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)
We calculate the oxidation state ($x$) for the central metal atom in each compound, assuming standard oxidation states for the halogens and oxygen:
|
Compound |
Calculation |
Oxidation State (x) |
|
(A) $\text{ScBr}_3$ |
$x + 3(-1) = 0 ⇒x = \mathbf{+3}$ |
$\text{Sc}^{3+}$ |
|
(B) $\text{V}_2\text{O}_5$ |
$2x + 5(-2) = 0 ⇒2x = +10 ⇒x = \mathbf{+5}$ |
$\text{V}^{5+}$ |
|
(C) $\text{CrF}_6$ |
$x + 6(-1) = 0 ⇒x = \mathbf{+6}$ |
$\text{Cr}^{6+}$ |
|
(D) $\text{Tc}_2\text{O}_7$ |
$2x + 7(-2) = 0 ⇒2x = +14 ⇒x = \mathbf{+7}$ |
$\text{Tc}^{7+}$ |
Increasing Order
Arranging the calculated oxidation states in increasing order:
$\mathbf{+3} < \mathbf{+5} < \mathbf{+6} < \mathbf{+7}$
Which corresponds to the order: $\mathbf{(A)} < \mathbf{(B)} < \mathbf{(C)} < \mathbf{(D)}$