A green complex, `K_(2)[Cr(NO)(NH_(3))(CN)_(4)]` is paramagnetic and has `mu_(eff)=1.73 BM` . Write the IUPAC name of the complex and draw the structure of anion and find out the the hybridisation of metal ion.

Explanation

IUPAC name is: Potassium amminetatracyanidonitrosoniumchromate(I)
or Potassium amminetatracyanidonitrocyliumchromate(I)
Let n is the number of unpaired electron in the chromium ion.
Since `mu=sqrt(n(n+2)) or 1.73=sqrt(n(n+2))BM or 1.73xx1.73=n^(2)+2n`
Hence n=1
As the `CN^(-) and NH_(3)` are strong fields ligands, they compel for pairing of electrons. So, `[Cr(NO)(CN)_(4)(NH_(3))]^(2-)=`
Hence, the oxidation state of chromium is +1 (having `3d^(5)` configuration ). So according to charge on the complex NO should be `NO^(+)` and the structure of this complex is octahedral with `d^(2)sp^(3)` hybridisation as given below