For identifying odd-electron species:
• Calculate the total number of valence electrons, considering charges on ions.
• Species with an odd total number of electrons will have unpaired electrons.
1. Odd-Electron Species: Odd-electron species have an unpaired electron in their structure, resulting in an odd total number of electrons.
2. Electron Count for Each Species
\(\text{NO}_2\):
\[\text{N (5) + O (6) + O (6) = 17~(odd~electrons).}\]
\(\text{ICl}_4^-\):
\(I (7) + 4Cl (4 \times 7) + 1 (charge) = 36~(even~electrons).\)
\(\text{BrF}_3\):
\(Br (7) + 3F (3 \times 7) = 28~(even~electrons).\)
\(\text{ClO}_2\):
\(\text{Cl (7) + O (6) + O (6) = 19~(odd~electrons).}\)
\(\text{NO}_2^+\):
\[\text{N (5) + O (6) + O (6) - 1 (charge) = 16~(even~electrons).}\]
\(\text{NO}\):
\[\text{N (5) + O (6) = 11~(odd~electrons).}\]
3. Species Without Odd Electrons: The species with an even number of electrons are:
\[\text{ICl}_4^-, \text{BrF}_3, \text{and } \text{NO}_2^+.\]
4. Count: Total number of species without odd electrons: 3.
Final Answer: \(3\).
0.1 mole of compound S will weigh ...... g, (given the molar mass in g mol\(^{-1}\) C = 12, H = 1, O = 16) 
Among $ 10^{-10} $ g (each) of the following elements, which one will have the highest number of atoms?
Element : Pb, Po, Pr and Pt
The molar mass of the water insoluble product formed from the fusion of chromite ore \(FeCr_2\text{O}_4\) with \(Na_2\text{CO}_3\) in presence of \(O_2\) is ....... g mol\(^{-1}\):
0.1 mol of the following given antiviral compound (P) will weigh .........x $ 10^{-1} $ g. 


For the circuit shown above, the equivalent gate is:
Let \( f : \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \[ (\sin x \cos y)(f(2x + 2y) - f(2x - 2y)) = (\cos x \sin y)(f(2x + 2y) + f(2x - 2y)), \] for all \( x, y \in \mathbb{R}. \)
If \( f'(0) = \frac{1}{2} \), then the value of \( 24f''\left( \frac{5\pi}{3} \right) \) is: