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Chapter 2: Problem 107

Arrange the following groups of atoms in order of increasing size. a. \(\mathrm{Te}, \mathrm{S}, \mathrm{Se}\) b. \(\mathrm{K}, \mathrm{Br}, \mathrm{Ni}\) c. \(\mathrm{Ba}, \mathrm{Si}, \mathrm{F}\)

Short Answer

Expert verified
The order of increasing atomic size for each group is: - Group a: \(S < Se < Te\) - Group b: \(Br < Ni < K\) - Group c: \(F < Si < Ba\)

Step by step solution

01

Group a: \(\mathrm{Te}, \mathrm{S}, \mathrm{Se}\)

All three elements belong to group 16, and they are all Chalcogens. As we move down a group in the periodic table, the atomic size increases due to the addition of electron shells. Thus, in this group, the order of increasing size will be: S < Se < Te
02

Group b: \(\mathrm{K}, \mathrm{Br}, \mathrm{Ni}\)

These elements belong to different groups and periods. K is in group 1 and period 4, Br is in group 17 and period 4, and Ni is in group 10 and period 4. Generally, as we move across a period from left to right, the atomic size decreases because of the increasing effective nuclear charge. In this case, we have: K > Ni > Br
03

Group c: \(\mathrm{Ba}, \mathrm{Si}, \mathrm{F}\)

Ba is in group 2 and period 6, Si is in group 14 and period 3, and F is in group 17 and period 2. As mentioned earlier, atomic size increases as we move down a group (due to the addition of electron shells) and decreases as we move from left to right across a period (due to increased effective nuclear charge). Considering these periodic trends, the order of increasing size in this group is: F < Si < Ba In conclusion, for each group, the order of increasing atomic size is as follows: - Group a: S < Se < Te - Group b: Br < Ni < K - Group c: F < Si < Ba

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Most popular questions from this chapter

For each of the following pairs of elements $$ (\mathrm{C} \text { and } \mathrm{N}) \quad(\mathrm{Ar} \text { and } \mathrm{Br}) $$ pick the atom with a. more favorable (more negative) electron affinity. b. higher ionization energy. c. larger size.

In defining the sizes of orbitals, why must we use an arbitrary value, such as \(90 \%\) of the probability of finding an electron in that region?

One of the emission spectral lines for \(\mathrm{Be}^{3+}\) has a wavelength of \(253.4 \mathrm{nm}\) for an electronic transition that begins in the state with \(n=5 .\) What is the principal quantum number of the lower-energy state corresponding to this emission? (Hint: The Bohr model can be applied to one- electron ions. Don't forget the \(Z\) factor: \(Z=\) nuclear charge \(=\) atomic number.

Answer the following questions based on the given electron configurations and identify the elements. a. Arrange these atoms in order of increasing size: \([\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{6} ;[\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{1} ;[\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{3}\). b. Arrange these atoms in order of decreasing first ionization energy: \([\mathrm{Ne}] 3 s^{2} 3 p^{5} ;[\mathrm{Ar}] 4 s^{2} 3 d^{10} 4 p^{3} ;[\mathrm{Ar}] 4 s^{2} 3 d^{10} 4 p^{5}\).

Calculate, to four significant figures, the longest and shortest wavelengths of light emitted by electrons in the hydrogen atom that begin in the \(n=5\) state and then fall to states with smaller values of \(n\).
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