From the given data regarding enthalpy changes in joule, find CORRECT statement(s) :-
Ca(s) $40$ Ca(g) $20$ Ca⁺(g) $25$ Ca²⁺(g) $H_{2} O - 20$ Ca²⁺(aq)
$\frac{1}{2}$ Cl₂(g) $10$ Cl(g) $\displaylines - 15$ Cl⁻(g) $H_{2} O - 5$ Cl⁻(aq)

Question

From the given data regarding enthalpy changes in joule, find CORRECT statement(s) :-

Ca(s) $40$ Ca(g) $20$ Ca⁺(g) $25$ Ca²⁺(g) $H_{2} O - 20$ Ca²⁺(aq)

$\frac{1}{2}$ Cl₂(g) $10$ Cl(g) $\displaylines - 15$ Cl⁻(g) $H_{2} O - 5$ Cl⁻(aq)

TestHub · Accepted Answer

(A) Lattice energy of CaCl₂ = –90 J/mol (Wrong)

Lattice energy ( $U$ ) formula:

$Δ H_{so l u t i o n} = Δ H_{s u b l ima t i o n} + I E_{1} + I E_{2} + \frac{1}{2} D (C l_{2}) + EGE (Cl) - Hy d r a t i o n e n er g i es$

Sum of given enthalpies:

$40 + 20 + 25 + 10 - 15 = 80 J (b e f ore h y d r a t i o n)$

Hydration energies subtract: –20 (Ca²⁺) –5 (Cl⁻) = –25

Net: $80 - 25 = 55 J$

Lattice energy is not –90 J

(B) |Hydration energy of Ca²⁺| > |Hydration energy of Cl⁻|

Ca²⁺ = –20 J (magnitude 20)

Cl⁻ = –5 J (magnitude 5)

True

(C) Bond dissociation energy of Cl₂ = 20 J/mol

½ Cl₂ → Cl = 10 J

So Cl₂ → 2Cl = 20 J

True

(D) IE₁(Ca) + EGE₁(Cl) = +5 J/mol

IE₁(Ca) = 20 J

EGE₁(Cl) = –15 J

Sum = $20 - 15 = + 5 J$

True

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