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  • The first ionisation potential of Na is 5.1 eV .The value of electron gain enthalpy (in eV) of Na+ will be :Option: 1 -5.1<d

The first ionisation potential of Na is 5.1 eV .The value of electron gain enthalpy (in eV) of Na+ will be :

Option: 1

-5.1


Option: 2

- 2.55


Option: 3

+ 2.55 


Option: 4

-10. 2


Answers (1)

best_answer

As discussed in

Electron gain enthalpy -

When an electron is added to a neutral gaseous atom (x) to convert it into a negative ion, the enthalpy change accompanying the process is defined as the electron gain enthalpy 
(\Delta_{eg}H^\circleddash )

-

 

 and, 

 

Ionization enthalpy -

It is defined as the minimum amount of energy required to remove the outer most shell electron from an isolated gaseous atom (X) to form the gaseous ion.

- wherein

X_{(g)}\rightarrow X^{+}_{(g)}  +e^{-}

 

 First IE is represented by:

Na\rightarrow Na^{+}+e^{-}\:\:\:\:\:\:\:\:5.1eV

and, Electron gain enthalpy is:

Na^{+}+e^{-}\rightarrow Na

Since the equation is reversed, so:

\Delta H_(eg)= -5.1eV

 

Posted by

avinash.dongre

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