To determine the number of paramagnetic species with a bond order of one, we analyze each given species:
Step 1: Identify Paramagnetic Species
Step 2: Calculate Bond Order
Bond order is calculated as: (Number of electrons in bonding orbitals-Number of electrons in antibonding orbitals)2
| Species | Electron Configuration | Paramagnetic? | Bond Order |
|---|---|---|---|
| H2 | (σ1s)² | No | 1 |
| He2+ | (σ1s)²(σ*1s)¹ | Yes | 0.5 |
| O2- | (σ2s)²(σ*2s)²(π2p)⁴(π*2p)³ | Yes | 1.5 |
| N2 | (σ2s)²(σ*2s)²(π2p)⁴(σ2p)² | No | 3 |
| O22- | (σ2s)²(σ*2s)²(π2p)⁴(π*2p)⁴ | No | 1 |
| F2 | (σ2s)²(σ*2s)²(π2p)⁴(π*2p)⁴(σ2p)² | No | 1 |
| Ne2+ | (σ2s)²(σ*2s)²(π2p)⁴(π*2p)⁴(σ2p)¹ | Yes | 0.5 |
| B2 | (σ2s)²(σ*2s)²(π2p)¹(π2p)¹ | Yes | 1 |
Final Results
The total number of species matching the criteria is: 1. This satisfies the range (1,1).
We analyze the magnetic behaviour and bond order for each species:
Among these, the only species that is both paramagnetic and has a bond order of 1 is: B2
Thus, the correct count is: 1.
From the given following (A to D) cyclic structures, those which will not react with Tollen's reagent are : 
Compound 'P' undergoes the following sequence of reactions : (i) NH₃ (ii) $\Delta$ $\rightarrow$ Q (i) KOH, Br₂ (ii) CHCl₃, KOH (alc), $\Delta$ $\rightarrow$ NC-CH₃. 'P' 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 ___________.
Consider the following two reactions A and B: 
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