A recently reported synthesis of the transuranium element bohrium (Bh) involved the bombardment of berkelium-249 with neon-22 to produce bohrium-267. Write a nuclear reaction for this synthesis. The half-life of bohrium-267 is \(15.0\) seconds. If 199 atoms of bohrium- 267 could be synthesized, how much time would elapse before only 11 atoms of bohrium- 267 remain? What is the expected electron configuration of elemental bohrium?

To write the nuclear reaction, we can use the following notation: $$ ^{_A}{_Z}X + ^{_B}{_Y}W \to ^{_{A+B}}{_{Z+Y}}V + Q, $$ where ('\(_X\)' and '\(_W\)') are the reactants, ('\(_V\)') is the product, and ('\(_Q\)') represents any other particles released during the reaction. Using the atomic numbers (Z) and mass numbers (A) of berkelium-249 (Bk) and neon-22 (Ne), we can write the nuclear reaction representing the synthesis process: $$ ^{249}_{97}Bk + ^{22}_{10}Ne \to ^{267}_{107}Bh + X $$ Now, we need to find what \(X\) is. The atomic number must be conserved, so we have: $$Z_{X} = 97 + 10 - 107 = 0.$$ The mass number must also be conserved: $$A_{X} = 249 + 22 - 267 = 4.$$ This means the particle is an alpha particle, which we can denote by "\( ^4_2He \)": $$ ^{249}_{97}Bk + ^{22}_{10}Ne \to ^{267}_{107}Bh + ^4_2He $$

We are given that the half-life of bohrium-267 is 15 seconds. We need to find the time it takes for the number of bohrium atoms to decrease from 199 to 11. We can use the half-life formula: $$N = N_0 \cdot \left(\frac{1}{2}\right)^{t/T_{1/2}}, $$ where '\(N\)' is the final number of atoms, '\(N_0\)' is the initial number of atoms, '\(t\)' is the time elapsed, and '\(T_{1/2}\)' is the half-life. Plugging in the values, we get: $$11 = 199 \cdot \left(\frac{1}{2}\right)^{t/15}.$$ To solve for '\(t\)', we will first divide both sides by 199: $$\frac{11}{199} = \left(\frac{1}{2}\right)^{t/15}.$$ Next, we will take the logarithm base 2 of both sides: $$\log_2{\frac{11}{199}} = \frac{t}{15}.$$ Now, we can solve for '\(t\)': $$t = 15 \cdot \log_2{\frac{11}{199}} \approx 55.4 \text{ seconds}.$$ So, it would take approximately 55.4 seconds for the number of bohrium-267 atoms to decrease from 199 to 11.

To determine the electron configuration of elemental bohrium (Bh), we need to know its atomic number, which is 107. The electron configuration can be obtained by following the pattern of the periodic table and filling up electron orbitals in the order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, ... Since the atomic number of Bh is 107, its electron configuration should account for the 107 electrons. Following the pattern, we get: $$[Rn]\ 5f^{14}\ 6d^{5}\ 7s^{2},$$ where '\([Rn]\)' represents the noble gas core of radon (86 electrons). Elemental bohrium has the electron configuration \(1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^2\ 3d^{10}\ 4p^6\ 5s^2\ 4d^{10}\ 5p^6\ 6s^2\ 4f^{14}\ 5d^{10}\ 6p^6\ 7s^2\ 5f^{14}\ 6d^{5}\).