What is the oxidation state of chromium in the complex [Cr(NH3)4Cl2]Cl?
Practice Questions
1 question
Q1
What is the oxidation state of chromium in the complex [Cr(NH3)4Cl2]Cl?
+2
+3
+4
+6
Let the oxidation state of Cr be x. The equation is x + 4(0) + 2(-1) = +1 (from Cl). Solving gives x = +3.
Questions & Step-by-step Solutions
1 item
Q
Q: What is the oxidation state of chromium in the complex [Cr(NH3)4Cl2]Cl?
Solution: Let the oxidation state of Cr be x. The equation is x + 4(0) + 2(-1) = +1 (from Cl). Solving gives x = +3.
Steps: 7
Step 1: Identify the complex and its components. The complex is [Cr(NH3)4Cl2]Cl, which contains chromium (Cr), ammonia (NH3), and chloride (Cl).
Step 2: Determine the oxidation state of each component. Ammonia (NH3) is a neutral ligand, so it has an oxidation state of 0. Each chloride ion (Cl) has an oxidation state of -1.
Step 3: Set the oxidation state of chromium (Cr) as 'x'.
Step 4: Write the equation for the total charge of the complex. The overall charge of the complex is +1 (from the outer Cl). The equation is: x + 4(0) + 2(-1) = +1.
Step 5: Simplify the equation. This becomes: x - 2 = +1.
Step 6: Solve for x. Add 2 to both sides: x = +1 + 2, which gives x = +3.
Step 7: Conclude that the oxidation state of chromium in the complex [Cr(NH3)4Cl2]Cl is +3.
