Q24.55 P - Chemistry: Molecular Nature Of Matter And Change

Question

Elements $\mathbf{\text{104, 105, and 106}}$ have been named rutherfordium (Rf), dubnium (Db), and seaborgium (Sg), respectively. These elements are synthesized from californium-  $\mathbf{\text{249}}$by bombarding with carbon- $\mathbf{\text{12}}$, nitrogen- , and oxygen- $\mathbf{\text{18}}$ nuclei, respectively. Four neutrons are formed in each reaction as well. 

(a) Write balanced nuclear equations for the formation of these elements. 

(b) Write the equations in shorthand notation.

Step-by-Step Solution

Answer

a.$\begin{array}{l}{}_{\text{98}}^{\text{249}}{\text{Cf+}}_{\text{6}}^{\text{12}}\text{C}\to {\text{4}}_{\text{0}}^{\text{1}}{\text{n+}}_{\text{104}}^{\text{257}}\text{Rf}\\ {}_{\text{98}}^{\text{249}}{\text{Cf+}}_{\text{7}}^{\text{15}}\text{ N}\to {\text{4}}_{\text{0}}^{\text{1}}{\text{n+}}_{\text{105}}^{\text{260}}\text{Db}\\ {}_{\text{98}}^{\text{249}}{\text{Cf+}}_{\text{8}}^{\text{18}}\text{O}\to {\text{4}}_{\text{0}}^{\text{1}}{\text{n+}}_{\text{106}}^{\text{263}}\text{Sg}\end{array}$

b.

$\begin{array}{l}{}^{\text{249}}\text{Cf}{{(}^{\text{12}}\text{C,4n})}^{\text{257}}\text{Rf}\\ {}^{\text{249}}\text{Cf}{{(}^{\text{15}}\text{ N,4n})}^{\text{260}}\text{Db}\\ {}^{\text{249}}\text{Cf}{{(}^{\text{18}}\text{O,4n})}^{\text{263}}\text{Sg}\end{array}$

Step 1: Subpart (a)

- Californium-  is bombarded with carbon-$12$ , and the product is rutherfordium (Rf, atomic number $104$ ) and  4 neutrons.

- Atomic number of californium is 98

- Atomic number of carbon is 6

- Neutron:  _{${}_{\text{0}}^{\text{1}}\text{n}$}

 _{${}_{98}^{249}\text{Cf}{+}_6^{12}\text{C}\to 4_0^1\text{n}{+}_{104}^{\text{A}}\text{Rf}$}

The value of A is

 _{$\begin{array}{c}\text{249+12 =4×1+A}\\ \text{A =257}\end{array}$}

So, the balanced nuclear equations is

 _{${}_{\text{98}}^{\text{249}}{\text{Cf+}}_{\text{6}}^{\text{12}}\text{C}\to {\text{4}}_{\text{0}}^{\text{1}}{\text{n+}}_{\text{104}}^{\text{257}}\text{Rf}$}

- Californium- _$249$ is bombarded with nitrogen-$15$ , and the product is dubnium (Db, atomic number  ) and 4  neutrons.

- Atomic number of nitrogen is 7

_{${}_{98}^{249}\text{Cf}{+}_7^{15} \text{N}\to 4_0^1\text{n}{+}_{105}^{\text{A}}\text{Db}$} 

The value of A is

 _{$\begin{array}{c}\text{249+15 =4×1+A}\\ \text{ A =260}\end{array}$}

So, the balanced nuclear equations is

 _{${}_{98}^{249}\text{Cf}{+}_7^{15} \text{N}\to 4_0^1\text{n}{+}_{105}^{260}\text{Db}$}

- Californium-249  is bombarded with oxygen-18, and the product is seaborgium (Sg, atomic number$106$  ) and 4  neutrons.

- Atomic number of oxygen is8

 _{${}_{98}^{249}\text{Cf}{+}_8^{18}\text{O}\to 4_0^1\text{n}{+}_{106}^{\text{A}}\text{Sg}$}

The value of A is

 _{$\begin{array}{c}\text{249+18 =4×1+A}\\ \text{ A =263}\end{array}$}

So, the balanced nuclear equations is

 _{${}_{98}^{249}\text{Cf}{+}_8^{18}\text{O}\to 4_0^1\text{n}{+}_{106}^{263}\text{Sg}$}

Step 2: Subpart (b)

The equations in shorthand notation

$\begin{array}{l}{}_{\text{98}}^{\text{249}}{\text{Cf+}}_{\text{6}}^{\text{12}}\text{C}\to {\text{4}}_{\text{0}}^{\text{1}}{\text{n+}}_{\text{104}}^{\text{257}}\text{Rf}\\ {}^{\text{249}}\text{Cf}{{(}^{\text{12}}\text{C,4n})}^{\text{257}}\text{Rf•}\\ {}_{\text{98}}^{\text{249}}{\text{Cf+}}_{\text{7}}^{\text{15}}\text{ N}\to {\text{4}}_{\text{0}}^{\text{1}}{\text{n+}}_{\text{105}}^{\text{260}}\text{Db}\\ {}^{\text{249}}\text{Cf}{{(}^{\text{15}}\text{ N,4n})}^{\text{260}}\text{Db•}\\ {}_{\text{98}}^{\text{249}}{\text{Cf+}}_{\text{8}}^{\text{18}}\text{O}\to {\text{4}}_{\text{0}}^{\text{1}}{\text{n+}}_{\text{106}}^{\text{263}}\text{Sg}\\ {}^{\text{249}}\text{Cf}{{(}^{\text{18}}\text{O,4n})}^{\text{263}}\text{Sg}\end{array}$