Arrange these species in order of increasing oxidation state of chlorine. ClO2 ClO2- Cl2 Cl2O5
The Correct Answer and Explanation is:
To arrange these species in order of increasing oxidation state of chlorine, we need to calculate the oxidation states of chlorine in each species:
- ClO2 (Chlorine dioxide):
In ClO2, the oxygen atoms are more electronegative than chlorine, so oxygen is typically assigned an oxidation state of -2.
The overall charge of the molecule is neutral, so we can assign the oxidation state of chlorine (x) as follows: x+2(−2)=0⇒x−4=0⇒x=+4x + 2(-2) = 0 \quad \Rightarrow \quad x – 4 = 0 \quad \Rightarrow \quad x = +4x+2(−2)=0⇒x−4=0⇒x=+4 Therefore, the oxidation state of chlorine in ClO2 is +4. - ClO2- (Chlorite ion):
In ClO2-, the oxygen atoms are assigned an oxidation state of -2. Since the overall charge is -1, we can assign the oxidation state of chlorine (x) as follows: x+2(−2)=−1⇒x−4=−1⇒x=+3x + 2(-2) = -1 \quad \Rightarrow \quad x – 4 = -1 \quad \Rightarrow \quad x = +3x+2(−2)=−1⇒x−4=−1⇒x=+3 Therefore, the oxidation state of chlorine in ClO2- is +3. - Cl2 (Chlorine molecule):
In Cl2, both chlorine atoms are in their elemental form, so their oxidation states are both 0.
Therefore, the oxidation state of chlorine in Cl2 is 0. - Cl2O5 (Dichlorine pentoxide):
In Cl2O5, the oxygen atoms are assigned an oxidation state of -2. The overall charge of the molecule is neutral. Let the oxidation state of chlorine be x. For the two chlorine atoms: 2x+5(−2)=0⇒2x−10=0⇒2x=10⇒x=+52x + 5(-2) = 0 \quad \Rightarrow \quad 2x – 10 = 0 \quad \Rightarrow \quad 2x = 10 \quad \Rightarrow \quad x = +52x+5(−2)=0⇒2x−10=0⇒2x=10⇒x=+5 Therefore, the oxidation state of chlorine in Cl2O5 is +5.
Final order of increasing oxidation state of chlorine:
Cl2 (0) < ClO2- (+3) < ClO2 (+4) < Cl2O5 (+5)
Thus, the correct order is:Cl2<ClO2-<ClO2<Cl2O5\text{Cl2} < \text{ClO2-} < \text{ClO2} < \text{Cl2O5}Cl2<ClO2-<ClO2<Cl2O5
