What is the increasing order of the oxidation state of the metal ion in following compounds?
(A). \( ScBr_3 \)
(B). \( V_2O_5 \)
(C). \( CrF_6 \)
(D). \( Tc_2O_7 \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (B), (C), (D)
The oxidation state of the metal in a neutral compound is found from the charge balance: the oxidation states of all the atoms add up to zero. Bromine and fluorine are taken as −1 in these halides and oxygen as −2 in the oxides.
(A) \( \ce{ScBr3} \): \( x + 3(-1) = 0 \Rightarrow x = +3 \)
(B) \( \ce{V2O5} \): \( 2x + 5(-2) = 0 \Rightarrow x = +5 \)
(C) \( \ce{CrF6} \): \( x + 6(-1) = 0 \Rightarrow x = +6 \)
(D) \( \ce{Tc2O7} \): \( 2x + 7(-2) = 0 \Rightarrow x = +7 \)
Each metal here shows its highest oxidation state, equal to its group number: Sc (group 3), V (group 5), Cr (group 6) and Tc (group 7, below manganese). The highest states are reached in fluorides and oxides because F and O are the most electronegative elements and oxygen can also form multiple bonds to the metal.
Shortcut: Sc has the lowest and Tc the highest state, so the order starts with (A) and ends with (D): Options (1) and (2). Since V (+5) is below Cr (+6), (B) comes before (C), which leaves Option (1).