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Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
ISBN: 9781305079243

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BuyFindarrow_forward

Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
ISBN: 9781305079243
Textbook Problem

Write electron configurations for the following ions.

a. Ni2+

b. Cd2+

c. Zr3+ and Zr4+

d. Os2+ and Os3+

Interpretation Introduction

Interpretation: The electron configuration of each given ion is to be stated.

Concept introduction: Valence electrons are filled in four principal orbitals according to the energy levels. The four orbitals are s,p,d and f .

The metals from group 3-12 are known as first row transition metals. The general form of valence electron configuration of these metals is,

[Nobelgas]nfxndynsz

When an element loses an electron, a positive charge is accumulated on it, whereas when an element gains an electron, negative charge is accumulated on it.

Explanation

Explanation

(a)

To write: The electron configuration of Ni2+ .

Valence electrons are filled in four principal orbitals according to the energy levels. The four orbitals are s,p,d and f .

The metals from group 3-12 are known as first row transition metals. The general form of valence electron configuration of these metals is,

[Nobelgas]nfxndynsz

Where,

  • n is the shell of principal orbital.
  • x is the number of electrons in f orbital.
  • y is the number of electrons in d orbital.
  • z is the number of electrons in s orbital.

Nickel is a group 10 element. The atomic number of Nickel (Ni) is 28 and its electronic configuration is,

[Ar]3d84s2

It loses two electrons to acquire two positive charges and becomes Ni2+ The electronic configuration of Ni2+ is,

Ni2+=Ni2e

Substitute the electron configuration of (Ni) in the above equation.

Ni2+=Ni2e=[Ar]3d84s22e=[Ar]3d8

(b)

To write: The electron configuration of Zr3+ and Zr4+

Zirconium is a group 4 element. The atomic number of zirconium (Zr) is 40 and its electronic configuration is,

[Kr]4d25s2

It loses three electrons to acquire three positive charges and becomes Zr3+ . The electronic configuration of Zr3+ is,

Zr3+=Zr3e

Substitute the electron configuration of (Zr) in the above equation.

Zr3+=Zr3e=[Kr]4d25s23e=[Kr]4d1

Zirconium loses four electrons to acquire four positive charges and becomes Zr4+ . The electronic configuration of Zr4+ is,

Zr4+=Zr4e

Substitute the electron configuration of (Zr) in the above equation.

Zr4+=Zr4e=[Kr]4d25s24e=[Kr]4d0

(c)

To write: The electron configuration of Cd2+

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