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Standard XII

Chemistry

Covalent Bond vs Ionic Bond

Calculate the...

Question

Calculate the electron affinity of oxygen atom in gaining two electrons to give the oxide ion $O^{- 2}$ from the following data.
Standard heat of formation of $M g O_{(s)} = - 600 k J m o l^{- 1}$
Lattice energy of $M g O_{(s)} = - 3860 k J m o l^{- 1}$
lonization energy of $M g_{(g)}$ to give $M g^{+ 2}$ is +2170 $k J m o l^{- 1}$
Dissociation energy of $O_{2 (g)}$ is +494 $k J m o l^{- 1}$
Heat of sublimation of $M g_{(s)}$ is +150 $k J m o l^{- 1}$

- 493 k J m o l^{- 1}

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+ 493 k J m o l^{- 1}

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- 693 k J m o l^{- 1}

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+ 693 k J m o l^{- 1}

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Solution

The correct option is D $+ 693 k J m o l^{- 1}$
Given, $Δ_{f} H_{M g O} = - 600 k J m o l^{- 1}$
$M g_{(s)} \to M g_{(g)} Δ_{s} H = + 150 k J m o l^{- 1}$
$M g_{(g)} \to M g_{(g)}^{2 +} I . E . = + 2170 k J m o l^{- 1}$
$\frac{1}{2} O_{2_{(} g)} \to O_{(g)} B . E . = + 494 k J m o l^{- 1}$ $Lattice energy of M g O_{(s)} = - 3860 k J m o l^{- 1}$
$Δ_{f} H_{M g O} = Δ_{s} H + I . E . + \frac{1}{2} B . E . + E . A . + Lattice energy$
$- 600 = 150 + 2170 + 247 + E . A . - 3860$
$E . A . = - 600 - 150 - 2170 - 247 + 3860$
$∴ E A = + 693 k J m o l^{- 1}$

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Similar questions

Q. Calculate the electron affinity of oxygen atom in gaining two electrons to give the oxide ion

O^{2 -}

from the following data.
Standard heat of formation of

M g O_{(s)} = - 600 k J m o l^{- 1}

Lattice energy of

M g O_{(s)} = - 3860 k J m o l^{- 1}

lonization energy of

M g_{(g)}

to give

M g^{2 +}

= +2170

k J m o l^{- 1}

Dissociation energy of

O_{2 (g)}

= +494

k J m o l^{- 1}

Heat of sublimation of

M g_{(s)}

= +150

k J m o l^{- 1}

Q. Use the following data to calculate second electron affinity of oxygen for the process

O^{-} (g) + e^{-} (g) \to O^{2 -} (g)

.
Heat of sublimation of

M g (s) = 147.7 k J / m o l

Ionisation energy of

M g (g)

to form

M g^{2 +} (g) = 2189 k J / m o l

Bond dissociation energy for

O_{2}

= 498.4 k J / m o l

First electron affinity of

O (g) = - 141 k J / m o l

Heat formation of

M g O (s) = - 601.7 k J / m o l

lattice energy of

M g O = - 3791 k J / m o l

Q. Use the following data to calculate second electron affinity of oxygen for the process

O^{-} (g) + e^{-} (g) \to O^{2 -} (g)

.
Heat of sublimation of

M g (s) = 147.7 kJ/mol

Ionisation energy of

M g (g)

to form

M g^{2 +} (g) = 2189 kJ/mol

Bond dissociation energy for

O_{2}

= 498.4 kJ/mol

First electron affinity of

O (g) = - 141 kJ/mol

Heat formation of

M g O (s) = - 601.7 kJ/mol

lattice energy of

M g O = - 3791 kJ/mol

Q. Calculate the enthalpy of formation of

M g F_{2}

from thefollowing data-
Enthalpy of sublimation for

M g (s) \to M g (g) = 146.4 k J m o l^{- 1}

Enthalpy of dissociation of

F_{2} (g) \to 2 F (g) = 155.8 k J m o l^{- 1}

Ionisation energy of

M g (g) \to M g^{2 +} (g) = 2186 k J m o l^{- 1}

Electron gain enthalpy of

2 F \to 2 F^{- 1} = 2 \times - 322.6 = - 645.2 k J m o l^{- 1}

Lattice enthalpy of

M g F_{2} = - 2922.5 k J m o l^{- 1}

Q. Use the following data to calculate the lattice energy of caesium oxide.
Enthalpy of formation of caesium oxide =

- 233 k J m o l^{- 1}

Enthalpy of sublimation of

C s = + 78 k J m o l^{- 1}

First ionization energy of

C s = + 375 k J m o l^{- 1}

Enthalpy of dissociation of

O_{2} (g) = 494 k J m o l^{- 1} o f O_{2} m o l e c u l e s

First electron affinity of

O = - 141 k J m o l^{- 1} o f O a t o m s

Second electron affinity of

O = + 845 k J m o l^{- 1} o f O^{1 -} i o n s

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Covalent Bond vs Ionic Bond

Standard XII Chemistry

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