If the ionisation potential is IP, electron affinity is EA and electronegativity is x then which of the following relation is correct.

2X-EA-IP=0

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2EA-X-IP=0

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2IP-X-EA=0

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All of the above

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Solution

The correct option is A 2X-EA-IP=0
Electron affinity: The amount of energy released when an electron is added to the outermost shell of one mole of an isolated gaseous atom in its lower energy state.

Ionisation energy: Minimum energy required to remove most loosly held outer most shell e – in ground state from an isolated gaseous atom is known as ionisation potential.

$X = \frac{I P + E A}{2}$

$2 X - I P - E A = 0$

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Similar questions

X

I P = 1681

E A = - 333

(IE = Ionisation energy in eV/atom, EA = Electron affinity in eV/atom, EN = Electronegativity in Mulliken scale)

(IE = Ionisation energy in eV/atom, EA = Electron affinity in eV/atom, EN = Electronegativity in Mulliken scale)

I) | I P_{1} of ion M^{2 +} | > | E A_{1} o f M^{3 +} |

(IP = ionization potential, and EA = electron affinity)

I I) S > S e > T e > O (order of EA)

I I I) L i < B e < B < C (order of electronegativity)

I V) A l^{3 +} < M g^{2 +} < N a^{+} < F^{-} (order of ionic size)

V) L i^{+} > N a^{+} < K^{+} (order of hydrated size)

V I) N a C l > M g C l_{2} > A l C l_{3} (order of lattice energy)

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