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Chapter 21: Problem 25

Write electron configurations for the following transition metals and their ions. a. \(\mathrm{Co}, \mathrm{Co}^{2+}, \mathrm{Co}^{3+}\) b. \(\mathrm{Pt}, \mathrm{Pt}^{2+}, \mathrm{Pt}^{4+}\) c. \(\mathrm{Fe}, \mathrm{Fe}^{2+}, \mathrm{Fe}^{3+}\)

Short Answer

Expert verified
a. Co: \(\mathrm{1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^7}\); Co²⁺: \(\mathrm{1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^7}\); Co³⁺: \(\mathrm{1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^6}\) b. Pt: \(\mathrm{1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^{10}, 4p^6, 5s^2, 4d^{10}, 5p^6, 6s^2, 4f^{14}, 5d^{9}}\); Pt²⁺: \(\mathrm{1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^{10}, 4p^6, 5s^2, 4d^{10}, 5p^6, 4f^{14}, 5d^{8}}\); Pt⁴⁺: \(\mathrm{1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^{10}, 4p^6, 5s^2, 4d^{10}, 5p^6, 4f^{14}, 5d^{6}}\) c. Fe: \(\mathrm{1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^6}\); Fe²⁺: \(\mathrm{1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^6}\); Fe³⁺: \(\mathrm{1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^5}\)

Step by step solution

01

Neutral Cobalt (Co) electron configuration

To write the electron configuration for the neutral Cobalt atom, we first note its atomic number (27). Using the periodic table, we determine the filling order of orbitals. The electron configuration can be written as: \[ \mathrm{1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^7} \]
02

Cobalt(II) (Co²⁺) electron configuration

To write the electron configuration for Co²⁺, we need to remove two electrons from the neutral Co atom configuration. Electrons are removed first from the 4s orbital. So, Co²⁺ configuration is: \[ \mathrm{1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^7} \]
03

Cobalt(III) (Co³⁺) electron configuration

To write the electron configuration for Co³⁺, we need to remove three electrons from the neutral Co atom configuration. After removing 2 electrons from the 4s orbital, we remove the third electron from the 3d orbital. So, Co³⁺ configuration is: \[ \mathrm{1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^6} \] b. Platinum (Pt), Platinum(II) (Pt²⁺), Platinum(IV) (Pt⁴⁺)
04

Neutral Platinum (Pt) electron configuration

To write the electron configuration for the neutral Platinum atom, we first note its atomic number (78). Using the periodic table, we determine the filling order of orbitals. The electron configuration can be written as: \[ \mathrm{1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^{10}, 4p^6, 5s^2, 4d^{10}, 5p^6, 6s^2, 4f^{14}, 5d^{9}} \]
05

Platinum(II) (Pt²⁺) electron configuration

To write the electron configuration for Pt²⁺, we need to remove two electrons from the neutral Pt atom configuration. Electrons are removed first from the 6s and then from 5d orbital. So, Pt²⁺ configuration is: \[ \mathrm{1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^{10}, 4p^6, 5s^2, 4d^{10}, 5p^6, 4f^{14}, 5d^{8}} \]
06

Platinum(IV) (Pt⁴⁺) electron configuration

To write the electron configuration for Pt⁴⁺, we need to remove four electrons from the neutral Pt atom configuration. After removing 2 electrons from 6s and 5d orbitals, we remove another 2 electrons from the 5d orbital. So, Pt⁴⁺ configuration is: \[ \mathrm{1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^{10}, 4p^6, 5s^2, 4d^{10}, 5p^6, 4f^{14}, 5d^{6}} \] c. Iron (Fe), Iron(II) (Fe²⁺), Iron(III) (Fe³⁺)
07

Neutral Iron (Fe) electron configuration

To write the electron configuration for the neutral Iron atom, we first note its atomic number (26). Using the periodic table, we determine the filling order of orbitals. The electron configuration can be written as: \[ \mathrm{1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^6} \]
08

Iron(II) (Fe²⁺) electron configuration

To write the electron configuration for Fe²⁺, we need to remove two electrons from the neutral Fe atom configuration. Electrons are removed first from the 4s orbital. So, Fe²⁺ configuration is: \[ \mathrm{1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^6} \]
09

Iron(III) (Fe³⁺) electron configuration

To write the electron configuration for Fe³⁺, we need to remove three electrons from the neutral Fe atom configuration. After removing 2 electrons from the 4s orbital, we remove the third electron from the 3d orbital. So, Fe³⁺ configuration is: \[ \mathrm{1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^5} \]

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Most popular questions from this chapter

Ethylenediaminetetraacetate (EDTA \(^{4-} )\) is used as a complexing agent in chemical analysis with the structure shown in Fig. \(21.7 .\) Solutions of EDTA \(^{4-}\) are used to treat heavy metal poisoning by removing the heavy metal in the form of a soluble complex ion. The complex ion virtually prevents the heavy metal ions from reacting with biochemical systems. The reaction of EDTA \(^{4-}\) with \(\mathrm{Pb}^{2+}\) is $$\mathrm{Pb}^{2+}(a q)+\mathrm{EDTA}^{4-(a q)} \rightleftharpoons \mathrm{PbEDTA}^{2-}(a q) \\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad K=1.1 \times 10^{18}$$ Consider a solution with 0.010 mol of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) added to 1.0 \(\mathrm{L}\) of an aqueous solution buffered at \(\mathrm{pH}=13.00\) and containing 0.050$M \mathrm{Na}_{4} \mathrm{EDTA} .\( Does \)\mathrm{Pb}(\mathrm{OH})_{2}$ precipitate from this solution? $\left[K_{\mathrm{sp}} \text { for } \mathrm{Pb}(\mathrm{OH})_{2}=1.2 \times 10^{-15}\right]$

Qualitatively draw the crystal field splitting for a trigonal bi- pyramidal complex ion. (Let the \(z\) axis be perpendicular to the trigonal plane.)

Consider the pseudo-octahedral complex ion of \(\mathrm{Cr}^{3+}\) , where A and \(\mathrm{B}\) represent ligands. Ligand A produces a stronger crystal field than ligand B. Draw an appropriate crystal field diagram for this complex ion (assume the A ligands are on the \(z\) -axis).

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 thiosul-fate. The equation for the reaction is: $$\operatorname{AgBr}(s)+2 \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(a q) \longrightarrow \\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \mathrm{Na}_{3}\left[\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}\right](a q)+\mathrm{NaBr}(a q)$$ 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: $$\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}(a q)+4 \mathrm{NH}_{3}(a q)+\mathrm{Cu}(s) \longrightarrow \\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad 2 \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}(a q)$$ Assume these copper complex ions have tetrahedral geometry.

Write electron configurations for the following transition metal ions. a. \(\mathrm{Sc}^{3+}\) b. \(\mathrm{Ru}^{2+}, \mathrm{Ru}^{3+}\) c. \(\operatorname{Ir}^{+}, \operatorname{Ir}^{3+}\) d. \(M n^{2+}\)
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