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Give the electron configurations of \(\mathrm{Ni}^{2+}\) and \(\mathrm{Ni}^{3+}\).

Short Answer

Expert verified
\(\mathrm{Ni}^{2+}\): \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^8\); \(\mathrm{Ni}^{3+}\): \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^7\).

Step by step solution

01

Determine the Electron Configuration of Neutral Nickel (Ni)

First, determine the electron configuration of a neutral nickel (Ni) atom. Nickel has an atomic number of 28, which means it has 28 electrons in its neutral state. The electron configuration is obtained by filling electrons in the orbitals according to the Aufbau principle:1s虏 2s虏 2p鈦 3s虏 3p鈦 4s虏 3d鈦.Thus, the electron configuration for neutral nickel is:\[ \text{Ni: } 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^8 \]
02

Remove Electrons to Form \\mathrm{Ni}^{2+}

For a \mathrm{Ni}^{2+} ion, 2 electrons need to be removed from the neutral nickel atom. Electrons are typically removed from the highest energy level first (n=4 in this case), specifically from the 4s orbital.The original configuration is:\[ 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^8 \]After removing two electrons from the 4s orbital, the configuration for \mathrm{Ni}^{2+} becomes:\[ \mathrm{Ni}^{2+}: 1s^2 2s^2 2p^6 3s^2 3p^6 3d^{8} \]
03

Remove Electrons to Form \\mathrm{Ni}^{3+}

For a \mathrm{Ni}^{3+} ion, 3 electrons need to be removed from the neutral nickel atom. Following the pattern, remove the 2 electrons from the 4s orbital first, and then one electron from the 3d orbital.Starting with the 4s and 3d electrons of neutral nickel:\[ 4s^2 3d^8 \]Removing two from 4s and one from 3d, the configuration for \mathrm{Ni}^{3+} is:\[ \mathrm{Ni}^{3+}: 1s^2 2s^2 2p^6 3s^2 3p^6 3d^{7} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Aufbau Principle
The Aufbau Principle is a fundamental guideline in understanding how electrons are configured within an atom. The word "aufbau" comes from German, meaning "building up." This principle helps us determine the sequence in which different atomic orbitals are filled with electrons. It is crucial for predicting how electrons are distributed among the orbitals of an atom in its ground state. According to the Aufbau Principle:
  • Electrons occupy the lowest energy orbitals first before moving to higher ones.
  • The order of filling is based on their energy levels, starting from 1s, 2s, 2p, 3s, 3p, and so on.
  • Hund's Rule and the Pauli Exclusion Principle complement the Aufbau Principle to ensure the most stable electron configuration is achieved.
For example, in the case of Nickel (Ni), the atomic number is 28, meaning there are 28 electrons to place in orbitals. Following the Aufbau Principle, these electrons are arranged in the following order: 1s虏, 2s虏, 2p鈦, 3s虏, 3p鈦, 4s虏, then finally 3d鈦. This helps achieve the most stable configuration for the nickel atom.
Nickel Ion
When discussing ions such as \mathrm{Ni^{2+}}\ and \mathrm{Ni^{3+}}, it becomes important to consider how electrons are removed according to their energy levels. Neutral nickel (Ni) has an electron configuration of \[ 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^8 \]. However, when it forms ions, electrons are removed in a specific order.To form the \(\mathrm{Ni^{2+}}\) ion:
  • Two electrons are taken from the highest energy level that nickel's electrons occupy, which is the 4s orbital.
  • This results in the electron configuration: \( 1s^2 2s^2 2p^6 3s^2 3p^6 3d^8 \).
For the \(\mathrm{Ni^{3+}}\) ion:
  • The process involves removing two electrons from the 4s orbital first and then one electron from the 3d orbital.
  • Thus, the configuration becomes: \( 1s^2 2s^2 2p^6 3s^2 3p^6 3d^7 \).
This method of removing electrons from the outer subshells allows nickel to mediate varying oxidation states, a key feature in many of its chemical reactions.
Orbitals
Orbitals are regions around the nucleus of an atom where electrons are likely to be found. Each orbital can hold up to two electrons with opposite spins, and these orbitals are designated as s, p, d, and f based on their shape and the number of electrons they can accommodate. To understand nickel's electron configuration, consider the following orbital types:
  • s orbitals: These spherical orbitals can hold up to two electrons. In the case of nickel, both the 1s and 2s orbitals are fully filled with two electrons each.
  • p orbitals: Shaped like dumbbells, each set of p orbitals (px, py, pz) can hold six electrons. For nickel, the 2p and 3p orbitals are filled with six electrons.
  • d orbitals: More complex in shape, these orbitals can hold up to ten electrons. Nickel's 3d orbital gains eight of such electrons, underlining its place in the d-block of the periodic table.
Orbitals not only dictate how electrons are arranged but also influence the chemical and physical properties of an element. The distribution and arrangement of electrons within these orbitals explain many intrinsic characteristics of elements, such as reactivity and ionization energy.

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Most popular questions from this chapter

Write Lewis formulas for the following molecules: a. \(\mathrm{CO}\) b. \(\mathrm{BrCN}\) c. \(\mathrm{N}_{2} \mathrm{~F}_{2}\)

Assuming that the atoms form the normal number of covalent bonds, give the molecular formula of the simplest compound of arsenic and bromine atoms.

Which has the larger radius, \(\mathrm{N}^{3-}\) or \(\mathrm{P}^{3-} ?\) Explain. (You may use a periodic table.)

For each of the following, use formal charges to choose the Lewis formula that gives the best description of the electron distribution: A. \(\mathrm{ClO}_{2} \mathrm{~F}\) b. \(\mathrm{SO}_{2}\) c. \(\mathrm{ClO}_{3}^{-}\)

When atoms of the hypothetical element \(\mathrm{X}\) are placed together, they rapidly undergo reaction to form the \(\mathrm{X}_{2}\) molecule: \(\mathrm{X}(g)+\mathrm{X}(g) \longrightarrow \mathrm{X}_{2}(g)\) A. Would you predict that this reaction is exothermic or endothermic? Explain. Is the bond enthalpy of \(\mathrm{X}_{2}\) a positive or a negative B. Is the bond enthalpy of \(\mathrm{X}_{2}\) a positive or a negative quantity? Why? C. Suppose \(\Delta H\) for the reaction is \(-500 \mathrm{~kJ} / \mathrm{mol}\). Estimate the bond enthalpy of the \(\mathrm{X}_{2}\) molecule. D. Another hypothetical molecular compound, \(\mathrm{Y}_{2}(g)\) has a bond enthalpy of \(750 \mathrm{~kJ} / \mathrm{mol}\), and the molecular compound \(\mathrm{XY}(g)\) has a bond enthalpy of \(1500 \mathrm{~kJ} / \mathrm{mol}\). Using bond enthalpy information, calculate \(\Delta H\) for the following reaction. $$\mathrm{X}_{2}(g)+\mathrm{Y}_{2}(g) \longrightarrow 2 \mathrm{XY}(g)$$ E, Given the following information, as well as the information previously presented, predict whether or not the hypothetical ionic compound \(\mathrm{AX}\) is likely to form. In this compound, A forms the \(\mathrm{A}^{+}\) cation, and \(\mathrm{X}\) forms the \(\mathrm{X}^{-}\) anion. Be sure to justify your answer. Reaction: \(\mathrm{A}(g)+\frac{1}{2} \mathrm{X}_{2}(g) \longrightarrow \mathrm{AX}(s)\) The first ionization energy of \(\mathrm{A}(g)\) is \(400 \mathrm{~kJ} / \mathrm{mol}\). The electron affinity of \(\mathrm{X}(g)\) is \(525 \mathrm{~kJ} / \mathrm{mol}\). The lattice energy of \(\mathrm{AX}(s)\) is \(100 \mathrm{~kJ} / \mathrm{mol}\). F. If you predicted that no ionic compound would form from the reaction in Part e, what minimum amount of \(\mathrm{AX}(s)\) lattice energy might lead to compound formation?

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