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Write electron configurations for the following metals. a. \(\mathrm{Ni}\) b. \(\mathrm{Cd}\) c. Zr d. \(\mathrm{Os}\)

Short Answer

Expert verified
The electron configurations for the given metals are: a. Nickel (Ni): \[1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^8\] b. Cadmium (Cd): \[1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^{10}\,4p^6\,5s^2\,4d^{10}\] c. Zirconium (Zr): \[1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^{10}\,4p^6\,5s^2\,4d^2\] d. Osmium (Os): \[1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^{10}\,4p^6\,5s^2\,4d^{10}\,5p^6\,6s^2\,4f^{14}\,5d^6\]

Step by step solution

01

To determine the electron configurations, first note the atomic numbers for each metal: Nickel \(\mathrm{(Ni)}\), Cadmium \(\mathrm{(Cd)}\), Zirconium \(\mathrm{(Zr)}\), Osmium \(\mathrm{(Os)}\). These atomic numbers are as follows: - Nickel: Z = 28 - Cadmium: Z = 48 - Zirconium: Z = 40 - Osmium: Z = 76 #Step 2: Determine Orbitals and Electron Number in Each Orbital#

Use the atomic numbers to fill orbitals in the order of increasing energy: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p. Here's an explanation of what this sequence means: - s orbitals can hold a maximum of 2 electrons - p orbitals can hold a maximum of 6 electrons - d orbitals can hold a maximum of 10 electrons - f orbitals can hold a maximum of 14 electrons Now let's determine the electron configurations for each metal: a. \(\mathrm{Ni}(28)\) electron configuration:
02

Nickel (Ni)

The electron configuration for Ni: \[1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^8\] b. \(\mathrm{Cd}(48)\) electron configuration:
03

Cadmium (Cd)

The electron configuration for Cd: \[1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^{10}\,4p^6\,5s^2\,4d^{10}\] c. \(\mathrm{Zr}(40)\) electron configuration:
04

Zirconium (Zr)

The electron configuration for Zr: \[1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^{10}\,4p^6\,5s^2\,4d^2\] d. \(\mathrm{Os}(76)\) electron configuration:
05

Osmium (Os)

The electron configuration for Os: \[1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^{10}\,4p^6\,5s^2\,4d^{10}\,5p^6\,6s^2\,4f^{14}\,5d^6\]

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Key Concepts

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

Atomic Orbitals
In chemistry, atomic orbitals are regions outside the nucleus where electrons are likely to be found. Each type of orbital shape can contain a certain number of electrons. There are several types of atomic orbitals, with the most common being s, p, d, and f. Each of these orbitals has a different shape and can hold a different number of electrons:
  • s orbitals: These are spherical shaped and can hold up to 2 electrons.
  • p orbitals: These have a dumbbell shape and can hold up to 6 electrons.
  • d orbitals: These can have more complex shapes, such as a four-leaf clover, and can accommodate 10 electrons.
  • f orbitals: These are even more complex in shape and can hold 14 electrons.
Understanding these shapes helps us predict and explain the behavior of electrons in atoms, which in turn affects how atoms bond and react chemically. By filling these orbitals with electrons in the correct order, we can represent the electron configuration of an element.
Electron Filling Order
The electron filling order, also known as the Aufbau principle, helps us determine how electrons are added to orbitals around an atom. Electrons fill the lowest energy orbitals first before moving to higher ones. This order is determined by the arrangement of orbitals in increasing energy levels. The sequence is as follows:
  • 1s, 2s, 2p, 3s, 3p
  • 4s, 3d, 4p, 5s
  • 4d, 5p, 6s, 4f
  • 5d, 6p, 7s
  • 5f, 6d, 7p
For example, the electron configuration of Nickel (Ni) with atomic number 28 would follow the pattern: 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d8. This pattern helps us write correct electronic configurations, allowing better understanding and prediction of chemical behavior.
Transition Metals
Transition metals are elements found in the center block of the periodic table, specifically in groups 3-12. These metals are unique because they often have electrons filling d orbitals. Key Features of Transition Metals:
  • They typically have multiple oxidation states.
  • Transition metals are known for forming colored compounds.
  • They tend to have high melting and boiling points due to the d-electrons contributing to metallic bonding.
  • They often exhibit good conductivity and malleability.
For example, the electron configuration of Zirconium (Zr), a transition metal with atomic number 40, is: 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6, 5s2, 4d2. Understanding transition metals is essential due to their wide application in industries, ranging from construction to electronics.
Atomic Number
The atomic number of an element is a fundamental concept in chemistry, which tells us the number of protons in an atom's nucleus. It equals the number of electrons in a neutral atom, which directly affects that element's chemical properties. Knowing an element's atomic number helps us determine how to fill its electron orbitals. Importance of Atomic Number:
  • Each element's unique atomic number helps define its identity on the periodic table.
  • It dictates how an element reacts with others.
  • The atomic number progresses sequentially as you move across the periodic table.
  • It also determines placement in the periodic table, giving insight into its behavior and characteristics.
Taking Osmium (Os) as an example, with atomic number 76: 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6, 5s2, 4d10, 5p6, 6s2, 4f14, 5d6. When dealing with electron configurations, always start by identifying the atomic number as it provides the roadmap for how electrons fill the available orbitals.

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Most popular questions from this chapter

The following statements discuss some coordination compounds. For each coordination compound, give the complex ion and the counterions, the electron configuration of the transition metal, and the geometry of the complex ion. a. \(\mathrm{CoCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) is a compound used in novelty devices that predict rain. b. During the developing process of black-and-white film, silver bromide is removed from photographic film by the fixer. The major component of the fixer is sodium thiosulfate. The equation for the reaction is: \(\begin{aligned} \mathrm{AgBr}(s)+2 \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}(a q) & \longrightarrow \\\& \quad\quad \mathrm{Na}_{3}\left[\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}\right](a q)+\mathrm{NaBr}(a q) \end{aligned}\) c. In the production of printed circuit boards for the electronics industry, a thin layer of copper is laminated onto an insulating plastic board. Next, a circuit pattern made of a chemically resistant polymer is printed on the board. The unwanted copper is removed by chemical etching, and the protective polymer is finally removed by solvents. One etching reaction is: $$\mathrm{Cu}\left(\mathrm{NH}_{3}\right){ }_{4} \mathrm{Cl}_{2}(a q)+4 \mathrm{NH}_{3}(a q)+\mathrm{Cu}(s) \longrightarrow \\\ \mathrm{Cu}\left(\mathrm{NH}_{3}\right){ }_{4} \mathrm{Cl}(a q)$$

The compound cisplatin, \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\), has been studied extensively as an antitumor agent. The reaction for the synthesis of cisplatin is: \(\mathrm{K}_{2} \mathrm{PtCl}_{4}(a q)+2 \mathrm{NH}_{3}(a q) \longrightarrow \mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}(s)+2 \mathrm{KCl}(a q)\) Write the electron configuration for platinum ion in cisplatin. Most \(d^{8}\) transition metal ions exhibit square planar geometry. With this and the name in mind, draw the structure of cisplatin.

Give formulas for the following complex ions. a. tetrachloroferrate(III) ion b. pentaammineaquaruthenium(III) ion c. tetracarbonyldihydroxochromium(III) ion d. amminetrichloroplatinate(II) ion

A coordination compound of cobalt(III) contains four ammonia molecules, one sulfate ion, and one chloride ion. Addition of aqueous \(\mathrm{BaCl}_{2}\) solution to an aqueous solution of the compound gives no precipitate. Addition of aqueous \(\mathrm{AgNO}_{3}\) to an aqueous solution of the compound produces a white precipitate. Propose a structure for this coordination compound.

A certain first-row transition metal ion forms many different colored solutions. When four coordination compounds of this metal, each having the same coordination number, are dissolved in water, the colors of the solutions are red, yellow, green, and blue. Further experiments reveal that two of the complex ions are paramagnetic with four unpaired electrons and the other two are diamagnetic. What can be deduced from this information about the four coordination compounds?

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