Correct Order of Electronegativity of B, C, N, O and F

Correct Answer: (B) F > O > N > C > B

The correct order of electronegativity among boron, carbon, nitrogen, oxygen, and fluorine is F > O > N > C > B. All five elements belong to the second period of the periodic table and are arranged consecutively from left to right as B, C, N, O, and F. Electronegativity generally increases from left to right across a period because effective nuclear charge increases while the electrons are added to the same principal energy level.

As we move from boron toward fluorine, the atomic nucleus attracts the bonding electron pair with progressively greater strength. Therefore, boron has the lowest electronegativity among the given elements, while fluorine has the highest. Reversing the left-to-right periodic sequence gives the decreasing order F > O > N > C > B, which corresponds to option (B).

Understanding Electronegativity

What Is Electronegativity?

Electronegativity is the tendency of an atom in a chemical bond to attract the shared pair of electrons toward itself. When two atoms with different electronegativities form a covalent bond, the bonding electrons are drawn more strongly toward the more electronegative atom.

Electronegativity is a relative property rather than a directly measurable physical quantity with a conventional unit. Several electronegativity scales have been developed, but the Pauling scale is the most widely used in introductory and competitive chemistry. On this scale, fluorine has the highest electronegativity value of approximately 3.98.

The greater the electronegativity of an atom, the stronger its ability to attract bonding electrons. Therefore, comparing electronegativity requires an understanding of nuclear charge, atomic size, shielding effect, and the position of an element in the periodic table.

Periodic Trend of Electronegativity Across Period 2

The elements given in the question are all members of Period 2:

B → C → N → O → F

Moving from left to right across this period, the atomic number increases by one at each step. This means that the number of protons in the nucleus increases progressively. At the same time, the additional electrons enter the same second principal shell.

Because no new principal electron shell is added from boron to fluorine, the increase in shielding is relatively small compared with the increase in nuclear charge. The effective nuclear attraction experienced by the valence electrons therefore increases across the period.

As a result, the atomic radius decreases and the ability of the atom to attract shared bonding electrons increases. Therefore, the electronegativity trend is:

B < C < N < O < F

Since the options are written in decreasing order of electronegativity, the required order becomes:

F > O > N > C > B

Step-by-Step Electronegativity Comparison of the Given Elements

Boron (B)

Boron is the leftmost element among the five elements given in the question. It has atomic number 5 and the electronic configuration 1s²2s²2p¹. Since it has the lowest effective nuclear attraction and the largest atomic size among these five Period 2 elements, it has the lowest electronegativity in the given series.

The Pauling electronegativity of boron is approximately 2.04. Therefore, boron appears at the end of the decreasing electronegativity order.

Carbon (C)

Carbon lies immediately to the right of boron in Period 2. Its atomic number is 6, and its electronic configuration is 1s²2s²2p². The increased nuclear charge causes carbon to attract bonding electrons more strongly than boron.

The Pauling electronegativity of carbon is approximately 2.55. Therefore:

C > B

Nitrogen (N)

Nitrogen is located to the right of carbon and has atomic number 7. Its electronic configuration is 1s²2s²2p³. The greater effective nuclear charge and smaller atomic radius allow nitrogen to attract shared electrons more strongly than carbon.

The Pauling electronegativity of nitrogen is approximately 3.04. Therefore:

N > C > B

Oxygen (O)

Oxygen lies to the right of nitrogen in Period 2 and has atomic number 8. Its electronic configuration is 1s²2s²2p⁴. Oxygen has a greater effective nuclear charge and a smaller atomic radius than nitrogen, so it attracts a bonding electron pair more strongly.

The Pauling electronegativity of oxygen is approximately 3.44. Therefore:

O > N > C > B

Fluorine (F)

Fluorine is the rightmost element among the given Period 2 elements and has atomic number 9. Its electronic configuration is 1s²2s²2p⁵. It has a very small atomic radius and a high effective nuclear charge.

Fluorine attracts shared bonding electrons more strongly than any other element in the periodic table. Its Pauling electronegativity is approximately 3.98, the highest value among all elements.

Therefore, fluorine must occupy the first position in the decreasing order of electronegativity.

Pauling Electronegativity Values of B, C, N, O and F

The approximate Pauling electronegativity values are:

B = 2.04
C = 2.55
N = 3.04
O = 3.44
F = 3.98

Arranging these values from highest to lowest gives:

3.98 > 3.44 > 3.04 > 2.55 > 2.04

Therefore:

F > O > N > C > B

Why Electronegativity Increases from Boron to Fluorine

The increase in electronegativity from boron to fluorine is mainly caused by increasing effective nuclear charge and decreasing atomic size. As the atomic number increases across Period 2, more protons are added to the nucleus. The additional electrons are added to the same principal shell, so the shielding effect does not increase enough to cancel the increasing nuclear attraction.

Consequently, the valence shell is pulled closer to the nucleus. A smaller atom with a greater effective nuclear charge can attract a shared pair of electrons more strongly. This explains the steady increase in electronegativity from B to C to N to O to F.

Relationship Between Atomic Size and Electronegativity

Atomic size and electronegativity generally show opposite trends across a period. As atomic radius decreases, the bonding electrons can approach the nucleus more closely. The electrostatic attraction between the nucleus and the shared electrons therefore becomes stronger.

Among the given elements, boron has the largest atomic size and the lowest electronegativity, while fluorine has the smallest atomic size and the highest electronegativity. This relationship supports the order:

F > O > N > C > B

Explanation of All Options

Option (A): B > C > N > O > F

This option is incorrect because it gives the exact reverse of the actual electronegativity trend across Period 2. Electronegativity increases from left to right, so boron cannot be the most electronegative and fluorine cannot be the least electronegative among the given elements.

Option (B): F > O > N > C > B

This is the correct option. Fluorine is the most electronegative element, followed by oxygen, nitrogen, carbon, and boron. This order agrees with the periodic trend and the Pauling electronegativity values of the five elements.

Option (C): F > N > O > C > B

This option correctly places fluorine first and boron last, but it incorrectly places nitrogen above oxygen. Oxygen lies to the right of nitrogen in Period 2 and has a higher electronegativity. Therefore, the correct relationship is O > N, not N > O.

Option (D): O > F > N > C > B

This option is incorrect because it places oxygen above fluorine. Fluorine is the most electronegative element in the periodic table and must have a greater electronegativity than oxygen. Therefore, the correct relationship is F > O.

Complete Electronegativity Order

Starting from the lowest electronegativity and moving toward the highest:

B < C < N < O < F

Writing the same trend in decreasing order:

F > O > N > C > B

Final Answer

Boron, carbon, nitrogen, oxygen, and fluorine are consecutive elements of Period 2. Electronegativity increases from left to right across a period because effective nuclear charge increases and atomic size decreases. Therefore, fluorine attracts bonding electrons most strongly, followed by oxygen, nitrogen, carbon, and boron.

Hence, the correct decreasing order of electronegativity is F > O > N > C > B.

Correct Answer: (B) F > O > N > C > B