0.5 moles of gaseous non-metallic $X^{-}$ anions (having positive electron affinity) requires 806.4 kJ energy to get completely converted into gaseous x+ ions. Calculate Pauling's electronegativity of the element X. Use Avogadro's no, = $6 \times 10^{23}$ and $1 eV =$ $1.6 \times 10^{- 19} J$
[Use the fact that Pauling's electronegativity
$= \frac{Mulliken' s electronegativity}{2.8}$ and
Mulliken's electronegativity
$\frac{Ionisation energy + Electron affinity}{2}$ ]

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

answer is 3.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Let $X ⟶ X^{+} e^{-}$

$∴ IE$ (energy absorbed) $= a eV$ per atom and $X + e^{-} ⟶ X^{-}$

$∴ EA$ (energy released) $= - b eV$ per atom

Now $\frac{N}{2}$ of $X^{-}$ will lose two electrons to give $\frac{N}{2} X^{+}$

$∴ X^{-} ⟶ X + e^{-}$

$∴$ energy absorbed $= + a eV$ per atom.

$X ⟶ X^{+} + e^{-}$

$∴ a \times \frac{N}{2} + b \times \frac{N}{2} = \frac{806.4 \times 1000}{1.6 \times 10^{- 19}} eV$

or $(a + b) = \frac{806.4 \times 1000 \times 2}{1.6 \times 10^{- 19} \times 6 \times 10^{23}} = \frac{806.4 \times 2}{1.6 \times 60} = 16.8$

$χ_{m} = \frac{I E + E A}{2} = \frac{16.8}{2} = 8.4$

$∴$ Mulliken's electronegativity = 8.4

Pauling's electronegativity $= \frac{8.4}{2.8} = 3.0$

Watch 3-min video & get full concept clarity

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test