Ionisation of energy of F -is 320 Mol -1A- a-320 KJ mol-1B- b -160 KJ mol-1C- c-320 KJ mol-1D- d 160 KJ mol-1

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Ionisation Energy

Ionisation of...

Question

Ionisation of energy of$ F^-$ is 320 Mol$^{-1}$

(a)-320 KJ mol

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(b) -160 KJ mol

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(c)+320 KJ mol

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(d)160 KJ mol

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Solution

The correct option is A
(a)-320 KJ mol

Let's look at this equation.

$F^{-}$ → F + $e^{-}$ $Δ$ H = 320 kJ

F + $e^{-}$ → $F^{-}$

$Δ$ H will be -320 kJ only

Second equation will represent electron gain enthalpy because an electron is added.

Therefore,

Electron gain enthalpy = -320kJ

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Ionisation of energy of\( F^-\) is 320 Mol\(^{-1}\)

The correct option is A (a)-320 KJ mol Let's look at this equation. F− → F + e− ΔH = 320 kJ F + e− → F− ΔH will be -320 kJ only Second equation will represent electron gain enthalpy because an electron is added. Therefore, Electron gain enthalpy = -320kJ

The correct option is A
(a)-320 KJ mol

Let's look at this equation.

$F^{-}$ → F + $e^{-}$ $Δ$ H = 320 kJ

F + $e^{-}$ → $F^{-}$

$Δ$ H will be -320 kJ only

Second equation will represent electron gain enthalpy because an electron is added.

Therefore,

Electron gain enthalpy = -320kJ