Q.Arrange the following in order of increasing molar conductivity

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 $-1$.

Step 3 — solve for Pt's oxidation number ($x$): $x + (-1) + 0 + 2(0) = +1$ $x - 1 = +1 \;\Rightarrow\; x = +2$

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.

In $K_4[Fe(CN)_6]$: the complex ion is $[Fe(CN)_6]^{4-}$; with 6 $CN^-$ ligands (total charge $-6$) and overall ion charge $-4$, Fe must be $+2$. In $[Fe(CO)_5]$: CO is a neutral ligand, and the whole molecule is neutral, so with 5 neutral ligands contributing zero charge, Fe must be $0$. In $[Fe(CN)_3(NH_3)_3]$: $NH_3$ is neutral but the three $CN^-$ ligands contribute $-3$ total; since the molecule overall is neutral, Fe must be $+3$ to balance that $-3$. So only option (c) has the metal at oxidation state zero.

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: two Cl and two $NH_3$, so coordination number = 4, not 2 (which would ignore the ammines), 5, or 6 (both too high for this formula).

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

$K_4[Fe(CN)_6]$ — six $CN^-$ ligands around $Fe^{2+}$, CN = 6. $[Co(en)_3]Cl_3$ — three bidentate ethylenediamine ligands, each donating 2 N atoms, giving CN = 6. $[Fe(H_2O)_6]Cl_3$ — six water ligands around $Fe^{3+}$, CN = 6. $[Cu(NH_3)_4]^{2+}$ — four $NH_3$ ligands bonded to $Cu^{2+}$ in a square-planar arrangement, CN = 4, matching the question.