The 'Spin only' Magnetic moment for $\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}$ is \_\_\_\_ $\times 10^{-1} \mathrm{BM}$. (given $=$ Atomic number of $\mathrm{Ni}: 28$ )

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · d and f Block Elements

The 'Spin only' Magnetic moment for

[Ni (NH_{3})_{6}]^{2 +}

is ____

\times 1 0^{- 1} BM

.
(given

=

Atomic number of

Ni : 28

)

Answer: 28

Ni

(Z = 28) has configuration

[Ar] 3 d^{8} 4 s^{2}

, so

N i^{2 +} = 3 d^{8}

N H_{3}

is a weak-field ligand, so it does not force pairing of the

d^{8}

electrons in this octahedral complex: there remain 2 unpaired electrons.

Spin-only magnetic moment:

μ = n (n + 2) = 2 \times 4 = 8 = 2.83 BM = 28.3 \times 1 0^{- 1} BM

So the value is

28 \times 1 0^{- 1} BM

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Integer answer