Write electron configurations for the following ions. a. \(\mathrm{Ni}^{2+}\) b. \(\mathrm{Cd}^{2+}\) c. \(\mathrm{Zr}^{3+}\) and \(\mathrm{Zr}^{4+}\) d. \(\mathrm{Os}^{2+}\) and \(\mathrm{Os}^{3+}\)

Short Answer

Expert verified
The electron configurations of the ions are: a. \(\mathrm{Ni}^{2+}\): \(1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^8\) b. \(\mathrm{Cd}^{2+}\): \(1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^{10}\, 4p^6\, 5s^2\, 4d^{10}\) c. \(\mathrm{Zr}^{3+}\): \(1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^{10}\, 4p^6\, 5s^2\, 4d^{7}\) and \(\mathrm{Zr}^{4+}\): \(1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^{10}\, 4p^6\, 5s^2\, 4d^{6}\) d. \(\mathrm{Os}^{2+}\): \(1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^{10}\, 4p^6\, 5s^2\, 4d^{10}\, 5p^6\, 6s^2 \, 4f^{14} \, 5d^{6}\) and \(\mathrm{Os}^{3+}\): \(1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^{10}\, 4p^6\, 5s^2\, 4d^{10}\, 5p^6\, 6s^2 \, 4f^{14} \, 5d^{5}\)

Step by step solution

01

a. Electron Configuration of \(\mathrm{Ni}^{2+}\)

The atomic number of nickel (Ni) is 28. In its neutral state, it has 28 electrons. As \(\mathrm{Ni}^{2+}\), it has lost 2 electrons, leaving it with 26 electrons. The electron configuration for 26 electrons is: \[1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^8\]
02

b. Electron Configuration of \(\mathrm{Cd}^{2+}\)

The atomic number of cadmium (Cd) is 48. In its neutral state, it has 48 electrons. As \(\mathrm{Cd}^{2+}\), it has lost 2 electrons, leaving it with 46 electrons. The electron configuration for 46 electrons is: \[1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^{10}\, 4p^6\, 5s^2\, 4d^{10}\]
03

c. Electron Configuration of \(\mathrm{Zr}^{3+}\) and \(\mathrm{Zr}^{4+}\)

The atomic number of zirconium (Zr) is 40. In its neutral state, it has 40 electrons. For \(\mathrm{Zr}^{3+}\), it has lost 3 electrons, leaving it with 37 electrons. The electron configuration for 37 electrons is: \[1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^{10}\, 4p^6\, 5s^2\, 4d^{7}\] For \(\mathrm{Zr}^{4+}\), it has lost 4 electrons, leaving it with 36 electrons. The electron configuration for 36 electrons is: \[1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^{10}\, 4p^6\, 5s^2\, 4d^{6}\]
04

d. Electron Configuration of \(\mathrm{Os}^{2+}\) and \(\mathrm{Os}^{3+}\)

The atomic number of osmium (Os) is 76. In its neutral state, it has 76 electrons. For \(\mathrm{Os}^{2+}\), it has lost 2 electrons, leaving it with 74 electrons. The electron configuration for 74 electrons is: \[1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^{10}\, 4p^6\, 5s^2\, 4d^{10}\, 5p^6\, 6s^2 \, 4f^{14} \, 5d^{6}\] For \(\mathrm{Os}^{3+}\), it has lost 3 electrons, leaving it with 73 electrons. The electron configuration for 73 electrons is: \[1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^{10}\, 4p^6\, 5s^2\, 4d^{10}\, 5p^6\, 6s^2 \, 4f^{14} \, 5d^{5}\]

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Most popular questions from this chapter

Write electron configurations for each of the following. a. \(\mathrm{Cr}, \mathrm{Cr}^{2+}, \mathrm{Cr}^{3+}\) b. \(\mathrm{Cu}, \mathrm{Cu}^{+}, \mathrm{Cu}^{2+} \quad\) c. \(\mathrm{V}, \mathrm{V}^{2+}, \mathrm{V}^{3+}\)

A compound related to acetylacetone is 1,1,1 -trifluoroacetylacetone (abbreviated Htfa): Htfa forms complexes in a manner similar to acetylacetone. (See Exercise \(47 .\) Both \(\mathrm{Be}^{2+}\) and \(\mathrm{Cu}^{2+}\) form complexes with tfa \(^{-}\) having the formula \(\mathrm{M}\) (tfa) \(_{2}\). Two isomers are formed for each metal complex. a. The \(\mathrm{Be}^{2+}\) complexes are tetrahedral. Draw the two isomers of Be(tfa)_2. What type of isomerism is exhibited by \(\mathrm{Be}(\mathrm{tfa})_{2} ?\) b. The \(\mathrm{Cu}^{2+}\) complexes are square planar. Draw the two isomers of \(\mathrm{Cu}(\mathrm{tfa})_{2} .\) What type of isomerism is exhibited by $\mathrm{Cu}(\mathrm{tfa})_{2} ?

Qualitatively draw the crystal field splitting of the \(d\) orbitals in a trigonal planar complex ion. (Let the \(z\) axis be perpendicular to the plane of the complex.)

Draw geometrical isomers of each of the following complex ions. a. \(\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}^{-}\) b. \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{2}^{2+}\) c. \(\operatorname{Ir}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}_{3}\) d. \(\operatorname{Cr}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{2} \mathrm{I}_{2}^{+}\)

Molybdenum is obtained as a by-product of copper mining or is mined directly (primary deposits are in the Rocky Mountains in Colorado). In both cases it is obtained as \(\mathrm{MoS}_{2},\) which is then converted to \(\mathrm{MoO}_{3}\). The \(\mathrm{MoO}_{3}\) can be used directly in the production of stainless steel for high-speed tools (which accounts for about \(85 \%\) of the molybdenum used). Molybdenum can be purified by dissolving \(\mathrm{MoO}_{3}\) in aqueous ammonia and crystallizing ammonium molybdate. Depending on conditions, either \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Mo}_{2} \mathrm{O}_{7}\) or \(\left(\mathrm{NH}_{4}\right)_{6} \mathrm{Mo}_{7} \mathrm{O}_{24} \cdot 4 \mathrm{H}_{2} \mathrm{O}\) is obtained. a. Give names for \(\mathrm{MoS}_{2}\) and \(\mathrm{MoO}_{3}\) b. What is the oxidation state of Mo in each of the compounds mentioned above?

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