The correct order of the number of unpaired electrons in the given complexes is A. $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ B. $\left[\mathrm{FeF}_{6}\right]^{3-}$ C. $\left[\mathrm{CoF}_{6}\right]^{3-}$ D. $\left[\mathrm{Cr}(\text{oxalate})_{3}\right]^{3-}$ E. $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ Choose the correct answer from the options given below:

(B) E $<$ A $<$ D $<$ C $<$ B

(A) A $<$ E $<$ D $<$ C $<$ B

(C) E $<$ A $<$ B $<$ D $<$ C

(D) A $<$ E $<$ C $<$ B $<$ D

JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Coordination Compounds

The correct order of the number of unpaired electrons in the given complexes is
A.

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

[FeF_{6}]^{3 -}

[CoF_{6}]^{3 -}

[Cr (oxalate)_{3}]^{3 -}

[Ni (CO)_{4}]

Choose the correct answer from the options given below:

Answer: (B)

Count the unpaired electrons in each complex.

F e^{3 +}

d^{5}

; with the strong-field

C N^{-}

the complex is low spin

t_{2 g}^{5}

, giving 1 unpaired electron. With weak-field

F^{-}

[Fe F_{6}]^{3 -}

is high spin

d^{5}

with 5 unpaired electrons.

C o^{3 +}

d^{6}

; with weak-field

F^{-}

it is high spin

t_{2 g}^{4} e_{g}^{2}

, giving 4.

C r^{3 +}

d^{3}

, always 3 unpaired. In

Ni (CO)_{4}

the metal is

Ni (0)

d^{10}

with strong-field CO, so none are unpaired. Arranging by increasing number of unpaired electrons gives

E (0) < A (1) < D (3) < C (4) < B (5)

JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Single correct option