Chapter 14: Problem 168
The decreasing values of bond angles from \(\mathrm{NH}_{3}\) \(\left(106^{\circ}\right)\) to \(\mathrm{SbH}_{3}\left(101^{\circ}\right)\) down group 15 of the periodic table is due to (a) increasing bp-bp repulsion (b) increasing p orbital character in sp \(^{3}\) (c) decreasing lp-bp repulsion (d) decreasing electronegativity
Short Answer
Step by step solution
Understanding Bond Angles in Group 15 Hydrides
Assessing Electronegativity Down Group 15
Examining Lone Pair-Bond Pair Repulsion
Analyzing Orbital Hybridization
Evaluating Options
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.