What is the oxidation state of chromium in [Cr(NH3)4Cl2]Cl?
Practice Questions
1 question
Q1
What is the oxidation state of chromium in [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. Solving gives x = +3.
Questions & Step-by-step Solutions
1 item
Q
Q: What is the oxidation state of chromium in [Cr(NH3)4Cl2]Cl?
Solution: Let the oxidation state of Cr be x. The equation is x + 4(0) + 2(-1) = +1. Solving gives x = +3.
Steps: 6
Step 1: Identify the complex ion. The compound is [Cr(NH3)4Cl2]Cl, which contains a chromium ion (Cr) surrounded by ammonia (NH3) and chloride (Cl) ligands.
Step 2: Determine the oxidation states of the ligands. Ammonia (NH3) has an oxidation state of 0, and each chloride (Cl) has an oxidation state of -1.
Step 3: Set up the equation for the oxidation state of chromium (Cr). Let the oxidation state of Cr be x. The equation will be: x + 4(0) + 2(-1) = +1.
Step 4: Simplify the equation. This becomes: x + 0 - 2 = +1.
Step 5: Solve for x. Rearranging gives: x - 2 = +1, so x = +1 + 2, which means x = +3.
Step 6: Conclude that the oxidation state of chromium in [Cr(NH3)4Cl2]Cl is +3.
