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: \(3\).
Which one of the following graphs accurately represents the plot of partial pressure of CS₂ vs its mole fraction in a mixture of acetone and CS₂ at constant temperature?

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 ___________.