The number of paramagnetic species from the following is ______. $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$, $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$, $\left[\mathrm{NiCl}_{4}\right]^{2-}$ $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}$, $\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}$ $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ and $\left[\mathrm{Fe}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}\right]^{2+}$

JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Coordination Compounds

The number of paramagnetic species from the following is ______.

[Ni (CN)_{4}]^{2 -}

[Ni (CO)_{4}]

[NiCl_{4}]^{2 -}

[Fe (CN)_{6}]^{4 -}

[Cu (NH_{3})_{4}]^{2 +}

[Fe (CN)_{6}]^{3 -}

and

[Fe (H_{2} O)_{6}]^{2 +}

Answer: 4

A complex is paramagnetic only if the metal keeps unpaired

d

electrons. In

[Ni (CN)_{4}]^{2 -}

the

d^{8}

nickel(II) with strong-field cyanide is square planar with all electrons paired, in

[Ni (CO)_{4}]

the nickel is

d^{10}

, and in

[Fe (CN)_{6}]^{4 -}

the

d^{6}

iron(II) with cyanide is low spin and fully paired, so these three are diamagnetic. The weak-field chloride leaves

[NiC l_{4}]^{2 -}

tetrahedral with two unpaired electrons,

[Cu (N H_{3})_{4}]^{2 +}

has one unpaired electron on

d^{9}

copper(II),

[Fe (CN)_{6}]^{3 -}

has one unpaired electron on low-spin

d^{5}

iron(III), and

[Fe (H_{2} O)_{6}]^{2 +}

has four unpaired electrons on high-spin

d^{6}

iron(II). The number of paramagnetic species is four.

JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Integer answer