In which of the following process highest energy is absorbed?

C u \to C u^{+}

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B r \to B r^{-}

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I \to I^{-}

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L i \to L i^{+}

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Solution

The correct option is C $I \to I^{-}$
Copper and lithium are electropositive element hence they readily donate electron to form corresponding cation, hence it requires less ionization energy. In case of bromine and iodine electron gain enthalpy becomes less negative as we go down a group. Larger the size of atom, larger will be there distance between added electron and the nucleus. The attraction between them will be weaker and hence electron gain enthalpy will be less negative.
Therefore, iodine will absorb highest energy to form iodine anion as compared to bromine, copper and lithium.

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Similar questions

$E^{⊖}$ values of some redox couples are given below. On the basis of these values choose the coorect option.
$E^{⊖}$ values :

$\frac{B r_{2}}{B r^{-}} = + 1.90 \frac{A g^{+}}{A g (s)} = + 0.80 \frac{C u^{2 +}}{C u (s)} = + 0.34 \frac{I_{2} (S)}{I^{-}} = + 0.54;$

(a) Cu will reduce $B r^{-}$
(b) Cu will reduce $A g$
(c) Cu will reduce $I^{-}$
(d) Cu will reduce $B r_{2}$

I^{-}

L i^{+}

Divide the following redox reactions into oxidation and reduction half reactions.

(i) $Z n + P b^{2} \to Z n^{2} + P b$

(ii) $Z n + C u^{2 +} \to Z n^{2} + C u$

(iii) $C l_{2} + 2 B r^{-} \to B r_{2} + 2 C l^{-}$

L i^{+}

L i X \cdot 3 H_{2} O [X = C l, B r, I, C l O_{2}, C l o_{4}, M n O_{4}, N O_{3}, B F_{4})

(L i^{+})

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