Step 1: Identify the paramagnetic species using Molecular Orbital Theory.& nbsp;
The given species are classified as follows:
\[ \begin{aligned} \text{Paramagnetic:}\quad & amp;\mathrm{O_2,\ NO,\ O_2^-,\ N_2^+,\ NO^-},\\[4pt] \text{Diamagnetic:}\quad & amp;\mathrm{N_2,\ F_2,\ CO,\ O_2^{2-},\ NO^+,\ H_2}. \end{aligned} \]
Thus, the number of paramagnetic species is
\[ 5. \]
Step 2: Write the answer.
Hence,
\[ \boxed{5}. \]
Thus, the correct option is
\[ \boxed{(B)}. \]
| Molisch's lest | Barfoed Test | Biuret Test | |
|---|---|---|---|
| A | Positive | Negative | Negativde |
| B | Positive | Positive | Negative |
| C | Negative | Negative | Positive |
According to MO theory, the molecule which contain only π-Bonds between the atoms is
In which if the following changes there is no change in hybridization of the central atom?
For the reaction at 25° C, X2O4 (l) → 2XO2(g), U and S are 2.1 K.Cal and 20 Cal/K respectively. what is G for the reaction at the same temperature? (R = 2 CAL K-1MOL-1)
The hybrids of which group elements of the periodic table form electron precise hybrids?