Coordination Compounds

$\mathrm{CN}^{-}$is a strong field ligand; it causes pairing in $\mathrm{Ni}^{2+}\left(d^8\right)$.

The $d s p^2$ hybridisation forms a square planar geometry with no unpaired electrons.

The correct match is A-II, B-I, C-IV, D-III.

A. $\mathrm{Ni}(\mathrm{CO})_4$, all ligands neutral. So, oxidation state, $\mathrm{Ni}=0$

B. $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{NO}^{2+}, \mathrm{NO}^{+}=+1, \mathrm{H}_2 \mathrm{O}=0\right.$. So oxidation state, $\mathrm{Fe}=+1$

C. $\left[\mathrm{Co}(\mathrm{CO})_5\right]^{2-}, x+5(0)=-2$, So, oxidation state, $\mathrm{Co}=-2$

D. $\left[\mathrm{Cr}_2(\mathrm{CO})_{10}\right]^{2-}$, each $\mathrm{Cr}=-1($ since total $=-2)$

$ \operatorname{In}\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}, \mathrm{Ni}^{2+}=3 d^{8} 4 s^{0} $

As $ \mathrm{CN}^{-} $is a strong ligand, pairing of electrons take place, Therefore,

$ d s p^{2} $ hybridisation gives square planar geometry.

Let’s determine the oxidation number of the metal in each complex step by step.

For A. $\mathrm{Ni}(\mathrm{CO})_4$

Each $CO$ ligand is neutral.

The overall complex is neutral.

Therefore, the oxidation number on Ni is $0$.

This matches List II item: II. Zero.

For B. $\left[\mathrm{Fe}\left(\mathrm{H}_2\mathrm{O}\right)_5 \mathrm{NO}\right]^{2+}$

Water ($H_2O$) is neutral.

In many nitrosyl complexes, when the $NO$ ligand binds linearly, it is considered to act as $NO^+$ (nitrosonium).

If we assign $+1$ to the $NO$ ligand, let the oxidation state of Fe be $x$:

$x + (+1)= +2$

So, $x = +1$.

This matches List II item: I. +1.

For C. $\left[\mathrm{Co}(\mathrm{CO})_5\right]^{2-}$

Again, all carbonyl ($CO$) ligands are neutral.

The overall complex carries a charge of $2-$.

Hence, the oxidation number on Co is $-2$.

This matches List II item: IV. -2.

For D. $\left[\mathrm{Cr}_2(\mathrm{CO})_{10}\right]^{2-}$

There are two chromium atoms and all $CO$ ligands are neutral.

Let the oxidation number on each Cr be $x$. Since the overall charge is $2-$:

$2x = -2$

So, $x = -1$.

This matches List II item: III. -1.

Thus, the correct matching is:

A-II, B-I, C-IV, D-III

Among the options given, this corresponds to Option A.

$\mathrm{Ni}: 3 d^8, 4 s^2\quad$ $\mathrm{Ni}^{2+}: 3 d^8$

Since, $\mathrm{Cl}$ is a weak field ligand, it doesn't cause pairing of electron.

Number of unpaired electrons, $n=2$

$\mu=\sqrt{n(n+2)}=\sqrt{2(2+2)} B M=\sqrt{8} ~B M=2.82 ~B M$

To determine the correct statement with respect to the complexes $\mathrm{Ni}(\mathrm{CO})_4$ and $[\mathrm{Ni}(\mathrm{CN})_4]^{2-}$, we need to analyze their coordination, oxidation states, and molecular geometries.

Option A: Nickel is in the same oxidation state in both.

To find the oxidation state of Ni in $\mathrm{Ni}(\mathrm{CO})_4$, we consider that CO is a neutral ligand. Therefore, the oxidation state of Ni here is +0.

In $[\mathrm{Ni}(\mathrm{CN})_4]^{2-}$, since CN is a -1 ligand and the overall charge on the complex is -2, the oxidation state of Ni is +2.

So, nickel is not in the same oxidation state in both complexes.

Option B: Both have square planar geometry.

To determine the geometry, we need to consider the electronic configuration and the nature of the ligands:

• $\mathrm{Ni}(\mathrm{CO})_4$: This complex has Ni in the 0 oxidation state with an electronic configuration of $3d^8 4s^2$. CO is a strong field ligand but with a $d^10$ configuration (Ni(0)), leading to a tetrahedral geometry.
• $[\mathrm{Ni}(\mathrm{CN})_4]^{2-}$: This complex has Ni in the +2 oxidation state with an electronic configuration of $3d^8$. CN is a strong field ligand, causing a square planar geometry due to the dsp² hybridization to minimize repulsion.

Thus, both do not have square planar geometry.

Option C: Have square planar and tetrahedral geometry, respectively.

Based on our analysis of the electronic configuration and hybridization, it is clear that $\mathrm{Ni}(\mathrm{CO})_4$ has a tetrahedral geometry and $[\mathrm{Ni}(\mathrm{CN})_4]^{2-}$ has a square planar geometry.

Hence, this statement is incorrect because it is the opposite of our findings.

Option D: Have tetrahedral and square planar geometry, respectively.

From the above analysis:

• $\mathrm{Ni}(\mathrm{CO})_4$: Tetrahedral geometry.
• $[\mathrm{Ni}(\mathrm{CN})_4]^{2-}$: Square planar geometry.

This statement is correct based on our conclusions.

Therefore, the correct answer is Option D: Have tetrahedral and square planar geometry, respectively.

The electronic configuration of Cr is [Ar] 3d⁵ 4s¹.

[CrF₆]^3$-$ has d²sp³-hybridisation and octahedral geometry.

In MnO$_4^ - $

${}_{25}$Mn⁺⁷ = [Ar], no unpaired elecrons.

Thus, it will now show d-d transition. It is dark purple coloured due to charge transfer from ligand to metal.

The oxidation state of Ni in [Ni(CO)₄] is 0.

The electronic configuration of Ni is 3d⁸ 4s².

As CO is a strong ligand, pairing of electrons occur.

As, it has sp³-hybridisation, so the geometry is tetrahedral. Since, it has no unpaired electrons. So, the complex is diamgnetic.

Oxidation number of Co,

x + ($-$1 $\times$ 6) = $-$2

$\therefore$ x = 4

$\therefore$ Electronic configuration of Co⁴⁺,

Co : [Ar] 3d⁵

$\therefore$ Number of unpaired electrons = 5

$\therefore$ Magnetic moment = $\sqrt {n(n + 2)} = \sqrt {5 + 7} = \sqrt {35} $ BM

IUPAC name of sodiumnitroprusside Na₂[Fe(CN)₅NO] is sodiumpentacyanonitrosyl ferrate (III) because in it NO is neutral ligand and the oxidation number of Fe is III.

On passing H₂S gas in the aqueous solution of salt in the presence of dilute HCl, black ppt of CuS is formed.

$3CuS + 8HN{O_3} \to 3Cu{(N{O_3})_2} + 2NO + 3S + 4{H_2}O$

$C{u^{2 + }} + 4N{H_4}OH \to \mathop {{{[Cu{{(N{H_3})}_4}]}^{2 + }}}\limits_{Deep\,blue\,complex} + 4{H_2}O$

$\underbrace {[Cr{{(N{H_3})}_6}]}_{Coordination\,sphere}\,\,\underbrace {C{l_3}}_{Ionisable}$ $\rightleftharpoons$ ${[Cr{(N{H_3})_6}]^{3 + }} + 3C{l^ - }$

$[Cr{(N{H_3})_6}]C{l_3}\buildrel {AgN{O_3}} \over \longrightarrow \mathop {AgCl \downarrow }\limits_{(White\,ppt.)} $

Config. of Cr (24) = 3d⁵ 4s¹

Config. Cr³⁺ = 3d³ 4s⁰

Thus option (d) is correct.

(a) d²sp³ hybridisation, octahedral. Thus, option (a) is correct.

(b) here are three unpaired electrons, hence it is paramagnetic. Thus, option (b) is correct.

(c) d²sp³ inner orbital complex. Thus option (c) is as incorrect.

(d) Due to ionisable Cl^$-$ ions, white ppt with AgNO₃. Thus, option (d) is correct.

So, the incorrect statement is (c).