KCET · Chemistry · Chemical Bonding and Molecular Structure
Bond angle is \(\mathrm{PH}_{4}^{+}\)is more than that of \(\mathrm{PH}_{3}\). This is because
- A lone pair - bond pair repulsion exists in \(\mathrm{PH}_{3}\)
- B \(\mathrm{PH}_{4}^{+}\)has square planar structure
- C \(\mathrm{PH}_{3}\) has planar trigonal structure
- D hybridisation of \(\mathrm{P}\) changes when \(\mathrm{PH}_{3}\) is converted to \(\mathrm{PH}_{4}^{+}\)
Answer & Solution
Correct Answer
(A) lone pair - bond pair repulsion exists in \(\mathrm{PH}_{3}\)
Step-by-step Solution
Detailed explanation
Bond angle of \(\mathrm{PH}_{4}^{+}\)is more than that of \(\mathrm{PH}_{3}\) as in \(\mathrm{PH}_{3}\) due to greater lone pair-bond pair repulsions than the bond pair-bond pair repulsions, the tetrahedral angle decreases from \(109^{\circ} 28^{\prime}\) to \(93.6^{\circ}\) and has a pyramidal structure. In \(\mathrm{PH}_{4}^{+}\), there are four bond pairs and no lone pair and due to the absence of lone pair-bond pair repulsions and presence of four identical bond pair-bond pair interactions \(\mathrm{PH}_{4}^{+}\)has tetrahedral geometry with bond angle \(109^{\circ}\).
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