d
$\text { Degree of freedom }(f)=5+2(3 N-5)$
$\qquad f=5+2(3 \times 2-1)=7$
$\text { energy of one molecule }=\frac{f}{2} K_B T$
energy of 10 molecules
$=10\left(\frac{\mathrm{f}}{2} \mathrm{~K}_{\mathrm{B}} \mathrm{T}\right)=10\left(\frac{7}{2} \mathrm{~K}_{\mathrm{B}} \mathrm{T}\right)=35 \mathrm{~K}_{\mathrm{B}} \mathrm{T}$