Question

6 Among the second period elements the a
order Li < B < Be <C<O<N<F < Ne.
Explain why
(i) Be has higher A, H than B
(ii) O has lower A, H than N and F?

​

Answer 1

Answer:

due to lathanoid contraction and higher EN

Answer 2

$\huge \mathfrak \red { \underline {Good morning}}$

________________________________________

$\huge \mathfrak \pink { \underline { Question}}$

Among the second period elements the actual ionization enthalpies are in the order

Li < B< Be < C< O < N < F < Ne.

Explain why

( a ) Be has higher ∆iH than B

( b ) O has lower ∆iH than N and F?

$\huge \mathfrak \blue { \underline {Answer}}$

a )

Be has elctrnic configuration 1s² 2s² whereas

B has electronic configuration 1s² 2s² 2p¹..as we can see that Be has its last orbital cmpltly filled therefore it is more stable than B...

Hence, more energy will be required to remove n electron from Be than from B and it will have high ionisation enthlpy.

b )

EC of O is 1s² 2s² 2p⁴

EC of N is 1s² 2s² 2p³

EC of F is 1s² 2s² ${2p}^{5}$

if you remembered

lonisation enthalpy increases when we move left to right due to decreases atomic size.

but here you see N has higher lonisation enthalpy than O.it is because of more stable f half filled or full filled orbitals shows more stablity } electronic configuration of N.so, it is difficult to remove an electron from N than O.

That's why Oxygen has lower lonisation enthalpy than Nitrogen and fluorine.

_____________________________________

I hope my answer helps you....

$\huge \mathfrak \purple { \underline {Stay positive}}$