Answer
(a) Tin(IV)chloride.
(b) Copper(I)oxide.
(c)Cobalt(II)nitrate.
(d) Sodiumdichromate.
Work Step by Step
(a) Tin(IV) chloride.
Tin, Sn exists in two oxidation states, +2 and +4. In $SnCl_{4}$ Sn is in the +4 oxidation states. So it must be mentioned in the formula.
(b) Copper(I) oxide.
Copper, Cu exists in two oxidation states, +1 and +2. In $Cu_{2}O$, Cu is in the +1 oxidation state. So it must be mentioned in the formula.
(c)Cobalt(II) nitrate
Cobalt, Co exists in two oxidation states, +2 and +4. In $Co(NO_{3})_{2}$, Co is in the +2 oxidation state. So it must be mentioned in the formula.
(d) Sodium dichromate – Dichromate ion is $(Cr_{2}O_{7})^{2-}$. So the formula is $Na_{2} Cr_{2}O_{7}$
