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Central Atom and Ligands

The change in...

Question

The change in enthalpy for hydration of ( i ) the chloride ion ; ( ii ) the iodide ion are :
Given :
* enthalpy change of solution of NaCl(s) $= - 2$ kJ / mol.
* enthalpy change of solution of NaI(s) $= + 2$ kJ / mol.
* enthalpy change of hydration of $N a^{+} (g) = - 390$ kJ / mol.
* lattice enthalpy of NaCl $= - 772$ kJ / mol.
* lattice enthalpy of NaI $= - 699$ kJ / mol.

for

C l^{-} - 741 k J m o l^{- 1},

for

I^{-} - 527 k J m o l^{- 1}

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for

C l^{-} - 384 k J m o l^{- 1},

for

I^{-} - 307 k J m o l^{- 1}

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for

C l^{-} - 454 k J m o l^{- 1},

for

I^{-} 963 k J m o l^{- 1}

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for

C l^{-} - 385 k J m o l^{- 1},

for

I^{-} - 332 k J m o l^{- 1}

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Solution

The correct option is D for $C l^{-} - 384 k J m o l^{- 1},$ for $I^{-} - 307 k J m o l^{- 1}$
$1)$ Enthalpy of hydration of $C l = Δ_{h} H_{C l} = ?$
Given,
$N a C l (s) + a q \to N a^{+} + {C l}^{-}$ $Δ H_{s o l} = - 2 k J / m o l$ ...(i)
$N a^{+} (g) \to N a^{+} (a q)$ $Δ_{L} H_{{N a}^{+}} = 390 k J / m o l$ ...(ii)
${C l}^{-} (g) \to {C l}^{-} (a q)$ $Δ_{h} H_{C l} = ?$ ...(iii)
$N a^{+} (g) + {C l}^{-} (g) \to N a C l (s) + a q$ $Δ_{L} H = - 772 k J / m o l$
or $N a C l (s) + a q \to N a^{+} (g) + {C l}^{-} (g)$ $Δ_{L} H = + 772 k J / m o l$ ...(iv)
So, (i),(ii),(iii),(iv) gives
$Δ H_{s o l} = + 772 - 390 + Δ_{h} H_{C l}$
$\Rightarrow - 2 = 772 - 390 + Δ_{h} H_{C l} \Rightarrow Δ_{h} H_{C l} = - 384 k J / m o l$
$2)$ Enthalpy of hydration of $I = Δ_{h} H_{I} = ?$
Given,
$N a I (s) + a q \to N a^{+} + I^{-}$ $Δ H_{s o l} = + 2 k J / m o l$ ...(i)
$N a^{+} (g) \to N a^{+} (a q)$ $Δ_{L} H_{{N a}^{+}} = 390 k J / m o l$ ...(ii)
$I^{-} (g) \to I^{-} (a q)$ $Δ_{h} H_{I} = ?$ ...(iii)
$N a^{+} (g) + I^{-} (g) \to N a I (s) + a q$ $Δ_{L} H = - 699 k J / m o l$
or $N a I (s) + a q \to N a^{+} (g) + I^{-} (g)$ $Δ_{L} H = + 699 k J / m o l$ ...(iv)
So, (i),(ii),(iii),(iv) gives
$Δ H_{s o l} = + 699 - 390 + Δ_{h} H_{I}$
$\Rightarrow + 2 = 699 - 390 + Δ_{h} H_{I} \Rightarrow Δ_{h} H_{I} = - 307 k J / m o l$

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