The correct order of the calculated spin-only magnetic moments of complexes (A) to (D) is : (a) $\mathrm{Ni}(\mathrm{CO})_{4}$ (b) $\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$ (c) $\mathrm{Na}_{2}\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]$ (d) $\mathrm{PdCl}_{2}\left(\mathrm{PPh}_{3}\right)_{2}$

(C) $(\mathrm{a}) \approx(\mathrm{c}) \approx(\mathrm{d})<(\mathrm{b})$

(A) (c) $\approx$ (d) $<$ (b) $<$ (a)

(B) $(\mathrm{a}) \approx(\mathrm{c})<(\mathrm{b}) \approx(\mathrm{d})$

(D) (c) $<$ (d) $<$ (b) $<$ (a)

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Coordination Compounds

The correct order of the calculated spin-only magnetic moments of complexes (A) to (D) is :
(a)

Ni (CO)_{4}

(b)

[Ni (H_{2} O)_{6}] Cl_{2}

(c)

Na_{2} [Ni (CN)_{4}]

(d)

PdCl_{2} (PPh_{3})_{2}

Answer: (C)

The spin-only moment

μ = n (n + 2)

BM depends on the number of unpaired electrons. In

Ni (CO) X_{4}

nickel is zero-valent and the strong-field

CO

pairs every electron, so

n = 0

. In

[Ni (H X_{2} O) X_{6}] X^{2 +}

the

Ni X^{2 +}

(

3 d^{8}

) sits in weak-field water with two unpaired electrons,

n = 2

and

μ = 8 \approx 2.83

BM. In

Na X_{2} [Ni (CN) X_{4}]

the strong-field

CN X^{-}

makes

Ni X^{2 +}

square planar and diamagnetic,

n = 0

, and

PdCl X_{2} (PPh X_{3}) X_{2}

with

4 d^{8}

Pd X^{2 +}

is square planar and diamagnetic too,

n = 0

. The three diamagnetic complexes therefore lie together below the aqua complex, which alone has a non-zero moment.

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Single correct option