State the number of valence electrons, bonding pairs, and lone pairs in each of the following Lewis structures: (6.6)
c.
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1
Identify the central atom in the Lewis structure, which is phosphorus (P) in this case.
Count the total number of valence electrons for the molecule. Phosphorus (P) has 5 valence electrons, and each hydrogen (H) has 1 valence electron. Since there are three hydrogens, the total is 5 + 3 = 8 valence electrons.
Determine the number of bonding pairs. Each P-H bond represents a bonding pair. There are three P-H bonds, so there are 3 bonding pairs.
Calculate the number of lone pairs on the central atom. After forming 3 bonds, phosphorus will have used 3 of its 5 valence electrons, leaving 2 electrons as a lone pair.
Summarize: The molecule has 8 valence electrons, 3 bonding pairs, and 1 lone pair on the phosphorus atom.
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이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
5m
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주요 개념
질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.
Valence Electrons
Valence electrons are the outermost electrons of an atom and are crucial for determining how an atom can bond with others. In the case of phosphine (PH3), phosphorus has five valence electrons, while each hydrogen atom has one, leading to a total of eight valence electrons in the molecule. Understanding valence electrons helps predict the bonding behavior and molecular structure.
Valence Electrons of Elements (Simplified) Concept 1
Bonding Pairs
Bonding pairs refer to pairs of electrons that are shared between atoms to form covalent bonds. In phosphine, there are three bonding pairs formed between the phosphorus atom and the three hydrogen atoms. Recognizing bonding pairs is essential for drawing Lewis structures and understanding molecular geometry.
Lone pairs are pairs of valence electrons that are not involved in bonding and are localized on a single atom. In the case of phosphine, phosphorus has one lone pair of electrons. Identifying lone pairs is important for predicting the shape of the molecule and its reactivity, as they can influence the overall electron distribution.