If M g - M - g - e - - - H -100 eVM g - M 2- g -2 e - - - H -250 eVWhich is-are correct statements- IE - ionization energy- - H ge - electron gain enthalpyA- IE M g is 100 eVB- IE 1 of M - g is 150 eVC- IE 2 of M g is 250 eVD- - H ge of M 2- g is -150 eV

The correct options are A IE_1 of M(g) is 100 eV B IE_1 of M^+(g) is 150 eV D Δ H_ge of nbsp; M^2+(g) is 150 eVIE_1 for M^+ is actually IE_2 M^+(g) → ...

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Byju's Answer

Standard XII

Chemistry

Ionization Enthalpy

If M g → M + ...

Question

If $M (g) \to M^{+} (g) + e^{-}; Δ H = 100 e V M (g) \to M^{2 +} (g) + 2 e^{-}; Δ H = 250 e V$
Which is/are correct statement(s)?
( $I E =$ ionization energy, $Δ H_{g e} =$ electron gain enthalpy)

I E_{1}

M (g)

is 100 eV

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I E_{1}

M^{+} (g)

is 150 eV

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I E_{2}

M (g)

is 250 eV

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Δ H_{g e}

M^{2 +} (g)

is -150 eV

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Solution

The correct options are
A $I E_{1}$ of $M (g)$ is 100 eV
B $I E_{1}$ of $M^{+} (g)$ is 150 eV
D $Δ H_{g e}$ of $M^{2 +} (g)$ is -150 eV
$I E_{1} f o r M^{+} i s a c t u a l l y I E_{2}$

$M^{+} (g) \to M^{+ 2} (g) + e^{-}$

$I E_{2} = (250 - 100) e V = 150 e V$
Since, $I E = - E A$
Hence,
$Δ H_{g e} (M^{2 +}) = - I E_{2} = - 150 e V$

Suggest Corrections

Similar questions

Q. If

M (g) \to M^{+} (g) + e^{-}; Δ H = 100 e V M (g) \to M^{2 +} (g) + 2 e^{-}; Δ H = 250 e V

Which is/are correct statement(s)?
(

I E =

ionization energy,

Δ H_{g e} =

electron gain enthalpy)

Q. If

M (g) \to M^{+} (g) + e^{-}; Δ H = 90 e V

M (g) \to M^{2 +} (g) + 2 e^{-}; Δ H = 260 e V

.
What is electron gain enthalpy of

M^{2 +} (g)

M (g) \to M^{+} (g) + e^{-}, Δ H = 100 e V, M (g) \to M^{2 +} (g) + 2 e^{-}, Δ H = 250 e V

which is/are incorrect statement(s).

M (g) \to M^{+} (g) + e^{-}, Δ H = 100 e V

M (g) \to M^{2 +} (g) + 2 e^{-}, Δ H = 250 e V

which is incorrect statement?

Q. Consider following ionisation steps:

M (g) \to M^{+} (g) + e^{-}; Δ H = 100 e V

M (g) \to M^{2 +} (g) + 2 e^{-}; Δ H = 250 e V

Select correct statement(s):

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If M(g) → M+(g)+e−;ΔH=100eVM(g) → M2+(g)+2e−;ΔH=250 eV Which is/are correct statement(s)? (IE= ionization energy, ΔHge= electron gain enthalpy)

The correct options are A IE1 of M(g) is 100 eV B IE1 of M+(g) is 150 eV D ΔHge of M2+(g) is -150 eVIE1 for M+ is actually IE2 M+(g)→M+2(g)+e− IE2=(250−100) eV=150 eV Since, IE=−EA Hence, ΔHge(M2+)=−IE2=−150 eV

If $M (g) \to M^{+} (g) + e^{-}; Δ H = 100 e V M (g) \to M^{2 +} (g) + 2 e^{-}; Δ H = 250 e V$
Which is/are correct statement(s)?
( $I E =$ ionization energy, $Δ H_{g e} =$ electron gain enthalpy)