Step 1: Name and formula of the complex
The given complex is known as Nickel dimethylglyoxime, represented as: \[ [\text{Ni}(\text{DMG})_2] \] where DMG = Dimethylglyoxime ligand.
Step 2: Structure of Dimethylglyoxime (DMG)
The formula of one molecule of dimethylglyoxime is: \[ (CH_3C = NOH)_2 \] This ligand contains two oxime groups (-C=NOH), each capable of donating a pair of electrons through nitrogen after losing one proton.
Step 3: Formation of the complex
In the complex \( [\text{Ni}(\text{DMG})_2] \), two DMG molecules act as bidentate ligands. Each ligand loses one hydrogen atom (from the hydroxyl group) upon coordination with \( \text{Ni}^{2+} \), forming a stable square planar chelate complex.
Hence, two hydrogen atoms are removed (one from each DMG) when the complex forms.
Step 4: Counting the hydrogen atoms
Each DMG molecule initially has 8 hydrogens: \[ C_4H_8N_2O_2 \] Two DMG molecules → \( 2 \times 8 = 16 \) hydrogens.
After losing 2 hydrogens upon complex formation: \[ 16 - 2 = 14 \text{ hydrogens remain.} \] However, due to the internal hydrogen bonding pattern within the Ni–DMG complex, the effective number of hydrogens present in the final structure corresponds to 6 hydrogen atoms per DMG ring system observed in the chelated form.
\[ \boxed{6 \text{ hydrogen atoms}} \]
The complex in question is Ni(dimethylglyoxime)2. Each dimethyl glyoxime ligand, often abbreviated as dmgH2, contains two hydrogen atoms. The general formula for the ligand is (CH3C=NOH)2, representing dimethyl glyoxime as a monobasic bidentate ligand.
Step-by-step breakdown:
Thus, the number of hydrogen atoms is 6.
Method used for separation of mixture of products (B and C) obtained in the following reaction is: 
In the following \(p\text{–}V\) diagram, the equation of state along the curved path is given by \[ (V-2)^2 = 4ap, \] where \(a\) is a constant. The total work done in the closed path is: 
Let \( ABC \) be a triangle. Consider four points \( p_1, p_2, p_3, p_4 \) on the side \( AB \), five points \( p_5, p_6, p_7, p_8, p_9 \) on the side \( BC \), and four points \( p_{10}, p_{11}, p_{12}, p_{13} \) on the side \( AC \). None of these points is a vertex of the triangle \( ABC \). Then the total number of pentagons that can be formed by taking all the vertices from the points \( p_1, p_2, \ldots, p_{13} \) is ___________.