For the reaction 2A + B $$\rightarrow$$ C, the values of initial rate at different reactant concentrations are given in the table below.

[A] (mol L$$^{-1}$$)    [B] (mol L$$^{-1}$$)    Initial Rate (mol L$$^{-1}$$s$$^{-1}$$)
0.05            0.05                0.045
0.10                0.05                0.090
0.20                0.10                0.72

The rate law for the reactions is:

For any reaction of the general type $$aA+bB\rightarrow \text{products}$$ we write the empirical rate law as

$$\text{Rate}=k[A]^m[B]^n$$

where $$k$$ is the rate constant and $$m,n$$ are the (experimental) orders with respect to the reactants A and B, respectively. Our task is to determine the values of $$m$$ and $$n$$ from the given initial-rate data.

First, label the three experiments for easy reference:

Experiment 1: $$[A]_1=0.05\ \text{mol L}^{-1},\quad [B]_1=0.05\ \text{mol L}^{-1},\quad R_1=0.045\ \text{mol L}^{-1}\text{s}^{-1}$$
Experiment 2: $$[A]_2=0.10\ \text{mol L}^{-1},\quad [B]_2=0.05\ \text{mol L}^{-1},\quad R_2=0.090\ \text{mol L}^{-1}\text{s}^{-1}$$
Experiment 3: $$[A]_3=0.20\ \text{mol L}^{-1},\quad [B]_3=0.10\ \text{mol L}^{-1},\quad R_3=0.72\ \text{mol L}^{-1}\text{s}^{-1}$$

We now compare pairs of experiments so that only one concentration changes at a time.

Step 1: Find $$m$$ by keeping $$[B]$$ constant.

Between Experiments 1 and 2, $$[B]$$ is the same (0.05 mol L−1), so the rate ratio depends only on $$[A]$$:

$$\frac{R_2}{R_1}= \frac{k[A]_2^m[B]_2^n}{k[A]_1^m[B]_1^n} =\left(\frac{[A]_2}{[A]_1}\right)^m\left(\frac{[B]_2}{[B]_1}\right)^n =\left(\frac{[A]_2}{[A]_1}\right)^m(1)^n =\left(\frac{0.10}{0.05}\right)^m.$$

The numerical ratio of rates is $$\dfrac{0.090}{0.045}=2$$, so

$$2=\left(\frac{0.10}{0.05}\right)^m=2^m.$$

Hence $$m=1$$.

Step 2: Find $$n$$ using $$m=1$$.

Now compare Experiments 2 and 3; both $$[A]$$ and $$[B]$$ double. Using the rate law with $$m=1$$, we have

$$\frac{R_3}{R_2}= \frac{k[A]_3^1[B]_3^n}{k[A]_2^1[B]_2^n} =\left(\frac{[A]_3}{[A]_2}\right)^1\left(\frac{[B]_3}{[B]_2}\right)^n =\left(\frac{0.20}{0.10}\right)\left(\frac{0.10}{0.05}\right)^n =2\cdot 2^n.$$

The experimental rate ratio is $$\dfrac{0.72}{0.090}=8$$, so

$$8=2\cdot 2^n \quad\Longrightarrow\quad 4=2^n \quad\Longrightarrow\quad n=2.$$

Step 3: Write the complete rate law.

With $$m=1$$ and $$n=2$$, the empirical rate expression becomes

$$\text{Rate}=k[A]^1[B]^2=k[A][B]^2.$$

Step 4: (Optional) Verify $$k$$ is constant.

Using Experiment 1 data:

$$k=\frac{R_1}{[A]_1[B]_1^2} =\frac{0.045}{(0.05)(0.05)^2} =\frac{0.045}{0.000125}=360.$$

Using Experiment 2:

$$k=\frac{0.090}{(0.10)(0.05)^2} =\frac{0.090}{0.00025}=360,$$

confirming that $$k$$ is indeed constant and our orders are correct.

Therefore, the reaction follows the rate law $$\boxed{\text{Rate}=k[A][B]^2}$$, which corresponds to Option D.

Hence, the correct answer is Option D.