Total number of species from the following with central atom utilising $$\mathrm{sp}^2$$ hybrid orbitals for bonding is ________.

$$\mathrm{NH}_3, \mathrm{SO}_2, \mathrm{SiO}_2, \mathrm{BeCl}_2, \mathrm{C}_2 \mathrm{H}_2, \mathrm{C}_2 \mathrm{H}_4, \mathrm{BCl}_3, \mathrm{HCHO}, \mathrm{C}_6 \mathrm{H}_6, \mathrm{BF}_3, \mathrm{C}_2 \mathrm{H}_4 \mathrm{Cl}_2$$

If the function $$f(x)=\left(\frac{1}{x}\right)^{2 x} ; x>0$$ attains the maximum value at $$x=\frac{1}{\mathrm{e}}$$ then :

If $$z_1, z_2$$ are two distinct complex number such that $$\left|\frac{z_1-2 z_2}{\frac{1}{2}-z_1 \bar{z}_2}\right|=2$$, then

Let $$\mathrm{A}=\{1,2,3,4,5\}$$. Let $$\mathrm{R}$$ be a relation on $$\mathrm{A}$$ defined by $$x \mathrm{R} y$$ if and only if $$4 x \leq 5 \mathrm{y}$$. Let $$\mathrm{m}$$ be the number of elements in $$\mathrm{R}$$ and $$\mathrm{n}$$ be the minimum number of elements from $$\mathrm{A} \times \mathrm{A}$$ that are required to be added to R to make it a symmetric relation. Then m + n is equal to :