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

Short Answer

Expert verified
a. \(\mathrm{Ni}\): 1s虏 2s虏 2p鈦 3s虏 3p鈦 4s虏 3d鈦 b. \(\mathrm{Cd}\): 1s虏 2s虏 2p鈦 3s虏 3p鈦 4s虏 3d鹿鈦 4p鈦 5s虏 4d鹿鈦 c. \(\mathrm{Zr}\): 1s虏 2s虏 2p鈦 3s虏 3p鈦 4s虏 3d鹿鈦 4p鈦 5s虏 4d虏 d. \(\mathrm{Os}\): 1s虏 2s虏 2p鈦 3s虏 3p鈦 4s虏 3d鹿鈦 4p鈦 5s虏 4d鹿鈦 5p鈦 6s虏 4f鹿鈦 5d鈦

Step by step solution

01

Find the atomic number.

Note down the atomic numbers of each element, which are as follows: a. \(\mathrm{Ni}\): Atomic number 28 b. \(\mathrm{Cd}\): Atomic number 48 c. \(\mathrm{Zr}\): Atomic number 40 d. \(\mathrm{Os}\): Atomic number 76
02

Determine the electron configurations.

Follow the Aufbau principle to fill the electron orbitals, and use Hund's rule and the Pauli Exclusion Principle to guide the electron distribution. This leads to the following electron configurations for the given elements: a. \(\mathrm{Ni}\) (Atomic number 28): 1s虏 2s虏 2p鈦 3s虏 3p鈦 4s虏 3d鈦 b. \(\mathrm{Cd}\) (Atomic number 48): 1s虏 2s虏 2p鈦 3s虏 3p鈦 4s虏 3d鹿鈦 4p鈦 5s虏 4d鹿鈦 c. \(\mathrm{Zr}\) (Atomic number 40): 1s虏 2s虏 2p鈦 3s虏 3p鈦 4s虏 3d鹿鈦 4p鈦 5s虏 4d虏 d. \(\mathrm{Os}\) (Atomic number 76): 1s虏 2s虏 2p鈦 3s虏 3p鈦 4s虏 3d鹿鈦 4p鈦 5s虏 4d鹿鈦 5p鈦 6s虏 4f鹿鈦 5d鈦 Using the above electron configurations, you can further study the properties and behaviors of these elements in more detail.

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Key Concepts

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

Aufbau Principle
When looking at electron configurations, the Aufbau principle is like a roadmap telling electrons how to fill up the seats in an atom's orbitals. Imagine a theater with multiple levels鈥攖he electrons would fill the seats starting from the lower levels before moving up to the balcony.

The Aufbau principle states that electrons occupy the lowest energy orbitals first before moving to higher energy orbitals. Using the noble gas configuration as a starting point can simplify the process, which is evidenced by the electron configurations for metals like nickel (Ni) with 28 electrons, filling up to the 3d level, and cadmium (Cd) with 48 electrons, going all the way up to the 4d level.
Hund's Rule
Imagine you're in a bus with lots of open seats. You wouldn't sit right next to someone if there are empty rows available, right? That's essentially what Hund's rule is about for electrons.

Bus Seat Analogy

Electrons will fill up orbitals in a way where they stay as unpaired as possible until they have to pair up. A great illustration is the electron configuration of zirconium (Zr) that has electrons populating the 4d orbitals: they don't start pairing until they absolutely must. This 'bus seat' approach minimizes electron repulsion and keeps energy low.
Pauli Exclusion Principle
Now, imagine that the bus seats are tiny and can only fit one person comfortably. The Pauli Exclusion Principle is like this rule for electrons鈥攊t tells us that no two electrons in an atom can have the exact same set of quantum numbers.

Think of quantum numbers as a unique address for each electron, consisting of the street (orbital type), the house number (orbital energy level), and even specifics of each room in the house (which direction the electron spins). Therefore, when an orbital is shared by two electrons, they have to 'spin' differently. This ensures configurations like that of osmium (Os) don't break any 'housing regulations'.
Atomic Number
The atomic number is like a person's unique ID number. It tells us exactly how many protons (and therefore, electrons) an element has.

Knowing the atomic number immediately tells us the element's position in the periodic table and is the jumping-off point for writing electron configurations. For instance, nickel (Ni) has an atomic number of 28, so we know it has 28 electrons to place in orbitals, while osmium (Os) with an atomic number of 76 has many more electrons to accommodate.
Electron Orbitals
Electron orbitals are like specific neighborhoods where electrons hang out around the nucleus of an atom. They come in different shapes (like s, p, d, f) that tell us about the electron's probable location.

The number in front of the orbital letter represents the energy level or 'distance' from the nucleus. For example, the configuration 1s虏 tells us there are two electrons in the s orbital at the first energy level. As we progress to larger atoms with more electrons, we see more complex 'neighborhoods' emerge, like in the electron configurations for metals such as cadmium (Cd) and zirconium (Zr).

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Most popular questions from this chapter

Which of the following ligands are capable of linkage isomerism? Explain your answer. $$\mathrm{SCN}^{-}, \mathrm{N}_{3}^{-}, \mathrm{NO}_{2}^{-}, \mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}, \mathrm{OCN}^{-}, \mathrm{I}^{-}$$

Compounds of copper(II) are generally colored, but compounds of copper(I) are not. Explain. Would you expect \(\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) to be colored? Explain.

Consider aqueous solutions of the following coordination compounds: \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{I}_{3}, \mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{4}, \mathrm{Na}_{2} \mathrm{Pt}_{6}\), and \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{3} .\) If aqueous \(\mathrm{AgNO}_{3}\) is added to separate beakers containing solutions of each coordination compound, how many moles of AgI will precipitate per mole of transition metal present? Assume that each transition metal ion forms an octahedral complex.

Both \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) and \(\mathrm{Ni}(\mathrm{SCN})_{4}^{2-}\) have four ligands. The first is paramagnetic, and the second is diamagnetic. Are the complex ions tetrahedral or square planar? Explain.

Silver is sometimes found in nature as large nuggets; more often it is found mixed with other metals and their ores. Cyanide ion is often used to extract the silver by the following reaction that occurs in basic solution: $$\mathrm{Ag}(s)+\mathrm{CN}^{-}(a q)+\mathrm{O}_{2}(g) \stackrel{\text { Basis }}{\longrightarrow} \mathrm{Ag}(\mathrm{CN})_{2}^{-}(a q)$$ Balance this equation by using the half-reaction method.

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