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.
Match List - I with List - II:
List - I:
(A) \([ \text{MnBr}_4]^{2-}\)
(B) \([ \text{FeF}_6]^{3-}\)
(C) \([ \text{Co(C}_2\text{O}_4)_3]^{3-}\)
(D) \([ \text{Ni(CO)}_4]\)
List - II:
(I) d²sp³ diamagnetic
(II) sp²d² paramagnetic
(III) sp³ diamagnetic
(IV) sp³ paramagnetic
If $ \theta \in [-2\pi,\ 2\pi] $, then the number of solutions of $$ 2\sqrt{2} \cos^2\theta + (2 - \sqrt{6}) \cos\theta - \sqrt{3} = 0 $$ is:
A thin transparent film with refractive index 1.4 is held on a circular ring of radius 1.8 cm. The fluid in the film evaporates such that transmission through the film at wavelength 560 nm goes to a minimum every 12 seconds. Assuming that the film is flat on its two sides, the rate of evaporation is:
The major product (A) formed in the following reaction sequence is
