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The correct order of bond angles (smallest first ) in H_{2}S,NH_{3},BF_{3},and \: SiH_{4} is

  • Option 1)

    H_{2}S< SiH_{4}< NH_{3}< BF_{3}

  • Option 2)

    NH_{3}<H_{2}S< SiH_{4}< BF_{3}

  • Option 3)

    H_{2}S< NH_{3}< SiH_{4}< BF_{3}

  • Option 4)

    H_{2}S< NH_{3}< BF_{3}< SiH_{4}

 

Answers (1)

best_answer

As we learnt in 

The correct order of bond angle ( smallest first )is

H_{2}S< NH_{3}< SiH_{4}< BF_{3}

                    92.6^{\circ}< 107^{\circ}< 109^{\circ}{28}'< 120^{\circ}

As we discussed in the concept

VSEPR Theory -

1.  The shape of the molecule is determined by repulsions between all of the electron pair present in valence shell.

2.  Order of repulsion 

             lone \:pair-Lone\: pair> Lone\:pair-Bond\:pair> Bond\:pair-bond\:pair

3  Repulsion among the bond pair is directly proportional to the bond  order and electronegativity difference between the central atom and the other atom.

-

 

 Molecule                      Bond angle 

H2S                                92.5^{o}

NH3                                      107^{o}

SiH4                                                 109.28'

BF_{3}                                120^{o}

Bond angle order = H2S < NH3 < SiH4 < BF3


Option 1)

H_{2}S< SiH_{4}< NH_{3}< BF_{3}

The option is incorrect 

Option 2)

NH_{3}<H_{2}S< SiH_{4}< BF_{3}

The option is incorrect 

Option 3)

H_{2}S< NH_{3}< SiH_{4}< BF_{3}

The option is correct 

Option 4)

H_{2}S< NH_{3}< BF_{3}< SiH_{4}

The option is incorrect 

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Plabita

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