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lf the ionisation energy of Na is 495.8 kJ mol^-1, the electron affinity of Cl is 349.4 kJ mol^-1 and lattice energy is 776 kJ mol^-1, the true statement for the formation of ionic bond
in NaCI is

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a

total energy released per mole of NaCl formed is 628.6 kJ

b

the ionic bond formed is unstable

c

the total energy absorbed per mole of NaCl formed is 1125.4 kJ

d

net energy released in this process is 629.6kJ

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detailed solution

Correct option is D

Na(g)+495.8⟶Na+(g)+e− . . . (i) Cl(g)+e−⟶Cl−(g)+349.4 Na+(g)+Cl−(g)⟶NaCl(s)+776 kJ mol−1Total energy released per mole of NaCl formed = 776 + 349.4 = 1125.4 kJTotal energy absorbed in step (i) = 495.8 kJNet energy released = 629.6 kJThus, a stable ionic bond is formed between NaCl.

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Similar Questions

Two elements A and B have following electronic configuration

$\begin{array}{l} A = 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} \\ B = 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{3} \end{array}$

$It is expected that they form a compound A_{x} B_{y} then sum of x and y is :$ (Ignore charge only observe the value)

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