Match the LIST-I with LIST-II.
| LIST-I | LIST-II | ||
| A. | Pnicogen (group 15) | I. | Ts |
| B. | Chalcogen (group 16) | II. | Og |
| C. | Halogen (group 17) | III. | Lv |
| D. | Noble gas (group 18) | IV. | Mc |
Choose the correct answer from the options given below :
To solve this matching problem, we need to align each group from LIST-I with its respective element from LIST-II. Let's break it down:
Now, let's match them with LIST-II:
Thus, the correct answer is: A-IV, B-III, C-I, D-II
Let's match the families of elements with their corresponding symbols from the given lists, focusing on the most recently named elements in these groups:
A. Pnicogen (group 15): The elements in Group 15 are Nitrogen (N), Phosphorus (P), Arsenic (As), Antimony (Sb), Bismuth (Bi), and Moscovium (Mc). Therefore, Pnicogen matches with Mc. A - IV
B. Chalcogen (group 16): The elements in Group 16 are Oxygen (O), Sulfur (S), Selenium (Se), Tellurium (Te), Polonium (Po), and Livermorium (Lv). Therefore, Chalcogen matches with Lv. B - III
C. Halogen (group 17): The elements in Group 17 are Fluorine (F), Chlorine (Cl), Bromine (Br), Iodine (I), Astatine (At), and Tennessine (Ts). Therefore, Halogen matches with Ts. C - I
D. Noble gas (group 18): The elements in Group 18 are Helium (He), Neon (Ne), Argon (Ar), Krypton (Kr), Xenon (Xe), Radon (Rn), and Oganesson (Og). Therefore, Noble gas matches with Og. D - II
The correct matching is A-IV, B-III, C-I, D-II, which corresponds to option (2).

The correct sequence of reagents for the above conversion of X to Y is :
Match List-I with List-II 
Choose the correct answer from the options given below:
Let \( \alpha = \dfrac{-1 + i\sqrt{3}}{2} \) and \( \beta = \dfrac{-1 - i\sqrt{3}}{2} \), where \( i = \sqrt{-1} \). If
\[ (7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}, \] then the value of \( m \) is ___________.