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STD 11 - 3. Classification of elements and periodicity in properties — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 3. Classification of elements and periodicity in properties4 Marks
MCQ
$EN$ of element $A$ is $E_1$ and $IP$ is $E_2$ hence $EA$ will be
  • ✓
    $2E_1 -E_2$
  • B
    $E_1 -E_2$
  • C
    $E_1 -2E_2$
  • D
    $(E_1 + E_2)/2$
✓

Answer

Correct option: A.
$2E_1 -E_2$
a
$EN = \frac{{IP + EA}}{2}$

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