The species that has a spin - only magnetic moment of 5.9 BM , is: ( $\mathrm{T}_{\mathrm{d}}=$ tetrahedral)

(D) $\left[\mathrm{MnBr}_{4}\right]^{2}-\left(\mathrm{T}_{\mathrm{d}}\right)$

(A) $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2}$ - (square planar)

(B) $\left[\mathrm{N}_{\mathrm{i}} \mathrm{Cl}_{4}\right]^{2}-\left(\mathrm{T}_{\mathrm{d}}\right)$

(C) $\mathrm{Ni}(\mathrm{CO})_{4}\left(\mathrm{~T}_{\mathrm{d}}\right)$

JEE Main · 2020 · 06 Sep 2020 · Shift 1 · d and f Block Elements

The species that has a spin - only magnetic moment of 5.9 BM , is:
(

T_{d} =

tetrahedral)

Answer: (D)

A spin-only moment of

5.9 BM

corresponds, through

μ = n (n + 2)

, to

n = 5

unpaired electrons. Among the choices,

[MnBr X_{4}] X^{2 -}

has manganese as

Mn X^{2 +}

, a

d^{5}

ion, and in a tetrahedral bromide field all five

d

electrons stay unpaired, giving exactly five unpaired electrons. The nickel species, whether

Ni X^{2 +}

(

d^{8}

) or

Ni (0)

(

d^{10}

), cannot reach five unpaired electrons, so the

5.9 BM

moment belongs to

[MnBr X_{4}] X^{2 -}

JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Single correct option