The lattice energy of NaCl is -780 , kJ , mol | vedclass

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(B) The enthalpy of solution $(\Delta H_{sol})$ is calculated using the Born-Haber cycle approach,where $\Delta H_{sol} = \Delta H_{lattice} + \Delta

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(B) The enthalpy of solution $(\Delta H_{sol})$ is calculated using the Born-Haber cycle approach,where $\Delta H_{sol} = \Delta H_{lattice} + \Delta H_{hyd(Na^+)} + \Delta H_{hyd(Cl^-)}$.
Step $1$: The lattice energy is given as $-780 \, kJ \, mol^{-1}$ (energy released during formation). The lattice enthalpy $(\Delta H_{lattice})$ is the energy required to break the lattice,which is the negative of the lattice energy: $\Delta H_{lattice} = -(-780 \, kJ \, mol^{-1}) = 780 \, kJ \, mol^{-1}$.
Step $2$: The hydration enthalpies are $\Delta H_{hyd(Na^+)} = -406 \, kJ \, mol^{-1}$ and $\Delta H_{hyd(Cl^-)} = -364 \, kJ \, mol^{-1}$.
Step $3$: Calculate the enthalpy of solution:
$\Delta H_{sol} = 780 \, kJ \, mol^{-1} + (-406 \, kJ \, mol^{-1}) + (-364 \, kJ \, mol^{-1})$
$\Delta H_{sol} = 780 - 770 = 10 \, kJ \, mol^{-1}$.

List-$I$	List-$II$
$A. PCl_5$	$I. \text{Tetrahedral}$
$B. BrF_5$	$II. \text{Square Planar}$
$C. BF_4^-$	$III. \text{Trigonal bipyramidal}$
$D. [Ni(CN)_4]^{2-}$	$IV. \text{Square pyramidal}$

The lattice energy of $NaCl$ is $-780 \, kJ \, mol^{-1}$. The enthalpies of hydration of $Na^{+}_{(g)}$ and $Cl^{-}_{(g)}$ ions are $-406 \, kJ \, mol^{-1}$ and $-364 \, kJ \, mol^{-1}$ respectively. The enthalpy of solution of $NaCl_{(s)}$ is $..... \, kJ \, mol^{-1}$.

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