards the bottom
♦ ‘Electron gain enthalpy’ becomes less and less negative towards the bottom
Are these two, related to each other?
• We can easily see the relation. It can be explained in 2 steps:
(i) The electron gain enthalpy becomes less negative because, the elements are less happier to add a new electron
♦ That is., the newly added electron will be held less tightly towards the nucleus
(ii) In the same way, the ‘shared pair of electrons’ will also be attracted less and less tightly towards the nucleus
6. So we have seen four items along a group:
(i) Atomic radii (ii) Ionization energy (iii) Electron gain enthalpy (iv) Electronegativity
• We saw that, all the four are closely related
• Since we compared their trends along groups (vertical columns), we will use four vertical arrows to show their increasing and decreasing trends
• This is shown in fig.3.15 below:
♦ This image is obtained from Wikimedia commons
♦ The link is given below:
♦ Link to image
 |
| Fig.3.15 |
• Next we will see the following relation:
1. Non-metallic elements have a strong tendency to gain electrons
2. We have seen that ‘electronegativity’ indicates the tendency to gain electrons
• So elements having ‘greater electronegativity’ will be ‘more non-metallic' in nature
3. Now we will bring ‘periodic trends in electronegativity’ into this discussion:
(i) Consider the ‘horizontal arrow of electronegativity’ in fig.3.14 above
• We see that, it points towards the right
♦ That means, electronegativity increases towards the right
♦ That means, elements towards the right will be more non-metallic
(ii) Consider the ‘vertical arrow of electronegativity’ in fig.3.15 above
• We see that, it points upwards
♦ That means, electronegativity increases towards the top
♦ That means, elements towards the top will be more non-metallic
4. So we have two arrows
• They are shown in figs.3.16 (a) and (b) below
♦ Fig.a indicates that, the elements towards the right will be more non-metallic
♦ Fig.b indicates that, the elements towards the top will be more non-metallic
 |
| Fig.3.16 |
■ Elements towards the top-right will be more non-metallic
• So we can draw an arrow towards the top-right corner of the periodic table
6. We can also write it in this way:
• A horizontal force acts towards the right as indicated by fig.3.16(a)
• A vertical force acts towards the top as indicated by fig.3.16(b)
■ The resultant of the two forces will obviously be a slanting force acting towards the top-right. This is shown in fig.3.16(c)
7. Now we know the periodic trend in non-metallic character
• We can show it in the periodic table. This is shown as the brown arrow in fig.3.17 below:
 |
| Fig.3.17 |
• Next we will see the following relation:
1. Metallic elements have a strong tendency to lose electrons
2. We have seen that ‘electronegativity’ indicates the tendency to gain electrons
• So elements having ‘lesser electronegativity’ will be ‘more metallic' in nature
3. Now we will bring ‘periodic trends in electronegativity’ into this discussion:
(i) Consider the ‘horizontal arrow of electronegativity’ in fig.3.14 above
• We see that, it points towards the right
♦ That means, electronegativity increases towards the right
♦ That means, electronegativity decreases towards the left
♦ That means, elements towards the left will be more metallic
(ii) Consider the ‘vertical arrow of electronegativity’ in fig.3.15 above
• We see that, it points upwards
♦ That means, electronegativity increases towards the top
♦ That means, electronegativity decreases towards the bottom
♦ That means, elements towards the bottom will be more metallic
4. So we have two arrows
• They are shown in figs.3.18 (a) and (b) below
♦ Fig.a indicates that, the elements towards the left will be more metallic
♦ Fig.b indicates that, the elements towards the bottom will be more metallic
 |
| Fig.3.18 |
■ Elements towards the bottom-left will be more metallic
• So we can draw an arrow towards the bottom-left corner of the periodic table
6. We can also write it in this way:
• A horizontal force acts towards the left as indicated by fig.3.18(a)
• A vertical force acts towards the bottom as indicated by fig.3.18(b)
■ The resultant of the two forces will obviously be a slanting force acting towards the bottom-left. This is shown in fig.3.18(c)
7. Now we know the periodic trend in metallic character
We can show it in the periodic table. This is shown as the magenta arrow in fig.3.17 above
Now we will see some solved examples
Considering the elements B, Al, Mg, and K, the correct order of their metallic character is :
(a) B > Al > Mg > K (b) Al > Mg > B > K
(c) Mg > Al > K > B (d) K > Mg > Al > B
Solution:
1. All the given four options are in the decreasing order
• That means:
♦ The most metallic element is written first
♦ The next most metallic element is written second
♦ so on . . .
2. We know that, the metallic character increases towards the bottom-left
• Out of the four given elements, the 'left-most' and 'bottom-most' is K
• So, out of the four given elements, K is the most metallic. It should be written first
3. Out of the four given elements, the 'right-most' and 'top-most' will be the least metallic
• We can see that, it is B
• So, out of the four given elements, B is the least metallic. It should be written last
4. Now, two elements remain: Al and Mg
• They belong to the same period
♦ So there is no question of 'being on top or bottom' of the other
• The element on the left will be more metallic
• So we get: Mg is more metallic. It should be written before Al
5. So the fourth option gives the correct order: K > Mg > Al > B
Solved example 3.20
Considering the elements B, C, N, F, and Si, the correct order of their non-metallic character is :
(a) B > C > Si > N > F (b) Si > C > B > N > F
(c) F > N > C > B > Si (d) F > N > C > Si > B
Solution:
1. All the given four options are in the decreasing order
• That means:
♦ The most non-metallic element is written first
♦ The next most non-metallic element is written second
♦ so on . . .
2. We know that, the non-metallic character increases towards the top-right
• Out of the five given elements, the 'right-most' and 'top-most' is F
• So, out of the five given elements, F is the most non-metallic. It should be written first
3. Out of the five given elements, the 'left-most' and 'bottom-most' will be the least non-metallic
♦ We see that, the left-most is B
♦ But the 'bottom-most' is Si
• So we have to pick the correct one from B and Si
• Which one is least non-metallic?
• To find the answer, we compare the electronegativities
♦ The electronegativity of B is 2.0
♦ The electronegativity of Si is 1.8
• B has greater electronegativity
♦ So B is more non-metallic than Si
♦ So the element with least non-metallic character is Si
♦ It should be written last
■ Note: To compare B and Si, we cannot seek the help of C. This is because, C is more non-metallic than both B and Si
4. Three elements remain: N, C and Si
• Out of them, the 'right-most' and 'top-most' is N
• So N should be written next to F
5. Two elements remain: C and Si
• They both belong to the same group
♦ So there is no question of being on left or right of the other
• The element on the top will be more non-metallic
• So C is more non-metallic than Si
• C must be written before Si
6. Thus the third option gives the correct order: F > N > C > Si > B
• We have completed a discussion on the basics of seven physical properties:
(i) Atomic radius
(ii) Ionic radius
(iii) Ionization energy
(iv) Electron gain enthalpy
(v) Electronegativity
(vi) Metallic character
(vii) Non-metallic character
• We also saw the basics about their periodic trends
• In the next section, we will see some chemical properties
(i) Atomic radius
(ii) Ionic radius
(iii) Ionization energy
(iv) Electron gain enthalpy
(v) Electronegativity
(vi) Metallic character
(vii) Non-metallic character
• We also saw the basics about their periodic trends
• In the next section, we will see some chemical properties
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