IE 1 and IE 2 of Mg are 178 and 348 kcal respectively- The energy required for the following reaction is-Mg - Mg 2-2 e -Where IE - ionization energyA- -170 kcalB- -526 kcalC- -170 kcalD- -526 kcal

The correct option is B +526 kcalAccording to given conditions Mg→ Mg^++e^ IE_1=178 kcal Mg^+→ Mg^2++e^ IE_2=348 kcal Therefore, Mg→ Mg^2++2e^ IE_3=IE ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Ionization Enthalpy

IE 1 and IE 2...

Question

$I E_{1}$ and $I E_{2}$ of $M g$ are $178$ and $348 kcal$ respectively. The energy required for the following reaction is:
$M g \to M g^{2 +} + 2 e^{-}$
(Where $I E$ = ionization energy)

+ 170 kcal

No worries! We‘ve got your back. Try BYJU‘S free classes today!

+ 526 kcal

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

- 170 kcal

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 526 kcal

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $+ 526 kcal$
According to given conditions
$M g \to M g^{+} + e^{-} I E_{1} = 178 kcal$
$M g^{+} \to M g^{2 +} + e^{-} I E_{2} = 348 kcal$
Therefore,
$M g \to M g^{2 +} + 2 e^{-} I E_{3} = I E_{1} + I E_{2}$
Hence, ionization energy of the given reaction is:
$I E = 178 kcal + 348 kcal$
$= 526 kcal$

Suggest Corrections

Similar questions

I E_{1}

and

I E_{2}

M g

are

178

and

348 kcal

respectively. The energy required for the following reaction is:

M g \to M g^{2 +} + 2 e^{-}

(Where

I E

= ionization energy)

$I E_{1}$ and $I E_{2}$ of Mg are 178 and 348 Kcal ${m o l}^{- 1}$ respectively. The energy required for the reaction

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

I E_{1}

and

I E_{2}

M g

are

178

and

348 kcal

respectively. The energy required (in kcal) for the following reaction is:

M g \to M g^{2 +} + 2 e^{-}

(Where

I E

= ionization energy)

I E_{1}

and

I E_{2}

M g

are

178

and

348 kcal

respectively. The energy required (in kcal) for the following reaction is:

M g \to M g^{2 +} + 2 e^{-}

(Where

I E

= ionization energy)

I E_{1}

and

I E_{2}

C a

are

141

and

274 kcal

respectively. The energy required for the following reaction is:

C a \to C a^{2 +} + 2 e^{-}

(Where

I E

= ionization energy)

Join BYJU'S Learning Program

Explore more

Ionization Enthalpy

Standard XII Chemistry

Join BYJU'S Learning Program

IE1 and IE2 of Mg are 178 and 348 kcal respectively. The energy required for the following reaction is: Mg→Mg2++2e− (Where IE = ionization energy)

The correct option is B +526 kcalAccording to given conditions Mg→Mg++e− IE1=178 kcal Mg+→Mg2++e− IE2=348 kcal Therefore, Mg→Mg2++2e− IE3=IE1+IE2 Hence, ionization energy of the given reaction is: IE=178 kcal+348 kcal =526 kcal

$I E_{1}$ and $I E_{2}$ of $M g$ are $178$ and $348 kcal$ respectively. The energy required for the following reaction is:
$M g \to M g^{2 +} + 2 e^{-}$
(Where $I E$ = ionization energy)