Chemistry, asked by anishakhatridk, 9 months ago

second ionization energy of nitrogen is less than second ionization energy of oxygen why? ​

Answers

Answered by zanwarpratham
0

Answer:

Well, I would use NIST to check these. I looked up the first two ionization energies (by typing the atomic symbol) and got...

IE

1

(

N

)

=

14.534 eV

IE

1

(

O

)

=

13.618 eV

IE

2

(

N

)

=

29.601 eV

IE

2

(

O

)

=

35.121 eV

or...

IE

1

(

N

)

1402 kJ/mol

IE

1

(

O

)

1314 kJ/mol

IE

2

(

N

)

2856 kJ/mol

IE

2

(

O

)

3389 kJ/mol

Here, you can see that the first ionization energy of oxygen atom is lower. Consider the electron configurations.

N

:

[

H

e

]

2

s

2

2

p

3

−−−−−

−−−−−

−−−−−



.

2

p

−−−−−

.

2

s

O

:

[

H

e

]

2

s

2

2

p

4

−−−−−

−−−−−

−−−−−



.

2

p

−−−−−

.

2

s

Since oxygen has a paired

2

p

electron, that one will repel the others, making it easier to remove than any of the others.

Easier to remove

lower ionization energy.

The second ionization energy of oxygen is higher than for nitrogen. In principle, they should be the same for two seemingly identical

2

p

electrons, BUT we would neglect an important factor...

It is because oxygen atom is smaller due to a higher effective nuclear charge

Z

e

f

f

=

Z

S

, where

S

is approximated to be the number of core electrons and

Z

is the atomic number.

Z

e

f

f

,

N

7 protons

2 core e

5

Z

e

f

f

,

O

8 protons

2 core e

6

So, it is more difficult to remove the electron from the smaller atom, whose electrons are more tightly bound to the nucleus.

Harder to remove

higher ionization energy.

Answered by xRapMonster1994x
0

Easier to remove → lower ionization energy. The second ionization energy of oxygen is higher than for nitrogen. It is because oxygen atom is smaller due to a higher effective nuclear charge Zeff=Z−S , where S is approximated to be the number of core electrons and Z is the atomic number

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