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Hydrides of Nitrogen

Bond dissocia...

Question

Bond dissociation enthalpy of EH (E = element) bonds is given below. Which of the compounds will act as strongest reducing agent?
Compound $N H_{3}$ $P H_{3}$ $A s H_{3}$ $S b H_{3}$
$Δ_{d i s s}$ $(E - H) / k J m o l^{- 1}$ 389 322 297 255

N H_{3}

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P H_{3}

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A s H_{3}

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S b H_{3}

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Solution

The correct option is C $S b H_{3}$
As the bond dissociation energy decreases reducing nature of the substance increases because the bond can be broken down easily and can donate electron easily and make the other substance to be reduced.
So the bond dissociation energy is inversely proportional to reducing nature.
The order of bond dissociation energy of given molecules is $N H_{3} > P H_{3} > A s H_{3} > S b H_{3}$ due to increase in size of the central atom.
So the strongest reducing agent is $S b H_{3}$ .
Hence option D is correct.

Suggest Corrections

Similar questions

1. S b H_{3} > N H_{3} > A s H_{3} > P H_{3}

2. N H_{3} > P H_{3} > A s H_{3} > S b H_{3}

3. N H_{3} > P H_{3} > A s H_{3} > S b H_{3}

4. N H_{3} > P H_{3} > A s H_{3} > S b H_{3}

1. S b H_{3} > N H_{3} > A s H_{3} > P H_{3}

2. N H_{3} > P H_{3} > A s H_{3} > S b H_{3}

3. N H_{3} > P H_{3} > A s H_{3} > S b H_{3}

4. N H_{3} > P H_{3} > A s H_{3} > S b H_{3}

E - H

E =

\begin{matrix} C o m p o u n d & = & N H_{3} & P H_{3} & A s H_{3} & S b H_{3} Δ_{d i s s {(E - H)}^{H} (k J {m o l}^{- 1})} & = & 389 & 322 & 297 & 255 \end{matrix}

{N H}_{3}, {P H}_{3}, A s H_{3}

S b H_{3}

N H_{3}, P H_{3}, A s H_{3}, S b H_{3}

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Compound	$N H_{3}$	$P H_{3}$	$A s H_{3}$	$S b H_{3}$
$Δ_{d i s s}$ $(E - H) / k J m o l^{- 1}$	389	322	297	255