According to VSEPR theory, what are the bond angles in NH₃, H₂O and SF₆ respectively?

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$107^{â} 48^{â²}; 104^{â} 27^{â²}; 90^{â}$

$109^{â} 28^{â²}; 107^{â} 48^{â²}; 90^{â}$

$104^{0} 27^{â²} : 109^{â} 28^{â²}; 84^{0} 30^{â²}$

$104^{â} 27^{â²}; 107^{â} 48^{â²}; 84^{â} 30^{â²}$

answer is D.

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Bond Angles in NH₃, H₂O, and SF₆:

Ammonia (NH₃):

Molecular geometry: Trigonal pyramidal
Electron pair arrangement: Tetrahedral (3 bond pairs and 1 lone pair on nitrogen)
The lone pair-bond pair repulsion reduces the ideal bond angle of 109°28′ (tetrahedral angle) to approximately 107°48′.

Water (H₂O):

Molecular geometry: Bent or V-shaped
Electron pair arrangement: Tetrahedral (2 bond pairs and 2 lone pairs on oxygen)
The lone pair-lone pair repulsion further reduces the bond angle to approximately 104°27′.

Sulfur Hexafluoride (SF₆):

Molecular geometry: Octahedral
Electron pair arrangement: Octahedral (6 bond pairs and no lone pairs on sulfur)
All bond angles in an octahedral geometry are 90°, but slight distortions reduce the bond angle to approximately 84°30′.

Answer:

The bond angles in NH₃, H₂O, and SF₆ are: