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Standard XII

Chemistry

Ionic Bond

Which is the ...

Question

Which is the correct order of formation of $M g C l_{2}$ by Born-Haber cycle ?
1. Sublimation of $M g$ ; 2. Bond dissociation of $C l_{2}$
3. Ionisation of $M g$ ; 4. Electron gained by $C l$

1, 3, 2, 4

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1.3, 4, 2

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1, 2, 3, 4

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None of these.

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Solution

The correct option is A $1, 3, 2, 4$
Sublimation: $M g (s) \to M g (g)$
Ionisation of : $M g (g) \to M g^{2 +} (g)$
Bond dissociation: $\frac{1}{2} C l_{2} \to C l (g)$
Electron gained: $C l (g) \to C l^{2 -} (g)$
$M g + {C l}_{2} \to M g {C l}_{2}$
Hence, 1, 3,2,4 is the correct order.

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Similar questions

Q. The correct folrmula of Q by using Born-Haber cycle for the preparation of

M g B r_{2} (s)

will be:
S

=

heat of sublimation of Mg(s)
I

=

ionisation energy of Mg(g)
D

=

dissociation energy of Br

_{2}

(g)
-E

=

electron affinity of Br(g)
-U

=

lattice energy of MgBr

_{2}

(s)
Q

=

heat of formation of MgBr

_{2}

(s)

Q. During the formation of

M g C l_{2}

by Born-Haber cycle, which of the following occurs ?

Q. Calculate the enthalpy of formation of

M g F_{2}

from the following 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. 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. (a) Which metal exists in the liquid state at room temperature?
(b) Give the electronic configuration of

M g

and

C l

atoms.
(c) Show the formation of

M g C l_{2}

by the transfer of electrons in between

M g

and

C l

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Which is the correct order of formation of MgCl2 by Born-Haber cycle ?1. Sublimation of Mg ; 2. Bond dissociation of Cl23. Ionisation of Mg ; 4. Electron gained by Cl

The correct option is A 1,3,2,4Sublimation: Mg(s)→Mg(g)Ionisation of : Mg(g)→Mg2+(g)Bond dissociation: 12Cl2→Cl(g)Electron gained: Cl(g)→Cl2−(g)Mg+Cl2→MgCl2Hence, 1, 3,2,4 is the correct order.

Which is the correct order of formation of $M g C l_{2}$ by Born-Haber cycle ?
1. Sublimation of $M g$ ; 2. Bond dissociation of $C l_{2}$
3. Ionisation of $M g$ ; 4. Electron gained by $C l$

The correct option is A $1, 3, 2, 4$
Sublimation: $M g (s) \to M g (g)$
Ionisation of : $M g (g) \to M g^{2 +} (g)$
Bond dissociation: $\frac{1}{2} C l_{2} \to C l (g)$
Electron gained: $C l (g) \to C l^{2 -} (g)$
$M g + {C l}_{2} \to M g {C l}_{2}$
Hence, 1, 3,2,4 is the correct order.