Amongst the following the lowest degree of paramagnetism per mol of the compound at 298 K will be shown by:

F e S O_{4} .6 H_{2} O

No worries! We‘ve got your back. Try BYJU‘S free classes today!

N i S O_{4} .6 H_{2} O

No worries! We‘ve got your back. Try BYJU‘S free classes today!

M n S O_{4} .4 H_{2} O

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C u S O_{4} .5 H_{2} O

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $C u S O_{4} .5 H_{2} O$
Amongst the given compounds $C u S O_{4} .5 H_{2} O$ has less number of unpaired electrons,so it has lowest degree of paramagnetism.
Hence option D is correct.

Suggest Corrections

Similar questions

The increasing order of paramagnetism of the following compound is:

(I) $M n S O_{4} .4 H_{2} O$

(II) $F e S O_{4} .7 H_{2} O$

(III) $N i S O_{4} .6 H_{2} O$

(IV) $C u S O_{4} .5 H_{2} O$

Among the following the lowest degree of paramagnetism per mole of the compound at 298 K will be shown by:

M n S O_{4} .4 H_{2} O

F e S O_{4} .7 H_{2} O

N i S O_{4} .6 H_{2} O

C u S O_{4} .5 H_{2} O

Join BYJU'S Learning Program