Question

If we have taken Carbon in the Diamond form (1 mol) , then how many C-C single bond will be present?

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{In}\mathrm{diamond},\mathrm{each}\mathrm{C}\mathrm{atom}\mathrm{forms}4\mathrm{sigma}\mathrm{bonds}\mathrm{with}\mathrm{other}\mathrm{C}\mathrm{atoms} \\ 1\mathrm{mol}\mathrm{of}\mathrm{diamond}=6.022\times {10}^{23}\mathrm{atoms} \\ \mathrm{So},\mathrm{No}.\mathrm{of}\mathrm{C}-\mathrm{C}\mathrm{bond}=4\times 6.022\times {10}^{23} \\ =2.4088\times {10}^{24}\mathrm{bonds} \end{gathered}$$