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Chapter 23: Problem 11

The lanthanide contraction explains which of the following periodic trends? (a) The atomic radii of the transition metals first decrease and then increase when moving horizontally across each period. (b) When forming ions the period 4 transition metals lose their \(4 s\) electrons before their \(3 d\) electrons. (c) The radii of the period 5 transition metals (Y-Cd) are very similar to the radii of the period 6 transition metals (Lu-Hg).

Short Answer

Expert verified
The lanthanide contraction explains the periodic trend given in option (c), where the radii of the period 5 transition metals (Y-Cd) are very similar to the radii of the period 6 transition metals (Lu-Hg).

Step by step solution

01

Option (a) Analysis

Here, we are given a trend regarding the atomic radii of transition metals decreasing and then increasing when moving across each period. The lanthanide contraction specifically affects elements in the lanthanide series and, as a result, neighboring elements in the d-block. This trend doesn't appear to directly involve the lanthanide contraction.
02

Option (b) Analysis

This option states that when forming ions, period 4 transition metals lose their 4s electrons before their 3d electrons. This trend is related to electron configurations and ionization energies, not the lanthanide contraction. Thus, this trend cannot be explained by the lanthanide contraction.
03

Option (c) Analysis

This option presents a trend where the radii of period 5 transition metals (Y-Cd) are very similar to the radii of the period 6 transition metals (Lu-Hg). We know that the lanthanide contraction causes a decrease in atomic radius across the lanthanide series. Consequently, the elements immediately following the lanthanides (in period 6) are expected to have a smaller atomic radius than one would expect without the lanthanide contraction. This trend involves the decrease in atomic radii across periods due to the lanthanide contraction and seems to be directly related.
04

Conclusion

Based on the analysis, we can conclude that the lanthanide contraction best explains the periodic trend given in option (c), where the radii of the period 5 transition metals (Y-Cd) are very similar to the radii of the period 6 transition metals (Lu-Hg).

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Most popular questions from this chapter

Which of the following objects is chiral: \((\mathbf{a})\) a pencil, \((\mathbf{b})\) a computer keyboard, (c) a grand piano, (d) a molecular model of \(c i s-\mathrm{Fe}(\text { bipy })_{2} \mathrm{Cl}_{2},(\mathbf{e})\) a piece of plane \(\mathrm{A} 4\) paper?

Which species are more likely to act as ligands? (a) Positively charged ions or negatively charged ions? (b) Neutral molecules that are polar or those that are nonpolar?

Determine if each of the following metal complexes is chiral and therefore has an optical isomer: (a) square planar \(\left[\mathrm{Pd}(\mathrm{en})(\mathrm{CN})_{2}\right],(\mathbf{b})\) octahedral $\left[\mathrm{Ni (\mathrm{en})\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+},$ (c) octahedral cis-[V(en) \(\left._{2} \mathrm{ClBr}\right]\).

When Alfred Werner was developing the field of coordination chemistry, it was argued by some that the optical activity he observed in the chiral complexes he had prepared was due to the presence of carbon atoms in the molecule. To disprove this argument, Werner synthesized a chiral complex of cobalt that had no carbon atoms in it, and he was able to resolve it into its enantiomers. Design a cobalt(III) complex that would be chiral if it could be synthesized and that contains no carbon atoms. (It may not be possible to synthesize the complex you design, but we will not worry about that for now.)

Give the number of (valence) \(d\) electrons associated with the central metal ion in each of the following complexes: (a) $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right] \mathrm{Cl}_{2}$, (b) $\mathrm{K}_{2}\left[\mathrm{Cu}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}\right]$, (c) \(\left[\mathrm{Os}(\mathrm{en})_{3}\right] \mathrm{Cl}_{3}\), (d) \([\mathrm{Cr}(\mathrm{EDTA})] \mathrm{SO}_{4}\), , (e) $\left[\mathrm{Cd}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$
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