JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Chemical Bonding and Molecular Structure

Among

H_{2}, He_{2}^{+}, Li_{2}, Be_{2}, B_{2}, C_{2}, N_{2}, O_{2}^{-}

, and

F_{2}

, the number of diamagnetic species is (Atomic numbers:

H = 1, He = 2, Li = 3, Be = 4, B = 5, C = 6, N = 7, O = 8, F = 9

)

Solution

Using molecular orbital theory:

$H_{2}$ : all electrons paired → diamagnetic
$He_{2}^{+}$ : 3 electrons, 1 unpaired → paramagnetic
$Li_{2}$ : all electrons paired → diamagnetic
$Be_{2}$ : all electrons paired (though unstable) → diamagnetic
$B_{2}$ : 2 unpaired electrons in $π$ orbitals → paramagnetic
$C_{2}$ : all electrons paired → diamagnetic
$N_{2}$ : all electrons paired → diamagnetic
$O_{2}^{-}$ : 17 electrons, 1 unpaired → paramagnetic
$F_{2}$ : all electrons paired → diamagnetic

Diamagnetic species:

H_{2}, Li_{2}, Be_{2}, C_{2}, N_{2}, F_{2}

= 6

However,

Be_{2}

does not exist as a stable molecule, so the answer could be 5.

Note: The official answer key states this question was dropped and +3 marks awarded to all candidates due to printing inconsistency.

JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Integer answer

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JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Chemical Bonding and Molecular Structure

Solution

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