To determine which pair of reagents yields paramagnetic species, we need to analyze the products of each reaction and check for the presence of unpaired electrons. A species is paramagnetic if it has one or more unpaired electrons.

Option A: $Na$ and excess of $NH_3$

1. Reaction: When sodium metal is dissolved in liquid ammonia, it forms a deep blue solution. The reaction is represented as: $Na + (x+y)NH_3 \rightarrow [Na(NH_3)_x]^+ + [e(NH_3)_y]^-$
2. Products: The products are the solvated sodium ion, $[Na(NH_3)_x]^+$, and the solvated electron, $[e(NH_3)_y]^-$.
3. Paramagnetism: The solvated electron is an electron surrounded by ammonia molecules. This species contains a single, unpaired electron. The presence of this unpaired electron makes the solution paramagnetic.
4. Conclusion: This reaction yields a paramagnetic species. Thus, option A is correct.

Option B: $K$ and excess of $O_2$

1. Reaction: Potassium is an alkali metal that reacts with an excess of oxygen to form potassium superoxide. $K + O_2 (\text{excess}) \rightarrow KO_2$
2. Product: The product is potassium superoxide ($KO_2$), which is an ionic compound composed of $K^+$ and $O_2^-$ ions.
3. Paramagnetism: We need to examine the superoxide ion ($O_2^-$). The molecular orbital (MO) configuration of an $O_2$ molecule has two unpaired electrons in its $\pi^*_{2p}$ orbitals. The superoxide ion, $O_2^-$, is formed by adding one electron to an $O_2$ molecule. This electron pairs up with one of the existing unpaired electrons, but one unpaired electron still remains. The MO configuration of $O_2^-$ is $(\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\sigma_{2p_z})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\pi^*_{2p_x})^2 (\pi^*_{2p_y})^1$. The single electron in the $\pi^*_{2p_y}$ orbital is unpaired.
4. Conclusion: Since the superoxide ion ($O_2^-$) is paramagnetic, potassium superoxide ($KO_2$) is also paramagnetic. Thus, option B is correct.

Option C: $Cu$ and dilute $HNO_3$

1. Reaction: Copper reacts with dilute nitric acid in a redox reaction. Nitric acid acts as an oxidizing agent. $3Cu(s) + 8HNO_3(\text{dilute}) \rightarrow 3Cu(NO_3)_2(aq) + 2NO(g) + 4H_2O(l)$
2. Products: The products are copper(II) nitrate ($Cu(NO_3)_2$), nitric oxide ($NO$), and water ($H_2O$).
3. Paramagnetism:
   • Copper(II) nitrate ($Cu(NO_3)_2$): This compound contains the $Cu^{2+}$ ion. The electronic configuration of Cu (Z=29) is $[Ar] 3d^{10} 4s^1$. The configuration of $Cu^{2+}$ is $[Ar] 3d^9$. A $d^9$ configuration has one unpaired electron in the 3d orbitals, making $Cu^{2+}$ paramagnetic.
   • Nitric oxide ($NO$): This molecule has a total of $7(N) + 8(O) = 15$ electrons. Since it has an odd number of electrons, it must be paramagnetic. Its MO configuration shows one unpaired electron in a $\pi^*$ antibonding orbital.
4. Conclusion: The reaction produces two paramagnetic species, $Cu(NO_3)_2$ and $NO$. Thus, option C is correct.

Option D: $O_2$ and 2-ethylanthraquinol

1. Reaction: This is the key step in the industrial production of hydrogen peroxide ($H_2O_2$) via the anthraquinone process. 2-ethylanthraquinol is oxidized by oxygen to produce 2-ethylanthraquinone and hydrogen peroxide. $\text{2-ethylanthraquinol} + O_2 \rightarrow \text{2-ethylanthraquinone} + H_2O_2$
2. Products: The products are 2-ethylanthraquinone and hydrogen peroxide ($H_2O_2$).
3. Paramagnetism:
   • Hydrogen peroxide ($H_2O_2$): It has an even number of electrons (18 total), and all electrons are paired up in covalent bonds or as lone pairs. It is diamagnetic.
   • 2-ethylanthraquinone: This is a stable organic molecule with a system of conjugated double bonds. All electrons are paired. It is diamagnetic.
4. Conclusion: Neither of the products is paramagnetic. Thus, option D is incorrect.

Final summary: Options A, B, and C all yield paramagnetic species.