In each of the following pairs,state which atom/ion has higher ionization enthalpy and more negative electron gain enthalpy?

I, I^{-}

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B r, {B r}^{-}

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L i, {L i}^{-}

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B r, I

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Solution

The correct option is A $I, I^{-}$
$I E = I > I^{-}, E A = I > I^{-}$
$I E = B r > {B r}^{-}, E A = B r > {B r}^{-}$
$I E = {L i}^{-} > L i, E A = L i > {L i}^{-}$
$I E = B r > I, E A = B r > I$
Negatively charged species gives $e^{-}$ easily due to more electron density on it $,$ and in group $I E$ or $E A$ decreases down the group due to less $Z_{e f f} .$

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Similar questions

Among halogens, the correct order of amount of energy released in electron gain (electron gain enthalpy ) is
(a) F > Cl > Br > I (b) F < Cl < Br < I (c) F < Cl > Br > I (d) F < Cl < Br < I

Q. Arrange the following in increasing order of property indicated:
(i) Li, Be and B (ionization energy)
(ii) F, Cl, Br, I (electron affinity)
(iii) SiO₂, P₂O₅, SO₃, Cl₂O₇ (acidic nature)
(iv) I, I⁺, I⁻ (atomic size)

Which of the following pairs of elements would have a more negative electron gain enthalpy?
(i) O or F

(ii) F or Cl

Q. Which of the following options does not give to the correct trends indicated against it?
(I)

F_{2} > C l_{2} > B r_{2} > I_{2}

: Bond dissociation energy
(II)

F_{2} > C l_{2} > B r_{2} > I_{2}

: Oxidizing nature
(III)

C l > F > B r > I

: Electron gain enthalpy
(IV)

F > C l > B r > I

: Electronegativity

Q. In which of the following arrangements, the order is according to the property indicated against it.

(I) I < B r < F < C l increasing electron gain enthalpy

(I I) L i < N a < K < R b increasing metallic radius

(I I I) B < C < N < O increasing first ionization energy

(I V) A l^{3 +} < M g^{2 +} < N a^{+} < F^{-} increasing ionic size

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In each of the following pairs,state which atom/ion has higher ionization enthalpy and more negative electron gain enthalpy?

The correct option is A I,I−IE=I>I−,EA=I>I−IE=Br>Br−,EA=Br>Br−IE=Li−>Li,EA=Li>Li−IE=Br>I,EA=Br>INegatively charged species gives e− easily due to more electron density on it , and in group IE or EA decreases down the group due to less Zeff.