Total number of molecules/species from following which will be paramagnetic is \_\_\_\_. $\mathrm{O}_{2}, \mathrm{O}_{2}^{+}, \mathrm{O}_{2}^{-}, \mathrm{NO}, \mathrm{NO}_{2}, \mathrm{CO}, \mathrm{K}_{2}[\mathrm{NiCl}_{4}]$, $[\mathrm{Co}(\mathrm{NH}_{3})_{6}]\mathrm{Cl}_{3}, \mathrm{~K}_{2}[\mathrm{Ni}(\mathrm{CN})_{4}]$

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Coordination Compounds

Total number of molecules/species from following which will be paramagnetic is ____.

O_{2}, O_{2}^{+}, O_{2}^{-}, NO, NO_{2}, CO, K_{2} [NiCl_{4}]

[Co (NH_{3})_{6}] Cl_{3}, K_{2} [Ni (CN)_{4}]

Answer: 6

Checking each species for unpaired electrons:

O_{2}

has two unpaired electrons in its

π^{*}

orbitals (paramagnetic). Removing an electron to give

O_{2}^{+}

leaves one unpaired electron (paramagnetic); adding one to give

O_{2}^{-}

also leaves one unpaired electron (paramagnetic). NO and

N O_{2}

are odd-electron molecules and so are necessarily paramagnetic. CO has all electrons paired (diamagnetic). In

K_{2} [N i C l_{4}]

N i^{2 +}

3 d^{8}

, and the weak-field

C l^{-}

gives a tetrahedral, high-spin arrangement with two unpaired electrons (paramagnetic). In

[C o (N H_{3})_{6}] C l_{3}

C o^{3 +}

3 d^{6}

but strong-field

N H_{3}

forces all electrons to pair (diamagnetic). In

K_{2} [N i (C N)_{4}]

N i^{2 +}

3 d^{8}

but the strong-field

C N^{-}

drives square-planar pairing (diamagnetic).

Paramagnetic species:

O_{2}

O_{2}^{+}

O_{2}^{-}

, NO,

N O_{2}

and

K_{2} [N i C l_{4}]

— six in total.

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Integer answer