Q.A complex in which the oxidation number of the metal is zero is

Since the compound's counter ion $Br^-$ carries charge $-1$, the complex ion itself must carry charge $+1$; subtracting the known charges of the neutral $NH_3$/$H_2O$ ligands and the $-1$ nitro ligand from that total isolates the platinum's oxidation number as $+2$.

Step 1 — charge on the complex ion: the compound $[Pt(NO_2)(H_2O)(NH_3)_2]Br$ is electrically neutral overall, and it has one $Br^-$ counter ion (charge $-1$) outside the coordination sphere. So the complex ion $[Pt(NO_2)(H_2O)(NH_3)_2]^{x}$ must carry a charge of $+1$ to balance it.

Step 2 — sum the ligand charges: within the coordination sphere, $NH_3$ (ammine) and $H_2O$ (aqua) are both neutral ligands (charge 0 each); $NO_2^-$ (nitro) is an anionic ligand carrying charge $-1$.

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Carbonyl ligands are neutral, so a homoleptic neutral carbonyl complex like $[Fe(CO)_5]$ has the metal in oxidation state zero; complexes with charged ligands like $CN^-$ require a nonzero metal oxidation state to balance charge.

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Coordination number = the total number of ligand donor atoms directly bonded to the central metal ion. In $[PtCl_2(NH_3)_2]$, the central $Pt^{2+}$ is surrounded by 2 chloride ligands and 2 ammonia ligands, i.e. $2+2=4$ donor atoms in total.

This complex is the classic example of cisplatin/transplatin — a square-planar $Pt(II)$ complex (consistent with $Pt^{2+}$'s $d^8$ configuration, which characteristically favours square-planar, 4-coordinate geometry). The formula itself directly tells us the ligand count: t …

Among the given complexes, only $[Cu(NH_3)_4]Cl_2$ has coordination number 4; the other three all have coordination number 6.

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