In the following sequence of reactions, the products (A) to (G) are formed:
i. `2CH_4(g)overset(Delta)underset(1773K)rarr(A)(g)+B(g)`
ii. `4 mol of (A) overset(Delta)underset(Ni(CN)_4+THF)rarr(C)overset(O_3//o x id.)rarr(D) only`
iii. `(A)overset((1) 1 mol of NaNH_2)underset((2) C_2H_5I)rarr(E)underset(Red hot tube)overset(3 mol of E)rarrunderset((G) only)underset(darr O_3//Red.)F`
Compound D is:

Options

A. Glyoxal
B. Glycol
C. Oxalic acid
D. Methylglyoxal

Explanation

To solve the problem step by step, we need to analyze each reaction and identify the compounds formed from the given reactions.

### Step 1: Identify Compound A
The first reaction is:
\[ 2CH_4(g) \overset{\Delta}{\longrightarrow} (A)(g) + B(g) \]
When methane (\(CH_4\)) is heated, it undergoes dehydrogenation to form acetylene (\(C_2H_2\)) and hydrogen gas (\(H_2\)). Therefore, compound A is:
\[ A = C_2H_2 \]

### Step 2: Identify Compound C
The second reaction is:
\[ 4 \text{ mol of } (A) \overset{\Delta}{\underset{Ni(CN)_4 + THF}{\longrightarrow}} (C) \]
When 4 moles of acetylene (\(C_2H_2\)) are treated with nickel in THF, cyclization occurs, leading to the formation of a cyclic compound. The product of this reaction is likely to be a cyclobutene (\(C_4H_6\)) or a similar cyclic structure. Thus, we can denote:
\[ C = C_4H_6 \]

### Step 3: Identify Compound D
The next part of the reaction is:
\[ C \overset{O_3 // oxid.}{\longrightarrow} (D) \]
Ozonolysis of cyclobutene will cleave the double bonds, leading to the formation of two molecules of carbonyl compounds. In this case, the ozonolysis of cyclobutene can yield oxalic acid (\(C_2H_2O_4\)). Therefore, we can conclude:
\[ D = C_2H_2O_4 \]

### Step 4: Identify Compound E
The next reaction is:
\[ (A) \overset{(1) 1 \text{ mol of } NaNH_2}{\underset{(2) C_2H_5I}{\longrightarrow}} (E) \]
Here, sodium amide (\(NaNH_2\)) acts as a strong base and deprotonates acetylene to form sodium acetylide, which then reacts with ethyl iodide (\(C_2H_5I\)) to form butyne (\(C_4H_6\)). Thus:
\[ E = C_4H_6 \]

### Step 5: Identify Compound G
The final reaction is:
\[ \underset{(G) \text{ only}}{(3 \text{ mol of } E) \overset{\text{Red hot tube}}{\longrightarrow}} \underset{(F)}{} \]
When butyne is heated in a red hot tube, it can undergo further reactions, possibly leading to the formation of smaller hydrocarbons or even carbon monoxide and hydrogen. However, the exact product is less relevant for identifying D.

### Conclusion
The compound D formed in the sequence of reactions is:
\[ D = C_2H_2O_4 \text{ (Oxalic Acid)} \]