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Most Upvoted AnswerHybridisation in Ni(NH3)4 2?**Hybridisation in Ni(NH3)4 2**The concept of hybridization is used to explain the molecular geometry and bonding in molecules and ions. In the case of Ni(NH3)4 2, where Ni represents the central metal ion and NH3 represents the ligand, hybridization plays a crucial role in determining the geometry and bonding.**1. Determining the total number of valence electrons**To understand the hybridization in Ni(NH3)4 2, we first need to determine the total number of valence electrons. Nickel (Ni) is a transition metal, and its valence electrons are distributed as 3d8 4s2. Each ammonia (NH3) ligand contributes one electron to the total count.Ni: 8 electrons (4 from 3d and 4 from 4s)NH3: 3 electrons (1 from N and 1 each from 3 hydrogens)Therefore, the total number of valence electrons in Ni(NH3)4 2 is 8 + (3 x 4) = 20.**2. Determining the central atom hybridization**In Ni(NH3)4 2, the central atom is nickel (Ni). To determine its hybridization, we need to consider the number of regions of electron density surrounding the central atom. This includes both the bonding pairs and lone pairs of electrons.In Ni(NH3)4 2, there are four ammonia ligands attached to the central nickel atom. Each ammonia ligand contributes a lone pair of electrons to the nickel atom. Additionally, the nickel atom has eight valence electrons of its own.Therefore, the total number of regions of electron density around the central atom is 4 (from ammonia ligands) + 8 (from nickel) = 12.**3. Applying the hybridization theory**Based on the number of regions of electron density, the central atom in Ni(NH3)4 2 undergoes sp3 hybridization. This means that the s orbital and three p orbitals of the nickel atom hybridize to form four equivalent sp3 hybrid orbitals.The four sp3 hybrid orbitals of the nickel atom then overlap with the nitrogen orbitals of the ammonia ligands, forming four sigma bonds. The remaining four sp3 hybrid orbitals accommodate the lone pairs of electrons on the nickel atom.**4. Molecular geometry**The hybridization of the central atom in Ni(NH3)4 2 is sp3, which leads to a tetrahedral molecular geometry. The four ammonia ligands are arranged symmetrically around the central nickel atom, forming bond angles of approximately 109.5 degrees.Overall, the hybridization in Ni(NH3)4 2 is sp3, and the molecular geometry is tetrahedral. This understanding helps explain the bonding and geometry in the compound.Answered by Infinity Academy · View profileCommunity AnswerHybridisation in Ni(NH3)4 2?Share your AnswerView all answers

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View all Question Description Hybridisation in Ni(NH3)4 2? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Hybridisation in Ni(NH3)4 2? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Hybridisation in Ni(NH3)4 2?.Solutions for Hybridisation in Ni(NH3)4 2? in English & in Hindi are available as part of our courses for Class 12. Download more important topics, notes, lectures and mock test series for Class 12 Exam by signing up for free.Here you can find the meaning of Hybridisation in Ni(NH3)4 2? defined & explained in the simplest way possible. Besides giving the explanation of Hybridisation in Ni(NH3)4 2?, a detailed solution for Hybridisation in Ni(NH3)4 2? has been provided alongside types of Hybridisation in Ni(NH3)4 2? theory, EduRev gives you an ample number of questions to practice Hybridisation in Ni(NH3)4 2? tests, examples and also practice Class 12 tests.

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