IE 1 and IE 2 of Ca are 141 and 274 kcal respectively- The energy required for the following reaction is-Ca - Ca 2-2 e -Where IE - ionization energyA- -415 kcalB- -415 kcalC- -175 kcalD- -175 kcal

The correct option is B +415 kcalAccording to given conditions Ca→ Ca^++e^ IE_1=141 kcal Ca^+→ Ca^2++e^ IE_2=274 kcal Therefore, Ca→ Ca^2++2e^ IE=IE_1 ...

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Byju's Answer

Standard XII

Chemistry

Atomic Radius

IE 1 and IE 2...

Question

$I E_{1}$ and $I E_{2}$ of $C a$ 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)

- 415 kcal

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+ 415 kcal

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- 175 kcal

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+ 175 kcal

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Solution

The correct option is B $+ 415 kcal$
According to given conditions
$C a \to C a^{+} + e^{-} I E_{1} = 141 kcal$
$C a^{+} \to C a^{2 +} + e^{-} I E_{2} = 274 kcal$
Therefore,
$C a \to C a^{2 +} + 2 e^{-} I E = I E_{1} + I E_{2}$
Hence, ionization energy of the given reaction is:
$I E = 141 kcal + 274 kcal$
$= 415 kcal$

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Similar questions

I E_{1}

and

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274 kcal

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and

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are

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and

348 kcal

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

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

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I E_{1}

and

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are

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Atomic Radius

Standard XII Chemistry

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IE1 and IE2 of Ca are 141 and 274 kcal respectively. The energy required for the following reaction is: Ca→Ca2++2e− (Where IE = ionization energy)

The correct option is B +415 kcalAccording to given conditions Ca→Ca++e− IE1=141 kcal Ca+→Ca2++e− IE2=274 kcal Therefore, Ca→Ca2++2e− IE=IE1+IE2 Hence, ionization energy of the given reaction is: IE=141 kcal+274 kcal =415 kcal

$I E_{1}$ and $I E_{2}$ of $C a$ 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)