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Explain why the transition metals in periods 5 and 6 have nearly identical radii in each group.

Short Answer

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The near-identical radii of transition metals in periods 5 and 6 can be explained through the similar energy levels and electron shielding provided by the 4d and 5d orbitals, as well as the influence of the lanthanide contraction in period 6. The combination of these factors results in minimal differences in the atomic radii of transition metals in these periods.

Step by step solution

01

Review Atomic Radius and Shielding Effect

The atomic radius is the distance between the nucleus of an atom and its outermost electron shell. In general, atomic radii decrease across a period due to an increase in the number of protons in the nucleus, which result in greater attraction between the nucleus and the electrons surrounding it. The shielding effect, on the other hand, describes the phenomenon in which inner electron shells shield the outer electron shells from the full positive charge of the nucleus. This diminishes the attractive force between the nucleus and the outermost electrons, causing the atomic radius to increase down a group.
02

Understand Periodic Trends for Transition Metals in Periods 5 and 6

Transition metals are found in the d-block of the periodic table. In periods 5 and 6, the transition metals fill their 4d and 5d orbitals, respectively. Since these elements have a similar electron configuration, they exhibit similar properties and trends in their atomic radii.
03

Explain the Role of Filling 4d and 5d Orbitals

As the transition metals in periods 5 and 6 are filling their 4d and 5d orbitals, the increase in nuclear charge (number of protons) is balanced by the increase in electron shielding. The 4d and 5d orbitals are quite similar in energy and are relatively far from the nucleus, which means they experience a significant shielding effect.
04

Discuss the Near-Identical Radii in Periods 5 and 6 Transition Metals

The near-identical radii of transition metals in periods 5 and 6 can be attributed to the following factors: 1. The 4d and 5d orbitals have similar energy levels and electron shielding, as mentioned earlier. This results in a small difference between the atomic radii as one moves from period 5 to period 6. 2. The lanthanide contraction: In period 6, the 4f orbitals are filled before the 5d orbitals. The 4f electrons are poor at shielding, allowing the nuclear charge to act more effectively on the outer 5d electrons. This causes a reduction in atomic radius, which counteracts the expected increase in atomic radius as one moves down the periodic table. In conclusion, the near-identical radii of transition metals in periods 5 and 6 can be explained through the similar energy levels and electron shielding provided by the 4d and 5d orbitals, as well as the influence of the lanthanide contraction in period 6. The combination of these factors results in minimal differences in the atomic radii of transition metals in these periods.

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Most popular questions from this chapter

Carbon monoxide is toxic because it binds more strongly to the iron in hemoglobin (Hb) than does \(\mathrm{O}_{2}\) , as indicated by these approximate standard free-energy changes in blood: $$\begin{array}{cl}{\mathrm{Hb}+\mathrm{O}_{2} \longrightarrow \mathrm{HbO}_{2}} & {\Delta G^{\circ}=-70 \mathrm{kJ}} \\\ {\mathrm{Hb}+\mathrm{CO} \longrightarrow \mathrm{HbCO}} & {\Delta G^{\circ}=-80 \mathrm{kJ}}\end{array}$$ Using these data, estimate the equilibrium constant at 298 K for the equilibrium $$\mathrm{HbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HbCO}+\mathrm{O}_{2}$$

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