If the ionization enthalpy and electron gain enthalpy of an element are 275 and86 kcal ${m o l}^{- 1}$ respectively, then the electro -negativity of the element on the Pauling scale is:

2.8

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2.6

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Solution

The correct option is D 2.8
$| X_{A} - X_{B} | = 0 \cdot 102 \times {Δ E}^{\frac{1}{2}} = 0 \cdot 102 \times \sqrt{{(275 - 86) \times 4 \cdot 18} k J / m o l} = 0 \cdot 102 \times 28 \cdot 1069 = 2 \cdot 86 \approx 2 \cdot 8$

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If the ionization enthalpy and electron gain enthalpy of an element are 275 and86 kcal mol−1 respectively, then the electro -negativity of the element on the Pauling scale is:

The correct option is D 2.8|XA−XB|=0⋅102×ΔE12=0⋅102×√{(275−86)×4⋅18}kJ/mol=0⋅102×28⋅1069=2⋅86≈2⋅8

If the ionization enthalpy and electron gain enthalpy of an element are 275 and86 kcal ${m o l}^{- 1}$ respectively, then the electro -negativity of the element on the Pauling scale is:

The correct option is D 2.8
$| X_{A} - X_{B} | = 0 \cdot 102 \times {Δ E}^{\frac{1}{2}} = 0 \cdot 102 \times \sqrt{{(275 - 86) \times 4 \cdot 18} k J / m o l} = 0 \cdot 102 \times 28 \cdot 1069 = 2 \cdot 86 \approx 2 \cdot 8$