Solid ammonium dichromate is heated strongly. The green solid residue is R and the gas evolved is G. G is passed over heated magnesium to give M. M…

Heating gives $\mathrm{(NH_4)_2Cr_2O_7\to Cr_2O_3+N_2+4H_2O}$, so $\mathbf{R}$ is green $\mathrm{Cr_2O_3}$ (Cr is $+3$) and $\mathbf{G}$ is $\mathrm{N_2}$. Nitrogen reacts with hot magnesium: $\mathrm{3Mg+N_2\to Mg_3N_2}$. Hydrolysis, $\mathrm{Mg_3N_2+6H_2O\to3Mg(OH)_2+2NH_3}$, releases pungent $\mathbf{P}=\mathrm{NH_3}$. Excess ammonia with copper(II) forms $\mathrm{[Cu(NH_3)_4]^{2+}}$, the deep blue ion $\mathbf{W}$. $\mathrm{Cu^{2+}}$ is $d^9$ with one unpaired electron, so the spin-only moment is $\sqrt{1(1+2)}=1.73\ \mathrm{BM}$. With four ammonia ligands the textbook description is $dsp^2$ hybridisation (one $3d$ electron promoted to $4p$) and square planar geometry. Chromium in $\mathbf{R}$ is not $+6$ since dichromate is reduced while ammonium is oxidised. A tetrahedral $sp^3$ ion or a diamagnetic $\mathbf{W}$ is inconsistent with the $d^9$ configuration. Only row (D), $+3$; $dsp^2$, square planar; $1.73\ \mathrm{BM}$, matches all three requirements.