Arrange $N - H, O - H$ and $F - H$ bonds in the decreasing order of bond energy.:

F - H > O - H > N - H

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O - H > N - H > F - H

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N - H > O - H > F - H

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F - H > N - H > O - H

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Solution

The correct option is A $F - H > O - H > N - H$
The bond energy of a molecule increases with increase in the diffference in the electronegativity difference between the bonding atom,
so, the increasing order of electronegativity is
$H < N < O < F$
so the decreasing order of bond energy as
$F - H > O - H > N - H$

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Question 43

Arrange the following bonds in order of increasing ionic character giving reason.

N - H, F - H, C - H and O - H

Δ {H_{f}}^{o} (N_{2} H_{4}, l) = 50 k J / m o l, Δ {H_{f}}^{o} ({N H}_{3}, g) = - 46 k J / m o l

B . E . (N - H) = 393 k J / m o l, B . E . (H - H) = 436 k J / m o l

Δ_{v a p} H (N_{2} H_{4}, l) = 18 k J / m o l

N - N

N_{2} H_{4}

F, O, N, C l

H

4.0, 3.5, 3.0, 3.2

2.1

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