The colour of [Ni(NO_2)_6]^4- complex is ______.

The correct answer is Option (1) → Brown-yellow

The colour of a complex comes from a d–d transition: the complex absorbs light whose energy equals the crystal field splitting \( \Delta_o \), and we see the complementary colour. A stronger field ligand gives a larger \( \Delta_o \), so the light absorbed moves to shorter wavelength.

In \( \ce{[Ni(NO2)6]^4-} \), the charge balance gives the oxidation state of nickel:

    \( \displaystyle x + 6(-1) = -4 \Rightarrow x = +2 \)

Ni (Z = 28) is \( \mathrm{[Ar]\,3d^{8}\,4s^{2}} \); losing the two 4s electrons gives \( \ce{Ni^2+} \): \( \mathrm{3d^{8}} \). For nickel(II) complexes the field strength of the ligands increases in the order \( \ce{H2O} \lt \ce{NH3} \lt \) en \( \lt \ce{NO2-} \). The nitrite ion is a strong field ligand, so \( \ce{[Ni(NO2)6]^4-} \) absorbs at the short-wavelength (blue–violet) end of the visible spectrum and appears brown-yellow. Along the same series, \( \ce{[Ni(H2O)6]^2+} \) is green, the ammonia and en complexes are blue, and \( \ce{[Ni(en)3]^2+} \) is violet.

Note: The question is from beyond the NCERT textbook.