The first ionization potential of Na is 5.1eV . The value of electron gain enthalpy of Na+ will be

Question

The first ionization potential of Na is 5.1eV . The value of electron gain enthalpy of Na+ will be (A) -5.1eV (B) -10.2eV (C) +2.55eV (D) -2.55eV

Tardigrade Admin · Accepted Answer

Ionization energy = 5.1 eV
Electron gain enthalpy = negative value of ionization energy = - 5.1 eV
As electron gain enthalpy is the energy released when an electron is added to the isolated gaseous atom so it is exothermic by nature.

Q. The first ionization potential of $N a$ is $5.1 e V$ . The value of electron gain enthalpy of $N a^{+}$ will be

$- 5.1 e V$

$- 10.2 e V$

$+ 2.55 e V$

$- 2.55 e V$

Ionization energy = 5.1 eV
Electron gain enthalpy = negative value of ionization energy = - 5.1 eV
As electron gain enthalpy is the energy released when an electron is added to the isolated gaseous atom so it is exothermic by nature.

Q. The first ionization potential of Na is 5.1eV . The value of electron gain enthalpy of Na+ will be

−5.1eV

−10.2eV

+2.55eV

−2.55eV

Ionization energy = 5.1 eV Electron gain enthalpy = negative value of ionization energy = - 5.1 eV As electron gain enthalpy is the energy released when an electron is added to the isolated gaseous atom so it is exothermic by nature.

Q. The first ionization potential of $N a$ is $5.1 e V$ . The value of electron gain enthalpy of $N a^{+}$ will be

$- 5.1 e V$

$- 10.2 e V$

$+ 2.55 e V$

$- 2.55 e V$

Ionization energy = 5.1 eV
Electron gain enthalpy = negative value of ionization energy = - 5.1 eV
As electron gain enthalpy is the energy released when an electron is added to the isolated gaseous atom so it is exothermic by nature.